AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 7th Lesson State Legislature Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 7th Lesson State Legislature

Long Answer Questions

Question 1.
Explain the composition, powers, and functions of the State Legislative Assembly.
Answer:
The Constitution provides for a Legislature for every State on the model of the Parliament. As per Article 168, the State Legislature consists of the Governor and one or two Houses. In India, while some States have Bicameral Legislatures, others have Unicameral Legislatures. Andhra Pradesh at present possesses Unicameral Legislature.

The Lower House of the State Legislature is known as the Legislative Assembly or Vidhana Sabha and the Upper House is the Legislative Council or Vidhana Parishad.

Composition of Legislative Council (Vidhana Parishad) :
The Upper House of the State Legislature is known as Legislative Council. The Constitution lays down that a Legislative Council shall have not less than 40 members and not more than \(\frac{1}{3}\)rd of the total membership of the State Assembly. The Legislative Council consists of both nominated and elected members.

The election is conducted through the indirect method by means of proportional representation with a single transferable vote.

Distribution of Seats :

  1. \(\frac{1}{3}\)rd are elected by the members of the State Assembly.
  2. \(\frac{1}{3}\)rd are elected by members of local bodies.
  3. \(\frac{1}{12}\)th are elected by teachers.
  4. \(\frac{1}{12}\)th are elected by graduates.
  5. The remaining \(\frac{1}{6}\)th members are nominated by the Governor from among persons who have distinguished themselves in the fields of Literature, Science, Arts, Social Services etc.

Qualifications :
The members of the Council 1) must be citizens of India, 2) must have completed 30 years of age and 3) must possess such other qualifications as may be prescribed by the Legislature.

Term :
the members are elected for a period of 6 years. But \(\frac{1}{3}\)rd of them retire for every 2 years. The Council is a permanent body. It cannot be dissolved by the Governor.

Chairman and Deputy Chairman:
The Council has Chairman and a Deputy Chairman who are elected by the members of the Council from among themselves. The Chairman presides over the meetings of the Council.

Legislative Assembly (Vidhana Sabha) :
Composition of the Legislative Assembly :
The Legislative Assembly is the popular and powerful chamber of the State Legislature. It is the lower house and resembles more or less the Lok Sabha at the Centre. It consists of representatives directly elected by the people of the State on the basis of universal adult franchise. It’s maximum strength is fixed at 500 and minimum strength at 60. Only the Legislative Assembly of Sikkim has less than 60 because of it’s small population. Those who become members of State Legislative Assembly must be citizens of India and must be above 25 years of age.

Term of Office :
The normal term of Assembly is 5 years. It may be dissolved earlier by the Governor on the advice of the Chief Minister. The Parliament may extend it’s term by one year, when National Emergency is in force.

Presiding Officers :
The Presiding officer of Assembly is known as Speaker. He is elected by the members of the Assembly. The Assembly also elects a Deputy Speaker to conduct the business of the House in the absence of the Speaker.

Powers and Functions of the State Legislature :
The State Legislature has the following powers.

1) Legislative Powers and Functions :
The State Legislature has the power to make laws on all the subjects included in the State List. It has also the power to make laws in respect of subjects included in the Concurrent List. However, such a law should not disagree with a law already made by the Parliament on the same subject. In the making of laws, Legislative Assembly has been given more powers than the Legislative Council. The Legislative Council at the most may delay the legislation for a period of 4 months. Later the Assembly sends the bill to the Governor for his assent.

2) Constitutional Powers and Functions :
Even though, the Legislature has no powers to move the Constitution amendment bills, it’s consent is required for amending certain provisions of the Constitution. Such bills have to be referred to it after they are approved by the Parliament.

3) Executive Powers and Functions :
The State Legislature exercises control over the Council of Ministers. It’s members make the Ministers individually and collectively responsible to the Legislature. The Council of Ministers is collectively responsible to the State Legislative Assembly. The Legislature can expose the actions of Executive, through questions, debates and adjournment motions. In controlling the Executive, the Legislative Assembly has more powers than the Council. The Ministry has to resign when the Legislative Assembly passes no confidence motion against the Government.

4) Financial Powers and Functions:
The State Legislature exercises complete control over the finances of the State Government. It sanctions money to the State Government to enable it to run the administration. It may pass, reduce or reject the demands for grants presented to it by the Government. It may accept or reject proposals for taxation and borrowings presented to it by the Government. In financial matters the Assembly is more powerful than the Council. Because all money bills, including the Budget, shall be introduced first only in the Assembly. It can accept or reject any recommendations made by the Council.

5) Electoral Powers :
The elected members of the Assembly participate in the election of the President. They also elect the representatives of the State to the Rajya Sabha and l/3rd members of the Legislative Council if the State Legislature is bicameral. They also elect the presiding officers and deputy presiding officers of Assembly and Council.

Miscellaneous Powers : The state legislature :

  1. Safeguards the dignity and privileges of its members.
  2. Suspends, expels or terminates their membership.
  3. Examines the report of the State Public Service Commission and the Comptroller and Auditor General etc.

Conclusion:
The State Legislature plays an important role in the State Administration. It makes necessary laws for the welfare of the people of the State. It controls the Executive by making it responsible for their actions.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Question 2.
Write briefly the composition, powers and functions of the. State Legislative Council.
Answer:
The Upper House of the State Legislature is known as Legislative Council. The Constitution lays down that a Legislative Council shall have not less than 40 members and not more than \(\frac{1}{3}\)rd of the total membership of the State Assembly. The Legislative Council is a body of partly nominated and partly elected members.

The election is conducted through indirect method by means of proportional representation with a single transferable vote.
Composition of the Council:

  1. \(\frac{1}{3}\)rd members are elected by the Legislative Assembly.
  2. \(\frac{1}{3}\)rd members are elected by members of local bodies.
  3. \(\frac{1}{12}\)th members are elected by teachers.
  4. \(\frac{1}{12}\)th members are elected by graduates.
  5. The remaining \(\frac{1}{6}\)rd members are nominated by the Governor from among persons who have distinguished themselves in the fields of Literature, Science, Arts, Social Services etc.

Tenure :
The Legislative council is a quasipermanent House. l/3rd of the members of this House retire every two years. But the term of each member is six years. New members are elected in the place of retired members. All the members of the House do not retire at a time as it is a permanent House.

Powers and Functions of State Legislative council:
State Legislative council is the primary law making body along with the Legislative Assembly. The State Legislative Council has the following powers and Functions.

a) Legislative powers and Functions :
The Legislative Council does not possess equal powers and functions when compared to counterpart, State Legislative Assembly. It is said that the Legislative Council enjoys equal status and not power. However it exercises the following powers and functions. All the bills, other than money bills may be introduced in either of the House. They will be sent to the assent of the Governor only with the approval of both the Houses. The Council may reject any bill and spnt it back for the reconsideration of the Assembly. However, incase of a disagreement between two Houses, the decision of the Assembly will be supreme.

The Council must approve all the bills sent by the Assembly with in a period of three months or at the maximum of four months. It implies that the Council can withhold its assent over the bills sent by the Assembly for a maximum period of four months. Thus, the Legislative Council can only delay the bills but the Legislative Assembly can override it.

b) Executive Powers and Functions :
The State Legislative Council has very limited executive powers when compared to that of the Assembly. The council of Ministers headed by the Chief Minister is responsible for its acts only to the Assembly and not to the Council. The Council cannot decide the future of the Council of Ministers. However, the Council can influence the policies and programmes of the ministers by asking questions and supplementary questions by drawing the call attention motion etc., but they cannot force the Council of Ministers to resign.

c) Financial Powers and Functions :
The Legislative Council has only limited powers in the financial matters. Money bills cannot at first be introduced in the Legislative Council first. The Council must accept all money bills with or without recommendation within fourteen days of the receipt of the bill. The Assembly possess the discretion powers either to accept or reject these recommendations. If the Council does not return the Money Bill to the Assembly within 14 days, then the bill is deemed to have been passed by both the Houses. It is clear that in the financial field the Legislative Council has a subordinate status and that Legislative Assembly has dominant position.

d) Electoral Functions:
The Legislative Council elects a Chairman and Deputy Chairman to preside over its meetings in a dignified manner. Some of its members are elected to various legislative committees like Public Accounts Committee, Estimates Committee and Public Undertakings committee etc.

e) Other Functions :
The Legislative council acts as the best means for formulating and consolidating public opinion. It discusses technical and other contemporary matters, as there are experts in various fields.

Question 3.
Explain the role and responsibilities of the Speaker of Legislative Assembly.
Answer:
The members of State Legislative Assembly elect one among them as a speaker to conduct the business of the House. His term of office is five years.

The speaker is the guardian of the Rights and Liberties of the members of the House.

Role and Responsibilities (or) Powers and Functions of the Speaker :
The powers and functions of the speaker of State Legislative Assembly are almost the same as those of the Speaker of Lok Sabha. His powers and functions are as follows.

  1. The speaker preserves order and decorum in the House for conducting legislative business.
  2. He allocates time for different kinds of business in the House.
  3. He interprets the rules and procedure.
  4. He puts matters to vote and announces the results.
  5. He has the right of casting vote in case of a tie.
  6. He admits motions, resolutions and points of order.
  7. He is empowered to adjourn the meeting of the House in the absence of a quorum.
  8. He can order for removal of indecent and incriminatory references from the records.
  9. He allows the members to speak in the House.
  10. He may name a member and ask him to leave the House in case of disorderly behavior.
  11. He can adjourn the House in case of grave disorder or serious matter.
  12. He accepts and rejects the resignation of a member of the House after ascertaining whether it was submitted under due process or not.
  13. He appoints the Chairmen of all the committee of the assembly and supervises their functioning. He himself is a Chairman of Business Advisory Committee, Rules Committee and the General Purpose Committee.
  14. He decides where a bill is a Money Bill or not. His decision on this question is final.

Short Answer Questions

Question 1.
Write a note on the Legislative Assembly.
Answer:
Legislative Assembly is the Lower House of the State Legislature. The Members of Legislative Assembly are called M.L.As. According to Article 170of the Indian Constitution it consists of not more than 500 members and not less than 60 members. It means that it’s strength depends on the population and size of the state. But small states have been allowed to have less number of members. Thus Goa and Mizoram have only 40 members, while Sikkim has 32 Members.

Composition of the Legislative Assembly:
The Legislative Assembly is the popular and powerful chamber of the State Legislature. It is the lower house and resembles more or less the Lok Sabha at the Centre. It consists of representatives directly elected by the people of the State on the basis of universal adult franchise. Those who become members of State Legislative Assembly must be citizens of India and must be above 25 years of age.

Term of office :
The normal term of Assembly is 5 years. It may be dissolved earlier by the Governor on the advice of the Chief Minister. The Parliament may extend it’s term by one year, when National Emergency is in force.

Presiding officers :
The Presiding Officer of Assembly is known as Speaker. He is elected by the members of the Assembly. The Assembly also elects a Deputy Speaker to conduct the business of the House in the absence of the speaker.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Question 2.
Write a note on Estimates Committee.
Answer:
According to the Rules of Procedure and Conduct of Business in the State Legislature, the Estimates committee consists of 20 members. Among them 15 members belong to Assembly. The remaining 5 members belong to Legislative Council. The members hold office for a period of one year. They are elected through in indirect election.

Functions :
The functions of the Estimates Committee in the State Legislature are the same as that of Estimates Committee of Lok Sabha. These are given here under.

  1. Estimates Committee exercises control over public expenditure.
  2. It suggests fiscal reforms in organization, the efficiency or administration reforms consistent with the policy underlying estimates.
  3. It advises alternative policies for securing efficiency and economy in administration.
  4. It examines whether the money is well laid out within the limits of the policy implied in the estimates.
  5. It also suggests the form in which the estimates shall be presented to the Assembly.

Question 3.
What do you know about Public Accounts Committee? [Mar.-18, 17]
Answer:
Public Accounts Committee consists of 20 members out of which 15 members belong to Assembly and 5 members belong to Legislative Council. They are elected through indirect election by following the principle of proportional representation Jor a period of one year. The Chairman is normally the member of Opposition Party. The Ministers of Cabinet cannot be member of Public Accounts Committee.

Functions :
Public Account Committee performs the following functions.

  1. The committee examines the accounts showing the appropriation of sums granted by the house for expenditure of the state government.
  2. It scrutinizes the appropriation accounts of the state and the reports of the Comptroller and Auditor General.
  3. It shall be the duty of the Public Accounts Committee to examine such a trading, manufacturing, and profit and loss accounts and balance sheets and the accounts of the state government and also to consider the report of the Comptroller and Auditor General.
  4. The committee carefully considers the accounting and audit procedures.
  5. The committee is not concerned with the question of policy approved by the legislature.
  6. The committee investigates expenditure after it has already incurred. An overall, this committee is generally described as a ‘post-mortem committee’.

Question 4.
Write the powers and functions of Vidhcma Sabha Speaker. [Mar. 16]
Answer:
The powers and functions of the Speaker of State Legislative Assembly are almost the same as those of the Speaker of Lok Sabha. His powers and functions are as follows.

  1. The speaker preserves order and decorum in the House for conducting legislative business.
  2. He allocates time for different kinds of business in the House.
  3. He interprets the rules and procedure.
  4. He puts matters to vote and announces the results.
  5. He has the right of easting vote in case of a tie.
  6. He admits motions, resolutions and points of order.
  7. He is empowered to adjourn the meeting of the House in the absence of a quorum.
  8. He can order for removal of indecent arid incriminatory references from the records.
  9. He allows the members to speak in the House. ‘
  10. He may name a member and ask him to leave the House in case of disorderly behavior.
  11. He can adjourn the House in case of grave disorder or serious matter.
  12. He accepts and rejects the resignation of a member of the House after ascertaining whether it was submitted under due process or not.
  13. He appoints the Chairmen of all the committees of the assembly and supervises their functioning. He himself is a Chairman of Business Advisory Committee, Rules Committee and the General Purpose Committee.
  14. He decides where a bill is a Money Bill or not. His decision on this question is final.

Very Short Answer Questions

Question 1.
Qualifications of M.L.A. [Mar. 18, 16]
Answer:
A person who wishes to contest for the membership of the State Legislative Assembly must be possess the following qualifications.

  1. He should be a citizen of India.
  2. He should have completed the age of 25 years.
  3. He should possess such other qualifications as prescribed by any act of Parliament.
  4. However, no person can simultaneously be a member of any House of the Parliament and of a State Legislature.

Question 2.
Qualifications of M.L.C.
Answer:
A person who wishes to contest for the membership of the State Legislative Council must possess the following qualifications.
a) He should be a citizen of India.
b) He should have completed 30 years of age.
c) He should possess such other qualifications as laid down by an Act of Parliament.

Question 3.
Quorum.
Answer:
Quorum is the minimum number of members required to be present in the house before it can transact any business. According to Article 188 of the constitution, the Quorum for conducting,the State Legislative Assembly meeting was fixed at 1/10th of the total membership. However, in some states, where the strength of the State Legislative Assembly is very less, the quorum will be a minimum number of 10. The speaker decides whether there is a quorum or not on a particular day.

Question 4.
Salaries and Allowances of M.L.A.
Answer:
The salary of MLA is decided by the respective State Legislature as per the Article 164 of the Indian constitution. The members of Andhra Pradesh State Legislative Assembly receive a monthly salary of ₹ 90,000/- which includes a basic pay of ₹ 15,000/- and constituency allowance of ₹ 75,000/-. Those legislators who are not provided government accommodation will get an additional ₹ 10,000/- as H.R.A members also get daily allowance of ₹ 800/- when the state legislature is in session.

Question 5.
Privileges of State Legislature.
Answer:
Privileges of a State Legislature are a sum of special rights, immunities and exemptions enjoyed by the State Legislatures. They are necessary in order to secure independence and effectiveness of their actions. The Houses cannot maintain the authority, dignity and honour without these privileges. They can protect their members from any obstructions in the discharge of their legislative responsibilities.

  1. Collective privileges
  2. Individual privileges

i) Collective Privileges :
The legislature has the right to publish its reports, debates and proceedings and also to prohibit others publishing the same.

ii) Individual Privileges:
The privileges belonging to the members of state legislature individually. They can not be arrested during the session of the state legislature or 40 days before and after the end of the session.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Question 6.
Brief History of AP legislature.
Answer:
The Andhra state was formed on October 1, 1953 Andhra State legislature initially had 140 MLAs. Elections were held to the Andhra State Legislative Assembly for the first time in 1955.

As per the recommendations of states re-organization committee, Hyderabad State was merged with Andhra State on linguistic basic and formed into Andhra Predesh State which had 245 MLAs [Including 150 MLAs of Hyderabad State.] Elections were held to the Andhra Pradesh legislative Assembly in 1957.

The State Legislative Council was established on July 1,1958. Since then it continued to exist till June 1, 1985, before being abolished. Again on March 30, 2007 the Andhra Pradesh Legislature became again bicameral after the revival of the legislative council.

Question 7.
Chairman of Legislative council. [Mar. 16]
Answer:
There will be a chairman in the Legislative council for conducting the meetings. He is elected by the members of the Legislative council among themselves. Dr. A. Chakrapani Yadav is the Present Chairman of Legislative council of Andhra Pradesh.

Question 8.
Deputy Speaker.
Answer:
The members of State Legislative Assembly elect one among them as a Deputy Speaker. His term of office is 5 years. The deputy speaker performs the duties in the absence of the speaker.

Question 9.
Deputy Chairman of Legislative Council.
Answer:
The Members of State Legislative Council elect one among them as a Deputy Chairman. The Deputy Chairman performs the duties in the absence of the Chairman.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Question 10.
Types of committees. [Mar. 18]
Answer:
The committees are of two types i.e., Standing committees and Ad-hoc committees.
i) Standing Committees:
Standing committees deal with specific business (financial matters) Ex : Estimates committee, Public accounts committee and Committee on public undertakings.

ii) Ad-hoc Committees :
Ad-hoc committees are concerned with the matters of temporary nature. They cease to exist after completion of the work. They perform some specific functions assigned to them from time to time.

AP Inter 2nd Year Civics Study Material Chapter 6 State Executive

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 6th Lesson State Executive Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 6th Lesson State Executive

Long Answer Questions

Question 1.
Discuss the powers and functions of the Governor. [Mar. 18]
Answer:
The Governor is the constitutional head of the state government. Article 153 of the Indian constitution provides for the office of the governor in the states. The administration of a state is carried on in the name of the governor

Appointment:
The president appoints the governor on the advice of the Prime Minister. In this Contest, the president generally follows two conventions which are mentioned below for the appointment of the governor.

  1. Consulting the Chief Minister of the State Concerned.
  2. Choosing an eminent person not belonging to the state concerned.

Qualifications :
Article 157 of our constitution lays down the following qualifications for the appointment of a person as a governor.

  1. He shall be a citizen of India.
  2. He should have completed the age of 35 years.
  3. He should not hold any office of profit.
  4. He should not be a member of either union or state legislature.
  5. He should not be an insolvent declared by any court of Law.

Pay and Allowances :
The governor is entitled to receive a monthly salary of ₹ 1,10,000/-. He resides in official rent-free building “Raj Bhavan”. Besides, he is entitled to many other allowances and privileges.

Oath of Office :
The Chief Justice of the state High Court administers the oath of office to the governor.

Tenure :
The Governor holds office as a convention for a term of five years. However, He holds office during the pleasure of the president.

Powers and Functions of the Governor:
The Governor exercises six important powers and functions. They are explained as follows.

1) Legislative Powers and Functions :
Article 163 describe that the Governor is an intregral part of the Stage Legislature. In that capacity he exercises certain powers and performs functions related to the State Legislature.

  1. The Governor inaugurates the first sessions of the State Legislative Assembly after the general elections are over.
  2. He also addressing the first session of State Legislative Assembly every year i.e. budget session.
  3. He appoints Pro-tern Speaker of the State Legislative Assembly.
  4. He summons and prorogues the sessions of the two Houses of the State Legislature.
  5. He addresses the Members of the state Legislature and sends messages in relation to the state legislature.
  6. The Governor gives his assent to the bills passed by the state Legislature.
  7. He may return a bill sent by the state Legislature for its reconsideration.
  8. He dissolves the State Legislative Assembly when he feels no party is in a position to form a stable and viable Government and the advice of the Chief Minister.
  9. He may promulgate Ordinances to meet an emergency which require immediate action during the recess of the State Legislature.
  10. He nominates members of Anglo-indian community to the Legislative Assembly of the state if he feels that community is not represented in the house!
  11. The Governor nominates 1/6 of the total members of the State Legislative Council.

2) Executive powers and functions:
Article 154 of our constitution vests the governor with the executive powers of the state. The governor exercises these powers either directly of through officers subordinate to him. The governor has the following executive powers.

  1. The Governor appoints, the Chief Minister and the members of the Council of Ministers on the advice of the Chief Minister.
  2. He allocates Portfolios among the ministers and reshuffles their portfolios.
  3. He removes the Ministers on the advice of the Chief Minister.
  4. He appoints the Vice-Chancellors of the Universities in the State. He acts as the Chancellor of the universities.
  5. He appoints the Chief Secretary and Advocate General of the State Government.
  6. He appoints the Chairmen and other members of the State Commissions such as a) State Public Service Commission, b) State Election Commission, c) Official Language Commission, d) Commission for Women, e) Minorities Commission, f) Backward Classes Commission and g) SC & ST Commission.
  7. He regulates the postings and transfers of the All India Services personnel working in the state.

3) Judicial Powers and Functions :
The Governor also exercises the following judicial powers and functions.

  1. The Governor renders advice to the President of India in the appointment of Chief Justice and other judges of the High Court of the State.
  2. The Governor appoints the Advocate General of the State.
  3. He makes appointments, postings and promotions of the District Judges in consultation with the Chief Justice of High Court of the State.
  4. He also appoints persons to the judicial services of the state (Other than the district courts) in consultation with the Chief Justice of High Court and State Public Service Commission.
  5. He can grant pardon, retrieve, remit and commute the sentence of any person convicted of any offence against any law of the concerned state.

4) Financial Powers and Functions :
The Governor will have the following powers and functions of financial nature.

  1. The Governor sees that the Annual Financial Statement,(i.e. Budget) is laid before the State Legislature.
  2. No Money bill shall be introduced in the prior permission of the Governor.
  3. No Demand for Grant can be made except on his recommendation.
  4. He maintains the Contingency Fund of the State. He can make advances out of the Contingency Fund to meet any unforeseen expenditure.
  5. He constitutes a Finance Commission for every five years to review the financial position of village Panchayats and Municipalities.
  6. He sees that reports of various financial committees are laid before the State Legislature.

5) Miscellaneous Powers and Functions :
The Governor receives the Annual Report of the State Public Service Commission and passes it on to the Council of Ministers for comments. Thereafter, he passes on the report on these comments to the Speaker of the Assembly for placing it before the legislature. He receives the report of the Auditor-General regarding income and expenditure made by different departments working under the State Government. On behalf of the President, he runs the administration as the real head of the state through the enforcement of law and policies during the period of President’s Rule.

6) Discretionary Powers:
Under Article 163 (1) of the constitutions the Governor has some discretionary powers which are discharged by him. His decisions in this regard are final, These are mentioned as below.

  1. Playing a decisive role in appointing the new Chief Minister in a situation when no single party has a clear majority in the state Legislative Assembly.
  2. Dismissing a Minister When it refuses to resign even after losing majority support in the House.
  3. Dissolution of the Assembly on the advice of the Chief Minister who lost the majority members support.
  4. Rendering advice to the President for the imposition of the President’s Rule in the State.
  5. Reserving a Bill for the consideration and approval of the President.
  6. Seeking instructions from the president before promulgating ordinance on some important matters.
  7. Sending back a bill passed by the state legislature for its reconsideration, .except money bills.

AP Inter 2nd Year Civics Study Material Chapter 6 State Executive

Question 2.
Explain the powers and Functions of the Chief Minister.
Answer:
Articles 163 and 164 of our constitution deal with the office of the chief minister. The chief minister is the real executive head of the State Government. He plays a decisive role and occupies a key position in the State Government.

Appointment :
The Chief Minister is appointed by the governor under article 164. After general elections, the governor normally invites the leader of the majority party in the Legislative Assembly to form the government and appoints him as the Chief Minister.

Powers and Functions of Chief Minister:
The Chief Minister has high authority and heavy responsibility in discharging his powers and related functions. His powers and functions are related to the following heads.

i) Formation of the Ministry :
The first and foremost responsibility of the Chief Minister is the formation of Ministry of his choice. The CM has a free hand in the selection and appointment of Ministers. He chooses some members of his party (or coalition partners in the case of a coalition) and recommends their names to the Governor to be appointed as Ministers. He advises the Governor to allocate portfolios among the Ministers.

ii) Leader of the State Council of Ministers :
The Chief Minister is the head of the Council of Ministers. As such he occupies a position of exceptional authority. He is the Chairman of the State Council of Ministers. The Chief Minister decides the time, venue and the agenda of Gabinet meetings. The CM presides over such meetings, discussions are carried under his direction. He guides, directs, controls and co-ordinates the activities of the Ministers.

iii) Link between the Governor and the State Council of Ministers:
The Chief Minister is the principal channel of communication between the Governor and the State Council of Ministers. As part of his Constitutional duty he communicates all the administrative decisions and legislative proposals of the State Council of Ministers to the Governor. It is his responsibility to furnish any information related to the actives of the Ministers as the Governor may call for. No minister shall meet the Governor without the consent of the Chief Minister.

iv) Leader of the Legislative Assembly:
As the Chief Minister enjoys the confidence and support of the majority Legislators he acts as the leader of the Assembly. In that capacity he extends complete co-operation to the Presiding Officers for the smooth conduct of the Business of the House. He ensures discipline .of his party members in the Assembly. The CM helps other Ministers in case they are unable to satisfy the House with their replies or when a situation goes out of control in the Assembly. He announces the Government policies on the floor of the Legislative Assembly.

v) Chief Spokes Person :
The Chief Minister is the chief spokesperson of the Government. He announces the major policies and programs of the State Government. His statements in and outside of the State Legislature will carry much legitimacy and influence in the State. The Members in the State Legislature demand for clarification and statements on particular issues of the State from the Cheif Minister. So, he maintains much restrain without making controversial statements.

vi) Leader of the party in power:
The Chief Minister is the leader of party in power at the State level, he participates in the meetings organized by his party. He informs the party members about the policies and programs initiated by the State Government to fulfill the poll promises of his party. He seeks the co-operation and support of the party members for the effective implementation of the government policies and successful function pf the Government. He brings co-ordination between the party in power and the Government.

If he happens to be the President or the General Secretary of the party he gains control over his party. He utilizes the services of the senior, experienced and prominent party leaders in improving the image and efficiency of the State Government. He sees that his party members do not make controversial and embarrassing comments that may land the executive in the troubled waters. –

vii) Leader of the people :
He tries to know and understand the needs and interests, aspirations and expectations of the people in the State. For this purpose he frequently makes visits to different places and addresses the public gatherings. He invites petitions from the people and patiently listens to them. He informs the people about the welfare measures and developmental programs taken up by the Government. He motivates the people to take active participation in the implementation of various welfare schemes. He undertakes relief „ measures and consoles the people affected during the natural calamities. He maintains good rapport with the people and wins their confidence and trust, as their prominent leader of the people.

viii) Chief Advisor to the Governor :
It is the Constitutional obligation of the CM to render advice to the Governor on all matter of the State Government. His advice is binding over the Governor in the matter of appointment of ministers, allocation of portfolios, reshuffling of the Ministry and accepting the resignation of Ministers. It is a rate privilege and opportunity of the Chief Minister to advise the Governor to dissolve the State Legislative Assembly when he still has majority support of the members in the Assembly.

ix) Cordial relations with the Union Government:
The Chief Minister, being the real head of the. State administration, has the main responsibility of maintaining harmonious relations with the Union Government. He should develop cordial and amicable relations with the Prime Minister and the Union Ministers. He will have to interaction with several Union Ministers particularly of Home, Finance, Industry, Agriculture, Education, and Rural Development etc.

How much Union support a State gets in the form of financial grants to the centrally sponsored schemes depend on the Chief Minister’s influence on and the rapport with the Union Ministers. The State’s representation in the Union Cabinet also influences the quantum and quality of the support to the State.

x) Relations with Party in Opposition :
The Chief Minister maintains good relations with the Presidents, Floor Leaders, and MLAs of the Opposition Parties. Good contacts, healthy relations and cordial approach the Chief Minister in securing constructive co-operation from the Opposition. He takes the Opposition parties into confidence on crucial issues of the State. He organizes all party meetings and takes delegation of all parties to the Union Government for communicating issues of the state.

xi) Related to the Constitution:
The Indian Constitution confers all the powers of real executive on the Chief Minister. He owes his position to the Constitution. He has to exercise his authority and discharge his responsibilities in accordance with the provisions of Constitution. He must uphold the democratic norms and Constitutional principles in running the State administration.

Question 3.
Describe the powers and functions of the State Council of Ministers.
Answer:
Article 163(1) of the Indian Constitution provides for the State Council Ministers with the Chief Minister at its head, to aid and advise the Governor in the exercise of his powers and in running the State administration.

Composition :
The State Council of Ministers is generally a three-tier body. It consisting of.
1. Cabinet Ministers 2. Ministers of State and 3. Deputy Ministers. There will be some only Parliamentary Secretaries in some states on rare occasions. It constitutes the fourth wing of the hierarchy of the Council of Ministers.

Qualifications :

  1. They should be members of either House of the Legislature (if it is bi-cameral)
  2. If the Ministers are not the members of the State Legislature, they should be elected to the State Legislature within six months from the date of assuming their office. Otherwise they cease to hold their office.
  3. They must possess such other qualifications as*is determined by the Parliament from time to time.

Appointment:
All the Ministers are appointed by the governor (Article 164) on the advice and recommendation of the chief minister.

Powers and Functions of the Council of Ministers :
i) Policy Formulation :
The State Council of Ministers formulates policies suitable for the progress of the people and development of the State. It is an intellectual and laborious process. The Cabinet Ministers meet frequently under the leadership of the Chief Minister, discuss thoroughly various matters of the State adihinistration and finalize the policies along with the necessary decisions.

ii) Enactment of Laws :
The State Council of Ministers takes Legislative initiation on different matters of State Government. It is the Council Ministers that drafts and finalizes the public Bill and pilots them in the State Legislature at different stages in order to get them approved by the Legislature. Once the bills are approved by the Legislature, the Council of Ministers advises the Governor to assent them so that they become laws. The Council of Ministers may propose amendment to the existing laws or enactment Of new laws for the administrative convenience.

iii) Provision of Good Administration :
The State Council of Ministers i.e., the real executive is voted to power to provide good administration and promote the well-being of the .people of the State. The chief responsibility of the Council of Ministers is running the administration in accordance with the Constitutional cardinals and democratic doctrines. The.total administrative work is divided into different ministries. Each minister has one or more departments under his control and is responsible for the effective and the transparent administration of such departments. It formulates and implements different developmental programs and welfare schemes.

iv) Co-ordination of Governmental Activities :
The State Council of Ministers is responsible and the authority for coordinating the functions of different government departments. Without proper co-ordination among the departments the success of the State administration cannot be ensured. The Chief Minister guides and takes lead in coordinating the cabinet discussions and government activities.

v) Appointment Power:
The State Council of Ministers plays a key role in all important appointments to various offices in the State. It makes all appointments in the name of the Governor to various higher offices like the Chief secretary. Advocate General, D.G.R Principal Secretaries and other Heads of the Departments etc.

vi) Financial Fucntions :
The State Council of Ministers wields control over the Finances of the State. It determines fiscal policy and deals with the matters concerning the State Revenue, Expenditure, Investment and Audit of Accounts. It prepares the budget proposals of the State Government and places it before the State Legislature for its consideration and approval. It manages the Finances of the State according to the policy and budget as approved by the Legislature. Its role is that of a trustee.

vii) Miscellaneous Functions :
The State Council of Ministers finalizes strategies for the overall development of the State in the sphere of Agriculture, Irrigation, Industry, Transport, Education, Planning, IT etc. It proclaims ordinances in the name of the Governor during the recess of the State Legislature.

Short Answer Questions

Question 1.
Explain any three Powers and Functions of the Governor.
Answer:
1) Legislative Powers and Functions :
Article 168 describes that the Governor is an intregral part of the State Legislature. In that capacity he exercises certain powers and performs functions related to the State Legislature.

  1. The Governor inaugurates the first sessions of the State Legislative Assembly after the general elections are over.
  2. He also addressing the first session of State Legislative Assembly every year i.e. budget session.
  3. He appoints Pro-tem Speaker of the State Legislative Assembly.
  4. He summons and prorogues the sessions of the two houses of the State Legislature.
  5. He addresses the Members of the state legislature and sends messages in relation to the state legislature.
  6. The Governor gives his assent to the bills passed by the State Legislature.
  7. He may return a bill sent by the State Legislature for its reconsideration.
  8. He dissolves the State Legislative Assembly when he feels no party is in a position to form a stable and viable Government and the advice of the Chief Minister.
  9. He may promulgate Ordinances to meet an emergency which require immediate action during the recess of the State Legislature.
  10. He nominates members of Anglo-Indian community to the Legislative Assembly of the state if he feels that community is not represented in the house.
  11. The Governor nominates 1/6 of the total members of the State Legislative Council.

2) Executive Powers and Functions:
The Governor has the following executive powers.

  1. The Governor appoints, the Chief Minister and the members of the Council of Ministers on the advice of the Chief Minister.
  2. He allocates Portfolios among the ministers and reshuffles their portfolios.
  3. He removes the Ministers on the advice of the Chief Minister.
  4. He appoints the Vice-Chancellors of the Universities in the State. He acts as the Chancellor of the universities.
  5. He appoints the Chief Secretary and Advocate General of the State Government.
  6. He appoints the Chairmen and other members of the State Commissions such as a) State Public Service Commission, b) State Election Commission, c) Official Language Commission, d) Commission for Women, e) Minorities Commission, f) Backward Classes Commission and g) SC & ST Commission.
  7. He regulates the postings and transfers of the All India Services personal working in the state.

3) Judicial Powers and Functions :
The Governor also exercises the following judicial powers and functions.

  1. The Governor renders advice to the President of India in the appointment of Chief Justice and other judges of the High Court of the State.
  2. The Governor appoints the Advocate General of the State.
  3. He makes appointments, postings and promotions of the District Judges in consultation with the Chief Justice of High Court of the State.
  4. He also appoints persons to the judicial services of the state (other than the district courts) in consultation with the Chief Justice of High court and State Public Service Commission.
  5. He can grant pardon; retrieve, remit and commute the sentence of any person convicted of any offence against any law of the concerned state.

AP Inter 2nd Year Civics Study Material Chapter 6 State Executive

Question 2.
What are the Discretionary Functions of the Governor?
Answer:
Under Article 163(1) of the constitution the governor has been armed with certain discretionary powers. In the exercise of his discretionary powers, the governor is not bound by the advice of his ministers or even to seek such advice. His actions shall not be called into question on the ground that he ought or ought not to have acted in his discretion.

The discretionary powers of governor are mentioned below.

  1. Playing a decisive role in appointing the new Chief Minister in a situation when single party has a clear majority in the State Legislative Assembly.
  2. Dismissing a Ministry when it refuses to resign even after losing majority support in the House.
  3. Dissolution of the Assembly on the advice of the Chief Minister who lost the majority members support.
  4. Rendering advice to the President for the imposition of the President’s Rule in the State.
  5. Reserving a Bill for the consideration and approval of the President.
  6. Seeking instructions from the President before promulgating ordinance on some important matters.
  7. Sending back a bill passed by the state legislature for its reconsideration, except money bills.
  8. Seeking information from the Chief Minister with regard to the administrative and legislative matters of the State.

Question 3.
What are the Differences between the Governor of a State and the President of India?
Answer:

Governor of a State President of India
1) The Governor is a nominated person. 1) The President is an elected person.
2) He has no security of Tenure. His Tenure depends upon the Pleasure of the President. 2) The President has a fixed tenure of Office of five years in general.
3) The Governor can be removed easily by the President on the advice of the Union Council of Ministers headed the Prime Minister. 3) He can be removed only by the difficult process of Impeachment by the Parliament.
4) The Governor has discretionary power. 4) The President has no discretionary, powers.
5) The Governor does not have Military and diplomatic powers. 5) He has Military and Diplomatic powers.
6) Pardoning power of the Governor is limited. He cannot pardon death sentence and any sentence inflicted by the Martial Court. 6) Pardoning power of the president is absolute. He can pardon even the death sentence and sentence of Martial Court.
7) The Governor does not have emergency powers. He can only suggest for the imposition of President’s Rule. 7) The President can Promulgate orders for the declaration of all the three types of Emergencies.
8) The Governor has no power to remove the Chairman and Members of the State Public Service Commission though he appoints them. 8) The President can remove the Chairman and the Members of the Union Public Service Commission on the grounds stipulated by the constitution.
9) The Governor sometimes may reserve a bill for the consideration of the President. 9) The President need not reserve any bill for the consideration of any other authority before giving his assent.
10) The Governor cannot issue ordinance without instructions from the President on the matters (a) which might affect the powers of the Union (b) affecting powers of the High Court (c) Imposing reasonable restrictions upon Inter¬State Trade or Commerce. 10) The President can promulgate any ordinance on the advice of Council of Ministers of the Union.

Question 4.
What is the position and significance of the Governor in the State?
Answer:
The Constitution of India provides for the Parliamentary System of Government both at Centre and in the States. While the Governor is only a nominal executive, the real executive constitutes the Council of Ministers headed by the Chief Minister. The Constitution has assigned a dual role to the office of a Governor in the Indian federal system. He is the Constitutional head of the State Government as well as the representative of the Union Government.

As the Constitutional head of the state government, he must positively contribute to the progress and development of the State. He has to see that the political and administrative heads of the State Government strive for the promotion of the interests of the people. The Governor has to ensure that the ministers and bureaucrats must observe the constitutional and democratic norms. It is the responsibility of the Governor to see that the affairs of the government are carried on in accordance with Constitutional provisions. The Governor has to maintain close and harmonious relations with the real executive heads of the Union and State Governments.

The Governor is not supposed to run a parallel government in the State. His role is that of a good counselor, mediator and arbitrator than an active politician. He shall abide by the advice of the State Council of Ministers. This does not mean that he should accept all proposals immediately. He can reserve Bills for reconsideration and prevent hasty decisions. Great caution and restrainment must be exercised while reporting to the President under Article 356. Otherwise, his image as guardian of the State Government would tarnished. He should keep himself away from active politics. If he identifies himself with a political party, he cannot inspire the total trust of the people.

Being the representative of the Centre, the Governor has the responsibility of informing through reports whether the State is complying with the directives issued by the Union from time to time. It is his constitutional obligation to inform the Union whether the constitutional machinery is functioning smoothly in the state or not.

The Centre-State relations largely depend upon the action and performance of the Governor. He can make or mar the healthy relations between the Union and the State. The Constitution has given certain discretionary powers to the Governor. If the Governor makes use of these powers sparingly, judiciously and impartially, tensions between the Centre and the States would certainly be reduced. If he acts with bias and at the behest of the Central Government, the tensions between the Centre and State would undoubtedly be enhanced. The role of the Governor in the formation or dissolution of the Ministry or imposing of President’s Rule will have far reaching implications and consequences in the healthy and harmonious Centre-State relations.

Question 5.
Explain any three Powers and Functions of the Chief Minister. [Mar. 16]
Answer:
The following are the three important powers and functions of the Chief Minister.
i) Formation of the Ministry :
The first arid foremost responsibility of the Chief Minister is the formation of Ministry of his choice. The CM has a free hand in the selection and appointment of Ministers. He chooses some members of his party (or coalition partners in the case of a coalition) and recommends their names to the Governor to be appointed as Ministers. He advises the Governor to allocate portfolios among the Ministers.

ii) Leader of the State Council of Ministers :
The Chief Minister is the head of the Council of Ministers. As such he occupies a position of exceptional authority. He is the Chairman of the State Council of Ministers. The Chief Minister decides the time, venue and the agenda of Cabinet meetings. The CM presides over such meetings, discussions are carried under his direction. He guides, directs, controls and co-ordinates the activities of the Ministers.

iii) Link between the Governor and the State Council of Ministers :
The Chief Minister is the principal channel of communication between the Governor and the State Council of Ministers. As part of his Constitutional duty he communicates all the administrative decisions and legislative proposals of the State Council of Ministers to the Governor. It is his responsibility to furnish any information related to the actives of the Ministers as the Governor may call for. No minister shall meet the Governor without the consent of the Chief Minister.

Question 6.
Explain the Composition of the State Council of Ministers.
Answer:
Article 163(1) of the Indian Constitution provides for the State Council Ministers with the Chief Minister at its head, to aid and advise the Governor in the exercise of his powers and in rurining the State administration.

Composition :
The State Council of Ministers is generally a three-tier body. It consisting of.
1. Cabinet Ministers 2. Ministers of State and 3. Deputy Ministers. There will be some only Parliamentary Secretaries in some States on rare occasions. It constitutes the fourth wing of the hierarchy of the Council of Ministers.

i) Cabinent Ministers :
The cabinent is a small body consisting of ministers holding the most important portfolios such as Home, Finance, Planning and Industries etc. They enjoy independence in taking and implementing decisions concerning their ministry. They attend the Cabinet meetings, concerned by the Chief Minister. Some times the Ministers of state and deputy ministers may attend the cabinet meetings, in case their presence is needed during deliberations. They meet frequently and determine the policies of the State Government under the stewardship of the Chief Minister.

ii) Ministers of State :
The Ministers of State hold portfolios of less importance compared to the Cabinent Ministers. They may be attached to the individual Cabinent Ministers or might be given independent charge of crucial departments’in the major Ministries. In such a case they enjoy independence. They are answerable directly to the Chief Minister. They are not subject to the control of Cabinet Ministers.

iii) Deputy Ministers :
The Deputy Ministers are attached to the Cabinet Ministers. They performs such functions which are assigned by the Cabinet Ministers. His role is mainly to relieve the burden of the Cabinet Minister. He assists the Cabinent Minister in the administrative and legislative affairs of the Ministry the Constitution (91st Amendment) Act 2003 fixes a ceiling on the size of the Council of Ministers. The total number of the Ministers cannot be more than 15% of the total strength of the State Legislative Assembly.

AP Inter 2nd Year Civics Study Material Chapter 6 State Executive

Question 7.
Point out any three Powers of the State Council of Ministers. [Mar. 17]
Answer:
The three of the following are the important powers of the state council of Ministers.
i) Policy Formulation :
The State Council of Ministers formulates policies suitable for the progress of the people and development of the State. It is an intellectual and laborious process. The Cabinent Ministers meet frequently under the leadership of the Chief Minister, discuss throughly various matters of the State administration, and finalize the policies along with the necessary decisions.

ii) Enactment of Laws :
The State Council of Ministers takes Legislative initiation on different matters of State Government. It is the Council of Ministers that drafts and finalizes the public Bills and pilots them in the State Legislature at different stages in order to get them approved by the Legislature. Once the bills are approved by the Legislature, the Council of ministers advises the Governor to assent them so that they become laws. The Council of Ministers may propose amendment to the existing laws or enactment of new laws for the administrative convenience.

iii) Provision of Good Administration :
The State Council of Ministers i.e., the real executive is voted to power to provide good administration and promote the wellbeing of the people of the State. The chief responsibility of the Council of Ministers is running the administration in accordance with the Constitutional cardinals and democratic doctrines.

The total administrative work is divided into different ministries. Each minister has one or more departments under his control and is responsible for the effective and the transparent administration of such departments. It formulates and implements different developmental programs and welfare schemes.

Question 8.
Estimate the relationship between the Chief Minister and the Governor.
Answer:
In a Parliamentary Democracy like India the real executive of the state plays a pivotal role in the state administration. The Cheif Minister as the real executive head in the State is responsible ultimately to the state electorate. The Chief Minister has also the obligation to facilitate the exercise of powers of the Governor by providing necessary information about the affairs of the administration of the State. The Governor has a right to seek any information on administrative and legislative activities of the state Council of Ministers through the Chief Minister.

However, this right does not allow permit the Governor to become a parallel centre in this state. It may be noted that the nature of the power available to the Governor is persuasive and not authoritarian. So he cannot under the grab of this right start over riding or vetoing the decisions or proposals of the state Council of Ministers.

The founding fathers of our Constitution have laid great emphasis on the need for harmonious relations between the Governor and his Council of Ministers headed by the Chief Minister. This was the main idea behind abandoning the proposal for an elected Governor and adopting for his nomination by the President.

The Sarkaria Commission in its report emphasized that for the proper working of the Parliamentary system there needs to be a good personnel rapport between the Governor and the Chief Minister of a State. For fostering good personnel relationship, the Sarkaria Commission suggested that the Union Government has to consult the concerned Chief Minister before appointing the Governor of the State.

Pandit Jawaharlal Nehru during his speeches in the Constituent Assembly stated that the Governor should be acceptable to the Chief Minister. Both the Chief Minister and Governor must work together in mutual co-operation to promote the development of the State and safeguard the interests of the people of the State.

Very Short Answer Questions

Question 1.
Qualifications of Governor. [Mar. 18]
Answer:
Article 157 of our Constitution lays down the following qualifications for the appointment of a person as a Governor.

  1. He shall be a citizen of India.
  2. He should have completed the age of 35 years.
  3. He should not be a member of either house of Parliament or state legislature.
  4. He should not hold any other office of profit.
  5. He should not be an insolvent declared by any court of law.

Question 2.
Special responsibilities of the Governor. [Mar. 17]
Answer:
The Governor has certain special responsibilities to discharge according to the directives issued by the President under Articles 371 (Z) 371 (A) (1) b, 371 (C) in case special responsibility through the Governor is to constilt the Council of Ministers the final decision shall be in his individual judgement which no court can question.

The Governor of Assom. Maharashtra, Gujarat, Wougaland, Manipur and Sikkim have special responsibility on specific matters related to their respective states. For example : 1. The Governor of Assom shall in his discretion determine the amount payable by the state of Assom to district council as the royalty accruing from licences of minerals decides the amount of money received from mineral resources and which has to be allocated to the District Council.

Question 3.
State Executive. [Mar. 18]
Answer:
Articles 153 to 167 deal with the matters of the state executive. The state executive consists of (a) the Governor (b) the Chief Minister and (c) Members of the State Council of Ministers. In our Parliamentary system Governor is the titular or constitutional head of the state. The Chief Minister is the real executive head of the Government. The Chief Minister and the Ministers being represents the people.

Question 4.
Immunities of the Governor.
Answer:
Our Constitution provides certain legal Immunities to the office of the Governor to enable him to discharge his constitutional functions in a free and fair manner, to ensure the state government works constitutionally. He shall not be held responsible for any act done or purporting to have been done in his official capacity. No criminal proceedings can be initiated against the Governor during his term of office. No proceedings for his arrest or imprisonment can be taken by any court of law.

Question 5.
Any two executive powers of the Governor.
Answer:

  1. The Governor appoints, the Chief Minister and the members of the Council of Ministers on the advice of the Chief Minister.
  2. He allocates portfolios among the Ministers and reshuffles their portfolios.

Question 6.
Two discretionary powers of the Governor.
Answer:

  1. Playing a decisive role in appointing the new Chief Minister in a situation when no single party has a clear majority in the State Legislative Assembly.
  2. Dismissing a Ministry when it refuses to resign even after losing majority support in the house.

Question 7.
The Chief Minister. [Mar. 17]
Answer:
The Chief Minister is the centre of the real executive authority at the state level. He plays a decisive role and occupies a key position in the State Government. The progress of the people and development of the state largely depends upon the Cabinet, Personality, Preservance and political stature of the Chief Minister.

Question 8.
Cabinet Ministers. [Mar. 16]
Answer:
The Cabinet is a small boy consisting of Ministers holding the most important portfolios such as Home, Finance, Planning and Industries etc. They enjoy independence in taking and implementing decisions concerning their Ministry. They attend the Cabinet meetings, concerned by the Chief Minister. Some times the Ministers of state and Deputy Ministers may attend the Cabinet meetings, incase their presence is needed during deliberations. They met frequently and determine the policies of the State Government under the Stewardship of the Chief Minister.

Question 9.
Composition of the State Council of Ministers. [Mar. 18]
Answer:
The State Council of Ministers is generally a three-tier body. It consisting of.

  1. Cabinet Ministers
  2. Ministers of state and
  3. Deputy Ministers. There will be some only Parliamentary Secretaries in some states on rare occassions. It constitutes the fourth wing of the hierarchy of the Council of Ministers. .

Question 10.
Deputy Ministers.
Answer:
The Deputy Ministers are attached to the Cabinet Ministers. They performs such functions which are assigned by the Cabinet Ministers. His role is mainly to relieve the burden of the Cabinet Minister. He assists the Cabinet Minister in the .administrative and legislative affairs of the Ministry.

AP Inter 2nd Year Civics Study Material Chapter 6 State Executive

Question 11.
State Ministers.
Answer:
The Ministers of State hold portfolios of less importance compared to the Cabinet Ministers. They may be attached to the individual Cabinet Ministers or might be given independent charges of crucial departments in the major minorities. In such a case they enjoy independence. They are answerably directly to the Chief Minister. They are not subject to the control of Cabinet Ministers.

AP Inter 2nd Year Civics Study Material Chapter 5 Union Judiciary

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 5th Lesson Union Judiciary Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 5th Lesson Union Judiciary

Long Answer Questions

Question 1.
Write an essay on the Supreme Court of India.
Answer:
The Supreme Court of India is the Highest Court of Justice in India.

Part V of the Indian constitution from Articles 124 to 147 deals with the composition, Appointment, Qualifications of Judges, powers and Functions of the Supreme Court.

Article 124 provides for the establishment of the Superme Court.

The Supreme Court of India was established in New Delhi on January 26, 1950 with the inauguration of our constitution.

Composition :
The Supreme Court Consists of the Chief Justice and 30 other Judges. There may be some Ad-hoc Judges and retired Judges on temporary basis.

1) Seat of the Supreme Court:
The Head quarters of the Superme Court is situated at New Delhi. The Supreme Court ordinarily shall sits at New Delhi.

  • All general cases are adjudicated by a Division Bench Comprising two or more Judges.
  • Cases involving the constitutional matters are heard by a constitutional Bench consisting Five Judges.
  • For considering special cases larger benches consisting of Five or more than Five Judges are constituted.

2) Appointment:
The judges are appointed by the President. While appointing the Chief Justice, the President consults the retiring Chief Justice and the Prime Minister. He appoints the remaining judges on the advice of the Chief Justice.

3) Qualification :
A judge of the Supreme Court 1) must be a citizen of India. 2) must have worked as a judge of any High Court for atleast 5 years or must have 10 years of experience as an advocate either in the Superme court or any High court or must be a legal expert in the opinion of the President.

4) Salaries :
The Chief Justice gets a monthly salary of Rs. 1,00,000/- and each other judge gets 90,000 per month. They also get allowances. Their salaries should not be reduced to their disadvantage during their term of office.

5) Term of Office :
The judges hold office till they reach 65 years of age. However, they can be removed from office by Parliament through a resolution known as impeachment before the end of their term or they may submit their resignation.

Removal:
A Judge of the Supreme Court can be removed from his position only on the grounds of proved misbehaviour or in capacity by an order of the president after on an address by the Parliament by a Majority of not less than 2/3 votes.

Immunities of Judges :
According to Article 121 of the constitution, No discussion shall take place in the Parliament with respect to the conduct of any Judge of the Supreme Court in the discharge of his duties except upon a motion of impeachment.

Powers and Functions of Supreme Court:
1) Original Jurisdiction :
According to the original Jurisdiction, the Supreme Court hears directly any dispute (a) between the Government of India and one or more States, (b) between the Government of India and any State or States on one side and one or more States on the other (c) between two or more States. The Supreme Court also decides all disputes and doubts regarding the election of the President and Vice President. It protects the Fundamental Rights guaranteed to the citizens by issuing several writs.

2) Appellate jurisdiction:
The Supreme Court is the highest Court of appeal in India. All appeals from all other courts can be heard by Supreme Court. The appellate jurisdiction extends to four types of cases namely civil, criminal, constitutional and special. But, the High Court should give certificate for appealing in the Supreme Court in the first three kinds of cases. But in special cases the certificate of High Court is not required.

3) Advisory Jurisdiction :
The Supreme Court offers it’s advice to the President on those matters of legal or public importance which are referred to him (Art. 143). It’s opinion is purely advisory and not binding on the President. The Supreme Court may refuse to give its opinion to the President. In October 1994 when Dr.S.D.Sharma, the then President of India asked the Supreme Court to give it’s advisory opinion on the Ayodhya issue, the court refused to give it’s opinion.

4) Judicial Review:
The Supreme Court has the power of Judicial Review. It examines the validity of laws passed by the legislatures or the orders issued by the Executive and declares them as ultra vires or unconstitutional if they are against the provisions of the constitution.

5) Court of Record:
The Supreme Court acts as a court of record. All the judgements and interpretations of the Supreme Court are recorded for future references. .

6) Other Powers :
The Supreme Court.

  1. reviews it’s own decisions.
  2. supervises the working of the State High Courts and other Subordinate Courts.
  3. recruits it’s own personnel for its maintenance.
  4. interprets the constitution and acts as it’s guardian.
  5. initiates contempt proceedings against those who criticise or defy it’s judgements etc.

Conclusion :
From the above account it is evident that our Supreme Court unique position in the judicial system of the country.

AP Inter 2nd Year Civics Study Material Chapter 5 Union Judiciary

Question 2.
Explain the Powers and Functions of the Supreme Court of India.
Answer:
The supreme Court of India is the Highest Court of Justice in India. Article 124 of the Indian constitution provides for the establishment of the Supreme Court.

The Supreme Court of India was established in New Delhi on January 26, 1950. The Supreme Court consists of the Chief Justice and 30 other Judges.

Powers and Functions of Supreme Court:
1) Original Jurisdiction :
According to the original jurisdiction, the Supreme Court hears directly any dispute

  • between the Government of India and one or more States.
  • between the Government of India and any State or States on one side and one or more States on the other.
  • between two or more State. The Supreme Court also decides all disputes and doubts regarding the election of the President and Vice President. It protects the Fundamental Rights guaranteed to the citizens by issuing several writs.

2) Appellate Jurisdiction:
The Supreme Court is the highest Court of appeal in India. All appeals from all other courts can be heard by Supreme Court. The appellate jurisdiction extends to four types of cases namely civil, criminal, constitutional and special. But, the High Court should give certificate for appealing in the Supreme Court in the first three kinds of cases. But in special cases the certificate of High Court is not required.

3) Advisory Jurisdiction :
The Supreme Court offers it’s advice to the President on those matters of legal or public importance which are referred to him (Art. 143). It’s opinion is purely advisory and not binding on the President. The Supreme Court may refuse to give its opinion to the President. In October, 1994 when Dr. S.D. Sharma, the then President of India asked the Supreme Court to give it’s advisory opinion on the Ayodhya issue, the court refused to give it’s opinion.

4) Judicial Review:
The Supreme Court has the power of Judicial Review. It examines the validity of laws passed by the legislatures or the orders issued by the Executive and declares them as ultravires or unconstitutional if they are against the provisions of the constitution.

5) Court of Record:
The Supreme Court acts as a court of record. All the judgements and interpretations of the Supreme Court are recorded for future references.

6) Other Powers :
The Supreme Court.

  1. reviews it’s own decisions.
  2. supervises the working of the State High Courts and other Subordinate Courts.
  3. recruits its own personnel for its maintenance.
  4. Interprets the constitution and acts as it’s guardian.
  5. initiates contempt proceedings against those who criticise defy it’s judgements etc.-

7) Review of Judgement :
The Supreme court is empowered to review its own Judgements. It can uphold, modify or nullify its previous judgements. For instance it, while pronouncing its judgement in Golak nathcase vs. Punjab state case in 1967, declared the Parliament has no powers sb amend any of the provisions of fundamental rights of Indian Citizens.

Conclusion :
From the above account it is evident that our Supreme Court enjoys unique position.

Question 3.
Describe Judicial Review.
Answer:
Judicial Review is perhaps the most important power of the Supreme Court. The purpose of Judicial Review is to maintain the supremacy of the Constitution. There is no explicit mention of Judicial Review in the Indian Constitution. The higher courts derive this power from the provisions of Article 13 of the Constitution.

This Article empowers the Supreme Court to validate those laws and executive orders which infringe upon the Fundamental Rights. The makers of our constitution adopted this concept from the American Constitution keeping in view the written nature of the Indian Constitution and federal character of Indian polity.

Judicial Review means the power of the Supreme. Court or High Courts to examine the Constitutional validity of the legislative enactments and executive actions of both Central and State Governments and to declare them ‘null and void’ if found repugnant, of the provisions of the Constitution. As M.V Pylee stated, “Judicial Review is the competence of a court of law to declare the Constitutionality or otherwise of a legislative enactment”.

Article 13 declares all laws that are inconsistent with or direspectful of the fundamental rights, or void to the extent of their inconsistency. Hence, the Supreme Court being responsible for protecting fundamental rights, can declare any legislative act or executive decision that is inconsistent with provisions on fundamental rights as ultra vires or null and void, meaning unconstitutional and inapplicable. Besides, in case of federal relations, the Supreme Court can avail this power if a law is inconsistent with the provisions concerning the distribution of powers between the governments as laid down by the Constitution.

In this context the Supreme Court and High courts reviews legislations on the grounds that (a) they violate fundamental rights or (b) they violate the federal distribution of powers. The Supreme Court’s power of Judicial Review extends to the

  • Laws passed by the union and state legislatures,
  • Executive actions of the union and states,
  • Decisions of the public sector undertakings and
  • Constitutional Amendments. The Supreme Court for the first time utilized this power in 1950 itself by declaring Section 14 of the Preventive Detention Act as unconstitutional.

It may be noted that the Supreme Court of India is prominent in the world by exercising the power to determine the validity of the Constitutional Amendment Acts. However, Judicial Review is inevitable due to the following reasons.

  1. The Supreme Court has to uphold the supremacy of the Constitution.
  2. It has to maintain the federal equilibrium.
  3. It has to protect the fundamentals rights of the citizens.

Apart from the above, the power of Judicial Review is a resultant of the position of the Supreme Court as the; guardian of the Constitution. As such it has the final say in interpretation of the Constitution and by such an interpretation, the Supreme Court has extended its power of Judicial Review to almost all the provisions of the Constitution.

AP Inter 2nd Year Civics Study Material Chapter 5 Union Judiciary

Question 4.
What is Judicial Activism? What are its merits and demerits?
Answer:
Generally, Judicial Activism is construed to be the over willingness of the judiciary to jump into the arena of executive or legislative functions. Judicial Activism, in fact, is not a distinctly separate concept from usual judicial activities. In general parlance, the expression “activism” means “being active”, “doing things with decision” and the expression “activist” should mean “one who favours intensified activities”. In this sense every judge is an activist.

Judicial Activism is a policy making in competition with policy making by legislature and executive. The essence of true Judicial Activism is rendering of decisions which are in tune with the temper and tempo of the times. The nature of Judicial Activisim is that it furthers the cause of social change or articulates concepts like liberty, equality or justice.

Judiciary which is an institution that traditionally confined to responding to cases brought before it, began considering many cases merely on the basis of Newspaper reports and postal complaints received by the court. But most of the cases of Judicial Activism have occurred through Public Interest Litigatiqn in the sphere of Public Health, Child Labour, Environment, Corruption etc. Therefore, the Judicial Activism became the most popular description of the role of Judiciary.
Causes for Judicial Activism :
The following are the causes for the emergence of Judicial Activism in India.

  1. Expansion of the rights of hearing in the administrative process.
  2. Excessive delegation without limitation.
  3. Judicial Review over administration.
  4. Promotion of open government.
  5. Indiscriminate exercise of contempt of power.
  6. Exercise of jurisdiction when non – exist.
  7. Over extending the standard rules of interpretation in its search for socio-economic and educational objectives.
    8) Breakdown of other machinery of the government.

Merits of Judicial Activism :

  1. Judicial Activism has democratized the judicial system by giving, access to the courts not just to individuals but also to groups.
  2. It has enforced executive accountability.
  3. It made an attempt to make the electoral system more free and fair.
  4. It is due to the impact of Judicial Activism during elections the candidates who tender affidavits disclosing their assets, income, educational qualifications, criminal record etc. This enable the people to elect better candidates.

De – Merits:

  1. Judicial Activism has blurred the line of distinction between the executive and legislature on the one hand and the judiciary on the other hand.
  2. Some felt that Judicial Activism led to the worsening of relations and balance among the three organs of government.
  3. Democracy is based on the principle that each organ of government will respect the powers and jurisdiction of others. But Judicial Activism may negate this democratic principle.

Short Answer Questions

Question 1.
Write about the composition of the Supreme Court.
Answer:
The Supreme Court of India is the highest court of Justice in India part V of the India Constitution from Articles 124 to 147 deals with the composition, Appointment, Qualifications of Judges, powers and Functions of the Supreme Court.

The Supreme Court of India was established in New Delhi on January 26, 1950.

Composition :
The Supreme Court consists of the chief Justice of India and such number of other Judges as is provided by the law. The parliament is authorised to determine the number of Judges in the Supreme Court. At present, there are a chief Justice and 30 other Judges in the Supreme Court. There may be some Adhoc Judges and retired Judges on temporary basis in the Supreme Court. All general cases are adjudicated by division Bench comprising two or more judges.

Cases involving the constitutional matters are heard by a constitution bench consisting Five Judges.

For considering special cases larger benches consisting of Five or more than Five Judges are constituted.

Question 3.
Mention any two Jurisdictions of the Supreme Court.
Answer:
1. Original Jurisdiction :
The original jurisdiction of the Supreme Court is purely of federal in nature. This power is confined to disputes between (a) the Government of India and any of the States in India, (b) The Government of India and any State of States on one side and other States on the other side or (c) two or more States. This power exclusively belongs to the Supreme Court and no other court in India is empowered to entertain any such suit. However disputes arising out of any treaty agreement, convenant, engagement etc., do not come under this Jurisdiction unless referred to by the President for advisory opinion. The Supreme Court can directly hear the disputes concerning the election of the President and the Vice-President.

2. Appellate Jurisdiction:
The Supreme Court is the highest court of appeal in India. Its appellate Jurisdiction may be divided into three heads.

  1. Cases involving interpretation of constitution,
  2. Civil cases and.
  3. Criminal cases.

i) The Supreme Court hears cases involving a substantial question of Law as the interpretation of. the constitution. It requires the certificate of the High Court for hearing such a case. Sometimes the Supreme Court can take up the appeal if it is satisfied that the case has do with certain intepretation of the constitution.

ii) In case where no constitutional question is involved, the Supreme Court hears appeals on the basis of a certificate of the High Court. Such cases, in the opinion of High Court involve (a) a substantial question of law and (b) the decision of the Supreme Court,

iii) In case of the criminal matters the Supreme Court hears appeals from any Judgement, whether they are in the form of final order or sentence of the High Court. It hears two specified cases namely.
a) Where the High Court has on an appeal reversed on order of acquittal of an accused and sentenced him to death and b) Where the High Court has tried the appeals from any of its subordinate courts convicted the accused and sentenced him to death.

Question 3.
What are the powers of Appellate Jurisdiction of the Supreme Court?
Answer:
The Supreme Court is the highest court of appeal in India. Its appellate Jurisdiction may be divided into three heads.

  1. Cases involving interpretation of constitution,
  2. Civil cases and
  3. Criminal cases.

i) The Supreme Court hears cases involving a substantial question of Law as to the interpretation of the constitution. It requires the certificate of the High Court for hearing such a case. Sometimes the Supreme Court can take up the appeal if it is satisfied that the ease has do with certain intepretation of the constitution.

ii) In case where no constitutional question is involved, the Supreme Court hears appeals on the basis of the High Court. Such cases, in the opinion of High Court involve (a) a substantial question of law and (b) the decision of the Supreme Court.

iii) In case of the criminal matters the Supreme Court hears appeals from any Judgement, whether they are in the form of final order or sentence of the High Court. It hears two specified cases namely.

a) Where the High Court has on an appeal reversed on order of acquittal of an accused and sentenced him to death and b) Where the High Court has tried the appeals from any of its subordinate courts convicted the accused and sentenced him to death.

The Supreme Court also hears appeals by special leave on any judgement of the High Court when the latter certifies that the case is fit for hearing by the Supreme Court. Besides, the Supreme Court as per Article 136 hears appeals over the cases that remain outside the purview of the ordinary law.

Question 4.
Explain the Advisory Jurisdiction of the Supreme Court.
Answer:
Under Article 143, the Supreme Court has advisory jurisdiction. Accordingly, the supreme Court offers its advice to the President on those matter of. legal or public or constitutional importance, when the President seeks such advice. It also reports its opinion over the disputes referred to it by the President, arising out of any treaty, agreement which was made or executed before the commencement of the Constitution. So far the Supreme Court rendered its advice to the President on eight occasions. The president, in the recent past, sought the advice of the Supreme Court on the ‘Ayodhya Issue’.

These are excluded by Article 131. However, the Supreme Court is not bound to render advice on such matters and the president is not bound to accept such an advice.

One may immediately question about the utility of the advisory powers of the Supreme Court. The utility is twofold. In the first place, it allows the government to seek legal opinion on a matter of importance before taking, action on it. This may prevent unnecessary litigations later. Secondly, in the light of the advice of the Supreme Court, the government can make suitable changes in its action or legislations.

AP Inter 2nd Year Civics Study Material Chapter 5 Union Judiciary

Question 5.
Write about the Writ Jurisdiction.
Answer:
The word ‘writ’ literally means ‘order’ in written form. Article 32 of our Constitution confers authority upon the Supreme Court, to issue a constitutional writ for the enforcement of Fundamental Rights of the citizens. Any person, whose fundamental rights have been violated, can directly move the Supreme Court for remedy. The Supreme Court issues Habeas Corpus, Mandamus, Prohibitions, Quo-Warranto and Certiorari, for enforcing the Fundamental rights.

i) Habeas Corpus :
Literally means ‘to have the body of. It is issued by the court to affect the release of a person who has not been detained legally. Under this writ the court issues orders to the concerned authority to produce the person before-the court. The failure to abide by the writ order is met with punishment for contempt of court.

ii) Mandamus :
It means ‘we command’. The writ is a command issued by the court to a public official to do a duty which he has failed to do. This writ cannot be issued against private persons.

iii) Prohibition :
It means ‘to forbid’. It is issued by a higher court to a lower court to prevent the latter from exceeding its jurisdiction that it does not posses. This writ can be issued only against judicial and quasi-judical authorities.

iv) Certiorari:
It means ‘to be certified’ or ‘to be informed’. This writ is issued against the lower courts by the Supreme Court or High Courts, if the lower former courts violate their jurisdiction.

v) Quo Warranto :
It means ‘by what authority’ or ‘warrant’. If the court finds that a person is holding a public office which is not entitled to hold that office, it issues this writ for restricting that person from acting in that office. This writ is also not issued with respect to the private offices.

Besides the above, many other mechanism have been established for protecting of human rights. The National Commission for Women, the National Commission for Scheduled Castes & Scheduled Tribes, the National Human Rights Commission etc are some examples in this regard.

Question 6.
Describe Judicial Activism in India.
Answer:
It is the collective responsibility of the legislative, executive and the judiciary to accomplish the goals of the Constitution. Social Justice is the prime goal of the Constitution. Judiciary plays a vital role in achieving this goal. In order to meet the basic needs of the poor the oppressed and suppressed classes of the society, the Supreme Court entertains and also encourages the Public Interest Litigation (PIL) in its expanded, role of Judicial Activism.

As a result the apex court has evolved, developed new techinique, discovered and applied new remedies for violation of Fundamental Rights, and attempted to fill the vacuum arising out of executive and legislative inaction.

Due to the negligible attitude of the legislature and lack of edicts from the executive, the vulnerable classes of the society are sometimes denied social justice. In such circumstances, Social Action Groups, Civil Liberties Organizations, Voluntary Organizations etc., have come forward to their rescue through Public Interest Litigation.

As Chief Justice A.S. Ananad remarked that “the expanded concept of Public Interest Litigation by judicial interpretation from time to time has expanded the judicial limits of the courts exercising Judicial Review. This expanded role has been given the title of Judicial Activism by those who are critical of this expanded role of the Judiciary”.

Question 7.
What is meant Public Interest Litigation (PIL). [Mar. 17]
Answer:
The institution of Public Interest Litigation originated in USA during the mid 1960s. PIL or Social Action Litigation is an offshoot of liberalized rules of locus – standi. The traditional rule of locus-standi was based on the fact that judicial remedy can be sought only by those who have suffered an injury on account of violation of legal right by some public authority. The PIL choose liberalize this rule by making it clear that any person who suffer an injury but is unable to reach the court can take help of public minded citizens to reach the court to seek justice.

Public Interest Litigation Movement in India emerged during post-emergency years intending to make the judicial system accessible to the socially and economically lower sections of the society. In most of cases, Judicial Activism has occurred through public interest litigation. In public interest litigation any person or group can approach the Supreme Court and High Court for the redressal of grievances on behalf of the victim or victims who were incapable of approaching the court.

Under this new arrangement, a destitute citizen can file a writ petition even through a simple letter written on the post card. This derives authenticity from the “right be heard” as implied by Article 32 of the Constitution. But the court has to ensure that the petitioner who approaches the court with PIL, is acting bona- fide and not for personal gains private profit, political or other oblique considerations. The court should not allow this process to be abused by politicians and others to delay legitimate administrative action or to gain a political objective.

Question 8.
What is the Meaning of Independence of Judiciary? How is it ensured by the constitution?
Answer:
The Judiciary performs its Functions independently. The Legislature or the executive shall not interfere in the working of the Judiciary. The Judiciary carries on its obligations according to the constitutional norms and democratic principles.

The following measures have been taken by the constitution to ensure the independence of Judiciary in India.

Measures ensuring for Independence of Judiciary
1. The Legislature is not involved in the process of appointment of judges. Thus, it is believed that party politics would not play a role in the process of appoinments. In order to be appointed as judge, a person must have experience as a advocate and / or must be well versed in law. Political opinion of the person or his/her political loyalty should not be the criteria for appointments to judiciary.

2. The judge have a fixed tenure. They hold office till reaching the age of retirement. Only in exceptional cases judges may be removed as per the procedure prescribed in the Constitution. This measure ensures that judge could function without fear % or favor.

3. The Judiciary is not financially dependent on either the executive or legislature. The Constitution provides that the salaries and allowances of the judges are not subjected to the approval of the legislature.

4. The actions and decisions of the judges are immune from personal criticism. The judiciary has the power to penalize those who are found guilty of contempt of court. This authority of the court is seen as an effective protection to the judges from unfair criticism.

5. Judiciary in India is neither a branch Of the executive nor a hand-made of the legislature. It has an independent identity under the Constitution.

6. Our Constitution prescribes specific and high qualifications for the judges. Thus, only those persons who have specific qualifications and experience as prescribed by the Constitution can be appointed as Judges of Supreme Court.

7. Security of the service for the Judges is an essential quality for securing the independence of Judiciary. No Judge can be removed from the office except by impeachment and only on the grounds of proven misbehavior or incapacity.

8. The Judiciary in India enjoys the vast jurisdiction. It is no way subordinate to the other organs of the government. Its decisions bind all. Such a powerful position helps the Judiciary to maintain its independence.

AP Inter 2nd Year Civics Study Material Chapter 5 Union Judiciary

Question 9.
What are the Power and Functions of the Attorney-General of India? [Mar. 18, 16]
Answer:
Article 76 of our Constitution provided for the office of the Attorney General of India. The Attorney General is the highest law officer of the Union Government. He is appointed by the President. He holds the office during the pleasure of the President. He is entitled to all privileges and immunities allowed to a Member of Parliament. When he attends sessions of the House, he occupies a seat on the treasury government benches.

Qualifications :
The Attorney General of India Possess the same qualifications that are necessary for a judge of the Supreme Court. They are as follows.

  1. He must be a citizen of India.
  2. He must have served as a judge in some High Court for a period of at least five years.
  3. He must have served as an advocate in some High Court for a period of at least ten years.
  4. He must be a distinguished jurist in the opinion of the President.

Pay and Allowances :
The Attorney General of India is paid not a salary but a remuneration that is determined by the President. The remuneration of Attorney General is equal to salary of a judge of the Supreme Court.

Removal:
He may quit his office by submitting his resignation to the President. He can be removed by the President in case a special address is passed charging him with ‘proved misbehavior’ or ‘incapacity’ by each House of Parliament with its absolute majority and with two-thirds majority of the members present and voting.

Powers and functions :
The Constitution assigned some specific powers and functions to the Attorney General of India. They are mentioned as follows :

  1. The Attorney General of India render advice to the Union Government upon such legal matters which are referred to him by the President.
  2. He performs such other functions of legal character that are assigned to him by the President from time to time.
  3. He discharges the Functions conferred on him by the constitution or any other Laws.
  4. He appears in any court of Law on behalf of the union government in all. cases.
  5. He represents the government in any reference made by the president to the Supreme Court.
  6. He appears in any High, Court on behalf of the union government.

Very Short Answer Questions

Question 1.
Qualifications of Judges of Supreme Court.
Answer:
A person to be appointed as a judge of the supreme court shall possess the following qualifications :

  1. He should be a citizen of India.
  2. He should have continuously worked as a judge in one or more High Courts at least for a period of 5 years.
  3. He should have continuously worked as an advocate of one or more High Courts for not less than 10 years or
  4. He should be a distinguished jurist in the opinion of the president of India.

Question 2.
Removal of the Judges of Supreme Court.
Answer:
A judge of the Supreme Court can be removed from his position only on the grounds of proved misbehavior or incapacity. He can be removed from his office by an order of the president, after an address from each house of Parliament, supported by a majority of the total membership of that house and by a majority of not less than 2/3 notes of the members present and voting passed.

Question 3.
Judicial Review. [Mar. 16]
Answer:
Judicial Review means the power of the Supreme Court or High Court to Examine the constitutional validity of the legislative enactments and executive actions of both central and state governments and to declare them ‘null and void’ if found repugnant of the provisions of the constitution.

Question 4.
Court of Record.
Answer:
According to Article 141, Supreme Court acts as court of Record. Being the highest court of the land, its proceedings acts and judgements are kept in record for perpetual memory and further verification and reference.

Question 5.
Judicial Activism.
Answer:
Generally, Judicial Activism is construed to be the over willingness of the judiciary to jump into the arena of executive or legislative functions. In general parlance, the expression “activism” means “being active”, “doing things with decision” and the expression “activist” should mean “one who favours intensified activities”. Judicial Activism is a policy making in competition with policy making by legislature and executive.

The essence of true Judicial Activism is rendering of decisions which are in tune with the temper and tempo of the times. The nature of Judicial Activism is that it furthers the cause of social change or articulates concepts like liberty equality or justice.

Question 6.
PIL.
Answer:
In Public Interest Litigation any person or group can approach the Supreme Court and High Court for the redressal of grievances on behalf of the victim or victims who were incapable of approaching the court. Under this new arrangements a destitute citizen can file a writ petition even through a simple letter written on the post card.

Question 7.
Independence of Judiciary.
Answer:
The Judiciary perform its functions independently. The legislature or the executive shall not interfere in the working of the judiciary. The Judiciary carries on its obligations accordingly to the constitutional norms and democratic principles.

Question 8.
Habeas corpus.
Answer:
Literally means ’to have the body of. It is issued by the court to affect the release of a person who has not been detained legally. Under this writ the court issues orders to the concerned authority to produce the person before the court. The failure to abide by the writ order is. met with punishment for contempt of court.

Question 9.
Seat of the Supreme Court. [Mar. 18, 17]
Answer:
The Supreme Court ordinarily shall sit at New Delhi. The Supreme Court of India was inaugurated on January 28, 1950. All general cases are adjudicated by a division Bench comprising two or more judges. Cases involving the constitutional matters are heard by a constitution bench consisting five judges. For considering special causes larger benches consisting of five or more than five judges are constituted.

AP Inter 2nd Year Civics Study Material Chapter 5 Union Judiciary

Question 10.
Writs.
Answer:
The word ‘writ’ literally means ‘order’ in written form. Article 32 of our constitution, confers authority upon the supreme court, to issue a constitutional writ for the enforcement of fundamental rights of the citizens. Any person whose fundamental rights have been violated, can directly move the supreme court for remedy. The Supreme Court issues harbeas corpus mapdomus prohibition quo-warranto and certiorari for enforcing the fundamental rights.

AP Inter 2nd Year Civics Study Material Chapter 4 Union Legislature

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 4th Lesson Union Legislature Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 4th Lesson Union Legislature

Long Answer Questions

Question 1.
Describe the powers and functions of the speaker. [Mar. 18]
Answer:
Articles 93 to 97 of the Indian Constitution deal with the office of the Speaker of the Lok Sabha. The Speaker acts as head of the Lok Sabha, guardian of members, and principal spokesman of the house. He enjoys Supreme Authority and power on the floor of the house.

Election :
The members of the Lok Sabha elect the Speaker from among themselves. According to the Parliamentary convention the speaker is unanimously elected or chosen by the members on the request of the Prime Minister.

A person elected as the speaker must be a member of the Lok Sabha.

Tenure:
The speaker continues in office for five years. Though the Lok Sabha is dissolved the speaker continues in office until the new Lok Sabha elects its speaker. (Article 94).

Removal:
The speaker can be removed from office by a majority members, resolution, proceeded by a 14-day prior notice to that effect.

Salary and. allowances :
At present, the speaker receives a monthly salary of ₹ 1,40,000. Besides he is provided with rent free accommodation, Medical, travelling, and telephone facilities.

Powers and functions of the speaker :

  1. The speaker presides over the meetings of the Lok Sabha. He conducts the meetings with dignity, order and efficiency.
  2. He allots time to the members to express their views on the bills, conducts voting if necessary and announces the results.
  3. He sends bills to the Rajya Sabha after they are approved by the Lok Sabha. On the receipt of the Bills from the Rajya Sabha, he certifies and sends them to the President of India for his consent.
  4. He acts as the representative of the Lok Sabha. He sends messages and directives to the members on behalf of the Lok Sabha.
  5. He takes steps for safeguarding the rights and privileges of the members and for upholding the respect of the house.
  6. He has the privilege of determining whether a bill is money bill or not.
  7. He accords permission to the members for introducing various bills in the house. He gives his signature on the bill approved by the house.
  8. He is empowered to permit the members to move a No-confidence motion against the government, postpone the meetings of the house and decide the Quorum in the house.
  9. He constitutes various house committees and appoints their chairpersons.
  10. He presides over the joint session of the Parliament.
  11. He exercises his casting vote in case of a tie over a bill. ,
  12. He conducts the election of the Deputy Speaker in case of a vacancy.

AP Inter 2nd Year Civics Study Material Chapter 4 Union Legislature

Question 2.
Explain the powers and functions of the Union Legislature. [Mar. 16]
(Or)
Write about the composition powers and functions of the Indian Parliament.
Answer:
Introduction :
The Union Legislature (Parliament) is the highest legislative organ of the Union government. Articles 79 to 129 in part V of Indian Constitution deals with the composition, organization, powers, and functions of the Indian Union Legislature.

Composition :
Indian Parliament consists of the

  1. President
  2. Rajya Sabha (Council of states)
  3. Lok Sabha (House of people)

The upper house Rajya Sabha represents the states and union territories.

The lower house Lok Sabha represents the people.

The President of India has the power to summon or prorogue the two houses of Parliament though he is not a member of either house.

He can dissolve the Lok Sabha on the advice of the Union Council of Ministers headed by the Prime Minister.

Powers and functions of the Union Legislature (or) Indian Parliament :
The Parliament enjoys extensive powers and performs variety of functions. These powers and functions are under the following points.

1) Legislative Powers and Functions:
The main function of the Indian Parliament is law making. It makes laws on all the subjects mentioned in the Union List and Concurrent List. Under certain circumstances it also makes laws on the subjects mentioned in the State List. Further, it also makes laws on the matters that are not included in any of the three Lists i.e., on residuary matters.

2) Executive Powers and Functions :
Another important function of the Indian Parliament is controlling the Executive (Union Council of Ministers). Parliament controls the Executive through various ways, such as by asking questions, supplementary questions, and by introducing adjournment motions and no confidence resolutions against the Ministry. Hence the survival of the Government depends upon the will of the members in the Lower House. The executive remains in office so long as it enjoys the confidence of the Lok Sabha.

3) Financial Powers and Functions :
The Parliament controls the financial resources of the nation. It accepts the budget and other money bills required by the government. Its permission is needed for the government for imposing and collecting tax and for revising the existing tax rates. In this regard the Lok Sabha has more financial powers than the Rajya Sabha. All money, bills shall at first be introduced in the Lok Sabha. The Rajya Sabha has to accept all money bills sent by the Lok Sabha within 14 days. It approves the railway budget, appropriation bill, and other money bills.

4) Judicial Powers and Functions :
The Parliament has certain judicial powers and functions. It has the power to remove the President and Vice President. The procedure is called impeachment. It has also the power to recommend to the President the removal of the higher officials of the country such as the Chief justice and Judges of Supreme Court, High Court and the Chairman, and other members of U.P.S.C., Chief Election Commissioner etc., for violation of certain principles.

5) Constitutional Powers and Functions:
The Parliament takes initiative for changing the provisions of the Constitution according to the changing times. Bills relating to the Constitution amendments may be introduced in either House. The State legislatures also join with the Parliament in accepting certain important Constitutional amendment bills. There are three methods of amending the Constitution.

6) Electoral Powers:
The Parliament also serves as an electoral college. It participates in the election of the President and Vice President. The Speaker and Deputy Speaker who act as the presiding officers are elected by the members of Lok Sabha. The Deputy Chairman is elected by the members of the Rajya Sabha.

7) Deliberative Powers and Functions :
The Parliament acts as the highest forum and direct agency of public opinion. Its members discuss various issues of national and international significance. They demand the government to solve the people’s problems.

8) Miscellaneous Powers :
In addition to the above, the Indian Parliament has also the power to a) create or abolish Legislative Councils b) change the names and boundaries of the States, etc.

Conclusion:
A look at the powers and functions of the Indian Parliament it is the centre of legislative activity and political activity of our country.

Question 3.
Examine the role of financial committees in Parliament.
Answer:
The Financial committees of Parliament endeavour of undertake the task of detailed scrutiny of governmental spending and performance, there by securing the accountability of the administration to the Parliament in financial matters.
There are three financial committees in Indian Parliament. They are :

  1. Public Accounts committee
  2. Estimates committee and
  3. Committee on public undertakings

1) Public Accounts Committee :
Public Accounts Committee was set up on 1921. It consists of 22 members. Out of them, 15 members belong to the Lok Sabha and 7 members to the Rajya Sabha. Their term of office is one year. They are elected by means of proportional representation and single transferable vote. The speaker nominates one of the members as the chairman of the committee. It became a convention to appoint one of the members of the opposition in the Lok Sabha as its chairman since 1967-68. The Committee performs the following functions :

  1. The Committee examines the annual audit reports of the Comptroller and Auditor General (CAG) of India.
  2. It examines Public Expenditure not only from legal and formal point of view to discover technical irregularities but also from the point of view of economy prudence, wisdom, and property.
  3. It brings out the appropriation accounts and the finance accounts of the Union Government and any other accounts laid before the Lok Sabha.
  4. It examines whether the public funds are disbursed properly.
  5. It examines the accounts of Autonomous and Semi-Autonomous bodies, the audit of which is conducted by the CAG.
  6. It investigates the money spent on any service during the financial year in the excess of the amount granted by the Lok Sabha for that purpose.

The Comptroller and Auditor General renders assistance to this committee. The members of this committee carry out a country-wide tour and meet the concerned officers non-officials, people and receive petitions from them and it submits a final report to the Speaker of the Lok Sabha.

2) Estimates Committee :
The origin of this committee can be traced to the Standing Financial Committee setup in 1921. The Estimates Committee was at first constituted in April 1950 in free India. It consists of 30 members of the Lok Sabha. The Rajya Sabha has no representation in this Committee. These members are elected by the Lok Sabha every year from among its own members according to the principle of proportional representation by means of single transferrable vote. The members hold their office for a year.

The Speaker ha appoints the Chairman of the Committee. If the Duty Speaker is a member ttee, the Speaker appoints him as the Chairman of the Committee. One third of the total members belong to newly elected members. The Chairman of the Committee will be appointed invariably from the ruling party. Mr. M. Ananthasayanam Ayyangar acted the first chairman of this committee in the first Lok Sabha.

The members of this-committee may visit different projects and hold discussions with the officers, non-officials, business groups and receive suggestions from them. The committee functions on permanent basis.

The committee performs the following four important functions :

  1. It offers suggestions in regard to the economy in expenditure, improvement in organization and efficiency of the Union Government.
  2. It examines as to whether the public funds are disbursed as per the estimates.
  3. It also examines the matters assigned to it by the Speaker of the Lok Sabha.
  4. It examines whether the money is well laid out within the, limits of the policy implied in the estimates. Hence, it has been described as a ‘continuous economy committee’.

3) Committee on Public Undertakings :
The Committee on Public Undertakings was created in 1964 on the recommendations of Krishan Menon Committee. It consists of 22 members out of which 15 are from Lok Sabha and 7 from Rajya Sabha. The members of the committee are elected by the Parlianient every year from amongst its own members according to the principle of proportional representation by means of single transferable vote.

The purpose of the committee is to lighten the burden of Public Accounts Committee. The Chairman of the committee is appointed by the speaker from amongst its members who are drawn from the Lok Sabha only. The members of Rajya Sabha are not be appointed as a Chairman of the committee. The functions of the committee are :

  1. To examine the reports and accounts of Public Undertakings.
  2. To examine the reports of the Comptroller and Audit of General on undertakings.
  3. To examine whether the affairs of the public undertakings are being managed in accordance with sound business principles and practices.
  4. To exercise such other functions vested in the Public Accounts Committee and the Estimates Committee.

Short Answer Questions

Question 1.
Write about the composition of the Lok Sabha.
Answer:
The Lok Sabha or the House of the people is the lower house in Indian Parliament. Maximum strength of the Lok Sabha envisaged by the constitution is now 552 (530 members to represent states, 20 to represent union territories and 2 members of Anglo-Indian community, to be nominated by the President).

At present there are 545 members in the Lok Sabha out of them,

  • 530 members are elected from the states.
  • 13 members are elected from the union territories and the remaining.
  • 02 members are nominated by the President from the Anglo Indian community.

Out of 543 elected seats, 79 seats are reserved for the scheduled castes and 41 for the scheduled tribes.

The election is through direct franchise.
The tenure of the Lok Sabha is normally 5 years.

A person who wishes to contest as a candidate for the membership of the Lok Sabha must
A) Be an Indian citizen.
B) Have completed 25 years of age.
C) Not hold any office of profit in union, state or local governments.
D) Possess such other qualifications as prescribed by the Parliament.

AP Inter 2nd Year Civics Study Material Chapter 4 Union Legislature

Question 2.
Explain the election of the speaker of the Lok Sabha.
Answer:
The members of the Lok Sabha elect the Speaker from among themselves. According to the Parliamentary convention, the speaker is unanimously elected or chosen by the members on the request of the Prime Minister.

When no single party secures majority or when a coalition Ministry is formed, the coalition Ministry is formed, the coalition partners will make efforts for deciding the candidature for the office of the speaker. Sometimes coalition partners may hand over that office to a candidate selected by the parties that declare support from outside. A person elected as the speaker must be a member of the Lok Sabha.

Question 3.
What do you know about the composition and qualifications of members . of the Rajya Sabha?
Answer:
The Rajya Sabfya is the upper chamber in Indian Parliament. Article 80 stipulates that the Rajya Sabha shall consist of
A) 12 members nominated by the President and
B) Not more than 238 representatives of the states and of the union territories.

Thus its maximum membership shall be 250 only.
At present there are 245 members in the Rajya Sabha of them

  • 229 members belong to the elected from the 29 states.
  • 3 members belong to the National capital territory of Delhi.
  • 1 member represents the union territory of Pondicherry and the remaining.
  • 12 members nominated by the President having practical experience in respect of matters such as literature, science, arts and social service.

The members are elected in accordance with the proportional representation by means of single transferable vote system.

Qualifications :

  1. He shall be a citizen of India.
  2. He shall have completed 30 years of age.
  3. He should not hold any office of profit under union, state or local government.
  4. He shall possess such other qualifications as prescribed by the Parliament.

The Raja Sabha is a permanent house of which l/3rd of its total members shall retire for every two years.

Question 4.
Write a note on the chairman and Deputy chairman of the Rajya Sabha.
Answer:
Chairman of Rajya Sabha:
The Presiding Officer of Rajya Sabha is popularly known as the Chairman. The Vice President of India acts as the Ex-Chairman of the Rajya Sabha. He is not a member of the House. The members of Parliament elected him for every five years as the Vice-President of India. It implies that both the members of the Lok Sabha and the Rajya Sabha cast their vote in the Vice-Presidential election.

At present he is paid ₹ 1,40,000/- towards monthly salary and allowances. His salaries and allowances are charged on the Consolidated Fund of India. The Chairman of Rajya Sabha vacates his office only if he is removed from the office of the Vice President.

Deputy Chairman:
The Deputy Chairman of Rajya Sabha is elected by the members of the Rajya Sabha amongst the members. The Deputy Chairman receives a monthly salary of ₹ 90,000/-. In the absence of the Chairman, the Deputy Chairman acts as Chairman and presides over the meetings of the Rajya Sabha. Whenver the office of the Deputy Chairman falls vacant, the members of Rajya Sabha will elect another member to fill the vacancy.

Question 5.
Mention any three powers and functions of Indian Parliament.
Answer:
The Indian Parliament, the law body in our country, has extensive powers and performs a variety of functions. There are as follows : ‘

1) Legislative Powers :
The main function of the Indian Parliament is law making. It makes laws on all the subjects mentioned in the Union List and Concurrent List. Under certain circumstances it also makes laws on the subjects mentioned in the State List. Further, it also makes laws on the matters that are not included in any of the three Lists i.e., on residuary matters.

2) Executive Powers :
Another important function of the Indian Parliament is controlling the Executive (Union Council of Ministers) . The members exercise control over the Executive by asking questions, supplementary questions, and by introducing adjournment motions and no-confidence resolutions against the ministry. The ministers are collectively responsible for their actions, to the Lower House of the Parliament i.e., Lok Sabha. They will be in office as long as they enjoy the confidence of the majority of members in the Lok Sabha.

3) Financial Powers :
The Parliament controls the financial resources of the nation. It accepts the budget and other money bills required by the government. It’s permission is needed for imposing and collecting taxes and for revising the existing tax rates. In this regard the Lok Sabha has more financial powers than Rajya Sabha.

Question 6.
Write a note on the types of bills.
Answer:
A Bill is a proposed law under consideration by a legislature. A bill does not become law until it is passed by the legislature. Once a bill has been enacted into law, it is called on Act or statute.

Bills introduced in the Parliament are of two typek 1) Public bill (Government bill) and 2) Private bills (Private member’s bills) the public bills are introduced by the Ministers in the Parliament whereas private bills are introduced by any member of Parliament other than a Minister. The bills introduced in the Parliament may also be classified into four categories.

1) Ordinary bill:
Ordinary bills are concerned with any matter other than financial subjects.

2) Money bill:
Money bills are concerned with the financial matters like taxation, public expenditure etc.

3) Finance bill:
The financial bills are also concerned with financial matters but are different from money bills. These bills deal with fiscal matters i.e., revenue of, expenditure.

4) Constitutional Amendment Bill: Constitutional Amendment bills are concerned with the provisions of the constitution.

AP Inter 2nd Year Civics Study Material Chapter 4 Union Legislature

Question 7.
Explain briefly about the stages of law-making procedure in Indian Parliament.
Answer:
Law making is an important function of Indian Parliament. The Parliament has the power to pass all acts. A bill becomes an act after it receives the assent of the President. Every bill has to pass through different stages as described below.

1) First Reading:
A bill may be introduced by any member of Parliament. One has to ask for the leave of the House to introduce a bill. The title of the bill is to be read out. If the bill is voted for it is deemed to have been read first time and is published in the Gazette of India.

2) Second Reading:
The printed copies of the bill are distributed to all the members at this stage. It may be moved that the bill be referred to a select committee or that the bill be circulated for public opinion. There will be a general discussion on the main principles of the bill at this stage.

3) Committee Stage :
If the House approves the principles, the bill is referred to and examined by the select committee. Then it is thoroughly discussed clause by clause.

4) Report Stage :
The report of the committee with suggestions is presented to the House. If the House agrees to consider the bill as reported by the select committee the bill is taken up for clause by clause discussions and members may move amendments.

5) Third Reading:
The bill enters for the third reading. If it is approved by the House, only oral amendments are allowed at this stage. If the House accepts the bill it is deemed to have been passed by the House.

6) Consideration by the other House :
When the bill is passed by the House, it is sent to the other House for consideration. The procedure in one House is repeated in the other House. If the second House disagrees, a joint sitting is arranged to resolve the differences.

7) Assent by President:
If the bill is passed by both the Houses of Parliament, it is sent to the President for his assent. After the bill is given assent, it becomes an act which will be implemented by the Executive.

The President sometimes send a bill passed by the Parliament for reconsideration. The suggestions sent by the President along with the bill have to be taken up for discussion by the Parliament immediately. If the bill is passed second time by the Parliament, then the President has to give his assent to the bill.

Question 8.
What do you know about Public Accounts Committee.
Answer:
Public Account’s Committee was set up in 1921. It consists of 22 members. Out of them 15 members belong to the Lok Sabha and 7 members to the Rajya Sabha. Their term of office is one year. They are elected by means of proportional representation and single transferable vote. It became a convention to appoint one of the members of the Opposition in the Lok Sabha as its chairman since 1967-68. The Committee performs the following functions :

  1. The Committee examines the annual audit reports of the Comptroller and Auditor General (CAG) of India.
  2. It examines Public Expenditure not only from legal and formal point of view to discover technical irregularities but also from the point of view of economy prudence, wisdom and property.
  3. It brings out the appropriation accounts and the finance accounts of the Union Government and any other accounts laid before the Lok Sabha.
  4. It examines whether the public funds are disbursed properly.
  5. It examines the accounts of Autonomous and Semi-Autonomous bodies, the audit of which is conducted by the CAG.
  6. It investigates the money spent on any service during the financial year in the excess of the amount granted by the Lok Sabha for that purpose.

Question 9.
Describe the composition and functions of Estimates Committee.
Answer:
The Estimates Committee was setup in 1950. It consists of 30 members of the Lok Sabha. The members hold their office for a year. The Speaker of the Lok Sabha appoints the Chairman of the Committee. If the Duty Speaker is a member of this Committee, the

Speaker appoints him as the Chairman of the Committee. The members of this committee may visit different projects and hold discussions with the officers, non-officials, business groups and receive suggestions from them. The committee functions on permanent basis.

The committee performs the following four important functions :

  1. It offers suggestions in regard to the economy in expenditure, improvement in organization and efficiency of the Union Government.
  2. It examines as to whether the public funds are disbursed as per the estimates.
  3. It also examines the matters assigned to it by the Speaker of the Lok Sabha.
  4. It examines whether the money is well laid out within the limits of the policy implied in the estimates. Hence, it has been described as a “continuous economy committee”.

Very Short Answer Questions

Question 1.
Composition of Indian Parliament.
Answer:
Indian Parliament consists of the i) President ii) Rajya Sabha (Council of states ) iii) Lok Sabha (House of people)
The upper house, Rajya Sabha represents the states and union territories.

The lower house, Lok Sabha represents the people.

The President of India has the power to summon or prorogue the two houses of Parliament though he is not a member of either house.

He can dissolve the Lok Sabha on the advice of the Union Council of Ministers headed by the Prime Minister :

Question 2.
Qualifications of Rajya Sabha member.
Answer:

  1. He shall be a citizen of India.
  2. He shall have completed 30 years of age.
  3. He should not hold any office of profit under union, state or local government.
  4. He shall possess such other qualifications as prescribed by the Parliament.

Question 3.
Quorum of Lok Sabha. [Mar. 18, 16]
Answer:
Quorum implies minimum attendance of members required for conducting the meetings of the Lok Sabha. Quorum is fixed at l/10th of the total membership. The speaker determines whether there is Quorum on a particular day for conducting the meetings.

Whenever there is no Quorum, he postpones the meetings for an hour or two or for the next day. There are several instances where in the meetings of the Lok Sabha were deffered due to lack of Quorum.

AP Inter 2nd Year Civics Study Material Chapter 4 Union Legislature

Question 4.
The Speaker of Lok Sabha.
Answer:
The office of the Speaker has great dignity, honour and authority. As the speaker acts as the chairman of the Lok Sabha and as the member of the Lok Sabha represents the people directly. He represents the whole Nation. He acts as the friend, philosopher and guide to the members.

Question 5.
Deputy Speaker of Lok Sabha.
Answer:
There will be a Deputy Speaker for conducting the proceedings of the Lok Sabha in the absence of the Speaker. The Deputy Speaker is elected by the members of the Lok Sabha from among themselves. The Deputy Speaker while acting as the presiding officer, enjoys all the powers and priveleges of the Speaker.

Question 6.
Committee on Public Undertakings. [Mar. 17]
Answer:
The Committee on Public Undertakings was created in 1964. It consists of 22 members out of which 15 are from Lok Sabha and 7 from Rajya Sabha. It examines whether the autonomy and efficiency of public sector undertakings are being managed in accordance with Sound business principles and prudent commercial practices.

Question 7.
Panel of Speakers.
Answer:
The Speaker nominates some of the members of the Lok Sabha as panel speakers. Maximum strength of panel chairpersons will be 10. If both the Speaker and Deputy Speaker are absent at particular time, one of the members from the panel of chairpersons will act as the Speaker. ,

Question 8.
Pro-tem Speaker.
Answer:
The President appoints the pro-tem Speaker for presiding over the meeting of the] first session of the Parliament after general elections. The pro-tem Speaker administers the oath of office on the elected members. Election to the office of the Speaker is held later. Pro-tem Speaker post is dissolved soon after the election of the new Speaker.

Question 9.
Question Hour. [Mar. 18, 16]
Answer:
In both houses of Parliament first hour is allotted to question hour. The members, by giving notice to the presiding officer, can ask questions pertaining to public issues, administrative inefficiency or about the role of the government.

Question 10.
Adjournment Motion.
Answer:
Adjournment Motion is tabled in the Parliament to draw attention of the house to a definite matter of urgent public importance and needs the support of 50 members to be admitted. If any member wants to introduce adjournment motion he should give in writing to the speaker, the Minister concerned and the Secretary General of Parliament before 10 A.M. on that day.

Question 11.
Whip [Mar. 17]
Answer:
Every political party whether ruling or opposition has its own whip in the Parliament. He is appointed by the concerned party to serve as an assistant floor leader. He is charged with the responsibility of ensuring the attendance of his party members in large numbers. He regulates and monitors their behaviour in the Parliament. The members are supposed to follow the directives given by whip, otherwise, disciplinary action can be taken against those members.

AP Inter 2nd Year Civics Study Material Chapter 4 Union Legislature

Question 12.
No Confidence Motion.
Answer:
According to Article 75 of the constitution, No Confidence Motion can be tabled in the Parliament when the Cabinet behaves in an irresponsible manner and if the ruling party does not enjoy majority. No confidence motion is introduced by the opposition parties through written notice supported by atleast 50 members and there will be a discussion on the motion. After the discussion there will be a voting. If the No-Confidence Motion is passed or approved In the house the cabinet has to resign.

AP Inter 2nd Year Civics Study Material Chapter 3 Union Executive

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 3rd Lesson Union Executive Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 3rd Lesson Union Executive

Long Answer Questions

Question 1.
Explain the powers and functions of the President of India.
Answer:
Introduction:
The President of India is the constitutional head of the Indian Republic. He is the First citizen of India. He administers the affairs of the union Government either himself or through the officers subordinate to him. (Articles 52 and 53)

Qualifications :
A person to be eligible to contest the office of the President shall possess the following qualifications :

  1. He should be a citizen of India.
  2. He should have completed the age of 35 years.
  3. He should be qualified for election as a member of the Lok Sabha.
  4. He should not hold any office of profit under the union, state or local Governments (Article 59 (i))
  5. Possess such other qualifications as prescribed by the Parliament.

Election Procedure :
The President of India shall be elected indirectly by an electoral college consisting of the elected members of both Houses of Parliament. State Legislative Assemblies and elected members of Delhi and Pondicheri. The election is held in accordance with the system of proportional representation by means of a single transferable vote system and secret ballot.

Oath of office :
The person who is elected as President assumes office only after he takes oath of office and secrecy by the Chief Justice of India.

Term of office :
The President continues in office for five years from the date of his assumption of office.

Salary and Allowances :
The President now gets a monthly salary of ₹ 1,50,000/-. His official residence is Rashtrapathi Bhavan at New Delhi. On retirement, he will get a monthly pension of ₹ 75,000/-.

Removal (or) Impeachment :
The President can be removed from the office by a process of impeachment for violation of the constitution. Impeachment is a quasi-judicial procedure adopted by the Parliament.

Powers and Functions :
The President shall exercise his powers with the help of the Council of Ministers headed by the Prime Minister. His powers may be analysed under the following heads :

1. Executive Powers :
An executive action of the Union Government shall be expressed in the name of the President. The President appoints the Prime Minister and other Ministers, Attorney General, Comptroller and Auditor General of India, State Governors, Judges of the Supreme Court and State High Courts, Finance Commission, Chairman and members of U.P.S.C., Election Commission, and Chief Commissioners of Unit Territories. He allocates portfolios to the Ministers.

2. Legislative Powers :
The President is an integral part of Parliament (Art. 79) and as such enjoys extensive legislative powers.
They are :

  1. He summons from time to time each House of Parliament, adjours, and prorogues either or both the Houses.
  2. He addresses either House separately or both the Houses Jointly.
  3. He can dissolve the Lok Sabha on the advice of the Prime Minister.
  4. He opens the first session of Parliament after the General Elections and at the commencement of the first session of each year.
  5. He can send messages to the Parliament.
  6. He arranges a joint session of both the Houses when there is a dead-lock over an ordinary bill.
  7. All bills passed by Parliament require his assent for becoming in acts.
  8. He nominates 12 members to Rajya Sabha and two Anglo Indian members to Lok Sabha.
  9. He promulgates ordinances when the Parliament is not in session.
  10. He sends the annual reports of Finance Commission, U.P.S.C etc., for the consideration and approval of Parliament.

3. Financial Powers :
The President also enjoys some financial powers. They are :

  1. He recommends the financial bills to be introduced by the members in parliament. The Budget is caused to be laid down before the Parliament by the President.
  2. He operates the Consolidated Fund of India.
  3. He determines the shares of States in the proceeds of Income Tax.
  4. No Money Bill can be introduced in the Parliament except on his recommendations.
  5. He constitutes a Finance Commission for every five years etc.

4. Judicial Powers :

  1. The President can grant pardons, reprieves, respites or remission of punishments.
  2. He appoints the judges of the Supreme Court and State High Courts.
  3. He can also remove them on an address by the Parliament.

5. Military Powers:
The President is the Supreme Commander of the Defence Forces of the Union. He appoints the Chiefs on the Staff for Army, Navy, and Air Force. He can declare war and conclude peace. But he has to take the approval of Parliament.

6. Diplomatic Powers :
The president appoints Ambassadors to foreign countries to represent India. He receives the credentials of the Ambassadors appointed in India. He represents our Nation in International affairs. He makes treaties and agreements with other countries subject to the ratification by the Parliament.

7. Emergency Powers :
In extraordinary conditions, the President can proelaim emergency to safeguard the security, integrity, and independence of our country. They are of three types :

  • Emergency caused by war or external aggression or armed rebellion (Article 352).
  • Emergency due to failure of Constitutional machinery in the States (Article 356)
  • Emergency due to threat to the financial stability of India (Article 360).

AP Inter 2nd Year Civics Study Material Chapter 3 Union Executive

Question 2.
Write briefly the Emergency powers of the President of India. [Mar. 16]
Answer:
Articles 352 to 360 of part XVIII of Indian constitution deals with three types of emergency powers of the Indian President. They are :

  1. National Emergency,
  2. Constitutional Emergency,
  3. Financial Emergency

They may be explained as follows.
1) National Emergency : (Article 352)
The President exercises this power during the period of war, external aggression or armed rebellion. He declares emergency if he is satisfied that the sovereignty and security of India or any part thereof is threatened by external aggression.

When National Emergency is in force, the federal provisions of our constitution ceases to operate. So far, National Emergency was proclaimed on four occasions in our country. They are : 1. Chinese Aggression (1962), 2. Indo – Pak war (1965), 3. Indo – Pak war in the context of Bangladesh Liberation Movement (1971), 4. Opposition’s call for blocking Parliament (1975).

2) Constitutional Emergency : (Article 356)
Article 356 of Indian constitution empowers the President to proclaim the constitutional emergency. If the President, on receipt of a report from the Governor or other wise; is satisfied that a situation has arisen in which the government of a state cannot be carried on according to the constitutional provisions. He is empowered to proclaim this emergency. It is also called as the President’s Rule. So far this type of emergency was proclaimed for over 100 times.

3) Financial Emergency : (Article 360)
If the President is satisfied that a situation has arisen where by the financial stability or credit of India is threatended, then he may proclaim financial emergency in the country. During the period of financial emergency, the President enjoys the following powers.

  • The President may reserve all the money bills or other financial bills of the state after they are approved by the state legislature.
  • He may reduce the salaries and allowances of all or any person serving in the states.
  • The President can reduce the salaries allowances of the persons working at the union level including the judges of the Supreme Court and the State High Courts. But so far the financial emergency has not been yet imposed in the country.

Question 3.
Discuss the powers and functions of the Prime Minister of India. [Mar. 18, 16]
Answer:
The Prime Minister is the real executive head of the Union Government. He occupies an important position in the administration of our country. Since India has a Parliamentary form of Government the real power rests with him. He is the ‘uncrowned king’ and “the keystone of the Cabinet arch in the Union Government”.

Qualifications :

  1. He should be citizen of India.
  2. He should have completed the age of 25 ytears.
  3. He should be qualified for election as a member of the Lok Sabha.
  4. He should not hold any office of profit under the union or state or local governments.

Appointment:
The President appoints the Prime Minister. Generally the President has to summon the leader of the majority party in the Lok Sabha to form the Ministry. If no party gets an absolute majority, the President can use his discretion and summon the leader of the party, who in his opinion can manage to form a ministry. Afterwards the Prime Minister will be asked to prove his majority in the Lok Sabha.

Oath of Office :
The President of India will administers the oath of office of the Prime Minister.

Term of Office :
The Prime Minister shall remain in office during the pleasure of the President. But actually he assumes his powers as long as he retains the confidence of the majority members in the Lok Sabha. He resigns when the Lok Sabha accepts a no-confidence motion against his ministry.

Salary and Allowances:
The salary and allowances of Prime Minister are decided by the Parliament from time to time. He gets his salary and allowances that are payable to a member of Parliament. At present the Prime Minister gets a salary and allowances of ₹ 1,60,000/- per month.

Powers and Functions :
The Prime Minister is the head of the union government. He is the real executive. The Council of Ministers cannot exist without the Prime Minister. His powers are explained here under.

1) Leader of the Union Cabinet:
The Prime Minister is the leader of the Union Cabinet and Union Council of Ministers. He selects some eminent members of his party in parliament and sees that they are appointed as ministers by the President. He has a free choice of both allocating portfolios and reshuffling the ministry. All the ministers are personally and politically loyal to the Prime Minister. He decides the agenda, of the cabinet meetings. Further, he presides over the cabinet meetings.

2) Leader of the Union Government:
The Prime Minister acts as the leader of the union government. The union executive (Union Council of Ministers) initiates its business after the swearing in ceremony of the Prime Minister. All the ministers in the union ministry assume their office, owe their position and exercise their powers along with the Prime Minister. Infact, the Prime Minister influences the nature and working of the union government. He not only has a clear understanding but holds complete control over the affairs of the union government. All the high-level officers and the entire ministry in the union government behave and act according to the wishes of the Prime Minister.

3) Leader of the parliament:
The Prime Minister acts the leader of the Parliament in India. He is primarily a member of Parliament. He extends co-operation to the presiding officers in the smooth conduct of the two Houses. He wields complete control over his party members in the Parliament. He ensures that his party members maintain discipline during the sessions of the Parliament. He informs out the cabinet decisions to the Parliament. He communicates the major domestic and foreign policies of the union government to the members of Parliament. He maintains rapport with the opposition leaders and discusses the major issues confronted by the nation with them.

4) Link between the President and the Council of Ministers :
The Prime Minister acts as the main link between the President and the Union Council of Ministers. It is his duty to communicate to the President about the decisions of the Union Council of Ministers. He furnishes the every information required by the President concerning the affairs of union government. All the ministers shall formally meet the President only with the consent of the Prime Minister.

5) Leader of the Majority Party:
The Prime Minister acts as the leader of the majority party or group in the lower House of Parliament. He participates in the meetings of the party and acquaints his party members on various issues and steps taken by his ministry in implementing the party promises. He utilizes the services of the senior party leaders in running the government. He acts as the main link between the part and the government.

6) Leader of the Nation :
The Prime Minister is the leader of the nation. He takes initiative in finding solutions to several problems in the internal matters of the country. He plays an important role in the development of the nation.

7) Maker of Foreign Policy :
The Prime Minister plays a dominant role in shaping the foreign policy of the nation. He keeps in touch with the developments in all countries. He meets Heads of various countries and maintains friendly relations with them.

8) Chairman of NITI Aayog :
The Prime Minister heads the NITI Aayog (National Institution for Transforming India) NITI Aayog means policy commission. It is a policy think tank of government of India that replaces planning commission which aims to involve the states in economic policy making in India. It will provide strategic and technical advice to the central and state governments. It will have a governing council comprising Chief Ministers of all the states and it governors of Union Territories. Union government set up the NITI Aayog on January 1,-2015.

Question 4.
Explain the composition, powers and functions of the Union Council of Ministers. .
Answer:
Article 74 (1) of the Constitution provides for a Council of Ministers at the Centre. It’s main function is to aid and advice the President in the performance of his duties. It consists of Prime Minister and other Ministers. It is this body which runs the entire administration of our country. It is thefreal Executive authority of the country. It functions on the ‘Principle of Collective Responsibility1. It holds office till it continues to enjoy the confidence and support of the Lok Sabha.

Formation of Council of Ministers :
The formation of the Council of Ministers starts with the appointment of the Prime Minister. The President appoints the Prime Minister and on the advice of the Prime Minister, the other Ministers are appointed by the President.

Composition of Council of Ministers :
Our Constitution did not mention the exact size of the Union Council of Ministers. But there are three kinds of Ministers:

  1. Cabinet Ministers
  2. Ministers of State
  3. Deputy Ministers.

The Cabinet Ministers are entrusted with the maintenance of some important ministries. They enjoy independence and decision making powers.

The Ministers of State act as the heads of some important sections in the ministry. They are directly responsible to the Prime Minister for their activities.

The Deputy Ministers have no independent and discretionary powers. They assist the Cabinet Ministers and perform the functions assigned to them.

The Cabinet or the Council of Ministers is the pivot around which the entire administration of our country revolves. “It is the steering wheel of the ship of the State.” It is the hyphen that joins the Executive and Legislative organs of the Government.

Powers and Functions :
1. Executive Powers :
The Union Cabinet is a deliberative and policy formulating body. It discusses and decides all National and International policies of the country. The policies decided by the cabinet are carried out by the Ministers. It directs and leads the Parliament for action and gets its approval for all its policies. It coordinates and guides the activities of departments of the Government. It also plays an important role by suggesting persons for all major appointments. It considers the reports of various committees before they are presented to the Parliament.

2. Legislative Powers :
The Cabinet plans the legislative programme of the Government at the beginning of each session of Parliament. It drafts Bills on all important matters and introduces them in the Parliament. It also decides the time of summoning and prorogation of Parliament. The inaugural speech of the President to the Parliament is prepared by the Cabinet.

3. Financial Powers:
The Cabinet possess important financial powers. It has complete control over national finance. It prepares the Union Budget. It decides what taxes are to be imposed and how much of expenditure is to be incurred. Money Bills are always introduced in the Lok Sabha by the Finance Minister.

4. Foreign Relations :
In the field of foreign relations also the Cabinet plays an important role. It determines and formulates the foreign policy of the country and decides India’s relations with other countries. It considers and approves all international treaties and agreements.

Collective Responsibility :
Article 75(3) of Indian constitution stated that the union council of Ministers shall be collectively responsible to the Lok Sabha, for all their acts of commissions and commissions. They act as a team under the leadership of the Prime Minister. They sail together, they swim together and they sink together.

Conclusion:
It is thus clear that the Council of Minister or Cabinet enjoys far reaching powers both with regard to the internal and external policies of the country. Internally it maintains law and order within the country and externally protects the country from foreign aggression. The progress of the country largely depends upon the ability of the Cabinet.

Short Answer Questions

Question 1.
How is the President of India elected?
Answer:
The President of India shall be elected indirectly by an electoral college consisting of the elected members of both Houses of Parliament. State Legislative Assemblies and elected members of Delhi and Pondicheri. The election is held in accordance with the system of proportional representation by means of a single transferable vote system and secret ballot.

Each member of the Electoral College has one vote. But the value of the vote differes from State to State. The value of the vote of a Parliament member also differs from the value of the vote of a member of State Assembly.

The value of the vote of an M.L.A is worked out as follows. The total population of the State is divided by the total number of elected members of the Assembly. The quotient thus obtained is to be divided by 1,000. Fractions of half or more should be counted as one and added to the quotient. If it is less than half, it is ignored. This may be shown as follows.

  1. Value of vote of an M.L.A = Population of State / number of elected members of the Assembly / 1,000.
  2. Value of vote of M.P. = Total value of votes of all Assembly members / Total number of elected members of both Houses of Parliaments.

This method is followed to keep the election of the President above narrow political considerations.

AP Inter 2nd Year Civics Study Material Chapter 3 Union Executive

Question 2.
Write briefly about the procedure of impeachment of President
Answer:
The President may be removed from the office for violation of the constitution by a process of impeachment. Impeachment is a quasi-judicial procedure adopted by the Parliament. Either House of Parliament shall prefer the charge for removal of the President. The other House shall investigate into the charges itself or cause the charge to be investigated.

There are four stages in the impeachment of the President.

Firstly the impeachment resolution has to be moved with a 14 days prior notice in writing signed by not less than 1/4th of the total members of that House. Such a resolution has to be passed by a majority of not less than 2/3rds of the total members of the House.

Secondly, the resolution approved by the first House will be sent to the second House for consideration and approval.

Thirdly, the second House investigates into the charges directly or constitutes a committee to enquire into the charges. The President has the right to present his views directly or through a deputy during such enquiry.

Fourthly, if the charges against the President are established and adopted by the second House with 2/3rds majority of the total members, the President stands removed from the office. With regard to voting on the resolution for impeachment, only the elected members cast their vote. No president has so far been impeached in our country till today.

Question 3.
Mention any two Emergency powers of the Indian President.
Answer:
1) National Emergency : (Article 352)
The President exercises this power during the period of war, External aggression or armed rebellion. He declares emergency if he is satisfied that the sovereignty and security of India or any part thereof is threatened by external aggression.

When National Emergency is in force, the federal provisions of our constitution ceases to operate. So far, National Emergency was proclaimed on four occasions in our country. They are :

  1. Chinese Aggression (1962) ‘
  2. Indo – Pak war (1965) .
  3. Indo – Pak war in the context of Bangaldesh Liberation Movement (1971)
  4. Oppositions call for blocking Parliament (1975)

2) Constitutional Emergency : (Article 356)
Article 356 of Indian constitution Empowers the President to proclaim the constitutional emergency. If the President, on receipt of a report from the Governor or other wise, is satisfied that a situation has arisen in which the government of a state cannot be carried on according to the constitutional provisions. He is empowered to proclaim this emergency, It is also called as the President’s rule. So far this type of emergency was proclaimed for over 100 times.

Question 4.
Explain the role and position of the President in Union Government.
Answer:
The President of India is the head of the union executive. He is the first citizen of India. He could exercise many powers as enstined in the constitution in the following manner.

Position of the President:
There are different opinions on the actual position of the president of India in the administration of our country. The farmers of the constitution wanted him to be a nominal Head of the state.

Dr. Ambedkar compared his position to that of the British King Sri. M. C. Setalved, the farmer Attorney General of India mentioned that the position of the President of India is like the king in England and the Governor General in a Dominion. Sri Alladi Krishna Swamya Ayyar also said that it was perfectly clear that our presidents position was similar to that of the constitutional Monarch in England.

Jawaharlal Nehru, the first Prime Minister of India, said that “We have not given our President any real power but we have made his position one of great authority and dignity.” These opinions make it clear that our President is only a nominal figure head and he does not have any real powers. This is confirmed by the 42nd Amendment of the constitution of India.

However, the President exercises independent powers under some conditions. He utilises these powers in regard to the appointment of the Prime Minister, dissolving the Lok Sabha and ordering midterm poll to the Lok Sabha. Some presidents like Sanjiva Reddy, Zail Singh, R.Venkata Raman, Dr. S.D. Sharma etc., utilised their discretionary powers where there was political instability or Hung Parliament after the general elections in the country. Like the Monarch of England, he still enjoys three rights the right to be ! consulted the right to encourage and the right to warn.

Question 5.
Write about any two powers of the Vice – President of India.
Answer:
The Vice President of India occupies the second highest position in the union government. He is accorded a rank next to the President.

Qualifications :
A person to be eligible as vice-president should possess the following qualifications.

  1. He should be a citizen of India.
  2. He should have completed 35 years of age.
  3. He should be qualified for election as a member of the eousil of states.
  4. He should not hold any office of profit under the union, state or local Governments in India.

Election :
The election of the vice – president like that of the President shall be indirect and in accordance with the system of proportional representation by means of the single transferable vote system. He is elected by the members of an electoral college consisting of the members of both the houses of Parliament.

Term of Office :
The Vice-President holds office for a term of 5 years from the date on which he enters his office.

Removal:
The Vice President may be removed from his office by a resolution of the council of states passed by a majority of all the members of the council and agreed to by the house of the people.

Salary and Allowances :
The Vice President of India receives a monthly salary of ₹ 1,25,000/-in addition, he is entitled to daily allowance, free furnished residence, medical, travel and other facilities.

Powers and Functions :
The Vice – President is the ex-officio Chairman of the Rajya Sabha. As. such he enjoys the same powers like the Speaker of Lok Sabha, such as (1) presiding over the meetings of Rajya Sabha, (2) maintaining discipline, decency and decorum in the House, (3) exercising casting vote in case of a tie, (4) admitting visitors, (5) protecting the privileges and rights of the members. He has no right with regard to money bills.

Acting as President of India :
He discharges the functions of the President during the temporary absertce of the President. He may take over the office of the President under 4 situations like (1) Death of the President, (2) Resignation of the President, (3) Removal of the President, (4) Inability of the President due to absence, illness or any other cause.

Question 6.
How is the Prime Minister appointed?
Answer:
Article 75 (1) of the Indian constitution deals with the appointment of the Prime Minister of India.

Appointment:
The constitution simply lays down that the Prime Minister shall be appointed by the President. After the conduct of General Elections to the Lok Sabha, the President has to invite the majority party leader of the Ldk Sabha to form the Government.

When no single party is able to secure majority seats in Lok Sabha, the President invites the leader of a coalition to form the Government. The president uses his discretionary powers in this regard. The President appoints the leader of the coalition as the Prime Minister on the condition that he has to prove his majority in the Lok Sabha within a specified period. Being the leader of the majority in Lok Sabha to be the Prime Minister, the person has to be a member of Parliament. If he is not a member at the time of appointment, he has to acquire it within six months from the date of his appointment as Prime Minister.

The powers of the President in choosing, inviting and appointing the Prime Minister cannot be questioned in any court of Law.

AP Inter 2nd Year Civics Study Material Chapter 3 Union Executive

Question 7.
Explain the role of the Prime Minister in Union Government.
Answer:
The prime Minister plays a predominant role in the affairs of the union government. He will have an indelible impression on every one in the administration of the union government. He is described as the Primus Interparus (first among equals). His role as the leader of the Union Council of Ministers, Union Cabinet, Party in power, Lok Sabha, Nation and as the link between the President and the Union Council of Ministers is unique. He wields tremendous political power and patronage. He enjoys enormous powers and fulfils innumerable tasks. It all depends on the image, influence, stature, and personality of the Prime Minister in the union government.

Jawaharlal Nehru, Dr. Ambedkar, and other eminent leaders of the National Movement and members of the Constituent Assembly described the Prime Minister as the linchpin of the union government. It is in this context that Sir william Harcourt remarked that every one expects from the Prime Minister dignity and authority, firmness to control, tact, practice and firmness, an impartial mind, a tolerant temper, a kind and prudent counsellorship, and accessibility to the people.

Question 8.
Describe the composition and powers of the Union Council of Ministers.
Answer:
Composition of Council of Ministers :
Our Constitution did not mention the exact size of the Union Council of Ministers. But there are three kinds of Ministers: 1) Cabinet Ministers 2) Ministers of State 3) Deputy Ministers.

The Cabinet Ministers are entrusted with the maintenance of some important ministries. They enjoy independence and decision making powers.

The Ministers of State act as the heads of some important sections in the ministry. They are directly responsible to the Prime Minister for their activities.

The Deputy Ministers have no independent and discretionary powers. They a&ist the Cabinet Ministers and perform the functions assigned to them.

The Cabinet Or the Council of Ministers is the pivot around which the entire administration of our country revolves. “It is the steering wheel of the ship of the State”. It is the hyphen that joins the Executive and Legislative organs of the Government.

Powers and Functions :
1) Executive Powers :
The Union Cabinet is a deliberative and policy formulating body. It discusses and decides all National and International policies of the country. The policies decided by the cabinet are carried out by the Ministers. It directs and leads the Parliament for action and gets its approval for all its policies. It co-ordinates and guides the activities of departments of the Government. It also plays an important role by suggesting persons for all major appointments. It considers the reports of various committees before they are presented to the Parliament.

2) Legislative Powers :
The Cabinet plans the legislative programme of the Government at the beginning of each session of. Parliament. It drafts Bills on all important matters and introduces them in the Parliament. It also decides the time of summoning and prorogation of Parliament. The inaugural speech of the President to the Parliament is prepared by the Cabinet.

3) Financial Powers :
The Cabinet possesses important financial powers. It has complete control over national finance. It prepares the Union Budget. It decides what taxes are to be imposed and how much of expenditure is to be incurred. Money Bills are always introduced in the Lok Sabha by the Finance Minister.

4) Foreign Relations :
In the field of foreign relations also the Cabinet plays an important role. It determines and formulates the foreign policy of the country and decides India’s relations with other countries. It considers and approves all international treaties and agreements.

Very Short answers

Question 1.
Composition of the Union Executive.
Answer:
The constitution of India provides for the Union Executive. Articles 52 to 78 in part V of the constitution deal with the union executive. The Union Executive consists.
i) The President
ii) The Vice-President
iii) The Prime Minister and
iv) The Union Council of Ministers

Question 2.
Qualifications required for contesting the Presidential elections.
Answer:
A person to be eligible to contest the office of the president shall possess the following qualifications.

  1. He should be a citizen of India.
  2. He should have completed the age of 35 years.
  3. He should be qualified for election as a member of the Lok Sabha.
  4. He should not hold any office of profit under the Union, State or Local Governments.
  5. Possess such other qualifications as prescribed by the Parliament.

Question 3.
Election of President.
Answer:
The President of India shall be elected indirectly by an electoral college consisting of the elected members of both houses of Parliament, State Legislative Assemblies and elected members of Delhi and Pondicheri. The election is held in accordance with the system of proportional representation by means of a single transferable vote system and secret ballot.

Question 4.
Important appointments of President.
Answer:
The president has the power to appoint the following high dignitaries :

  • The Prime Minister of India.
  • Members of the union council of Ministers.
  • The Affomey General of India.
  • The Comptroller and Auditor general of India.
  • The judges of the Supreme Court and the High Court.
  • State Governors etc.

Question 5.
Judicial Powers of the President.
Answer:

  1. The President can grant pardons, reprieves, respites or remission of punishments.
  2. He appoints the Judges of the Supreme Court and State High Courts.
  3. He can also remove them on an address by the Parliament.

Question 6.
Article 352. [Mar. 18, 16]
Answer:
Article 352 of the Indian constitution empowers the President to impose National Emergency during the period of war, External Aggression, Armed Rebellion or internal disturbance. So far National Emergency was proclaimed on Four occasions. They are :

  1. Chinese Aggression (1962)
  2. Indo – Pak War (1965)
  3. Indo – Pak war in the context of Bangladesh Liberation movement (1971) and
  4. Opposition’s call for blocking Parliament (1975).

Question 7.
Article 356.
Answer:
Article 356 of Indian Constitution empowers the President to proclaim the constitutional emergency. If the President on receipt of a report from the governor or otherwise is satisfied that a situation has arisen in which the government of a state cannot be carried on according to the constitutional provisions. He is empowered to proclaim this emergency. It is also called as the President’s Rule.

Question 8.
Financial Emergency.
Answer:
If the President is satisfied that a situation has arisen where by the financial stability or credit of India is threatened then he may proclaim financial emergency in the country as per Article 360 of the Indian Constitution.

Question 9.
National Emergency.
Answer:
Article 352 of the Indian constitution empowers the president to impose National Erpergency during the period of war, External Aggression, Armed Rebellion or Internal disturbances. When National Emergency is in force, the Federal provisions of our constitution cease to operate. So far, National Emergency was imposed four times in 1962, 1965, 1971 and 1975.

Question 10.
Qualifications required for contesting as Vice-President. [Mar. 16]
Answer:
A person to be eligible for election as Vice-president should possess the following qualifications : ,

  1. He should be a citizen of India.
  2. He should have completed 35 years of age.
  3. He should be qualified for election as a member of the council of states.
  4. He should not hold any office of profit under the Union, State or Local Governments in India.

Question 11.
Chairman of Rajya Sabha,
Answer:
The Vice President is the ex-officio chairman of the Rajya Sabha. As such he enjoys the same powers like the speaker of Lok Sabha such as

  1. Presiding over the meetings of Rajya Sabha.
  2. Maintaining discipline, decency and decorum in the House.
  3. Exercising casting vote in case of a tie.
  4. Protecting the privileges and rights of the members.

Question 12.
Appointment of Prime Minister.
Answer:
After the conduct of General elections of the Lok Sabha, the President has to invite the majority party leader of the Lok Sabha to form the Government. When no single party is able to secure majority seats in Lok Sabha, the President invites the Leader of a coalition to form the Government. The president uses his discretionary powers in this regard.

Question 13.
Categories of Union Council of Ministers.
Answer:
There are three kinds of ministers in the union council of ministers. They are

  1. Cabinet Minister.
  2. Ministers of State.
  3. Deputy Ministers.

1) The Cabinet Ministers are entrusted with the maintenance of some important ministries like Finance, Home, Defence etc.

2) The Ministers of state act as the heads of some important sections in the Ministry. They are directly responsible to the Prime Minister for their activities.

3) The Deputy Ministers have no independent and discretionary powers. They assist the Cabinet Ministers.

Question 14.
Any two functions of the Union Cabinet.
Answer:

  1. The Union Cabinet formulates the policies of the union government. It finalizes the domestic as well as foreign policies of the nation after having serious deliberations.
  2. It pilots several bills in the Parliament at various stages and strives to secure the approval of the latter.

AP Inter 2nd Year Civics Study Material Chapter 3 Union Executive

Question 15.
Collective Responsibility. [Mar. 18]
Answer:
Article 75 (3) of Indian constitution stated that the union council of Ministers shall be collectively responsible to the Lok Sabha, for all their acts of omissions and commissions. They act as a team tinder the leadership of the Prime Minister. They sail together, they swim together and they sink together.

AP Inter 2nd Year Civics Study Material Chapter 2 Fundamental Rights and Directive Principles of State Policy

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 2nd Lesson Fundamental Rights and Directive Principles of State Policy Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 2nd Lesson Fundamental Rights and Directive Principles of State Policy

Long Answer Questions

Question 1.
Explain the characteristic Features of Fundamental Rights.
Answer:
Fundamental Rights :
Fundamental Rights are an important feature of Indian constitution. They are meant for Indian citizens realising the ideal of political democracy. These rights are assigned to the Indian citizens. They enable the citizens to realize their personality. Fundamental Rights will act as a means for leading a happy and honourable life by citizens their render strength and succor to the citizens. They serve as the main source for realising the ideals of political democracy in India.

The makers of Indian constitution have incorporated Fundamental Rights in Articles 12-35 in part III of the constitution.

Characteristics features of Fundamental Rights :
Fundamental Rights have the following characteristic features.

1) Some of the fundamental rights are granted to the ‘citizens’ alone for example, equality of opportunity in matters of public employment, protection against discrimination on any ground; freedom of speech, assembly, association, etc., and cultural and educational rights of the minorities. On the other hand, some of the fundamental rights are available to any person living in the country whether Indian of foreign. For example, equality before law and its equal protection, protection of life, freedom of religion etc.

2) Some of the fundamental rights are positive in nature. They provide scope for the citizens to enjoy some types of freedom. On the other hand, some of the fundamental rights are negative in nature. They impose some restrictions upon the activities of the state.

3) Fundamental Rights are not absolute. In this sense the state can impose reasonable restrictions on their utilisation and enjoyment in the interest of public order, morality, and friendly relations with foreign states.

4) State may impose some restrictions on all or some of the fundamental rights of the citizens during the emergency. The president of India can suspend all the fundamental rights except article 21 (Right to Life) during the national emergency. However, the freedoms guaranteed can’t be restricted by any body.

5) Fundamental Rights are component of the Indian constitution. So they can’t be altered through ordinary laws.

6) Fundamental Rights are comprehensive, integrative and detailed in nature. Some restraints have also been imposed against the utilization of these rights under specific conditions.

7) Fundamental Rights are protected by the judicial organizations in the country. Especially the Supreme court and state High courts play a crucial role in this regard. The ensure justice to those whose rights are infringed or confiscated by others including the state authorities. They issue several writes for the protection of fundamental rights.

8) Fundamental Rights serve as the main means for proper utilization of the capacities and intelligence hidden among the Indian citizens.

9) Though the constitution guarantees six categories of fundamental Rights, all are not of equal weight. That is three can be discovered a hierarchy of values. It becomes evident when justice M. Hidayatullah in the Golaknath case ruled that right to property is the Weakest of all rights’.

AP Inter 2nd Year Civics Study Material Chapter 2 Fundamental Rights and Directive Principles of State Policy

Question 2.
Explain the various types of Directive principles of state policy mentioned in Indian constitution. [Mar. 18, 17]
Answer:
Directive principles of state policy are enumerated in articles from 36 to 51 in past -IV of the Indian constitution. They are borrowed from the Irish constitution. They help in realizing the objectives mentioned in the preamble.

Types of Directive principles of state policy: Directive principles can be classified into three broad categories namely Socialistic, Liberal-intellectual and Gandhian principles.

Article 36 defines the term “State”.

Article 37 declares that the directive principles shall not be enforceable by any court.

1) Socialist Principles:
Articles 38,39,41,42,43 and 47 explains about the socialistic ideology of the directive principles of state policy.
1. Article 38 prescribes that the state shall strive to provide justice and promote welfare of the people by creating a proper economic, social and political atmosphere.

2. Article 39 directs the state to secure its citizens.

  • Adequate means of livelihood for all citizens.
  • Equitable distribution of wealth for sub-serving the common good.
  • Equal pay for equal work for all.
  • Protection of adult and child labour.
  • Decentralization of nation’s wealth.
  • Preserving the health and strength of workers, men and women.
  • Protecting childhood and youth against exploitation.

3. To secure right to work and education for all people* relief in the case of unemployment; old age; sickness and disablement and in dther cases of under served want. (Article 41).

4. To make provision for just and human conditions of work and maternity relief (Article 42).

5. To secure living wage and decent standard of life so as to ensure to the workers sufficient leisure and enjoyment of social and cultural Opportunities. (Article 43).

6. Raising the level of nutrition and standard of living of the people and the improvement of public health (Article 47).

2) Liberal-intellectual Principles:
The principles represent the ideology of liberalism and certain objective like provision of basic education, uniform civil code, independent judiciary and international peace. They are incorporated in Articles 44, 45, 50 and 51 of the Constitution.

  1. The State shall secure for the citizens uniform civil code throughout the country. (Article 44)
  2. The State shall provide free and compulsory education for all the children below 14 years of age. The Constitution (Eighty Sixth Amendment) Act, 2002 substituted, the following words in Article 45. “The State shall endeavor to provide early childhood care and education for all children until they complete the age of six years.” (Article 45)
  3. The state organize agriculture and animal husbandry on modem and scientific lines (Article 48)
  4. The state protect monuments which are declared to be of national importance (Article 49)
  5. The state protect and improve the environment and to safeguard forests and wild life. (Article 48 A)
  6. The State shall take steps to separate judiciary from executive in public services of the State. (Article 50)
  7. The State shall (a) promote international peace, justice and security, (b) Maintain just and honorable relations with other nations (c) protection of monuments and place of historical and cultural interest (d) respect for international laws and treaty obligations; and (e) encourage settlement of international dispute by arbitration. (Article 51)

3) Gandhian Principles :
These Principles are based on gandhian ideology. They represent the programme of reconstruction enunciated by Mahatma Gandhi during the national movement. These principles provide ideal rule in India. They are reflected in Articles 40, 43, 46 and 47. They may be enumerated as under.

  1. The State shall organize village Panchayats and endow them with adequate powers and authority so as to enable them to function as the units of self-government. (Article 40)
  2. The State shall strive for the promotion of cottage industries on individual or cooperative basis in rural areas. (Article 43)
  3. The State shall promote the educational and economic interests of the SCs, STs and BCs of society with special care, (Article 46)
  4. The State shall endeavour to bring about prohibition of intoxicating drinks and of drugs which are injurious to health (Article 47)

Other Principles :
The Constitution (Forty Second and Forty-Fourth Amendment) Acts of 1976 and 1978 added a few more subjects to the list of Directive Principles. While the Constitution (Forty Second Amendment) Act inserted Articles 39A, 43A, and 48 A, the Constitution (Forty Fourth Amendment) Act included Article 39 Clause (2). They cc the following provisions.

  1. Providing opportunities for healthy development of children.
  2. Promotion of equal justice and legal aid to the poor.
  3.  Securing participation of workers in the management of industries.
  4. Protecting the environment, forests and wild animals.

Question 3.
Describe the Fundamental Duties incorporated in Indian Constitution. [Mar. 16]
Answer:
Fundamental Duties are a significant feature of Indian Constitution. They are incorporated in our constitution by the constitution 42nd amendment act of 1976 in part – IV under article 51 A. They are borrowed from Russian Constitution. They are 11 in number as mentioned below :

  • To abide by the Constitution and respect the National Flag and the National Anthem.
  • To cherish and follow the noble ideals which inspired our national struggle for freedom.
  • To protect the sovereignty, unity and integrity of India.
  • To defend the country and render national service when called upon to do so.
  • To promote harmony and the spirit of common brotherhood amongst all the people of India and renounce practices derogatory to the dignity of women.
  • To value and preserve the rich heritage of our composite culture.
  • To protect and improve the natural environment including forests, rivers and wildlife and to have compassion for living creatures.
  • To develop the scientific temper, humanism and the spirit of inquiry and reform.
  • To safeguard public property and to abjure violence.
  • To strive towards excellence in all spheres of individual and collective activity.
  • To provide educational opportunities by the parent or guardian to his child or ward between the age of six and fourteen years.

Question 4.
Explain the differences between Fundamental Rights and Directive principles of State Policy.
Answer:
Fundamental Rights are incorporated in Part – III (Article 12 to 35) of our Constitution. Fundamental Rights are of great significance. They serve as the best means to safeguard the life, liberty and property of individuals. They act as the main instrument for releasing the inherent talents and capabilities of the individuals. Fundamental Rights are borrowed from the American Constitution. These are justiciable and can be enforced by the courts.

The Directive Principles of State Policy is a feature of Indian Constitution. They are included in Part -IV from Articles 36 to 51. The principles help to realize the objectives mentioned in the preamble of our Constitution. The makers of our Constitution drew them from the Irish Constitution. These are directives to different governments and agencies of our country. These principles aim at transforming our country into a Welfare State. The rulers have to respect them. They cannot be enforced through any Court of Law. The I governments have to implement them subject to availability of funds. They explain the I responsibilities of the State towards the people.

Differences between Fundamental Rights and Directive Principles of state policy :

Fundamental Rights Directive Principles of state policty
1) The concept was borrowed from the American Constitution. 1) The concept was borrowed from the Irish Constitution.
2) These are enumerated in Part – III of the Constitution covering articles from 12 to 35. 2) These are enumerated in Part – IV of the Constitution covering articles from 36 to 51.
3) These-are negative in character in the sense that they prohibit the Government from doing certain things. 3) These are positive in character in the sense that they direct the government to work for the attainment of certain objects.
4) These aim at establishing political democracy in the country. 4) These aim at establishing social and economic democracy in the country.
5) These have constitutional sanction and so their implementation needs no legislation. 5) These have no constitutional sanction and so their implementation needs legislation.
6) These are justicable and can be enforced by the courts. 6) These are non – justiciable and cannot be enforced by the courts.
7) A law violating a Fundamental Right can be declared by the court as unconstitutional. 7) A law violating a Directive Principle cannot be declared by the court as unconstitutional.
8) These are personal and individualistic in nature. 8) These are societarian and socialistic in nature.
9) These promote the welfare pf the individual. 9) These promote the welfare of the society.
10) These have been laid down in clear legal language. 10) These are stated in general terms.

Short Answer Questions

Question 1.
Write a note on the changing Relationship between Fundamental Rights and Directive Principles of State Policy.
Answer:
Fundamental Rights and Directive principles of state policy are the salient features of Indian Constitution.

Changing Relationship between Fundamental Rights and Directive Principles :
Although a distinction is mode, between fundamental rights and directive principles of sate policy by way of justiciable and non-justiciable nature. Yet over the years directive principles of state policy have become politically important and the relation between the two has undergone several changes.

The Supreme court consistently held the opinion that the directive principles of state policy should be subsidiary to the fundamental rights. Judgements in various disputes like Sajjan Singh Vs state of Rajasthan and Golaknath Vs state of Punjab, the Supreme court confirmed its stand and reiterated that it is the duty of Parliament to enforce the Directive principles without tampering the Fundamental Rights.

As a result of the invalidation of certain laws like the Nationalization of Banks, Abolition of privy purses, the Parliament enacted the (Twenty Fifth Amendment) constitution Act in 1971 which declated that the enforcement of the directive principles of state policy shall not be invalidated by any court on the grand that it violates the fundamental rights in articles 14, 19 and 31 of the constitution.

Again the (Forty Second Amendment) constitution Act passed by Parliament in 1976 declared that no law, giving effect to any of all directive principles, shall be invalid on the ground that it infringes on Fundamental Rights. However, in the Minerava Mills case, the Supreme court restricted the original supremacy and sanctity of the fundamental rights over the directive principles of state policy. Thus, the fundamental rights have primary over the directive principles.

In the case of Keshavananda Bharati Vs state of Kerala. The supreme court held that the Parliament cannot among the basic structure of the constitution. By implication the Supreme court considered fundamental rights as a part of the basic structure of the constitution.

AP Inter 2nd Year Civics Study Material Chapter 2 Fundamental Rights and Directive Principles of State Policy

Question 2.
Explain any three fundamental rights of a citizen. [Mar. 18]
Answer:
1) Right to freedom of Religion :
This right denotes the secular nature of Indian political system. It aims at transforming India into a secular state. Both the citizens and aliens of India enjoy this right.

Article 25 empowers every persons to profess, practice and propagate a religion of this liking.

Article 26 guarantees the following rights to every person.

  1. To establish and maintain religious and charitable institutions.
  2. To mange his their religious affairs.
  3. To own and acquire moveable and immovable properties and
  4. To maintain such properties in accordance with the provision of the law.

Article 27 prohibits the state to impose or collect taxes from individuals purely on religious grounds. It also prohibits the state to’impose and collect taxes for the benefit and maintenance of any particular religion or religious denominations.

Article 28 bans religious instructions in educational institutions wholly or partly maintained by the state funds.

2) Educational and Cultural Rights :
Indian constitution provided several cultural and educational opportunities for Indian citizens through this right. Article 29 enables every citizen to preserve and protect his own language and culture irrespective of one’s religion, language or region.

Article 30 prohibits special treatment to any citizen in the admission into educational institutions either wholly or partly funded by the state on the grounds of caste, religion, region, colour, language or sect. However, it allowed the minotities some special facilities for preserving and promoting their language and culture. The state can grant financial assistance to them in this regard.

3) Right to Constitutional Remedies :
This right enables the individuals to approach a high court under article 226 or the supreme court under article 32 to get any of the fundamental rights restored in case of their violation. The Supreme court and the state High courts issue various writs for the implementation of Fundamental Rights. Dr. Ambedkar described this right as the Heart and Soul of the constitution.

Question 3.
Describe the six Freedoms of a citizen. [Mar. 17]
Answer:
Our constitution in chapter III under Article 19 (clause 1) guarantees certain fundamental rights subject to certain restrictions. They are also known as fundamental freedoms. They are :

  1. Freedom of speech and expression.
  2. Freedom of peaceful Assembly without Arms.
  3. Freedom of Associations and Unions.
  4. Freedom of movement throughout the territory of India.
  5. Freedom of residence and settlement in any part of the Territory of India.
  6. Freedom of profession, trade occupation or business.

These freedoms would facilitate the progress of Indian citizens in social, political and economic spheres. These freedoms are not absolute. The state may, if necessary, impose certain reasonable restrictions on the enjoyment of the above freedoms by the Indian citizens. These restrictions relate to the maintenance and safe guarding of the independence, sovereignty, integrity, law and order.

Question 4.
Write briefly on the right to Constitutional Remedies.
Answer:
This right is the most significant of all the fundamental rights. It extends protection and relief to those whose Fundamental Rights were abridged, confiscated or infringed by others including the public, authorities. As this right gives a citizen the right to approach a rights restored in case of their violation. The supreme court and the High court can issue orders and give directions to the governments for the enforcement of the Fundamental Rights. The courts can issue various writs like habeas corpus, Mandamus, prohibition, Quowarranto and certiorari Dr. Ambedkar rightly described this right as Heart and Soul of the constitution.

Question 5.
Explain any five differences between Fundamental rights and Directive principles of state policy. [Mar. 16]
Answer:

Fundamental Rights Directive Principles
1) The concept was borrowed from American constitution. 1) The concept was borrowed from Irish Constitution.
2) These aim at establishing a political democracy in the country. 2) These aim at establishing a social and economical democracy in the society.
3) These are justicable and then enforced by the courts. 3) These are non-justicable and cannot be enforced by the courts.
4) These are personal and Individualistic in nature. 4) These are societarian and socialistic in nature.
5) These promote the welfare of the Individual. 5) These promote the welfare of the society.

Question 6.
Explain the important characteristics of Directive principles of State Policy.
Answer:
The following are the characteristics of the Directive principles :

  1. These are in the form of directives or instructions issued to the successive governments.
  2. These are positive in nature as they extend the jurisdiction of the powers and functions of the governments at various levels in India.
  3. Enforcement of these principles depends upon availability of financial resources.
  4. They are popular in nature as they aim at the establishment of egalitarian society.
  5. They are supposed to be implemented by any party in power irrespective of it’s policies and Ideology.
  6. Failure to implement these principles is not considered as a breach of law.
  7. They are non-justicable in nature as no one can force the governments to implement them immediately. The governments have discretion in implementing these principles.

Question 7.
Examine the implementation of Directive principles of state policy.
Answer:
Keeping in view the fact, Directive Principles of State Policy are fundamental in the governance of the country. The Union as well as the State Governments, since 1950, have been continuously taking various steps to implement them from to time. These are mentioned below:

  1. Abolition of Zamindari, Jagirdari and Inamdari systems.
  2. Introduction of Land Ceiling Acts.
  3. Abolition of Privy Purses.
  4. Nationalization of 14 leading commercial banks.
  5. Establishment of Khadi and Village Industries Board etc. .
  6. Organization of Village Panchayats.
  7. Reservation of seats are reserved for SCs and STs in educational institutions and representative bodies.
  8. Enactment of Ancient and Historical Monuments and Archeological Sites Remains Act 1951.
  9. Separating criminal procedure code from the executive.
  10. Prohibition of cow slaughter, calves and bullocks in some States.
  11. Establishment of primary health centers and hospitals throughout the country.
  12. Implementation of Non-Alignment and Panchasheel Principles.
  13. Initiation of old age pension schemes.
  14. Introduction of unemployment schemes.
  15. Enactment of Minimum Wages Act.
  16. Enactment of Wild Life Act.

Question 8.
Explain the significance of Directive principles of State Policy.
Answer:
Directive principles of state policy are considered as the most significant feature of the Indian constitution. They have great significance in the administration of our country. Though they lack legal sanction, they enjoy popular support and possess public sanction. If the party in power fails to implement these principles, it has to face resistance from the electorate in the coming elections.

So, no government can afford to ignore the implementation of these principles. The opposition, the press and the electorate would exert pressure on the government for implementing these principles. Hence the union government considered these principles as the basis for the functioning of National Planning Commission (or) NITI Ayog. They have helped the cburts in exercising their power of Judicial Review.

To conclude, directive principles of state policy are considered as an authoritative declaration of the aims and aspirations of Indians formulated by their representatives after solemn and mature deliberations as they are considered” as the goal of social Revolution”.

Very Short Questions

Question 1.
Fundamental Rights.
Answer:
Fundamental Rights are an important feature of Indian constitution. They are incorporated in part – III Articles from 12 to 35 in our constitution. They enable the citizens to realise their personality. They are :

  • Right to Equality
  • Right to Freedom
  • Right against exploitation
  • Right to Freedom of Religion
  • Educational and cultural rights
  • Right to constitutional remedies

Question 2.
Types of Directive principles.
Answer:
On the basis of their content and direction directive principles of state policy can be classified into three broad categories namely socialistic, Liberal – intellectual and Gandhian principles.

AP Inter 2nd Year Civics Study Material Chapter 2 Fundamental Rights and Directive Principles of State Policy

Question 3.
Habeas corpus. [Mar. 16]
Answer:
Habeas corpus literally means “To produce the body of”. It is in the nature of calling upon a person who has detained another to produce the latter before it. The court wants to know on what grounds a person has been detained. This writ frees a person whose detention has no legal justification.

Question 4.
Mandamus.
Answer:
Mandamus literally means “A mandate or command”. It is issued by a competent court for directing any person, corporation or inferior courts requiring him, it or them to do some particular, thing specified there in which appertains to his of their and is in the nature of public duty.

Question 5.
Cultural and Educational Rights.
Answer:
Article 29 guarantees to every citizen to protect his own language, script or culture. Article 30 provides that all minorities whether based on religion or language shall have the right to establish and maintain educational institutions of their choice.

Question 6.
Gandhian Ideas of Directive principles of state policy.
Answer:

  1. The state shall organize village panchayats and endow them with adequate powers and authority so as to enable them to function as the units of self government (Article 40).
  2. The state shall strive for the promotion of cottage industries on individual or cooperative basis in rural areas (Article 43).
  3. The state shall promote the educational and economic interests of the SCs, STs and BCs of society with special care (Article 46).
  4. The state shall endeavour to. bring about prohibition of intoxicating drinks and of drugs which are injurious to health (Article 47).

Question 7.
Significance of Fundamental Duties. [Mar. 17]
Answer:
Fundamental duties are considered most significant from the following view points.

  1. The Fundamental Duties act as a reminder to the citizens that while enjoying their rights, they should also be conscious of duties they owe to their country, their society and to their fellow citizens.
  2. They serve as warning against antinational and anti social activities.
  3. They serve as the source of inspiration for the citizens and promote sense of discipline and commitment among them.
  4. They help the courts in examining and determining the constitutional validity of a law.

Question 8.
Mention any three liberal principles.
Answer:

  1. The state shall secure for the citizens uniform civil code throughout the country (Article 44).
  2. The state organize Agriculture and Animal husbandry on modem and scientific lines, (Article 48).
  3. The state shall take steps to separate judiciary from executive in public services of the state (Article 50)

Question 9.
Quo – Warranto
Answer:
Quo – Warranto literally means ‘What Warrant or Authority”. It enables the competent court to enquire into the legality of the claim which a party assets to a public office and to oust him from its enjoyment if the claim is not well founded.

Question 10.
Right to Religion.
Answer:
This right denotes the secular nature of Indian political system. It aims at transforming India into a secular state. Both the citizens and aliens of India enjoy this right. Articles from 25 to 28 in part – III of Indian constitution deals with right to freedom of Religion. This right enables the individuals to profers, practice and propagate any religion according to their conscience.

AP Inter 2nd Year Civics Study Material Chapter 2 Fundamental Rights and Directive Principles of State Policy

Question 11.
Right against exploitation.
Answer:
In our country, there are millions of people who are underprivileged and deprived. They are subjected to exploitation by their fellow human beings. One such form of exploitations is ‘begar’ or ‘forced labour’ without payment. Another closely related form of exploitation is trafficking of human beings and using them as slaves. Both are prohibited under Article 23 of our constitutions.

Article 24 of the Indian constitution forbids all forms of child labour below the age of 14 years in factories, mines and other hazardous industries.

AP Inter 2nd Year Civics Study Material Chapter 1 The Constitution of India

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 1st Lesson The Constitution of India Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 1st Lesson The Constitution of India

Long Answer Questions

Question 1.
What is a constitution? Explain the historical background of the Indian Constitution.
Answer:
The term Constitution implies a written document embodying the provisions relating to the powers and functions of the government organs, the rights of the people, and their relations with the government. In simple,

Powers and Functions of the state + Rights and Duties of the Citizens = Constitution

Historical Background of the Indian constitution :
The constitution of India is the best constitution in the world.

It was formulated by Drafting Committee headed by Dr. B.R. Ambedkar and approved by the Constituent Assembly Chaired by Dr. Rajendra Prasad.

Earlier the Leaders of the Indian National Movement demanded several times for a separate constitution for Indians. They at first emphasized, during the second decade of the 20th century, that the British government shall grant self-rule or the right to formulate a constitution for Indians. It is in this context that leaders like Bala Gangadhara Tilak and Annie Besant launched the Home Rule movement. They requested the British government to apply the “right of self-determination”.

The Swaraj Party Leaders in 1922 made it clear that constitutional arrangements must be made for providing representation to Indians in Legislative Assembly.

Later Motilal Nehru moved a resolution in the Central Legislative Assembly requesting the British government for convencing a Round Table Conference for formulating a constitution for Indians.

The Three Round Table conferences held in London in 1930, 1931 and 1932 and their recommendations led to the passage of Government of India Act, 1935.

The Act provided for the introduction of Federal Polity and the establishment of provincial autonomy in the British India Provinces.

In the wake of Second World War, the Congress Ministers in the provincial legislative councils proposed a resolution meant for recognizing India as independent state and the power to make a constitution by the provincial legislative councils.

Viceroy Linlithgow through his August offer (1940) announced for the first time that Indians must Cooperate with Britain in the Second World War and the primary responsibility of making a constitution wholly vests with the Indians.

In 1942 Prime Minister Winston Churchill made efforts for resolving the Indian constitutional crisis by sending Stafford Cripps, a member of war cabinet to India for extensive discussions with Indian leaders.

The Cripps Mission (1942) was a failure. In this back drop, Indian National Congress under Gandhiji’s leadership adopted the famous “Quit India Revolution on 8th August, 1942 for the immediate ending of British rule in India.

After the Second World War, the labour party came to power in Britain under the leadership of element Attlee.

Then Viceroy Lord Wavell announced the latest policy of government in September, 1945. The British Government despatched a three member cabinet mission in February 1946 to India for making deliberations with Indian leaders on the issue of setting up of the Constituent Assembly and the provision of independence for Indians.

The Cabinet Mission made it clear that elections will be held to the Constituent Assembly and citizens having franchise will elect the members. Except Muslim League all the parties in India have agreed broadly the suggestions of the above team.

Lord Wavell formed on interion government with Jawaharlal Nehru as it’s head in 1946, with an increasing communal violence it seems there is no alternative to partition of the sub-continent.

The Mount Batten plan (1947) provided for the partition of the country. The Indian Independence Act of 1947 provided for setting up of a dominion of India and a dominion of Pakistan.

The Native states were given the choice of joining and becoming an integral part of either of the two dominions.

The vision and constructive statemanship of Sardar Vallabhai Patel the deputy Prime Minister and Home Minister facilitated the peaceful merger of most of the Native states into the Indian union.

Of course, the merger of Jammu and Kashmir, Junagadh and Hyderabad states into the Indian Union provided to be a difficult task.

AP Inter 2nd Year Civics Study Material Chapter 1 The Constitution of India

Question 2.
Explain in brief the salient features of Indian Constitution. [Mar. 18, 16]
Answer:
Introduction :
The Indian constitution was prepared and adopted by the Constituent Assembly which was set up in 1946. The Constituent Assembly took nearly three years (From 9th December, 1946 to 25th November, 1949) to complete the framing of the constitution.

The Constituent Assembly approved the Indian Constitution on 26th November, 1949. The Indian constitution came into force on 26th January, 1950. Which we have been celebrating as “The Republic Day”. The following are the salient or basic features of the Indian constitution.

1) Preamble :
The Indian constitution begins with a preamble. The preamble clearly defines the objectives of our constitution. It declares India as a Sovereign, Socialist, Secular, democratic Republic. It provides liberty, equality, fraternity and justice. It states that the people of India are the chief source of the political authority.

2) A lengthy Written Document:
The Constitution of India is the most written, lengthy and detailed document in the world. In 1950, the Indian Constitution had 22 parts, 8 schedules and 395 articles. After subsequent amendments, it contains 12 schedules and 444 articles for the present.

3) A combination of rigidity and flexibility :
The constitution of India is a blend of rigidity and flexibility. Our founding Fathers were programatic enough to provide for rigidity and flexibility as the situation demands. Article 368 provides the details of the amendment procedure.

i) Some of the provisions like admission of New states (Ex : Telangana), provisions relating to citizenship, salaries and allowances of the members of the constitutional bodies like President, Vice-President, Supreme Court and High Court judges etc. Can be amended by simple majority. It is said to be flexible.

ii) Some provisions can be amended by a special majority i.e., not less than Two – Thirds of the members of the House present and voting.
Ex : Fundamental Rights, Directive Principles of state policy etc.

It is said to be half rigid and half flexible.

iii) Some provisions can be amended by two-thirds majority of the Parliament and with the concurrence of half of the states.
Ex: Election of the President, executive powers of the union and the states, distribution of legislative powers between the union and the states etc. It is said to be rigid.

4) Quasi-Federal polity:
India is a states according to the constitution. Our constitution contains both the features of unitary and Federal Governments. It prescribed unitary system in emergencies and Federal system on ordinary occassions. Provisions of unitary state such as Single Citizenship, Single Integrated Judiciary, Single Election Commission, Role of All India Services Personnel etc., are found in our constitution. At the same time certain federal features like written, rigid constitution, Dual government, Bicameralism etc., are profoundly seen in our constitution. Thus it is a Quasi-Federal. Polity like Canada.

5) Republican government :
Unlike the colonial Master, the UK, India preferred a Republican government. Here all public offices right from World member to the top president of India are open to all eligible citizens and there is no place for hereditary principle.

6) Parliamentary government :
The constitution of India provided Parliamentary government of the British type but with an elected President of Irish Model. Accordingly, the features of Parliamentary government such as two executive heads, Ministerial accountability to the lower house of the legislature, Prime Minister leadership etc., are prevalent in our political system.

7) Fundamental Rights and Fundamental Duties:
Part – III of the constitution, Articles from 12 to 35, provides for a set of basic human rights to ail. They are justiceable and ensure basic freedoms. They are six in number. 1) Right to equality 2) Right to freedom 3) Right against exploitation 4) Right to religion 5) Cultural and educational right and 6) Right to constitutional remedies.

The 42nd Amendment to the constitution in 1976 incorporated the fundamental duties in Article 51A under part – IVA. Though they are not justiciable. But they put an obligation on the citizens to render certain duties in return for the protection they have been enjoying through fundamental rights. Fundamental duties relate to respecting the constitution, the National Flag and National Anthem, safeguarding public property etc.

8) Single citizenship :
Our Constitution provides for single citizenship for all persons who are born in India and who resided in India for a specific period. It enables the citizens to possess and enjoy identical rights and privileges. It also promotes unity, integrity and fraternity among the people.

9) Universal and Adult Franchise :
The makers of the Indian constitution provided for the universal adult Franchise for all citizens without any discrimination based on caste, colour, creed, community, language, religion, region, sex, property etc. At the beginning, adult Franchise was given to all the citizens who attained the age of 21 years. Later voting age was reduced to 18 years through the 61st constitution, Amendment Act in 1988.

10) Secular state :
Our constitution stands for a secular state. It does not uphold any particular religion as the official religion of. the Indian state. It ensures complete religious freedom to the people. It abolishes discrimination between individuals on religious grounds in the matters of employments education and legislation. It prohibits religious instructions in state owned or state – aided educational institutions.

11) Independent judiciary :
The Judiciary performs its functions independently. The legislature or the executive shall not interfere in the working of the Judiciary. The Judiciary carries on its obligations according to the constitutional norms and democratic principles.

12) Directive principles of state policy:
Our constitution hinted out certain directive principles as the policy of the state in part IV from Articles from 36 to 51. These principles are drawn from Irish constitution. These principles reflect the welfare state concept. They are the directions to be followed by the various governments. Equal pay for equal work, provision of employment opportunities, fair distribution of wealth, old age pension, protection of illhealth, provision of education, protection of women and children etc., are the some examples of these principles. Though these principles are non-justiciable, No responsible government can afford to ignore them.

13) Bi-cameralism:
The constitution of India introduced Bi-cameralism at the National level. Accordingly, The Indian Parliament consists of two houses namely the Rajya Sabha (upper house) and the Lok Sabha (lower house). While the Rajya Sabha represents the states, the Lok Sabha represents the people.

14) Panchayati Raj and Nagar Palikas Acts :
The Panchayati Raj and Nagar Palikas Acts are recent features of our constitution. The 73rd and 74th Constitutional Amendment Acts gave constitutional recognition to the Rural and Urban local governments. Which came into force in 1993 and 1994 respectively had become operative all over the territory of India. The ideals of democratic decentralisation or the grass roof democracy are realised by these acts.

These acts provides for adequate representation for women, scheduled castes, scheduled tribes and other weaker sections in the policy making bodies of the local governments.

Short Answer Questions

Question 1.
Write briefly the elements of the constitution.
Answer:
The term constitution implies a written document embodying the provisions relating to the powers and functions of the government organs, the rights of the people and their relations with the government.
Elements of the constitution :
1) The constitution prescribes a set of basic rules that ensure coordination amongst the members and groups of a society. The constitution specifies the basic allocation of power in a society. It decides who frames the laws. For example, in a democratic country like, the people through their elected representatives make the laws i.e., the Parliament at national level and state legislatures at state level are empowered to make laws.

2) It specifies the structure of the government and its limitations of the modern govemrttents are made up of three organs, viz.
i) Legislature which frames the laws within the limits set by the constitution.
ii) The executive, the President or the Governor, the Council of Ministers headed by the Prime Minister or Chief Minister would be taking policy decisions as per the guidelines provided by the Parliament or Legislature.
iii) The Judiciary by interpreting the laws would set limits on the powers of both the legislature and executive.

Through the Judiciary, the constitution ensures limited and responsible government.
For example : Article 13 of the Indian constitution establishes the Supremacy of the constitution.

3) The constitution establishes the relationship between the rulers and the ruled. Fundamental rights and Fundamental Duties spell out in detail the mutual obligations of the citizens and the state. Part III and Part IVA of the Indian constitution bind the state and the citizens towards each other.

4) Every society has certain aspirations and goals. The state came into existence to fulfill the bare needs of the people and continues to exist for the good life of all. It is the constitution which directs the state i.e., government to make certain policies for welfare of the people.
Ex : Directive principles of state policy which incorporated in part – IV of the Indian
1 Constitution.

5) The constitution, as the Supreme document, serves as a shock absorber in limiting the fluctuations of present and future generations. It is a living document that connects the past with the present and assures a predictable future. Modem societies cannot survive without a just constitution.

AP Inter 2nd Year Civics Study Material Chapter 1 The Constitution of India

Question 2.
Write about the making of Indian constitution.
Answer:
While Negotiations were going on about the modalities of transfer of power, a l Constituent Assembly was set up to draw the constitution for India. The Cabinet Mission and the major political parties reached an agreement over the constitution of Constituent Assembly in 1946.

Elections to the Constituent Assembly were held in July, 1946 in which 292 members from British provinces, 93 members from Native states and 4 members from central provinces were elected. The Constituent Assembly of India held its first meeting on December 9, 1946. It elected Sachchidanand Sinha as its protemporaiy chairman.

On December 11, 1946 it elected Dr. Babu Rajendra Prasad as its permanent chairman.

The making of the constitution really began at its third session held between April 22 and May 2, 1947.

The Fourth session of the Assembly was held on July 14 and continued till July 31, 1947. It held discussions on Model constitution, Adopted the National Flag. The Assembly met for the fifth time on the eve of the independence day.

On August 29, 1947, it set up a seven member Drafting committee with Dr. B.R. Ambedkar as its chairman.

The Drafting Committee presented the Draft constitution on February 21, 1948. The Constituent Assembly adopted the Draft constitution on November 26, 1949. The last session of the Assembly was held on January 24, 1950.

It elected Dr. Babu Rajendra Prasad as the first President of Indian Republic under the new constitution.

On January 26, 1950 the New constitution came into operation and India was declared as a Republic state.

Question 3.
Write a note on the sources of Indian constitution. [Mar. 18, 16]
Answer:
The constitution of India was formulated on the basis of various experiences. Almost all the noble features of the world constitutions have been incorporated in it. Similarly the peculiar political, social and cultural conditions present in India have been taken into account at the time of drafting the constitution. The makers of Indian constitution ignored those conditions prevalent in other countries which are contrary to the socio-economic and political background of India.

On the whole the following sources figure prominently in making the Indian constitution.

  1. Many provision of Indian constitution have been drawn on the basis of West Minister Model (british). These include parliamentary traditions, rule of law, cabinet government, legislative-executive relations, single citizenship’ nominal executive head etc.
  2. Some provisions like fundamental rights, judicial review, federal system, president’s election, impeaching the president etc., have been taken from the American constitution.
  3. Items relating to directive Principles of State Policy have been drawn from the constitution of Ireland.
  4. The emergency powers of the President have been taken basing on the German constitution.
  5. Matters such as Concurrent List, Business, Commerce, Inter State trade, Special privileges of legislators etc., have been added on the model of Australia.
  6. The Union-State relations mentioned in Indian constitution have been designed basing on the Canadian constitution.
  7. Matters of constitutional amendment procedure were drawn from South African constitution.
  8. The idea of republic and the ideals of liberty, equality and fraternity have been taken from the constitution of France.
  9. Most of provisions of Indian constitution were drawn from the Government of India Act, 1935.

Question 4.
Explain any three salient features of Indian constitution.
Answer:
1) A combination of rigidity and flexibility :
The constitution of India is a blend of rigidity and flexibility. Article 368 provides the details of the amendment procedure. Some Of the’provisions like Admission of New states (Ex : Telanganaj, provisions relating to citizenship, salaries and allowances of the President, Vice president, Supreme Court and High Court judges etc., can be amended by simple majority. It is said to be flexible.

Some provisions can be amended by a special majority i.e., not less than two-thirds of the members of the House present and voting. Ex : Fundamental Rights, Directive principles of state policy etc. It is said to be half rigid and half flexible. Some provisions can be amended by two-thirds majority of the Parliament and with the concurrence of half of the states.
Ex : Election of the President, Distribution of Legislative powers between the union and the states etc. It is said to be rigid.

2) Republican government:
Unlike the colonial master, the U.K., India preferred a Republican government. Here all public offices right from ward member to the top of India are open to all eligible citizens and there is no place for hereditary principle.

3) Single citizenship :
Our constitution provides for single citizenship for all persons who are born in India and who resided in India for a specific period. It enables the citizens to possess and enjoy identical rights and privileges. It also promotes unity, integrity and fraternity among the people.

Question 5.
“The preamble is the soul of the constitution”. Explain.
Answer:
Preamble is the most salient features of Indian constitution. It denotes the essential philosophy of Indian constitution. It reflects the aims, aspirations and objectives of the makers of Indian constitution. Jawaharlal Nehru described the Preamble as a declaration, a firm resolve, a pledge, an understanding and something more than a resolution. Preamble begins with the sentence “we, the people of India’ adopt, enact and give to ourselves this constitution”.

It declared India as a sovereign socialist, secular, democratic and republic. It announced that Indians enjoy liberty, equality, fraternity and justice. It clearly stated that sovereignty in India belongs the people of India. Justice Madholkar pronounced that Preamble is the essence of Indian constitution. Justice Hidayatullah praised the Preamble as “the soul of the constitution”.

The Preamble after the Constitution (Forty Second Amendment) Act, 19718 is as follows :

We, the people of India having solemnly resolved to constitute India into a Sovereign Socialist Secular Democratic Republic and to secure to all its citizens.

“Sovereign” – which means its authority within the country is undisputed and externally it is free from others contort. “Socialist” means a state that doesn’t, allow any kind of exploitation – social, economic and political.

“Secular” – It gives equal freedom to all religions.

“Democratic” – Stipulates that the sovereign power vests with the people. They exercise it periodically through universal Adult Franchise.

A “Republican” state assures that public offices are open to every citizen without any discrimination. There is no place for hereditary principle.

  1. Justice – social, economic and political.
  2. Liberty of-thought, expression, belief, faith and worship.
  3. Equality of status and of opportunity; and to promote among them all.
  4. Fraternity assuring the dignity of the individual and the unity and integrity of the nation.

In our Constituent Assembly this twenty-sixth day of November, 1949 do hereby adopt, enact and give to ourselves this Constitution”.

Very Short Answer Questions

Question 1.
Written constitutibn
Answer:
The constitution of India is the most written, lengthy and detailed document in the world. In 1950 the Indian constitution had 22 parts, 8 schedules and 395 articles. After subsequent amendments, it contains 12 schedules and 444 articles for the present.

AP Inter 2nd Year Civics Study Material Chapter 1 The Constitution of India

Question 2.
Rigid constitution
Answer:
Rigid constitution is one whose provisions cannot be charged easily. In this system the constitutional amendment methods are different from those of ordinary laws. There will be a special procedure for amending the provisions of the rigid constitution. The rigid constitution will have firmness due to its special procedures of amendment.
Ex : United states and India.

Question 3.
Parliamentary form of government
Answer:
The constitution of India provided Parliamentary Government of the british type but with an elected president of Irish model. According to the features of Parliamentary government such as two executive heads, ministerial accountability to the lower house, of the legislature, Prime Minister, leadership etc., are prevalent in our political system.

Question 4.
Fundamental Rights and Duties
Answer:
Fundamental Rights and Fundamental Duties are the salient features of Indian constitution.

Six Fundamental Rights were incorporated in Part – III articles from 12 to 35. They are justiciable and ensures basic freedoms to all Indians.

Eleven fundamental duties were incorporated in Part IV A under article 51 A. They put an Obligation on the citizens to render certain duties in return for the protection they have been enjoying through Fundamental Rights.

Question 5.
Secular state
Answer:
Our constitution stands for a secular state. It does not uphold any particular religion as the official religion of the Indian state. It ensures complete religious freedom to the people it abolishes discrimination between individuals on religious grounds in the matters of employment,’ education and legislation. It prohibits religious instructions’in state-owned or state-aided educational institutions.

Question 6.
Universal Adult Franchise
Answer:
The makers of the Indian constitution provided for the universal adult Franchise for all citizens without any discrimination based on caste, colour, creed, community, language, religion, region, sex, property etc. At the beginning, adult Franchise was given to all .the citizens who attained the age of 21 years. Later voting age was reduced to 18 years through the 61st constitution Amendment Act in 1988.

Question 7.
Bicameralism
Answer:
The constitution of India introduced Bi-cameralism at the national level. Accordingly, the Indian Parliament consist of two houses namely the Rajya Sabha (upper house) and the Lok Sabha (Lower house). While the Rajya Sabha represents the states, the Lok Sabha represents the people.

Question 8.
Directive principles of state policy
Answer:
Our constitution hinted out certain directive principles as the policy of the state in part IV from Articles from 36 to 51. These principles are drawn from Irish constitution. These principles reflect the welfare state concepts. They are the directions to be followed by the various governments. Though these principles are non-justiciable, No responsible government can afford to ignore them.

Question 9.
Independence of Judiciary
Answer:
The Judiciary performs its functions independently. The legislature of the executive shall not interfere in the working of the judiciary. The Judiciary carries on its obligations according to the constitutional norms and democratic principles.

AP Inter 2nd Year Civics Study Material Chapter 1 The Constitution of India

Question 10.
Preamble [Mar. 18, 17, 16]
Answer:
The Indian constitution begins with a preamble. The preamble clearly defines the objectives of our constitution. It declares India as a sovereign, as a socialist, Secular Democratic Republic. It provides liberty, equality, fraternity and justice. It states that the people of India are the chief sources of the political authority.

AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 16th Lesson Communication Systems Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 16th Lesson Communication Systems

Very Short Answer Questions

Question 1.
What are the basic blocks of a communication system?
Answer:
Basic blocks in a communication system are

  1. Transmitter
  2. Receiver
  3. Channel.

Question 2.
What is “World Wide Web” (WWW) ?
Answer:
Tern Berners -Lee invented the World Wide Web.
It is an encyclopedia of knowledge accessible to every one round the clock through out the year.

AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems

Question 3.
Mention the frequency range of speech signals.
Answer:
Speech signals frequency range is 300 Hz to 3100 Hz.

Question 4.
What is sky wave propagation ?
Answer:
In the frequency range from a MHz upto about 30 MHz, long distance communication can be achieved by ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave propagation.

Question 5.
Mention the various parts of the ionosphere ?
Answer:
Parts of ionosphere are

  1. D (Part of stratosphere (65-70 km day only),
  2. E (Part of stratosphere (100 km day only),
  3. F1 (Part of mesosphere (170 km – 190 km),
  4. F2 (Part of thermosphere [300 km at night 250 – 400 km during day time]).

Question 6.
Define modulation. Why is it necessary ? (T.S. Mar.’19, 16, 15; A.P. Mar. 16, 15) (Mar. ’14)
Answer:
Modulation : The process of combining low frequency audio signal with high frequency carrier wave is called modulation.

Necessary : Low frequency signals cannot transmit directly. To reduce size of the antenna and to avoid mixing up of signal from different transmitters modulation is necessary.

Question 7.
Mention the basic methods of modulation. (A.P. Mar. ’19, ’16, T.S. Mar. ’15)
Answer:
The basic methods of modulation are :

  1. Amplitude modulation (AM)
  2. Frequency modulation (FM)
  3. Phase modulation (PM)

AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems

Question 8.
Which type of communication is employed in Mobile Phones ? (A.P. Mar. ’15)
Answer:
Space wave mode of propagation is employed in mobile phones.

Short Answer Questions

Question 1.
Draw the block diagram of a generalized communication system and explain it briefly.
Answer:
Every communication system has three essential elements :

  1. Transmitter
  2. Medium / Channel
  3. Receiver.

The block diagram is shown in Fig.
AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems 1
In Communication system, the transmitter and the receiver are located at two different places separate from the transmitter.

The channel is the physical medium that connects. The purpose of transmitter is to convert the message signal produced by the source of information into a form, suitable for transmission through the channel. If the output of the information source is a non-electrical signal like a voice signal, a transducer converts it to electrical form before giving it as an input to the transmitter. When a transmitted signal propagates along the channel it may get distorted due to channel imperfection. Moreover, noise adds to the transmitted signal and the receiver receives a corrupted version of the transmitted signal. The receiver has the task of operating on the received signal. It reconstructs a recognizable form of the original message signal for delivering it to the user of information.

Question 2.
What is a Ground wave ? When is it used for communication ?
Answer:
Ground Wave : To radiate signals with high efficiency, The antennas should have a size comparable to the wavelength λ of the signal (at least ~ λ/4). At longer wavelengths (i.e., at lower frequencies), the antennas have large physical size and they are located on or very near to the ground. In standard AM broadcast, ground based vertical towers are generally used as transmitting antennas. For such antennas, ground has a strong influence on the propagation of the signal. The mode of propagation is called surface wave propagation and the wave glides over the surface of the earth. A wave induces current in the ground over which it passes and it is attenuated as a result of absorption of energy by the earth. The attenuation of surface waves increases very rapidly with increase in frequency. The maximum range of coverage depends on the transmitted power and frequency (less than 2 MHz). Ground waves will propagate long distances over sea water due to its high conductivity.

Question 3.
What are Sky Waves ? Explain Sky Wave propagation, briefly.
Answer:
Sky Waves : Long distance communication between two points on the earth is achieved through reflection of electromagnetic waves by Ionosphere, Such waves are called sky waves.

This mode of propagation is used by short wave broadcast services. The Ionosphere is so called because of the presence of a large number of ions or charged particles. It extends from a height of ~ 65 Km to about 400 km above the earth’s surface.

The degree of ionisation varies with the height. The density of atmosphere decreases with height. At greater heights the solar radiation is intense but there are few molecules to be ionized. Close to the earth, the radiation intensity is low so that the ionization is again low. However at some intermediate heights, there occurs a peak of ionization density. The ionospheric layer acts as a reflector for a certain range of frequencies (3 to 30 MHz).
Electromagnetic waves of frequencies higher than 30 MHz penetrate ionosphere and escape. This phenomenon is shown in the Fig.
AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems 2
The phenomenon of bending of em waves is so that they are diverted towards the earth which is similar to Total Internal Reflection in optics.

AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems

Question 4.
What is Space Wave Communication ? Explain.
Answer:
A spcae wave travels in a straight line from transmitting antenna to the receiving antenna. Space waves are used for line – of – sight (LOS) communication as well as satellite communication. At frequencies above 40 MHz, communication is essentially limited to line-of-sight paths. At these frequencies, the antennas are relatively smaller and can be placed at heights of many wavelengths above the ground. Because of line-of-sight nature of propagation, direct waves get blocked at some point by the curvature of the earth as illustrated in Fig. If the signal is to be received beyond the horizon then the receiving antenna must be high enough to intercept the line-of-sight waves.

If the transmitting antenna is at a height hT then it can be shown that the distance to the horizon dT is given as dT = \(\sqrt{\left(2 \mathrm{Rh}_{\mathrm{T}}\right)}\) where R is the radius of the earth (approximately 6400 km). Similarly if the receiving antenna is at a height hR, the distance to the horizon dR is dR = \(\sqrt{\left(2 \mathrm{Rh}_{\mathrm{R}}\right)}\). With reference to Fig. the maximum line-of sight distance dM between the two antennas having heights hT and hR above the earth is given by dM = \(\sqrt{\left(2 \mathrm{Rh}_{\mathrm{T}}\right)}\) = \(\sqrt{\left(2 \mathrm{Rh}_{\mathrm{R}}\right)}\)

Television broadcast, microwave links and satellite communication are some examples of communication systems that use space wave mode of propogation.
AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems 3

Question 5.
What do you understand by modulation ? Explain the need for modulation.
Answer:
The process of combining audio frequency signal with high frequency signal is called modulation.
To transmit an electronic signal in the audio frequency range (20 Hz to 20 KHz) over a long distance directly the following constraints limit the possibility :

  1. Size of the antenna
  2. Effective power radiated by the antenna
  3. Mixing up of signals from different transmitters.

For 20 KHz signal the height of antenna is about 4km, still a large height. This makes antenna length impractical. Even if we transmit the signal, they may combine with low frequency signals present in the atmosphere and it is impossible to distinguish the signals at the receiving end.

In order to avoid these problems a low frequency audio signal is combined with high frequency signal to translate the audio signal to high frequencies.

Question 6.
What should be the size of the antenna or aerial ? How the power radiated is related to length of the antenna and wavelength ?
Answer:
Size of antenna (or) aerial: For trasmitting a signal, we need an antenna. The size of the antenna comparable to the wavelength of the signal (at least λ/4). So that the antenna properly senses the time variation of the signal. For an e.m waves of frequency 20 kHz, the wavelength λ is 15 km. Obviously such a long antenna is not possible to construct and operate. There is a need of translating the information contained in our original low frequency base band signal into high frequencies before transmission.

Effective power radiated by an antenna : A linear antenna (length l) show that the power radiated is proportional to \(\frac{l}{\lambda^2}\). For the same antenna length, the power radiated increases with decreasing λ. i.e., increasing frequency. Hence the effective power radiated by a long wave length base band signal would be small.

Question 7.
Explain amplitude modulation.
Answer:
Amplitude modulation (AM.) : In amplitude modulation, the amplitude of carrier wave varies, but frequency and phase remains constant.
Here we can explain AM using a sinusoidal signal as a modulating signal.

Let C(t) = Ac sin ωc t represent carrier wave
m(t) = Am sin ωm t represent modulating signal.
The modulating Cm(t) can be written as
Cm(t) = (Ac + Am sin ωmt) sin ωCt
Cm(t) = Ac (1 + \(\frac{A_m}{A_c}\) sin ωmt) sin ωct —— (1)
Where ωm = 2πfm is the angular frequency of message signai
Note that the modulated signal now contains the message signal.
Cm(t) = Ac sin ωct + µ Ac sin ωmt sin ωc t —– (2)
Where µ = \(\frac{A_m}{A_c}\) = Modulation index.
To avoid distortion keep . µ ≤ 1
Cm(t) = Ac sin ωt + \(\frac{\mu A_c}{2}\) cos (ωc – ωm) t – \(\frac{\mu \mathrm{A}_{\mathrm{c}}}{2}\) cos (ωc + ωm)t —— (3)
Here (ωc – ωm) and (ωc + ωm) are lower side and upper side frequencies.
As long as the broadcast frequencies (camer wave) are sufficiently spaced out the side bands donot over lap.

Question 8.
How can an amplitude modulated wave be generated?
Answer:
AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems 4
The modulated signal Am sin ωmt is added to the camer wave signal Ac sin ωct to produce the signal x(t).
x(t) = Am sin ωm t + A sin ωct is passed through a square law device produces an output.
y(t) = B x(t) + Cx2 (t)
where B and C are constant.
This signal is passed through a band pass filter. The output of band pass filter produces AM wave, In band pass filter dc and the sinusoidal frequencies ωm, 2ωm and 2ωc rejects and retains the frequency ωc, (ωc – ωm) and (ωc + ωm)

AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems

Question 9.
How can an amplitude modulated wave be detected ?
Answer:
AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems 5
A block diagram of a typical receiver is shown in figure. Detection is the process of recovering the modulating signal from the modulated carrier wave.
We just saw that the modulated carrier wave contains the frequencies ωc and ωc ± ωm.. In order to obtain message signal m(t) of angular frequency ωm, a simple method is shown in figure.

The modulated signal is passed through a rectifier produces the output message signal. This message signal is passed through envelope detector (RC circuit).
AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems 6

Textual Exercises

Question 1.
Which of the following frequencies will be suitable for beyond the horizon communication using sky waves ?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Solution:
(b) 10 kHz frequencies cannot be radiated due to large antenna size, 1GHz and 1000 GHz will be generated. So option (b) is correct.

Question 2.
Frequencies in the UHF range normally propagate by means of:
(a) Ground waves
(b) Sky waves
(c) Surface waves
(d) Space waves
Answer:
(d). The frequencies in UHF range normally propagate by means.of space waves. The high frequency space does not bend with ground but are ideal for frequency modulation.

Question 3.
Digital signals
(i) do not provide a continuous set of values
(ii) represent values are discrete steps
(iii) can utilize binary system and
(iv) can utilize decimal as well as binary systems.
Which of the above statements are true ?
(a) (i) and (ii) only
(b) (ii) and (iii) only
(c) (i), (ii) and (iii) but not (iv)
(d) All of (i), (ii), (iii) and (iv).
Answer:
(c). A digital signal is a discontinuous function of time in contrast to an analogue signal. The digital signals can be stored as digital data and cannot be transmitted along the telephone lines. Digital signal cannot utilize decimal signals.

Question 4.
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication ? A TV transmitting antenna is 81m tall. How much service area can it cover if the receiving antenna is at the ground level ?
Answer:
No, it is not necessary for line of sight communication, the two antennas may not be at the same height.
Given, height of antenna h = 81 m.
Radius of earth R = 6.4 × 106 m.
Area = πd2; Range, d = \(\sqrt{2 \mathrm{hR}}\)
∴ Service area = π × 2πR = \(\frac{22}{7}\) × 81 × 2 × 6.4 × 106 = 3258.5 km2.

AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems

Question 5.
A carrier wave of peak voltage 12V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75% ?
Answer:
Given, Peak voltage V0 = 12V
Modulation index μ = 75% = \(\frac{75}{100}\)
We know that Modulation index (μ)
AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems 7
So, peak voltage of modulating signal Vm = μ × Peak voltage = \(\frac{75}{100}\) × 12 = 9V.

Question 6.
A modulating signal is a square wave, as shown in fig.
The carrier wave is given by c(t) = 2 sin (8πt) volts.
(i) Sketch the amplitude modulated waveform.
(ii) What is the modulation index ?
AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems 8
Answer:
Given, equation of carrier wave c(t) = 2 sin (8πt) —– (1)

(i) According to the diagram
Amplitude of modulating signal Am = 1V
Amplitude of carrier wave Ac = 2V
TM = 1s; Ωm = \(\frac{2 \pi}{T_m}\) = \(\frac{2 \pi}{1}\) = 2π rad/s —– (2)
From equation (1)
c(t) = 2 sin 8πt = Ac sin ωc
From equation (2)
so, ωc = 4ωm
Amplitude of modulated wave A = Am + Ac = 2 + 1 = 3V.
The sketch of amplitude modulated waveform is shown below.
AP Inter 2nd Year Physics Study Material Chapter 16 Communication Systems 9

(ii) Modulation index μ = \(\frac{A_m}{A_c}\) = 0.5

Question 7.
For an amplitude modulated wave, the maximum amplitude is found to be 10V while the ‘ minimum amplitude is found to be 2V. Determine the modulation index, μ. What would be the value of m if the minimum amplitude is zero volt ?
Answer:
Given, maximum amplitude Amax = 10V
Minimum amplitude Amin = 2V
Let Ac and Am be amplitudes of carrier wave and signal wave
Amax = Ac + Am = 10 —— (1)
and Amin = Ac – Am = 2 —– (2)
Adding the equations (1) & (2) we get 2 Ac = 12; Ac = 6V; Am = 10 – 6 = 4V
Modulation index μ = \(\frac{A_m}{A_c}\) = \(\frac{4}{6}\) = \(\frac{2}{3}\).
When the minimum amplitude is zero, then i.e., Amin = 0
Ac + Am = 10 —– (3)
Ac – Am = 0 —- (4)
By solving (3) & (4) we get 2 Am = 10; Am = 5; Ac = 5
Modulation index μ = \(\frac{A_m}{A_c}\) = \(\frac{5}{5}\) = 1.

Question 8.
Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.
Answer:
Let ωc be the angular frequency of carrier waves & ωm be the angular frequency of signal waves.
Let the signal received at the receiving station be e = E1. cos (ωc + ωm) t
Let the instantaneous voltage of carrier wave ec = E0 cos ωc t is available at receiving station.
Multiplying these two signals, we get
e × ec = E1Ec cos ωct. cos (ωc + ωm) t
E = \(\frac{E_1 E_c}{2}\) 2. cos ωct. cos (ωc + ωm) t. (Let e × ec = E)
= \(\frac{\mathrm{E}_1 \mathrm{E}_{\mathrm{c}}}{2}\) [cos (ωc + ωc + ωm) t + cos (ωc + ωm – ωc) t]
∵ [2 cos A cos B = cos(A + B) + cos (A – B)]
\(\frac{E_1 E_c}{2}\) = [cos (2 (ωc + ωm)t + cos ωmt]

Now, at the receiving end as the signal passes through filter, it will pass the high frequency (2ωc + ωm)t + cos ωm) but obstruct the frequency ωm. So we can record the modulating signal \(\frac{\mathrm{E}_1 \mathrm{E}_{\mathrm{c}}}{2}\) cos ωmt which is a signal of angular frequency to ωm.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits

Very Short Answer Questions

Question 1.
What is an n-type semiconductor? What are the majority and minority charge carriers in it?
Answer:
If a pentavalent impurity is added to pure tetravalent semiconductor, it is called n-type semiconductor.
In n-type semiconductor majority charge carriers are electrons and minority charge carriers are holes.

Question 2.
What are intrinsic and extrinsic semiconductors? (A.P. Mar. ’15)
Answer:
Pure form of semiconductors are called intrinsic semiconductors.
When impure atoms are added to increase their conductivity, they are called extrinsic semiconductors.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 3.
What is a p-type semiconductor? What are the majority and minority charge carriers in it?
Answer:
If a trivalent impurity is added to a tetravalent semiconductor, it is called p-type semi-conductor.
In p-type semiconductor majority charge carriers are holes and minority charge carriers are electrons.

Question 4.
What is a p-n junction diode ? Define depletion layer. (T.S. Mar. ’19)
Answer:
When an intrinsic semiconductor crystal is grown with one side doped with trivalent element and on the other side doped with pentavalent element, a junction is formed in the crystal. It is called p-n junction diode.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 1
A thin narrow region is formed on either side of the p-n junction, which is free from charge carriers is called depletion layer.

Question 5.
How is a battery connected to a junction diode in

  1. forward and
  2. reverse bias ?

Answer:

  1. In p-n junction diode, if p-side is connected to positive terminal of a cell and n-side to negative terminal, it is called forward bias.
    AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 2
  2. In a p-n junction diode, p-side is connected to negative terminal of a cell and n-side to positive terminal, it is called reverse bias.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 6.
What is the maximum percentage of rectification in half wave and full wave rectifiers ?
Answer:

  1. The percentage of rectification in half-wave rectifier is 40.6%.
  2. The percentage of rectification in full-wave rectifier is 81.2%.

Question 7.
What is Zener voltage (Vz) and how will a Zener diode be connected in circuits generally ?
Answer:

  1. When Zener diode is in reverse biased, at a particular voltage current increases suddenly is called Zener (or) break down voltage.
  2. Zener diode always connected in reverse bias.

Question 8.
Write the expressions for the efficiency of a full wave rectifier and a half wave rectifier.
Answer:

  1. Efficiency of half wave rectifier (η) = \(\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
  2. Efficiency of full wave rectifier (η) = \(\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)

Question 9.
What happens to the width of the depletion layer in a p-n junction diode when it is
i) forward-biased and
ii) reverse biased ?
Answer:
When a p-n junction diode is forward bias, thickness of depletion layer decreases and in reverse bias, thickness of depletion layer increases.

Question 10.
Draw the circuit symbols for p-n-p and n-p-n transistors. (Mar. 14) (A.P. Mar. 19; T.S. Mar. 16)
Answer:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 3

Question 11.
Deflne amplifier and amplification factor.
Answer:

  1. Rising the strength of a weak signal is known as amplification and the device is called amplifier.
  2. Amplification factor is the ratio between output voltage to the input voltage.
    A = \(\frac{\mathrm{V}_0}{\mathrm{~V}_{\mathrm{i}}}\)

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 12.
In which bias can a Zener diode be used as voltage regulator ?
Answer:
In reverse bias Zener diode can be used as voltage regulator.

Question 13.
Which gates ae called universal gates ?
Answer:
NAND gate and NOR gates are called universal gates.

Question 14.
Write the truth table of NAND gate. How does it differ from AND gate.
Answer:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 4

Short Answer Questions

Question 1.
What are n-type and p-type semiconductors? How is a semiconductor junction formed?
Answer:
n-type semiconductor : When a pure semiconductor is droped with pentavalent atoms like Arsenic, Antimony, Bismuth, then n-type semiconductor is formed.

p-type semiconductor: When a pure semiconductor is doped with trivalent atoms like Indium, Gallium. Al, p-type semiconductor is formed.

Formation of p-n junction diode: When p-type and n-type semiconductors are formed side by side at the junction, holes from p side diffuse to the n -side and electrons from n -side to p-side.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 5
Hence positive charge is built in n-side and negative charge is established at the p-side.
At the junction, the electrons and holes recombine and the region is free from charge carriers. This region which is free from charge carriers is called depletion layer.

The potential developed near the p-n junction is called barrier potential.
The potential barrier stops further diffusion of holes and electrons across the junction.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 2.
Discuss the behaviour of p-n junction. How does a potential barrier develop at the junction ?
Answer:
When p-n junction is formed, the free electrons on n-side diffuse over to p-side, they combine with holes and become neutral. Similarly holes on p-side diffuse over to n-side and combine with electrons become neutral.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 6
This results in a narrow region formed on either side of the junction. This region is called depletion layer. Depletion layer is free from charge carriers.

The n-type material near the junction becomes positively charged due to immobile donor ions and p-type material becomes negatively charged due to immobile acceptor ions. This creates an electric field near the junction directed from n-region to p-region and causes a potential barrier.

The potential barrier stops further diffusion of holes and electrons across the junction. The value of the potential barrier depends upon the nature of the crystal, its temperature and the amount of doping.

Question 3.
Draw and explain the current-voltage (I -V) characteristic curves of a junction diode in forward and reverse bias.
Answer:
A graph is drawn between the applied voltage (V) and the current (I) passing through the p-n junction diode is called characteristics of a diode.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 7
As forward bias voltage is increased potential barrier decreases, but initially increases in current is negligible (region OA). This is due to the potential barrier.

The voltage at which the current starts to increase rapidly is called knee voltage (or) cut in voltage.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 8
In this reverse bias, small current flows in the circuit due to minority charge carriers. 1f the reverse voltage is increased further, after some voltage, there will be a sudden rise of reverse current. This region is named as breakdown region.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 4.
Describe how a semiconductor diode is used as a half wave rectifier. (Mar. ’14) (T.S. & A.P. Mar. ’16)
Answer:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 9

  1. A half wave rectifier can be constructed with a single diode. The ac input signal is fed to the primary coil of a transformer. The output .signal is taken across the load resistance RL
  2. During positive half cycle, the diode is forward biased and current flows through the diode.
  3. During negative half cycle, diode is reverse biased and no current flows through the load resistance.
  4. This means current flows through the diode only during positive half cycles and negative half cycles are blocked. Hence in the output we get only positive half cycles.
  5. Rectifier efficiency is defined as the ratio of output dc power to the input ac power.
    η = \(\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{ac}}}\) = \(\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
    where rf = Forward resistance of a diode; RL = Load resistance
    The maximum efficiency of half wave rectifier is 40.6%.

Question 5.
What is rectification ? Explain the working of a full wave rectifier. (T. S. Mar. 19: A.P. & T.S. Mar. ’15)
Answer:
Rectification : The process of converting on alternating current into a direct current is called rectification.
The device used for this purpose is called rectifier.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 10

  1. A full wave rectifier can be constructed with the help of two diodes D1 and D2.
  2. The secondary transformer is centre tapped at C and its ends are connected to the p regions of two diodes D1 & D2.
  3. The output voltage is measured across the load resistance RL.
  4. During positive half cycles of ac, the diode D1 is forward biased and current flows through the load resistance RL. At this time D2 will be reverse biased and will be in switch off position.
  5. During negative half cycles of ac, the diode D2 is forward biased and the current flows through RL. At this time D1 will be reverse biased and will be in switch off position.
  6. Hence positive output is obtained for all the input ac signals.
  7. The efficiency of a rectifier is defined as the ratio between the output dc power to the input ac power.
    η = \(\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{ac}}}\) = \(\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
    The maximum efficiency of a full wave rectifier is 81.2%

Question 6.
Distinguish between half-wave and full-wave rectifiers. (A.P. Mar. ’19)
Answer:
Half wave rectifier

  1. Single diode is used.
  2. Only half wave is rectified.
  3. Rectifier efficiency η = \(\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
  4. Efficiency of half wave rectifier is 40.6%.
  5. Output is discontinuous and pulsative.

Full wave rectifier

  1. Two diodes are used.
  2. Full wave is rectified.
  3. Rectifier efficiency η = \(\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
  4. Efficiency of full wave rectifier is 81.2%.
  5. Output is continuous and pulsative.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 7.
Distinguish between zener breakdown and avalanche breakdown.
Answer:
Zèner break down

  1. Zener break down occurs at heavily doped diodes.
  2. This occurs at low reverse bias voltages.
  3. This occurs due to field emissîon.
  4. Width of depletion layer is small.

Avalanche break down

  1. Avalanche break down occurs at lightly doped diodes.
  2. This occurs at high reverse bias voltages.
  3. This occurs due to ionisation by collision.
  4. Width of depletion layer is also small.

Question 8.
Explain hole conduction in intrinsic semiconductors.
Answer:
Pure semiconductors are called intrinsic semiconductors. At low temperature, valency band is filled with electrons and conduction band is empty. Hence it acts as an insulator at low temperature.

As temperature increases electrons in valence band get energy and jumps into the conduction band crossing the forbidden band. At their places in the valency band a vacancy is created.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 11
This vacancy of electron in the valency band is called a hole. Hole has positive charge and move only in the valency band; giving hole current.

In this Fermi-energy level will be at the middle of the forbidden band.

Question 9.
What is a photodiode ? Explain its working with a circuit diagram and draw its I-V characteristics.
Answer:
Photodiode : Photodiode is an optoelectronic device in which current carriers are generated by photons through photo excitation.

Working : When visible light of energy greater than forbidden energy gap is incident on a reverse biased p-n juncton photodiode, additional electron-hole pairs are created in the depletion layer (near the junction). These charge carriers will be seperated by the junction field and made to flow across the junction. Creating a reverse current across the junction.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 12
The value of reverse saturation current increases with the increase in the intensity of incident light.
V-I characteristics of photodiode is shown in figure. It is found that reverse saturation current through the photodiode varies almost linearly with the light flux.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 13
Uses :

  1. It is used in switching the light ON and OFF.
  2. It is used in demodulation in optical signals.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 10.
Explain the working of LED and what are its advantages over conventional incandescent low power lamps.
Answer:
Light emitting diode (LED) : It is a photoelectronic device which converts electrical energy into light energy.
It is a heavily doped p-n junction diode which under forward bias emits spontaneous radiation. The diode is covered with a transparent cover so that the emitted light may be come out.

Working ; When p-n junction diode is forward biased, the movement of majority charge carriers takes place across the junction. The electrons move from n-side to p-side through the junction and holes move from p-side to n-side.
As a result of it, concentration of majority carriers increases rapidly at the junction.

When there is no bias across the junction, therefore there are excess minority carriers on either side of the junction, which recombine wito majority carriers near the junction.

On recombination of electrons and hole, die energy is given out in the form of heat and light.

Advantages of LED’S over incandescent lamp:

  1. LED is cheap and easy to handle.
  2. LED has less power and low operational voltage.
  3. LED has fast action and requires no warm up time.
  4. LED can be used in burglar alarm system.

Question 11.
Explain the working of a solar cell and draw Its I-V characteristics.
Answer:
Solar cell is a p-n junction device which converts solar energy into electric energy.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 14
It consists of a silicon (or) gallium – arsenic p-n junction diode packed in a can with glass window on top. The upper layer is of p-type semiconductor. It is very thin so that the incident light photons may easily reach the p-n junction.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 15

Working : When light (E = hv) falls at the junction, electron – hole pairs are generated near 13% junction. The electrons and holes produced move in opposite directions due to junction field. They will be collected at the two sides of the junction giving rise to a photo voltage between top and bottom metal electrodes. Top metal acts as positive electrode and bottom metal acts as a negative electrode. When an external load is connected across metal electrodes a photo current flows.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 16
I – V characteristics : I – V characteristics of solar cell is drawn in the fourth quadrant of the co-ordinate axes. Because it does not draw current.
Uses : They are used in calculators, wrist watches, artificial satellites etc.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 12.
Explain the different transistor configurations with diagrams.
Answer:
There are three configurations. They are

1) Common Base
2) Common emitter
3) Common collector.

1) Common Base configuration :
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 17
In this configuration base is earthed. Base is common both input and output. Input is given across base – emitter and output is taken across base-collector.

2) Common Emitter configuration :
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 18
In this configuration emitter is earthed. Emitter is common both input and output. Input is given across base – emitter and output is taken across collector – emitter.

3) Common collector configuration :
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 19
In this configuration collector is earthed. Collector is common both input and output. Input is given across base collector and output is taken across collector – emitter.

Question 13.
Explain how transistor can be used as a switch?
Answer:
To understand the operation of transistor as a switch.

  1. As long as Vi is low and unable to forward bias the transistor, V0 is high (at VCC).
  2. If Vi is high enough to drive the transistor into saturation then V0 is low, very near to zero.
    AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 20
  3. When the transistor is not conducting it is said to be switched off and when it is driven into saturation it is said to be switched on.
  4. We can say that a low input to the transistor gives a high output and high input gives a low output.
  5. When the transistor is used in the cutoff (or) saturation state it acts as a switch.
    AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 21

Question 14.
Explain how transistor can be used as an oscillator ?
Answer:

  1. In an oscillator, we get ac output without any external input signal.
    AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 22
  2. Here L – C circuit is inserted in emitter-base circuit of transistor winch is forward biased with battery VBB. The collector emitter circuit is reverse biased with battery VCC-.
  3. A coil L1 is inserted in collector emitter circuit. It is coupled with L.

Working:

  1. S we close the key (K), weak collector current start rising with time due to inductance L1. As a resell, increasing magnetic flux is finked with L1 and L.
  2. Due to mutual induction, an emf is induced in L which will charge the upper plate of capacitor (C), consequently there will be support to the forward biasing of emitter base circuit.
  3. This results in an increasing in the emitter current and hence an increase in the collector current.
  4. Due to it, more increasing magnetic flux is linked with L1 & L.
  5. The above process continues till the collector current becomes maximum (or) saturated,
  6. The resonant frequency of tuned circuit at which the oscillator will oscillate.
    v = \(\frac{1}{2 \pi \sqrt{L C}}\)

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 15.
Define WND and NOR gases. Give their truth tables.
Answer:
NAND gate : NAND gate is a combination of AND gate and NOT gate.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 23
NAND gate can be obtained by connecting a NOT gate in the output of an AND gate. NAND gates are called universal gates.

  1. If both inputs are low, output is high
    A = 0, B = 0, X = 1
    AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 24
  2. If any input is low, output is high.
    A = 0, B = 1, X = 1
    A = 1, B = 0, X = 1
  3. If both inputs are high, output is low.
    A = 1, B = 1, X = 0

NOR gate : NOR gate is a combination of OR gate and NOT gate when the output of OR gate is connected to NOT gate. It has two (or) more inputs and one output.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 25

  1. If both inputs are low, output is high
    A = 0, B = 0, X = 1
    AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 26
  2. If input is high, output is low.
    A = 0, B = 0, X = 1
  3. If both inputs are high, the output is low.
    A = 1, B = 1, X = 0
    NOR gate is also a universal gate.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 16.
Explain the operation of a NOT gate and give, its truth table.
Answer:
NOT gate: NOT gate is the basic gate. It has one input and one output. The NOT gate is also called an inverter. The circuit symbol of NOT gate is shown in figure.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 27

  1. If input is low, output is high.
    A = 0, X = \(\overline{0}\) = 1
    AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 28
  2. If input is high, output is low.
    A = 1, X = \(\overline{1}\) = 0

Long Answer Questions

Question 1.
What is a junction diode ? Explain the formation of depletion region at the junction. Explain the variation of depletion region in forward and reverse-biased condition.
Answer:
p-n junction diode: When a p-type semiconductor is suitably joined to n-type semiconductor, a p-n junction diode is formed.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 29
The circuit symbol of p-n junction diode is shown in figure.

Formation of depletion layer at the junction: When p-n junction is formed, the free electrons on n-side diffuse over to p-side, they combine with holes and become neutral. Similarly holes on p-side diffuse over to n-side and combine with electrons become neutral.

This results in a narrow region formed on either side of the junction. This region is called depletion layer. Depletion layer is free from charge carriers.

The n-type material near the junction becomes positively charged due to immobile donor ions and p-type material becomes negatively charged due to immobile acceptor ions. This creates an electric field near the junction directed from n-region to p-region and cause a potential barrier.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 30
The potential barrier stops further diffusion of holes and electrons across the junction. The value of the potential barrier depends upon the nature of the crystal, its temperature and the amount of doping.

Forward bias:
“When a positive terminal of a battery is connected to p-side and negative terminal is connected to n-side; then p-n junction diode is said to be forward bias”.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 31
The holes in the p-region are repelled by the positive polarity and move towards the junction.
Similarly electrons in the n-region are repelled by the negative polarity and move towards the junction.
As a result, the width of the depletion layer decreases. The charge carriers cross the junction and electric current flows in the circuit.

Hence in forward bias resistance of diode is low. This position is called switch on position.

Reverse bias:

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 32

‘When the negative terminal of the battery is connected to p-side and positive terminal of the battery is connected n-side of the p-n junction, then the diode is said to be reverse bias”.

The holes in the p-region are attracted towards negative polarity and move away from the junction. Similarly the electrons in the n-region are attracted towards positive polarity and move away from the junction.

So, width of the depletion layer and potential barrier increases. Hence resistance of p-n junction diode increases. Thus the reverse biased diode is called switch off position.

Question 2.
What is a rectifier ? Explain the working of half wave and full wave rectifiers with diagrams.
Answer:
Rectifier: It is a circuit which converts ac into d.c. A p-n junction diode is used as a rectifier.
Half-wave rectifier:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 33

  1. A half wave rectifier can be constructed with a single diode. The ac input signal is fed to the primary coil of a transformer. The output signal is taken across the load resistance RL.
  2. During positive half cycle, the diode is forward biased and current flows through the diode.
  3. During negative half cycle, diode is reverse biased and no current flows through the load resistance.
  4. This means current flows through the diode only during positive half cycles and negative half cycles are blocked. Hence in the output we get only positive half cycles.
  5. Rectifier efficiency is defined as the ratio of output dc power to the input ac power.
    η = \(\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{ac}}}\) = \(\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
    Where rf = Forward resistance of a diode; RL = Load resistance
    The maximum efficiency of half wave rectifier is 40.6%.

Full wave rectifier: The process of converting an alternating current into a direct current is called rectification.
The device used for this purpose is called rectifier.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 34

  1. A full wave rectifier can be constructed with the help of two diodes D1 and D2.
  2. The secondary transformer is centre tapped at C and its ends are connected to the p regions of two diodes D1 & D2.
  3. The output voltage is measured across the load resistance RL.
  4. During positive half cycles of ac, the diode D1 is forward biased and current flows through the load resistance RL. At this time D2 will be reverse biased and will be in switch off position.
  5. During negative half cycles of ac, the diode D2 is forward biased and the current flows through RL. At this time D1 will be reverse biased and will be in switch off position.
  6. Hence positive output is obtained for all the input ac signals.
  7. The efficiency of a rectifier is defined as the ratio between the output dc power to the input ac power.
    η = \(\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{ac}}}\) = \(\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
    The maximum efficiency of a full wave rectifier is 81.2%.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 3.
What is a Zener diode ? Explain how it is used as a voltage regulator.
Answer:
Zener diode : Zener diode is a heavily doped germanium (or) silicon p-n junction diode. It works on reverse bias break down region.

The circuit symbol of zener diode is shown in figure.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 35
Zener diode can be used as a voltage regulator. In general zener diode is connected in reverse bias in the circuits.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 36

  1. The zener diode is connected to a battery, through a resistance R. The battery reverse biases the zener diode.
  2. The load resistance RL is connected across the terminals of the zener diode.
  3. The value of R is selected in such away that in the absence of load RL maximum safe current flows in the diode.
  4. Now consider that load is connected across the diode. The load draws a current.
  5. The current through the diode falls by the same amount but the voltage drops across the load remains constant.
  6. The series resistance R absorbs the output voltage fluctuations, so as to maintain constant voltage across the load.
  7. The voltage across the zener diode remains constant even if the load RL varies. Thus, zener diode works as voltage regulator.
  8. If I is the input current, IZ and IL are zener and load currents.
    I = IZ + IL; Vin = IR + VZ
    But Vout= VZ
    ∴ Vout= Vin – IR

Question 4.
Describe a transistor and explain its wbking.
Answer:
Transistor: It is a device which consists of two p-n junctions joined back to back. Transistor means transfer of resistance.
Transistor has three regions. They are :

1) Emitter (E)
2) Base (B)
3) Collector (C)

1) Emitter (E) The section at one end of transistor is called emitter. It is heavily doped region. It emitts charge carriers.
2) Base (B) : The middle section off transistor is called base. This is lightly doped and very thin and almost of the charge carriers injected into it to flow into collector with out neutralised.
3) Collector (C) : The section at the other end is called collector. It is moderately doped. It collects the charge carriers. Physically it is large.

Note : Usually the emitter-base junction is forward biased and collector-base junction is reverse biased.

Working of p-n-p transistor : The base part is made of n-type, emitter and collector parts are ‘made of p-type. The circuit symbol of p-n-p transistor is shown in figure.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 37
The emitter junction is forward biased by connecting positive terminal off battery to the emitter and negative to the base. The collector junction is reverse biased by connecting negative to fee collector and positive to fee base.

The holes in p-region (emitter) are repelled by the positive terminal and crossing the emitter junction, enter the base region causing emitter current IE. Few holes are combined with electrons in base region, this constitute base current (IB). Majority of holes enters through the collector region. The negative terminal battery rapidly sweeps the holes in the collector region causing collector current (IC).
IE = IB – IC
In p-n-p transistor inside the circuit charge carriers are holes and outside the circuit charge carriers are electrons.

n-p-n transistor: If the base part is made of p-type, emitter and collector parts are made of n-type, we get n-p-n transistor. The circuit symbol of n-p-n transistor is shown in figure.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 38
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 39
The emitter junction is forward biased with negative end of a battery connected to emitter and positive to the base. The collector junction is reverse biased with collector positive end of a battery and negative to the base.

The electrons in the emitter (n-region) are repelled by the negative terminal of a battery and cross the base region constituting the emitter current IE. A small number of electrons may recombine with holes in base region constituting base current IB.

Majority charge carriers (electrons) enters into the collector region. The positive terminal of battery rapidly sweeps the electrons in collector, constituting collector current IC.
IE = IB + IC
In n-p-n transistor charge carriers inside and outside the circuit are electrons.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 5.
What is amplification ? Explain the working of a common emitter amplifier with necessary diagram.
Answer:
Amplification : The process of raising the strength of a weak signal is called amplification and the device which accomplishes this job is called an amplifier.
Amplifiers are of two types.

  1. Power amplifier
  2. Voltage amplifier.

Amplification factor: The ratio between output voltage to the input voltage is called amplification factor.
A = \(\frac{v_0}{v_i}\)
Common emitter transistor amplifier:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 40
The n-p-n common emitter amplifier circuit is shown in figure. In this circuit the battery VBB provides the biasing voltage VBE for the base emitter junction. The emitter junction is forward biased and the battery VCC provides the biasing voltage VCE for the emitter collector junction. The junction is reverse biased. Most of the electrons from emitter cross the base region and move into the collector.

The input signal to be amplified is connected in series with the biasing battery (VBB). A load resistance RL is connected in the collector circuit and output voltage is taken across RL.

As the base emitter voltage (VBE) changes due to input signal, the base current changes (IB) . This results in large change in collector current (ΔIC). The change in collector emitter voltage (ΔVCE) is taken across RL. Thus amplified output is obtained across RL.

Current gain (β) : The ratio of change in collector current to the change in base current is called current gain (or) current amplification factor.
β = \(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\)

Voltage gain (AV) : It is the ratio of change in output voltage (ΔVCE) to the change in input voltage (ΔVBE).

Power gain : It is defined as the product of current gain and voltage gain.
Power gain (Ap) = Current gain × Voltage gain.

Question 6.
Draw an OR gate using two diodes and explain its operation. Write the truth table and logic symbol of OR gate.
Answer:
OR Gate : It has two input terminals and one output terminal. If both inputs are low the output is low. If one of the inputs is high, or both the inputs are high then the output of the gate is high. The truth tables of OR gate.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 41
In truth table logic function is written as A or B ‘OR’ logic function is represented by the symbol ‘plus’
Q = A + B
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Logic gate ‘OR’ is shown given below.

Implementation of OR gate using diodes:
Let D1 and D2 be two diodes.
A potential of 5V represents the logical value 1
A potential of 0V represents the logical value 0
When A = 0, B = 0 both the diodes are reverse, biased and there no current through the resistance. So, the potential at Q is zero. i.e. Q = 0. When A = 0 or B = O and the other equal to a potential of 5V i.e., Q = 1. When both A and B which is 5V i.e., Q = 1. The output is same as that of the OR gate.

Question 7.
Sketch a basic AND circuit with two diodes and explain its operation. Explain how doping increases the conductivity in semiconductors?
Answer:
AND gate : It has two input terminals and one out terminal

  • If both the inputs are low (or) one of the inputs is low.
    ⇒ The out is low in an AND gate.
  • If both the inputs are high ⇒ The output of the gate is high

Note: If A and B are the inputs of the gate and the output is ‘Q’ then ‘Q‘ is a logical function of A and B.
AND gate Truth Tables
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 43
The logical function AND is represented by the symbol dot so that the output, Q = A.B and the circuit symbol used for the logic gate AND is shown in Fig.
The logical function AND is similar to the multiplication.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 44

Implementation of AND gate using diodes: Let D1 and D2 represents two diodes. A potential of 5 V represents the Logical value 1 and a potential of 0 V represents the logical value zero (O). When A = 0, B = 0 both the diodes D1 and D2 are forward-biased and they behave like closed switches. Hence, the output Q is same as that A or B(equal to zero.) When A = 0 or B = 0, D1 or D2 is forward – biased and Q is zero. When A = 1 and B = 1 both the diodes are reverse – biased and they behave like open switches. There is no current through the resistance R making the potential at Q equal to 5V i.e., Q = 1. The output is same as that of an AND gate.

Doping increases the conductivity in Semiconductors : If a pentavalent impurity (Arsenic) is added to a pure tetravalent semiconductor it is called n-type semiconductor. Arsenic has 5 valence electrons, but only 4 electrons are needed to form covalent bonds with the adjacent Germanium atoms. The fifth electron is very loosely bound and become a free electron. Therefore excess electrons are available for conduction and conductivity of semiconductor increases.

Similarly when a trivalent impurity Indium is added to pure Germanium it is called p-type semiconductor. In this excess holes in addition to those formed due to thermal energy are available for conduction in the valence bând and the conductivity of semiconductor increases.

Problems

Question 1.
In a half wave rectifier, a p-n junction diode with internal resistance 20 ohm is used. If the load resistance of 2 ohm is used in the circuit, then find the efficiency of this half wave rectifier.
Solution:
Internal resistance (rf) = 20Ω
RL = 2kΩ = 2000 Ω
η = \(\frac{0.406 R_L}{r_f+R_L}\) = \(\frac{0.406 \times 2000}{20+2000}\) × 100 = \(\frac{812 \times 100}{2020}\)
η = 40.2%

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 2.
A full wave p-n junction diode rectifier uses a load resistance of 1300 ohm. The internal resistance of each diode is 9 ohm. Find the efficiency of this full wave rectifier.
Solution:
Given that RL = 1300 Ω
rf = 9Ω
η = \(\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\) = \(\frac{0.812 \times 1300}{9+1300}\) × 100
η = \(\frac{8120 \times 13}{1309}\)
η = 80.64%

Question 3.
Calculate the current amplification factor β(beta) when change in collector current is 1mA and change in base current is 20μA.
Solution:
Change in collector current (ΔIC) = 1mA = 10-3 A
Change in base current (ΔIB) = 20μA = 20 × 10-6 A
β = \(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\) = \(\frac{10^{-3}}{20 \times 10^{-6}}\)
β = \(\frac{1000}{20}\)
β = 50

Question 4.
For a transistor amplifier, the collector load resistance RL = 2k ohm and the input resistance Ri = 1 k ohm. If the current gain is 50, calculate voltage gain of the amplifier.
Solution:
RL = 2kΩ = 2 × 103 Ω
Ri = 1kΩ = 1 × 103 Ω
β = 50.
Voltage gain (AV) = β × \(\frac{\mathrm{R}_{\mathrm{L}}}{\mathrm{R}_{\mathrm{i}}}\) = \(\frac{50 \times 2 \times 10^3}{1 \times 10^3}\)
AV = 100.

Textual Exercises

Question 1.
In an n-type silicon, which of the following statement is true :
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer:
(c). In an n-type semiconductor, it is obtained by doping the Ge or Si with pentavalent atoms. In n-type semiconductors, electrons are majority carriers and holes are minority carriers.

Question 2.
Which of the statements given in Exercise 1 is true for p-type semiconductors ?
Answer:
(b) & (d). p-type semiconductor is obtained by doping Ge or Si with trivalent atoms. In p-type semiconductor holes are majority carriers and electrons are minority carriers.

Question 3.
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)c, (Eg)Si and (g)Ge. Which of the following statements is true ?
(a) (Eg)Si < (Eg)Ge < (Eg)C
(b) (Eg)C < (Eg)Ge > (Eg)Si
(c) (Eg)C > (Eg)Si > (Eg)Ge
(d) (Eg)C = (Eg)Si = (Eg)Ge
Answer:
(c). The energy band – gap is largest for carbon, less for sillicon and least for Germanium.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 4.
In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
Answer:
(c). In an unbiased p-n junction, the diffusion of charge carriers across the junction takes place from higher concentration to lower concentration. Therefore, hole concentration in p-region is more as compared to n-region.

Question 5.
When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above.
Answer:
(c). When a forward bias is applied across the p-n junction, the applied voltage opposes the barrier voltage. Due to this, the potential barrier across the junction is lowered.

Question 6.
For transistor action, which of the following statements are correct:
(a) Base, emitter and collector regions should have similar size and doping concentrations.
(b) The base region must be very thin and lightly doped.
(c) The emitter junction is forward biased and collector junction is reverse biased.
(d) Both the emitter junction as well as the collector junction are forward biased.
Answer:
(b) and (c). For a transistor β = \(\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}\)
IB = \(\frac{I_C}{\beta}\) (Or) Rinput = \(\frac{V_{\text {input }}}{V_B}\) = \(\frac{V_{\text {input }}}{I_C}\) . β. i.e., Rinput ∝ \(\frac{1}{\mathrm{I}_{\mathrm{C}}}\)

Therefore Rinput is inversly proportional to the collector current. For high collector current, the Rinput should be small for which the base region must be very thin and lightly doped for a transistor action, the emitter junction is forward biased and collector junction is reverse biased.

Question 7.
For a transistor amplifier, the voltage gain
(a) remains constant for all frequencies.
(b) is high at high and low frequencies and constant in the middle frequency range.
(c) is low at high and low frequencies and Constant at mid frequencies.
(d) None of the above.
Answer:
(c). The voltage gain is low at high and low frequencies and constant at mid-frequency.

Question 8.
In half-wave rectification, what is the output frequency, if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency ?
Answer:
A half wave rectifier rectifies only the half of ac input i.e., it conducts once during an ac input cycle while a full wave rectifier rectifies both the half cycles of the ac input i.e., it conducts twice during a cycle.
The output frequency for half-wave is 50Hz.
The output frequency of a full-wave rectifier is 2 × 50 = 100Hz.

Question 9.
For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.
Solution:
Given, collector resistance Routput = 2KΩ = 2000Ω.
Current amplification factor of the transistor βAC = 100.
Audio signal voltage Voutput = 2V
Input (base) resistance Rinput = 1KΩ = 1000Ω
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 45

Question 10.
Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 26. If the input signal is 0.01 volt, calculate the output ac signal.
Solution:
Given, voltage gain of first Amplifier, \(\mathrm{A}_{\mathrm{V}_1}\) = 10
Voltage gain of second Amplifier, \(\mathrm{A}_{\mathrm{V}_2}\) = 20
Input voltage V1 = 0.01V
Total voltage gain Av = \(\frac{v_0}{v_i}\) = \(\mathrm{A}_{\mathrm{V}_1}\) × \(\mathrm{A}_{\mathrm{V}_2}\)
\(\frac{v_0}{0.01}\) = 10 × 20; V0 = 2V

Question 11.
A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm ?
Solution:
Energy(E) = \(\frac{h c}{\lambda}\) = \(\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-9} \times 1.6 \times 10^{-19}}\) eV = 2.06 eV.
The band – gap is 2.8 eV and energy E is less than band gap (E < Eg). So p-n junction cannot detect, file radiation of given wavelength 6000 nm.

Additional Exercises

Question 1.
The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1028 per m3 atoms of Indium. Calculate the number of electrons and boles. Gives that n,sub>i = 1.5 × 1016 m-3. Is the material n-type or p-type ?
Solution:
We know that for each atom doped of arsenic one free electron is received. Similarly, for each- atom doped of Indium a, vacancy is crested So, the number of free electrons introduced by pentavalent impurity added.
ne = NAs = 5 × 1022m3 —- (i)
The number of holes introduced by trivalent impurity added.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 46
As number of electrons ne (= 4.95 × 1022) is greater than number of holes nh (= 45 × 109). So, the material is n-type semiconductor.

Question 2.
In an intrinsic semiconductor the energy gap Eg is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300 K ? Assume that the temperature dependence of intrinsic carrier concentration ni is given by ni = n0 exp \(\left(-\frac{\mathbf{E}_{\mathbf{g}}}{2 \mathbf{K}_{\mathbf{B}} \mathbf{T}}\right)\) where n0 is a constant.
Solution:
Given, intrinsic carrier concentration ni = \(n_0 e^{-E g / 2 k_B T}\) and Energy gap Eg = 1.2 eV
KB = 8.62 × 10-5 eV/K
For T = 600k
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 47
Let the conductivities are σ600 and σ300 (∵ σ = e n µe)
\(\frac{\sigma_{600}}{\sigma_{300}}\) = \(\frac{n_{600}}{n_{300}}\) = 1.1 × 105.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 3.
In a p-n junction diode, the current I can be expressed as I = I0 exp \(\left(\frac{e V}{2 K_B T}-1\right)\) where I0 is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, kB is the Boltzmann constant (8.6 × 10-5 eV/K) and T is the absolute temperature. If for a given diode I0 = 5 × 10-12 A and T = 300 K, then
(a) What will be the forward current at a forward voltage of 0.6 V ?
(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?
(c) What is the dynamic resistance ?
(d) What will be the current if reverse bias voltage changes from IV to 2 V ?
Solution:
Given I0 = 5 × 10-12 A, T = 300K
KB = 8.6 × 10-5eV/K = 8.6 × 10-5 × 1.6 × 10-19 J/K
a) Given, voltage V = 0.6V
\(\frac{\mathrm{eV}}{\mathrm{K}_{\mathrm{B}} \mathrm{T}}\) = \(\frac{1.6 \times 10^{-19} \times 0.6}{8.6 \times 10^{-5} \times 1.6 \times 10^{-19} \times 300}\) = 23.26
The current I through a junction diode is given by
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 48
= 5 × 10-12(e23.26 – 1) = 5 × 10-12 (1.259 × 1010 – 1)
= 5 × 10-12 × 1.259 × 1010 = 0.063 A
Change in current ΔI = 3.035 – 0.063 = 2.9A

b) Given voltage V = 0.7 V
\(\frac{\mathrm{eV}}{\mathrm{K}_B T}\) = \(\frac{1.6 \times 10^{-19} \times 0.7}{8.6 \times 10^{-5} \times 1.6 \times 10^{-19} \times 300}\) = 27.14
Now, I = \(I_0 e^{\frac{e V}{K_B T}}-1\) = 5 × 10-12(e27.14 – 1)
= 5 × 10-12(6.07 × 1011 – 1)
= 5 × 10-12 × 5.07 × 1011 = 0.035 A
Change in current ΔI = 3.035 – 0.693 = 2.9 A

c) ΔI = 2.9A, voltage ΔV = 0.7 – 0.6 = 0.1 V
Dynamic retistance Rd = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}\) = \(\frac{0.1}{2.9}\) = 0.0336Ω

d) As the voltage changes from IV to 2V, the current I will be almost equal to
I0 = 5 × 10-12A.
It is due to that the diode possesses practically infinite resistance in the reverse bias.

Question 4.
You are given the two circuits as shown in Fig.Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 49
Solution:
a) Split the gate, OR gate
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 50
The truth table :
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 51
Here, for given A and B as inputs, C is the output of OR gate and input of NOT gate 1, D is the output of NOT gate 1 and input of NOT gate 2, then Y is finally output.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 52
This is same as OR gate. So, this circuit acts as OR gate.

b) Split the gate:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 53
The truth table:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 54
Here, for given A and B as inputs, C is the output of A and D is the output of B, E is the output of OR gate and input of NOT gate 3, then Y is finally output. .
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 55
This is same as AND gate, So the given circuit acts as AND gate

Question 5.
Write the truth table for a NAND gate connected as given in fig.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 56
Hence identify the exact logic operation carried out by the circuit.
Solution:
Split the gate:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 57
Truth table:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 58
B is the output of AND gate and input of NOT gate.
So, for input A and output Y, the table is
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 59
Here, it is same as NOT gate, so the logic operation is carried by this circuit as NOT gate

Question 6.
You are given two circuits as shown in fig. which consist of NAND gates. Identify the logic operation carried out by the two circuits.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 60
Solution:
a) Split the gate:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 61
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 62
C is the output of AND gate 1 and input of NOT gate 1, D is the output of NOT gate 1 and input of AND gate 2, E is the õut put of AND gate 2 and input of NOT gate 2, Y is finally output
Truth table:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 63
So, this logic operation is AND gate.

b)
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 64
C is the out put of AND gate 1
D is the out put of AND gate 2
E is the out put of NOT gate 1
F is the out put of NOT gate 2
G is the out put of AND gate 3 and input of NOT gate 3
Here, it is same as OR gate. A and B are inputs and Y is output.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 65
So, this logic operation resembles to OR gate.

Question 7.
Write the truth table for circuit given in fig. below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 66
Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y = 1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)
Solution:
Split the gate:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 67
C is the output of OR gate 1.
D is the output of NOT gate 1 and input of NOT gate 2. E is the output of OR gate 2 and input of the NOT gate 2. Here, it is same as OR gate. A and B are inputs and Y is output.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 68
So, this logic operation resembles to OR gate.

AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 8.
Write the truth table for the circuits given in fig. consisting of NOR.gates ònly. Identify the logic operations (OR, AND, NOT) performed by the two circuits.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 69
Solution:
a) Split the gate
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 70
B is the output of OR gate and input of NOT gate. So, the gate resembles to NOT gate as A is input and Y is output.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 71

b) a) Split the gate:
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 72
C is the output of OR gate 1 and input of NOT gate 1, D is the output of OR gate 2 and input of NOT gate 2, E is the output of NOT gate 1, F is the output of NOT gate 2.

G is the output of OR gate 3 and input of NOT gate 3. The truth table resembles to AND gate as A and B inputs and Y is output.
AP Inter 2nd Year Physics Study Material Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 73
This operation is AND gate.

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 14th Lesson Nuclei Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 14th Lesson Nuclei

Very Short Answer Questions

Question 1.
What are isotopes and isobars?
Answer:
Isotopes : The nuclei having the same atomic number (Z) but different mass numbers (A) are called isotopes.
E.g.:
\({ }_8^{16} \mathrm{O}\), \({ }_8^{17} \mathrm{O}\), \({ }_8^{18} \mathrm{O}\)

Isobars : The nuclei having the same neutron number (N) but different atomic number (Z) are called isobars.
E.g.: \({ }_6^{14} \mathrm{C}\), \({ }_7^{14} \mathrm{N}\)

Question 2.
What are isotones, and isomers?
Answer:
Isotones : The nuclei having same neutron number (N) but different atomic numbers (Z) are called isotones.
E.g. : \({ }_80^{198} \mathrm{Hg}\), \({ }_79^{197} \mathrm{N}\)

Isomers: Nuclei having the same atomic number (Z) and mass number (A) but with different nuclear properties such as radioactive decay and magnetic moments are called isomers.
E.g. : \({ }_{35}^{80} \mathrm{Br}^{\mathrm{m}}\), \({ }_{35}^{80} \mathrm{Br}^{\mathrm{g}}\)

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 3.
What is a.m.u. ? What is its equivalent energy ?
Answer:
The mass of \(\frac{1}{12}\)th of the mass of \({ }_6^{12} \mathrm{C}\) atom is called atomic mass unit (a.m.u)
1 a.m.u = \(\frac{1}{12}\) of mass of \({ }_6^{12} \mathrm{C}\) atom = 1.66 × 10-27 kg
Equivalent energy of a.m.u= 931.5 MeV

Question 4.
What will be the ratio of the radii of two nuclei of mass numbers A1 and A2 ?
Answer:
The ratio of the radii of two nuclei of mass numbers A1 and A2 will be \(\frac{R_1}{R_2}\) = \(\left[\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right]^{\frac{1}{3}}\) since R = R0 A1/3.

Question 5.
Natural radioactive nuclei are mostly nuclei of high mass number why ?
Answer:
Natural radioactivity is displayed by heavy nuclei beyond lead in the periodic table because of relatively low binding energy per nucleon as 7.6 MeV Hence to attain greater stability.

Question 6.
Does the ratio of neutrons to protons in a nucleus increase, decrease or remain the same after the emission of an α – particle ?
Answer:
The ratio of neutrons to protons in a nucleus, increases after the emission of an α – particle.
E.g. : Taking,
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 1
Before emission, the ratio of neutrons to protons
= \(\frac{\mathrm{A}-\mathrm{Z}}{\mathrm{Z}}\) = \(\frac{238-92}{92}\) = \(\frac{144}{92}\) = 1.57
After emission, the ratio of neutrons to protons
= \(\frac{234-90}{90}\) = 1.6

Question 7.
A nucleus contains no electrons but can emit them. How?
Answer:
When the nucleus disintegrates and radiates β-rays, it is said to be β-decay. β-particles are nothing but electrons. So the nucleus eject electrons.

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 8.
What are the units and dimensions of the disintegration constant ?
Answer:
λ = –\(\frac{0.693}{\mathrm{~T}}\)
Units = sec-1
Dimensions = -1

Question 9.
Why do all electrons emitted during β-decay not have the same energy ?
Answer:
When a neutron is converted into a proton, an electron and neutron are emitted along with it
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 2
In β – decay proton remains in the nucleus, but electron and neutron are emitted with constant energy. The energy of neutron is not constant. So, all electrons do not have same energy.

Question 10.
Neutrons are the best projectiles to produce nuclear reactions. Why ?
Answer:
Neutrons are uncharged particles. So they do not deflected by the electric and magnetic fields. Hence Neutrons are considered as best projectiles in nuclear reaction.

Question 11.
Neutrons cannot produce ionization. Why ?
Answer:
Because neutrons are uncharged particles and cannot produce ionization.

Question 12.
What are delayed neutrons ?
Answer:
Neutrons are emitted in the fission products after sometime are called delayed neutrons.

Question 13.
What are thermal neutrons ? What is their importance ?
Answer:
Neutrons having kinetic energies approximately 0.025 eV are called as slow neutrons or thermal neutrons. 235U undergoes fission only when bombarded with thermal neutrons.

Question 14.
What is the value of neutron multiplication factor in a controlled reaction and in an uncontrolled chain reaction ?
Answer:
In controlled chain reaction K = 1
In uncontrolled chain reaction K > 1

Question 15.
What is the role of controlling rods in a nuclear reactor ?
Answer:
In nuclear reactor controlling rods are used to absorb the neutrons. Cadmium, boron materials are used in the form of rods in reactor. These control the fission rate.

Question 16.
Why are nuclear fusion reactions called thermo nuclear reactions ?
Answer:
Nuclear fusion occurs at very high temperatures. So it is called as thermo nuclear reaction.

Question 17.
Define Becquerel and Curie.
Answer:
Becquerel: 1 disintegration or decay per second is called Becquerel. It is SI unit of activity.
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 3
Curie : 3.7 × 1010 decays per second is called Curie.
1 Curie = 1Ci = \(\frac{3.7 \times 10^{10} \text { decays }}{\text { second }}\) = 3.7 × 1010 Bq.

Question 18.
What is a chain reaction ?
Answer:
Chain reaction: The neutrons produced in the fission of a nucleus can cause fission in other neighbouring nuclei producing more and more neutrons to continue the fission until the whole fissionable material is disintegrated. This is called chain reaction.

Question 19.
What is the function of moderator in a nuclear reactor ?
Answer:
They are used to slow down the fast moving neutrons produced during the fission process.
e.g.: Heavy water, Berilium.

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 20.
What is the energy released in the fusion of four protons to form a helium nucleus ?
Answer:
26.7 MeV energy is released.

Short Answer Questions

Question 1.
Why is the density of the nucleus more than that of the atom ? Show that the density of nuclear matter is same for all nuclei.
Answer:

  1. Volume of the atom is greater than that of nucleus and it consists of nucleons
  2. Since density ∝ \(\frac{1}{\text { volume }}\)
    ∴ Density of the nucleus more than that of the atom.
  3. Mass of the nucleus = no.of nucleons (A) × mass of nucleon (m)
    = Am
  4. Volume of the nucleus V= \(\frac{4}{3} \pi \mathrm{R}^3\)
    = \(\frac{4}{3} \pi\left(R_0 A^{1 / 3}\right)^3\)
    = \(\frac{4 \pi \mathrm{R}_0^3 \mathrm{~A}}{3}\) = 1.2 × 10-45m3.A. [∵ R0 = 1.2 × 10-15m]
    i. e., the volume of the nucleus is proportional to the mass number A.
  5. Density of the nucleus (ρ)
    AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 4
  6. The above equation represents it clear that the density of the nucleus is independent of the mass number A and is same for all the nuclei.

Question 2.
Write a short note on the discovery of neutron.
Answer:

  1. Bothe and Becker found that when beryllium is bombarded with α – particles of energy 5 MeV, which emitted a highly penetrating radiation.
  2. The equation for above process can be written as
    AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 5
  3. The radiations are not effected by electric and magnetic fields.
  4. In 1932, James Chadwick, had subjected nitrogen and argon to the beryllium radiation. He interpreted the experimental results by assuming that the radiation is of a new kind of particles which has no charge and its mass is equal to proton. These neutral particles were named as ‘neutrons’. Thus the neutron was discovered.
  5. The experimental results can be represented by the following equation.
    AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 6

Question 3.
What are the properties of a neutron ?
Answer:

  1. Neutron is an uncharged particle and hence it is not deflected by the electric and magnetic fields.
  2. It has very high penetrating power and has very low ionization power.
  3. Inside the nucleus neutrons appear to be stable. The average life of an isolated neutron is about 1000 seconds. A free neutron is unstable and spontaneously decays into a proton, electron and an antineutrino \((\bar{v})\).
  4. If fast neutrons pass through substances like heavy water, paraffin wax, graphite etc., they are slowed down.
  5. Neutrons are diffracted by crystals.

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 4.
What are nuclear forces ? Write their properties.
Answer:
The forces which hold the nucleons together in nucleus are called nuclear forces. Properties of Nuclear forces :

  1. Nuclear forces are attractive forces between proton and proton (P – P), proton and neutron (P – N) and neutron and neutron (N – N).
  2. Nuclear forces are independent of charge. It was found that force between proton and proton is same as force between neutron and neutron.
  3. These forces are short range forces i.e., these forces will act upto a small distance only. Generally the range of nuclear forces is upto few Fermi (10-15 m).
  4. These forces are non central forces, i.e., they do not act along the line joining the two nucleons.
  5. These forces are exchange forces. The force between two nucleons is due to exchange of π-mesons.
  6. These forces are spin dependent. These forces are strong when the spin of two nucleons are in same direction and they are weak when they are in opposite direction.
  7. Nuclear forces are saturated forces i.e., the force between nucleons will extend upto the immediate neighbouring nucleons only.
  8. These are the strongest forces in nature. They are nearly 1038 times stronger than gravitational forces and nearly 100 times stronger than Coulombic forces.

Question 5.
For greater stability a nucleus should have greater value of binding energy per nucleon. Why?
Answer:

  1. Uranium has a relatively low binding energy per nucleon as 7.6 MeV Hence to attain greater stability Uranium breaks up into intermediate mass nuclei resulting in a phenomenon called nuclear fission.
  2. Lighter nuclei such as hydrogen combine to form heavy nucleus to form helium for greater stability, resulting in a phenomenon called nuclear fusion.
  3. Iron whose binding energy per nucleon stands maximum at 8.7 MeV is the most stable and will undergo neither fission nor fusion.

Question 6.
Explain α – decay ?
Answer:

  1. It is the phenomenon of emission of an a particle from a radioactive nucleus. When a nucleus emits an alpha particle, its mass number decreases by 4 and charge number decreases by 2.
  2. In general, alpha decay is represented as
    AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 7
    Where Q is the energy released in the decay.
  3. Thus the total mass energy of the decay products is less than the mass energy of the original nuclide.
  4. The difference between the initial mass energy and the total mass energy of decay products is called disintegration energy (Q) of the process.
  5. This can be calculated using Einstein’s mass energy equivalence relation, E = (Δm). c2
    i-e., Q = (mx – my – mHe) c2
    The energy released (Q) is shared by daughter nucleus y and alpha particle.

Question 7.
Explain β – decay ?
Answer:

  1.  It is the phenomenon of emission of an electron from a radioactive nucleus.
  2. When a parent nucleus emits a β-particle (i.e., an electron), mass number remains same because mass of electron is negligibly low. However, the loss of unit negative charge is equivalent to a gain of unit positive charge. Therefore, atomic number is increased by one.
  3. In general, we can write
    AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 8
    where Q is the energy released in β-decay.
  4. The basic nuclear process underlying p-decay is the conversion of neutron to proton.
    n → P + \(\overline{\mathrm{e}}\) + \(\overline{\mathrm{v}}\)
    While for β+ decay, it is the conversion of proton ino neutron.
    P → n + e+ + v
  5. The emission of electron in β-decay is accompained by the emission of an anti neutrino \((\bar{v})\) In β, decay instead, a neutrino (v) is generated. Neutrons are neutral particles with very small mass compared to electrons. They have only weak interactions with other particles.

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 8.
Explain γ – decay ?
Answer:

  1. It is the phenomenon of emission of gamma ray photon from a radioactive nucleus.
  2. Like an atom, a nucleus has discrete energy levels in the ground state and excited states.
  3. When a nucleus in an excited state spontaneously decays to its ground state (or to a lower energy state), a photon is emitted with energy equal to the difference in the two energy levels of the nucleus. This is the so called gamma-decay.
  4. The energy (MeV) corresponds to radiation of extremely short wave length, shorter than the hard X-ray region.
  5. A Gamma ray is emitted when a or p decay results in a daughter nucleus in an excited
    state.
  6. The \(\bar{\beta}\) -decay of 27CO60 transforms it into an excited 28Ni60 nucleus. This reaches the ground state by emission of γ-rays of energy 1.17 MeV and 1.33 MeV. This is shown in figure.
    AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 9

Question 9.
Define half life period and decay constant for a radioactive substance. Deduce the relation between them.
Answer:
Half life period (T) : Time taken for the number of radio active nuclei to disintegrate to half of its original number of nuclei is called Half life period.
Decay constant (λ) : The ratio of the rate of radioactive decay to the number of nuclei present at that instant.
It is a proportional constant and is denoted by ‘λ’.
λ = \(\frac{-\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)}{\mathrm{N}}\)

Relation between half the period and decay constant:

  1. The radioactive decay law N = N0 e-λt states that the number of radioactive-nuclei in a radioactive sample decreases exponentially with time. Here λ is called decay constant.
  2. If N0 is the number of nuclei at t = 0 and N is the radioactive nuclei at any instant of time’t’.
  3. Substituting N = \(\frac{\mathrm{N}_0}{2}\) at t = T in N = N0e-λt.
    Where T is half life of the radioactive substance.
  4.  \(\frac{\mathrm{N}_0}{2}\) = N0 e-λT
    eλT = 2
    λT = ln2
    T = \(\frac{\ln 2}{\lambda}\) = \(\frac{2.303 \log _{10}^2}{\lambda}\)
    ∴ T = \(\frac{0.693}{\lambda}\)

Question 10.
Define average life of a radioactive substance. Obtain the relation between decay constant and average life.
Answer:
Average life \((\tau)\) : It is equal to the total life time of all the N0 nuclei divided by the total number of original nuclei N0. It is denoted by \((\tau)\).

Relation between decay constant and average life :

  1. Let N0 be the radioactive nuclei that are present at t = 0 in the radioactive sample; The no’ of nuclei which decay between t and t + dt is dN.
  2. The total life time of these dN nuclei is t dN. The total life time of all the nuclei present initially in the sample = \(\int_0 \mathrm{t} \mathrm{dN}\)
  3. Average life time \((\tau)\) is equal to the total life time of all the N0 nuclei divided by the total number of original nuclei N0.
  4. Average \((\tau)\) = \(\frac{\int \mathrm{tdN}}{\mathrm{N}_0}\)
    But \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = -λN
    dN = -λNdt = N0e-λtdt [∵ N = N0e-λt]
  5. AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 10
    On integrating, we get \(\tau\) = \(\frac{1}{\lambda}\)
    \(\tau\) = \(\frac{T}{0.693}\) [∵ λ = \(\frac{0.693}{\mathrm{~T}}\)]
  6. From the above equation ‘the reciprocal of the decay constant gives us the average life of a radioactive sample.

Question 11.
Deduce the relation between half life and average life of a radioactive substance.
Answer:
Relation between half life (T) and average life (\(\tau\)) :

  1. We know, the radioactive decay law, N = N0 e-λt —– (1)
  2. Consider, ‘N0‘ is the number of nuclei present at t = 0 and after time T, only \(\frac{\mathrm{N}_0}{2}\) are left and after a time ‘2T’, only \(\frac{\mathrm{N}_0}{4}\) remain and soon.
  3. Substituting N = \(\frac{\mathrm{N}_0}{2}\) at t = T in eqn. (1) then
    \(\frac{\mathrm{N}_0}{2}\) = N0 e-λT ⇒ \(\frac{1}{2}\) = \(\frac{1}{\mathrm{e}^{\lambda \mathrm{T}}}\) ⇒ eλT = 2
    Taking loge on both sides, we get
    λT = \(\log _{\mathrm{e}}^2\) = 2.303 \(\log _{\mathrm{e}}^2\) = 0.693
    ∴ T = \(\frac{0.693}{\lambda}\) —— (2)
  4. Average life \(\tau\) = \(\frac{\int \mathrm{tdN}}{\mathrm{No}}\)
  5. But –\(\frac{\mathrm{dN}}{\mathrm{dt}}\) = λN dN = -λ.N0 e-λt dt [∵ from eqn. (1)]
  6. AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 11
    on integrating, we get \(\tau\) = \(\frac{1}{\lambda}\) —– (3)
  7. From equs (2) and (3) we get \(\tau\) = \(\frac{\mathrm{T}}{0.693}\)
    This is the relation between average life and half life of radioactive substance.

Question 12.
What is nuclear fission ? Give an example to illustrate it.
Answer:
Nuclear fission : The process of dividing a heavy nucleus into two or more smaller and stable nuclei due to nuclear reaction is called nuclear fission.
Ex : The fission reaction is
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 12
Where Q is the energy released.
Q = (Total mass of reactants – Total mass of product) C2
= [(Mass of \({ }_{92}^{235} \mathrm{U}\) + Mass of \({ }_0^1 n\)) – (Mass of \({ }_{56}^{141} \mathrm{Ba}\) + Mass of \({ }_{36}^{92} \mathrm{Kr}\) + Mass of three neutrons)]C2
= (235.043933 – 140.9177 – 91.895400 – 2 × 1.008665) amu × C2.
= 0.2135 × 931.5 MeV = 198.9 MeV = 200 MeV

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 13.
What is nuclear fusion ? Write the conditions for nuclear fusion to occur.
Answer:
Nuclear fusion : The process of combining lighter nuclei to produce a larger nucleus is known as nuclear fusion.
E.g : Hydrogen nuclei (1H1) are fused together to form heavy Helium (2He4) along with 25.71 MeV energy released.

Conditions for nuclear fusion :

  1. Nuclear fusion occurs at very high temperatures such as 107 kelvin and very high pressures. These are obtained under the explosion of an atom bomb.
  2. Higher density is also desirable so that collisions between light nuclei occur quite frequently.

Question 14.
Distinguish between nuclear fission and nuclear fusion.
Answer:
Nuclear fission

  1. In this process heavy nucleus is divided into two fragments along with few neutrons.
  2. These reactions will takes place even at room temperature.
  3. To start fission atleast one thermal neutron from out side is compulsory.
  4. Energy released per unit mass of participants is less.
  5. In this process neutrons are liberated.
  6. This reaction can be controlled.
    Ex: Nuclear reactor.
  7. Atom bomb works on principle of fission reaction.
  8. The energy released in fission çan be used for peaceful purpose.
    Ex : Nuclear reactor and Atomic power stations.

Nuclear fusion

  1. In this process lighter nuclei will join together to produce heavy nucleus.
  2. These reactions will takes place at very high temperature such as Kelwin.
  3. No necessary of external neutrons.
  4. Energy released per unit mass of participants is high. Nearly seven times more than fission reaction.
  5. In this process positrons are liberated.
  6. There is no control on fusion reaction.
  7. Hydrogen bomb works on the principle of fusion reaction.
  8. The energy released in fusion cannot be used for peaceful purpose.

Question 15.
Explain the terms tchain reaction’ and multiplication factor’. How is a chain reaction sustained?
Answer:
Chain reaction : In nuclear fission nearly three neutrons are produced when one uranium atom is destoryed. If they again participate in fission reaction nine neutrons are produced. In next generation the neutrons becomes 27. In this process the number of neutrons increases in geometric progression and the whole uranium is destroyed in few seconds. This type of self sustained fission reaction is called chain reaction.

Neutron multiplication factor (K) : Neutron multiplilcation factor is defined as the ratio of number of neutrons produced in one generation to the number of neutrons in previous generation.
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 13

Neutron multiplication factor is useful to understand the nature of nuclear reactions in a nuclear reactor.

To sustained chain reaction:
1. Neutron multiplication factor K ≥ 1.

Long Answer Questions

Question 1.
Define mass defect and binding energy. How does binding energy per nucleon vary with mass number ? What is its significance ?
Answer:

  1. Mass defect (ΔM) : The difference in mass of a nucleus and its constituents is called the mass defect. The nuclear mass M is always less than the total mass, Σm, of its constituents.
    Mass defect, (ΔM) = [Zmp + (A – Z)mn – M]
  2. Binding energy: The energy required to break the nucleus into its constituent nucleons is called the binding energy.
    Binding Energy, (Eb) = ΔMC2 = [Zmp + (A – Z)mn – M] 931.5 MeV
    Nuclear binding energy is an indication of the stability of the nucleus.
    Nuclear binding energy per nucleon Ebn = \(\frac{\mathrm{E}_{\mathrm{b}}}{\mathrm{A}}\).
  3. The following graph represents how the binding energy per nucleon varies with the mass number A.
    AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 14
  4. From the graph that the binding energy is highest in the range 28 < A < 138. The binding energy of these nuclei is very close to 8.7 MeV.
  5. With the increase in the mass number the binding energy per nucleon decreases and consequently for the heavy nuclei like Uranium it is 7.6 MeV
  6. In the region of smaller mass numbers, the binding energy per nucleon curve shows the characteristic minima and maxima.
  7. Minima are associated with nuclei containing an odd number of protons and neutrons such as \({ }_3^6 \mathrm{Li}\), \({ }_5^{10} \mathrm{~B}\), \({ }_7^{14} \mathrm{~N}\) and the maxima are associated with nuclei having an even number of protons and neutrons such as \({ }_2^4 \mathrm{He}\), \({ }_6^{12} \mathrm{C}\), \({ }_8^{16} \mathrm{O}\).
    Significance :
  8. The curve explains the relationship between binding energy per nucleon and stability of the nuclei.
  9. Uranium has a relatively low binding energy per nucleon as 7.6 MeV. Hence to attain greater stability Uranium breaks up into intermediate mass nuclei resulting in a phenomenon called fission.
  10. On the other hand light nuclei such as hydrogen combine to form heavy nucleus to form helium for greater stability, resulting in a phenomenon called fusion.
  11. Iron is the most stable having binding energy per nucleon 8.7 MeV and it neither undergoes fission per fusion.

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 2.
What is radioactivity ? State the law of radioactive decay. Show that radioactive decay is exponential in nature.
Answer:

  1. Radioactivity : .The nuclei of certain elements disintegrate spontaneously by emitting alpha (α), beta (β) and gamma (γ) rays. This phenomenon is called Radioactivity or Natural radioactivity.
  2. Law of radioactivity decay: The rate of radioactive decay \(\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)\) (or) the number of nuclei decaying per unit time at any instant, is directly proportional to the number of nuclei (N) present at that instant is called law of radioactivity decay’.
  3. Radioactive decay is exponential in nature : Consider a radioactive substance. Let the number of nuclei present in the sample at t = 0, be N0 and let N be the radioactive nuclei remain at an instant t.
    \(\frac{\mathrm{dN}}{\mathrm{dt}}\) ∝ N ⇒ \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = -λN
    dN = -λNdt ——— (1)
    The proportionality constant λ is called decay constant or disintegration constant. The negative sign indicates the decrease in the number of nuclei.
  4. From eq.(1) \(\frac{\mathrm{dN}}{\mathrm{N}}\) = -λ dt ——- (2)
  5. Integrating on both sides
    \(\int \frac{\mathrm{dN}}{\mathrm{N}}\) = \(-\lambda \int \mathrm{dt}\)
    In N = -λt + C ——- (3)
    Where C = Integration constant.
  6. At t = O; N = N0. Substituting in eq. (3), we get in ln N0 = C
    ∴ ln N – ln N0 = – λt
    ln \(\left(\frac{\mathrm{N}}{\mathrm{N}_0}\right)\) = – λt
    ∴ N = 0e-λt
    The above equation represents radioactive decay law.
  7. It states that the number of radioactive nuclei in a radioactive sample decreases exponentially with time.

Question 3.
Explain the principle and working of a nuclear reactor with the help of a labelled diagram. (A.P.Mar.’19,’16, ’15 & T.S. Mar. ‘15) (Mar. ’14)
Answer:
Principle : A nuclear reactor works on the principle of achieving controlled chain reaction in natural Uranium 238U enriched with 235U, consequently generating large amounts of heat.
A nucleàr reactor consists of
(1) Fuel
(2) Moderator
(3) Control rods
(4) Radiation shielding
(5) Coolant.
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 15

1. Fuel and clad : In reactor the nuclear fuel is fabricated in the form of thin and long cylindrical rods. These group of rods treated as a fuel assembly. These rods are surrounded by coolant, which is used to transfer of heat produced in them. A part of the nuclear reactor which use to store the nuclear fuel is called the core of the reactor. Natural uranium, enriched uranium, plutonium and uranium – 233 are used as nuclear fuels.

2. Moderator : The average energy of neutrons released in fission process is 2 MeV They are used to slow down the velocity of neutrons. Heavy water or graphite are used as moderating materials in reactor.

3. Control Rods : These are used to control the fission rate in reactor by absorbing the neutrons. Cadmium and boron are used as controlling the neutrons, in the form of rods.

4. Shielding : During fission reaction beta and gamma rays are emitted in addition to neutrons. Suitable shielding such as steel, lead, concrete etc are provided around the reactor to absorb and reduce the intensity of radiations to such low levels that do not harm the operating personnel.

5. Coolant: The heat generated in fuel elements is removed by using a suitable coolant to flow around them. The coolants used are water at high pressures, molten sodium etc.

Working: Uranium fuel rods are placed in the aluminium cylinders. The graphite moderator is placed in between the fuel cylinders. To control the number of neutrons, a number of control rods of cadmium or beryllium or boron are placed in the holes of graphite block: When a few 235U nuclei undergo fission fast neutrons are liberated. These neutrons pass through the surrounding graphite moderator and loose their energy to become thermal neutrons.

These thermal neutrons are captured by 235U. The heat generated here is used for heating suitable coolants which in turn heat water and produce steam. This steam is made to rotate steam turbine and there by drive a generator of production for electric power.

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 4.
Explain the source of stellar energy. Explain the carbon – nitrogen cycle, proton – proton cycle occurring in stars.
Answer:
Scientists proposed two types of cyclic processes for the sources of energy in the sun and stars. The first is known as carbon-nitrogen cycle and the second is proton-proton cycle.

1. Carbon-Nitrogen Cycle: According to Bethe carbon – nitrogen cycle is mainly responsible for the production of solar energy. This cycle consists of a chain of nuclear reactions in which hydrogen is converted into Helium, with the help of Carbon and Nitrogen as catalysts. The nuclear reactions are as given below.
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 16
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 17

2. Proton – Proton Cycle: A star is formed by the condensation of a large amount of matter at a point in space. Its temperature rises to 2,00,000°C as the matter contracts under the influence of gravitational attraction. At this temperature the thermal energy of the protons is sufficient to form a deuteron and a positron. The deuteron then combines with another proton to form lighter nuclei of helium \({ }_2^3 \mathrm{He}\). Two such helium nuclei combine to form a helium nucleus \({ }_2^4 \mathrm{He}\) and two protons releasing a total amount of energy 25.71 MeV The nuclear fusion reactions are given below.
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 18
The net result is
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 19

Problems

Question 1.
Show that the density of a nucleus does not depend upon its mass number (density is independent of mass)
Solution:
Density of nucleus matter =
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 20
No. of nucleons (A) × mass of nucleons (m)
Volume of nucleus V = \(\frac{4}{3} \pi R^3\)
But R = R0A1/3
∴ V = \(\frac{4}{3} \pi \mathrm{R}_0^3 \mathrm{~A}\)
∴ Density of nucleus matter = \(\frac{\mathrm{Am}}{\frac{4}{3} \pi \mathrm{R}_0^3 \mathrm{~A}}\) = \(\frac{3 \mathrm{~m}}{4 \pi \mathrm{R}_0^3}\)
∴ Density of nucleus is independent of mass.

Question 2.
Compare the radii of the nuclei of mass numbers 27 and 64.
Solution:
A1 = 27; A2 = 64
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}\) = \(\left[\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right]^{1 / 3}\) [∵ R = R0A1/3]
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}\) = \(\left[\frac{27}{64}\right]^{\frac{1}{3}}\) = \(\frac{3}{4}\)
∴ R1 : R2 = 3 : 4

Question 3.
The radius of the oxygen nucleus \({ }_8^{16} \mathrm{O}\) is 2.8 × 10-15m. Find the radius of lead nucleus \({ }_{82}^{205} \mathrm{~Pb}\).
Solution:
R0 = 2.8 × 10-15 m; A0 = 16
APb = 205; RPb = ?
\(\frac{R_{P b}}{R_0}\) = \(\left[\frac{A_{P b}}{A_0}\right]^{1 / 3}\) = \(\left[\frac{205}{16}\right]^{1 / 3}\)
[∵ R = R0A1/3]
\(\frac{R_{P b}}{2.8 \times 10^{-15}}\) = (12.82)1/3 = 2.34
RPb = (2.34) × (2.8 × 10-15)
= 6.55 × 10-15m

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 4.
Find the binding energy of \({ }_{26}^{56} \mathrm{Fe}\). Atomic mass of Fe is 55.9349u and that of Hydrogen is 1.00783u and mass of neutron is 1.00876u.
Solution:
Mass of hydrogen atom
mp = 1.00876u; mn = 1.00867u
Z = 26; A = 56
Mass of Iron atom M = 55.9349u

i) Mass defect Δm
= [Zmp + (A – Z) mn – M]
= [26 × 1.00876 + (56 – 26) (1.00867) – 55.9349] u
∴ Δm = 0.55296u

ii) BE of nucleus = ΔMC2
= ΔM × 931.5 MeV
= 0.55296 (931.5) MeV
= 515.08 MeV

Question 5.
How much energy is required to separate the typical middle mass nucleus \({ }_{50}^{120} \mathrm{Sn}\) into its constituent nucleons? (Mass of \({ }_{50}^{120} \mathrm{Sn}\) = 119.902199u, and mass of neutron = 1.008665u)
Solution:
mp = 1.007825u
mn = 1.008665u
For Sn, Z = 50;

A = 120; M = 119.902199u

i) Mass defect Δm
= [Zmp + (A – Z)mn – M]u
= 50 (1.007825) + (120 – 50) [(1.008665) – 119.902 199]
= [150 × 1.007825 + 70 × 1.008665 – 119.902199]u
= [50.39125 + 70.60655 – 119.902199]u
ΔM = [120.9978 – 119.902199]
= 1095601u

ii) Energy required to šeparate the nucleons = B.E of the nucleus
BE = ΔMc2 = ΔM × 931.5MeV
= 1.095601 × 931.5 MeV
= 1020.5 MeV

Question 6.
Calculate the binding energy of an α-particle. Given that mass of proton = 1.0073 u, mass of neutron = 1.0087u, and mass of α- particle = 4.0015u.
Solution:
For 2He4, A = 4, Z = 2, mp = 1.0073u
mn = 1.0087u, mn = 4.0015u

i) ΔM
= [Zmp + (A – Z)mn – M]
= [2(1.0073) + (4 – 2) (1.0087) – 4.00260]
= [2 × 1.0073 + 2 × 1.0087 – 4.00260]
= (2.0146 + 2.0174) – 4.0015
ΔM = [4.032 – 4.0015] = 0.0305 u

ii) BE = ΔM × c2 = ΔM × 931.5 MeV
= 0.0305 × 931.5
∴ BE = 28.41 MeV

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 7.
Find the energy required to split \({ }_8^{16} \mathrm{O}\) nucleus into four α-particles. The mass of an α-particle is 4.002603u and that of oxygen is 15.994915u.
Solution:
The energy required to split O =
[Total mass of the products – Total mass of the reactants] c2
Mass of four \({ }_2^4 \mathrm{He}\) – Mass of \({ }_8^{16} \mathrm{O}\)] × c2
= [(4 × 4.002603) – 15.994915] u × c2
= [16.010412 – 15.994915] u × c2
= (0.015497) 931.5 MeV
= 14.43 MeV

Question 8.
Calculate the binding energy per nucleon of \({ }_{17}^{35} \mathrm{Cl}\) nucleus. Given that mass of \({ }_{17}^{35} \mathrm{Cl}\) nucleus = 34.98000 u, mass of proton = 1.007825u, mass of neutron = 1.008665u and 1 is equivalent to 931 MeV.
Solution:
For \({ }_{17}^{36} \mathrm{Cl}\), A = 35, Z = 17;
mp = 1.007825 u
mn = 1.008665 u,
M = 34.98u

(i) ΔM = [Zmp + (A – Z) mn – M]
= [17 × 1.007825 + (35 – 17)(1.008665) – 34.98]
= [17.13303 + 18.15597 – 34.98]
ΔM = [35.289 – 34.98]
= 0.3089 u

(ii) BE = ΔMc2
= 0.3089 × 931 MeV = 287.5859 MeV
∴ BE per nucleon
= \(\frac{B \cdot E}{A}\) = \(\frac{287.5859}{35}\) = 8.21 MeV

Question 9.
Calculate the binding energy per nucleon of \({ }_{20}^{40} \mathrm{Ca}\). Given that mass of \({ }_{20}^{40} \mathrm{Ca}\) nucleus = 39.962589u, mass of a proton = 1.007825 u,; mass of Neutron = 1.008665u and 1u is equivalent to 931 MeV.
Solution:
For \({ }_{20}^{40} \mathrm{C}\), A = 40, Z = 20; mp = 1.007825 u
mn = 1.008665 u; M = 39.962589 u

(i) ΔM = [Zmp + (A – Z) mn – M]
= [(20) (1.007825) + (40 – 20)(1 .008665) – 39.962589]
[(20 × 1.007825) + (20 × 1.008665) – 39.962589]
= [40.3298 – 39.962589] = 0.3672 u

(ii) BE = ΔMc2 = 0.3672 × 931 MeV
= 341.86 MeV
B.E ‘341.86

(iii) B.E per nucleon = \(\frac{B . E}{A}\) = \(\frac{341.86}{40}\)
= 8.547 MeV

Question 10.
Calculate
(i) mass defect,
(ii) binding energy and
(iii) the binding energy per nucleon of \({ }_6^{12} \mathrm{C}\) nucleus. Nuclear mass of \({ }_6^{12} \mathrm{C}\) = 12.000000 u; mass of proton = 1.007825 u and mass of neutron = 1.008665 u.
Solution:
For \({ }_6^{12} \mathrm{C}\), A = 12; Z = 6; mp = 1.007825u
mn = 1.008665u; M = 12.000000u

(i) ΔM = [Zmp + (A – Z) mn – M]
= [6(1.007825) + (12 – 6)(1.008665) – 12,00]
= [6 × 1.007825 + 6 × 1.008665 – 12]
= [6.04695 + 6.05199 – 12.00]
ΔM = [12.09894 – 12.00] = 0.098944

(ii) BE = ΔM × c2 = 0.09894 × 931.5 MeV
= 92.16 MeV

(iii) BE per nucleon
= \(\frac{B . E}{A}\) = \(\frac{92.16}{12}\) = 7.68 MeV

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 11.
The binding energies per nucleon for deuterium and helium are 1.1 MeV and 7.0 MeV respectively. What energy in joules will be liberated when 106 deuterons take part in-the reaction.
Solution:
\(\left[\frac{B \cdot E}{A}\right]_D\) = 1.1 MeV; \(\left[\frac{B . E}{A}\right]_{H e}\) = 7.0 MeV
For deuterium \(\left({ }_1^2 \mathrm{H}\right)\),
A = 2
For He \(\left({ }_2^4 \mathrm{He}\right)\), A = 4
\(\left[\frac{B \cdot E}{2}\right]_D\) = 1.1 MeV ⇒ [B.E.]D
= 2 × 1.1 MeV = 2.2 MeV
\(\left[\frac{B . E}{4}\right]_{\mathrm{He}}\) = 70 MeV ⇒ [B.E.]He
= 4 × 7.0 MeV = 28 MeV
We know 1H2 + 1H2 → 2He4
Energy released = B.E of 106 deuterons – B.E of \(\frac{1}{2}\) × 106 Helium atoms
B.E = 2.2 × 106 × \(\frac{1}{2}\) × 106 × 28
= 106(2.2 – 14)
= -11.8 × 106 MeV
= -11.8 × 106 × 1.6 × 10-13J
= -18.88 × 10-7 J
(- ve sign indicates that energy is released)
∴ Energy released = 18.88 × 10-7 J

Question 12.
Bombardment of lithium with protons gives rise to the following reaction :
\({ }_3^7 \mathrm{Li}\) + \({ }_1^1 \mathrm{H}\) → 2 \(\left[{ }_2^4 \mathrm{He}\right]\) + Q. Find the Q-value of the reaction. The atomic masses of lithium, proton and helium are 7.016u, 1.0084 and 4.004u respectively.
Solution:
Mass of Lithium = 7.0 16 u
mp = 1.008 u
Mass of Helium = 4004 u;
u = 931.5 MeV
Q = [Total mass of the reactants – Total mass of the products] c2
= [Mass of Lithium + mp – (2 × mass of Helium)] × 931.5 MeV
= [7.016+ 1.008 – 2(4.004)] × 931.5MeV
= [18.024 – 8.008] × 931.5 MeV
∴ Energy Q = 0.016 × 931.5
= 14.904 MeV

Question 13.
The half life radium is 1600 years. How much time does lg of radium take to reduce to 0.125 g. (T.S. Mar. ’16)
Solution:
Half life of radium = 1600 years
Initial mass = 1g
Final mass 0.125 g = \(\frac{1}{8}\) g
The quantity remaining after ‘n’ half lifes is \(\frac{1}{2^{\mathrm{n}}}\) of the initial quantity.
In this problem,
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 21
∴ Time taken = ’n’ half-lifes = 3 × 1600
= 4,800 years

Question 14.
Plutonium decays with a half – life of 24,000 years. If plutonium is stored for 72,000 years, what fraction of it remains ?
Solution:
Half-life of plutonium = 24,000 years
The duration of the time = 72,000 years
Initial mass = Mg
Final mass = mg
Number of half lifes (n)
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 22
Fraction of plutonium that remains
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 23

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 15.
A certain substance decays to 1/232 of its initial activity in 25 days. Calculate its half-life.
Solution:
Fraction of substance decays
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 24
= \(\frac{1}{2^n}\) = \(\frac{1}{32}\) = \(\frac{1}{2^5}\)
∴ n = 5
Duration of time = 25 days
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 25

Question 16.
The half-life period of a radioactive substance is 20 days. What is the time taken for 7/8th of its original mass to disintegrate?
Solution:
Half life period = 20 days
In this problem,
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 26
∴ Time taken to disintegrate
= n × Half life time
= 3 × 20 = 60 days

Question 17.
How many disintegrations per second will Occur in one gram of \({ }_{92}^{238} \mathrm{U}\), if its half-life against α-decay is 1.42 × 10-17s?
Solution:
T = 1.42 × 1017 sec
Decay constant (λ) = \(\frac{0.693}{\mathrm{~T}}\) = \(\frac{0.693}{1.42 \times 10^7}\)
No. of disintegration (n) in 1 gm
= \(\frac{1}{238}\) × 6.023 × 1023
∴ Activity A = λN
= \(\frac{0.693}{1.42 \times 10^{17}}\) × \(\frac{1}{238}\) × 6.023 × 1023
= 1.235 × 104 disintegrations / sec

Question 18.
The half-life of a radioactive substance is 100 years. Calculate in how many years the activity will decay to 1/10th of its initial value.
Solution:
T = 100 years
λ = \(\frac{0.693}{\mathrm{~T}}\) = \(\frac{0.693}{\mathrm{~T}}\) = 0.693 × 10-2 years
N = N0e-λt ⇒ e-λt = \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = \(\frac{1}{10}\)
eλt = 10 ⇒ λt = \(\log _e^{10}\) = 2.303 × \(\log _{10}^{10}\)
t = \(\frac{2.303 \times 1}{0.693 \times 10^{-2}}\) = 3.323 × 102
= 332.3years

Question 19.
One gram of radium is reduced by 2 milligram in 5 years by a-decay. Calculate the half-life of radium.
Solution:
Initial (original) mass (N0) = 1 gram
Reduced mass = 2 mg = 0.002 grams
Final mass (N) = 1 – 0.002 = 0.998 grams
t = 5 years
e-λt = \(\frac{\mathrm{N}}{\mathrm{N}_0}\) ⇒ eλt = \(\frac{\mathrm{N}_0}{\mathrm{~N}}\) ⇒ λt = \(\log _{\mathrm{e}}\left[\frac{\mathrm{N}_0}{\mathrm{~N}}\right]\)
λt = 2.303 log\(\left[\frac{\mathrm{N}_0}{\mathrm{~N}}\right]\)
λt = 2.303 log \(\left[\frac{1}{0.998}\right]\)
= 2.303 × 0.000868
= 0.001999
λ = \(\frac{0.001999}{5}\) = 0.0003998
T = \(\frac{0.693}{\lambda}\) = \(\frac{0.693}{0.0003998}\) = 1733.3 years

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 20.
The half-life of a radioactive substance is 5000 years. In how many years, its activity will decay to 0.2 times a its initial value ? Given log105 = 0.6990.
Solution:
T = 5000 years; t = ?
Activity, A = Nλ = 0.2 times initial value
Initial activity A0 = N0λ
In radioactivity,
N = N0e-λt ⇒ \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = e-λt = 0.2
Put 0.2 = \(\frac{1}{5}\) ; \(\frac{1}{5}\) = e-λt (or) -λt = \(-\log _e^5\)
(or) t = \(\log _{\mathrm{e}}^5 \frac{5}{\lambda}\)
Radioactive decay constant A = \(\frac{\log _{\mathrm{e}} 2}{\mathrm{~T}}\)
= \(\frac{2.303 \log _{10}^2}{5000}\) = \(\frac{0.693}{5000}\)
Time taken to decay
t = \(\frac{\log _{\mathrm{e}}^5}{\lambda}\) × 5000
∴ t = \(\frac{2.303 \times 0.6990 \times 5000}{0.693}\) = \(\frac{8049}{0.693}\)
= 1.161 × 104 years

Question 21.
An explosion of atomic bomb releases an energy of 7.6 × 1013J. If 200 MeV energy is released of fission of one 235U atom calculate
(i) the number of uranium atoms undergoing fission,
(ii) the mass of uranium used in the bomb.
Solution:
Energy released (E) = 7.6 × 1013 J
Energy released on fissions (E) = 200 MeV
= 200 × 106 × 1.6 × 10-19 J
i) No. of fissions (n) = \(\frac{\mathrm{E}^{\prime}}{\mathrm{E}}\)
= \(\frac{7.6 \times 10^{13}}{200 \times 10^6 \times 1.6 \times 10^{-19}}\)
∴ n = 2.375 × 1024 atoms

But Avogadro number N = 6.023 × 1023 atoms
Mass of uranium (m) = \(\frac{\mathrm{n} \times 235}{\mathrm{~N}}\)
= \(\frac{2.375 \times 10^{24}}{6.023 \times 10^{23}}\)
= 926.66 g

Question 22.
If one microgram of \({ }_{92}^{235} \mathrm{U}\) is completely destroyed in an atom bomb, how much energy will be released ? (T.S. Mar. ’19)
Solution:
m = 1μg = 1 × 10-6 g = 1 × 10-6 × 10-3 kg
= 10-9 kg
c = 3 × 108 m/s
E = mc2 = 1 × 10-9 × 9 × 106 = 9 × 107 J

Question 23.
Calculate the energy released by fission from 2g of \({ }_{92}^{235} \mathrm{U}\) in kWh. Given that the energy released per fission is 200 MeV.
Solution:
Mass of uranium = 2g
Energy per fission = 200 MeV
No. of atoms in 2g, n = \(\frac{2 \times 6.023 \times 10^{23}}{235}\)
Total energy released (E’) = nE
= \(\frac{2 \times 6.023 \times 10^{23}}{235}\) × 200 × 106 × 1.6 × 10-19J
= \(\frac{2 \times 602.3}{235}\) × 32 × 109
= \(\frac{1640.3 \times 10^8}{36 \times 10^5}\)
∴ E’ = 45.56 × 103 kWh
= 4.556 × 104 kWh

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 24.
200 MeV energy is released when one nucleus of 235U undergoes fission. Find the number of fissions per second required for producing a power of 1 megawatt.
Solution:
E = 200 MeV
P = 1 × 106W
P = \(\frac{\mathrm{nE}}{\mathrm{t}}\) ⇒ \(\frac{\mathrm{n}}{\mathrm{t}}\) = \(\frac{\mathrm{P}}{\mathrm{E}}\) = \(\frac{10^6}{200 \times 10^6 \times 1.6 \times 10^{-19}}\)
= \(\frac{1}{32}\) × 1018
= 3.125 × 106

Question 25.
How much 235U is consumed in a day in an atomic power house operating at 400 MW, provided the whole of mass 235U is converted into energy ?
Solution:
P = 400 MW = 400 × 106 W, c = 3 × 108 m/s
t = 24 hours = 24 × 60 × 60 sec
E = mc2
\(\frac{\mathrm{Pt}}{\mathrm{c}^2}\) = m [∵ P = \(\frac{E}{t}\)]
m = \(\frac{400 \times 10^{-6} \times 24 \times 60 \times 60}{9 \times 10^6}\)
= 384 × 10-6 kg
∴ Mass required = 384 × 10-6 × 103 g = 0.384 g

Textual Exercises

Question 1.
a) Two stable isotopes of lithium \({ }_3^6 \mathrm{Li}\) and \({ }_3^7 \mathrm{Li}\) have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.
b) Boron has two stable isotopes, \({ }_5^{10} \mathrm{~B}\) and \({ }_5^{11} \mathrm{~B}\). Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811u. Find the abundances of \({ }_5^{10} B\) and \({ }_5^{11} \mathrm{~B}\).
Solution:
a) Atomic weight
= Weighted average of the isotopes.
= \(\frac{6.01512 \times 7.5+7.01600 \times 92.5}{(7.5+92.5)}\)
= \(\frac{45.1134+648.98}{100}\)
= 6.9414

b) Let relative abundance of 5B10 be x%
∴ Relative abundance 5B11 = (100 – x) %
Proceeding as above
10.811 = \(\frac{10.01294 x+i 1.00931 \times(100-x)}{100}\)
x = 19.9% and (100 – x) = 30.1%

Question 2.
The three stable isotopes of neon : \({ }_{10}^{20} \mathrm{Ne}\), \({ }_{10}^{21} \mathrm{Ne}\) and \({ }_{10}^{22} \mathrm{Ne}\) have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99u, 20.99 u and 21.99u, respectively. Obtain the average atomic mass of Neon.
Solution:
The masses of three isotopes are 19.99u, 20.99 u, 21.99u
Their relative abundances are 90.51%, 10.27% and 9.22%
∴ Average atomic mass of Neon is
m = \(\frac{90.51 \times 19.99+0.27 \times 20.99+9.22 \times 21.99}{(90.51+0.27+9.22)}\)
= \(\frac{1809.29+5.67+202.75}{100}\) = \(\frac{2017.7}{100}\) = 20.17u

Question 3.
Obtain the binding energy (in MeV. of a nitrogen nucleus \(\left({ }_7^{14} \mathrm{~N}\right)\), given m \(\left({ }_7^{14} \mathrm{~N}\right)\) = 14.00307 u.
Solution:
7N14 Nucleius contains 7 protons and 7 neutrons
Mass defect (ΔM) = 7mH + 7mn – mN
= 7 × 1.00783 + 7 × 1.00867 – 14.00307
= 7.05481 + 7.06069 – 14.00307
= 0.11243μ
Binding energy = 0.11243 × 931 MeV
= 104.67 MeV

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 4.
Obtain the binding energy of the nuclei \({ }_{26}^{56} \mathrm{~F}\) and \({ }_{83}^{209} \mathrm{Bi}\) in units of MeV from the following data : m\(\left({ }_{26}^{56} \mathrm{Fe}\right)\) = 55.934939 u, m\(\left({ }_{83}^{209} \mathrm{Bi}\right)\) = 208.980388 u
Solution:
(i) 26F56 nucleus contains 26 protons and (56 – 26) = 30 neutrons
Mass of 26 protons = 26 × 1.007825
= 26.26345 a.m.u
Mass of 30 neutrons = 30 × 1.008665
= 30.25995 amu
Total mass of 56 nucleons
= 56.46340 a.m.u
Mass of 26Fe56 Nucleus
= 55.934939 a.m.u
Mass defect Δm = 56.46340 – 55.934939
= 0.528461 a.m.u
Total binding energy = 0.524861 × 931.5 MeV
= 492.26 MeV
Average B.E per nucleon = \(\frac{492.26}{56}\)
= 8.790 MeV

(ii) 83Bi209 nucleus contains 83 protons and (209 – 83) = 126 neutrons
Mass of 83 protons = 83 × 1.007825
= 83.649475 a.m.u
Mass of 126 Neutrons = 126 × 1.008665
= 127.09170 a.m.u
Total mass of nucleons = 210.741260 a.m.u
Mass of 83Bi209 nucleus = 208.986388 a.m.u
Mass defect Δm = 210.741260 – 208.980388
= 1.760872
Total B.E = 1.760872 × 931.5 MeV
= 1640.26 MeV
Average B.E. per nucleon = \(\frac{1640.26}{209}\)
= 7.848 MeV
Hence 26Fe56 has greater B.E per nucleon than 83Bi209

Question 5.
A given coin has a mass of 3.0gi Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 29Cu atoms (of mass 62.92960 u.
Solution:
Number of atoms in 3g coin =
\(\frac{6.023 \times 10^{23} \times 3}{63}\)
= 2.868 × 1022
Each atom of copper contains 29 protons and 34 neutrons. Therefore, mass defect of each atom= [29 × 1.00783 + 34 × 1.00867] – 62.92960 = 0.59225 u
Total mass defect for all the atoms
= 0.59225 × 2.868 × 1022 u
ΔM = 1.6985 × 1022u
As, 1u = 931 MeV
Nuclear energy required
= 1.6985 × 1022 × 931 MeV
= 1.58 × 1025 MeV

Question 6.
Write nuclear reaction equations for
i. α-decay of \({ }_{88}^{226} \mathrm{Ra}\)
ii. α-decay of \({ }_{94}^{242} \mathrm{Pu}\)
iii. β-decay of \({ }_{15}^{32} \mathrm{P}\)
iv. β-decay of \({ }_{83}^{210} \mathrm{Bi}\)
v. β+-decay of \({ }_6^{11} \mathrm{C}\)
vi. β+-decay of \({ }_{43}^{97} \mathrm{Tc}\)
Solution:
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 27

Question 7.
A radio active isotope has a half-life of T years. How long will it take the activity to reduce to
a) 3.125%,
b) 1% of its original value ?
Solution:
a) Here \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = \(\frac{3.125}{100}\) = \(\frac{1}{32}\)
From \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = \(\left(\frac{1}{2}\right)^{\mathrm{n}}\) = \(\frac{1}{32}\left(\frac{1}{2}\right)^5\) ∴ n = 5
From t = nT = 5T

b) Here \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = \(\frac{1}{100}\)
From \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = e-λt = \(\frac{1}{100}\)
-λt = log 1 – \(\log _{\mathrm{e}} 100\)
= 0 – 2.303 \(\log _{10} 100\) = -2.203 × 2
= -4.606
t = \(\frac{4.606}{\lambda}\) = \(\frac{4.606}{0.693 / \mathrm{T}}\) = 6.65 T

Question 8.
The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive \({ }_6^{14} \mathrm{C}\) present with the stable carbon isotope \({ }_6^{12} \mathrm{C}\). When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity, ceases and its activity begins to drop. From the known half-life (5730 years. of \({ }_6^{14} \mathrm{C}\), and the measured activity, the age of the specimen can be approximately estimated. This is the principle of \({ }_6^{14} \mathrm{C}\) dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.
Solution:
Here normal activity R0 = 15 decays/min
Present activity R = 9 decays / min
T = 5730 yrs
Age t = ?
As activity is proportional to the number of radio active atoms, therefore
\(\frac{\mathrm{N}}{\mathrm{N}_0}\) = \(\frac{\mathrm{R}}{\mathrm{R}_0}\) = \(\frac{9}{15}\)
But \(\frac{N}{N_0}\) = e-λt
e-λt = \(\frac{9}{15}\) = \(\frac{3}{5}\)
e+λt = \(\frac{5}{3}\)
λt \(\log _{\mathrm{e}} \mathrm{e}\) = \(\log _e \frac{5}{3}\) = 2.3023 log 1.6667
λt = 2.3026 × 0.2218 = 0.5109
t = \(\frac{0.5109}{\lambda}\)
λ = \(\frac{0.693}{T}\) = \(\frac{0.693}{5730}\)yt-1
∴ t = \(\frac{0.5109}{0.693 / 5730}\) = \(\frac{0.5109 \times 5730}{0.693}\)
t = 4224.3 years

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 9.
Obtain the amount of \({ }_{27}^{60} \mathrm{Co}\) necessary to provide a radioactive source of 8.0 mCi strength. The half-life of \({ }_{27}^{60} \mathrm{Co}\) is 5.3 years.
Solution:
Here, mass of 27Co60 =?
Strength of source, \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = 8.0 mci
= 8.0 × 3.7 × 107 disint/sec
Half life, T = 5.3 years
= 5.3 × 365 × 24 × 60 × 60 sec
= 1.67 × 108 sec
λ = \(\frac{0.693}{T}\) = \(\frac{0.693}{1.67 \times 10^8}\) = 4.14 × 10-9 s-1
As \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = 2N
∴ N = \(\frac{\mathrm{dN} / \mathrm{dt}}{\lambda}\) = \(\frac{8 \times 3.7 \times 10^7}{4.14 \times 10^{-9}}\)
= 7.15 × 1016
By definition of Avogadros number, Mass of 6.023 × 1023 atoms of 27Co60 = 60 g
Mass of 7.15 × 1016 atoms of 27Co60
= \(\frac{60 \times 7.15 \times 10^{16}}{6.023 \times 10^{23}}\) = 7.12 × 10-6g

Question 10.
The half-life of \({ }_{38}^{90} \mathrm{~Sr}\) is 28 years. What is the disintegration rate of 15 mg of this isotope?
Solution:
Here T = 28 years = 28 × 3.154 × 107s
As number of atoms in 90 g of 38Sr90
= 6.023 × 1023
∴ Number of atoms in 15mg of 38Sr90
= \(\frac{6.023 \times 10^{23}}{90}\) × \(\frac{15}{1000}\)
i.e., N = 1.0038 × 1020
Rate of disintegration \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = λN
= \(\frac{0.693}{\mathrm{~T}} \mathrm{~N}\)
= \(\frac{0.693 \times 1.0038 \times 10^{20}}{28 \times 3.154 \times 10^7}\)
= 7.877 × 1010 Bq

Question 11.
Obtain approximately the ratio of the nuclear radii of the gold isotope \({ }_{79}^{197} \mathrm{Au}\) and the silver isotope \({ }_{47}^{107} \mathrm{Ag}\).
Solution:
Here A1 = 197 and A2 = 107
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}\) = \(\left(\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right)^{1 / 3}\) = \(\left(\frac{197}{107}\right)^{1 / 3}\) = 1.225

Question 12.
Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) \({ }_{88}^{226} \mathrm{Ra}\) and (b) \({ }_{86}^{220} \mathrm{Rn}\).
Given m \(\left({ }_{88}^{226} \mathrm{Ra}\right)\) = 226.02540 u,
m \(\left({ }_{86}^{222} R n\right)\) = 222.01750 u,
m \(\left({ }_{86}^{222} \mathrm{Rn}\right)\) =220.01137 u, m \(\left({ }_{84}^{216} \mathrm{Po}\right)\) = 216.00189 u.
Solution:
a) 88Ra226 → 85Rn222 + 2He4 Q value
[m(88Ra226) – m(86Rn222) – mα] × 931.5 MeV
= [226.02540 – 222.0 1750 – 4.00260] × 931.5 MeV
Q = 0.0053 × 931.5 MeV = 4.94 MeV
K.E of a particle =
\(\frac{(A-4) Q}{A}\) = \(\frac{226-4}{226}\) × 4.94 = 4.85 MeV

b) Proceeding as above, in case of
Q = 6.41 MeV
K.E of a particle
= \(\frac{(\mathrm{A}-4) \mathrm{Q}}{\mathrm{A}}\) = \(\frac{(220-4)}{220}\) × 6.41 = 6.29 MeV

Question 13.
The radionuclide 11C decays according to AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 28 The maximum energy of the emitted positron is 0.960 MeV. Given the mass values ; m\(\left({ }_6^{11} \mathrm{C}\right)\) = 11.011434u and \(\left({ }_6^{11} \mathrm{~B}\right)\) = 11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted.
Solution:
Mass defect in the given reaction is Δm = m(6C11)
= [m (5B11) + Me]
This is in terms of nuclear masses. If we express the Q value interms of atomic masses we have to subtract 6me from atomic mass of carbon and 5 me from that of boron to get the corresponding nuclear masses
Therefore, we have
Δm = [m(6C11) – 6 me – m(5B11) + 5me – me
= [m(6C11) – m(5B11) – 2 me]
= [11.011434 – 11.009305 – 2 × 0.000548] u
= 0.001033u
As, 1u = 931 MeV
Q = 0.001033 × 931 MeV = 0.961 MeV
Which is the maximum energy of emitted position.

Question 14.
The nucleus \({ }_{10}^{23} \mathrm{Ne}\) decays by β-emission. Write down the β – decay. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
\(m\left({ }_{10}^{23} \mathrm{Ne}\right)\) = 22.994466 u
\(\mathrm{m}\left({ }_{11}^{23} \mathrm{Na}\right)\) = 22.089770 u.
Solution:
The β decay of 10Ne23 may be represented as
10Ne23 → 11Na23 – 1e0 + v + Q
Ignoring the rest mass of antineutrino and v electron
Mass defect, Δm = m(10Ne23) – m (11Na23)
= 22.994466 – 22.989770
= 0.004696 amu
Q = 0.004696 × 931 MeV
= 4.372 MeV
As 11Na23 is very massive, this energy of 4.3792 MeV is shared by e–v pair. The max K.E of e– = 4.372 MeV when energy carried by v is zero.

Question 15.
The Q value of a nuclear reaction A + b → C + d is defined by Q = [mA + mb – mc – md] c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
i) \(\mathrm{H}_{\mathrm{1}}^{\mathrm{1}}\) + \({ }_1^3 \mathrm{H}\) → \({ }_1^2 \mathrm{H}\) + \({ }_1^2 \mathrm{H}\)
ii) \({ }_6^{12} \mathrm{C}\) + \({ }_6^{12} \mathrm{C}\) → \({ }_{10}^{20} \mathrm{Ne}\) + \({ }_2^4 \mathrm{He}\)
Atomic masses are given to be
\(\mathrm{m}\left({ }_{\mathrm{1}}^2 \mathrm{H}\right)\) = 2.014102u
\(\mathrm{m}\left({ }_1^3 \mathrm{H}\right)\) = 3.016049 u
\(\mathrm{m}\left({ }_6^{12} \mathrm{C}\right)\) = 12.000000u
\(\mathrm{m}\left({ }_{10}^{20} \mathrm{Ne}\right)\) = 19.992439 u
Solution:
i) \(\mathrm{H}_{\mathrm{1}}^{\mathrm{1}}\) + \({ }_1^3 \mathrm{H}\) → \({ }_1^2 \mathrm{H}\) + \({ }_1^2 \mathrm{H}\)
Q = Δm × 931 MeV =
[m (1H1 + m(1H3) – 2m (1H2)] × 931 MeV
= [1.007825 + 3.01604 – 2 × 2.014102] × 931 MeV
= -4.03 MeV
∴ This reaction is endothermic,

ii) 6C12 + 6C12 → 10Ne20 + 2He4
Q = Δm × 931 MeV =
[2m(6C12) – m(10Ne20) – m(2He4)] × 931 MeV
= [24.000000 – 19.992439 – 4.002603] × 931 MeV
= ±4.61 MeV
∴ The reaction is exothermic.

Question 16.
Suppose we think of fission of a \({ }_{26}^{56} \mathrm{Fe}\) nucleus into two equal fragments, \({ }_{13}^{28} \mathrm{Al}\), Is the fission energetically possible ? Argue by working out Q of the process. Given \(\mathrm{m}\left({ }_{26}^{56} \mathrm{Fe}\right)\) = 55.93494 u and \(\mathbf{m}\left({ }_{13}^{28} \mathrm{~A} l\right)\) = 27.98191 u.
Solution:
Q = [m(26Fe56 – 2m (13 Al28.] × 931.5 MeV
= [55.93494 – 2 × 27.9819] × 931.5 MeV
Q = – 0.2886 × 931.5 MeV = – 26.88 MeV
Which is negative.
This fission is not possible energetically.

Question 17.
The fission properties of \({ }_{94}^{239} \mathrm{Pu}\) are very similar to those of \({ }_{92}^{238} U\). The average-energy released per fission is 180 MeV. How much energy, in MeV is released if all the atoms in 1kg of pure \({ }_{94}^{239} \mathrm{Pu}\) undergo fission?
Solution:
No. of atoms in 1kg of pure .
UP = \(\frac{6.023 \times 10^{23}}{239}\) × 1000 = 2.52 × 1024
As average energy released / fission is 180 MeV, therefore total energy released
= 2.52 × 1024 × 180 MeV = 4.53 × 1026 MeV

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 18.
A 1000 MW fission reactor consumes half of its fuel In 5.00 y. How much \({ }_{92}^{235} \mathrm{U}\) did it contain initially ? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of \({ }_{92}^{235} \mathrm{U}\) and that this nuclide is consumed only by the fission process.
Solution:
In the fission of one nucleus of 92U235 energy generated is 200 MeV
∴ Energy generated in fission of 1 kg of
92U235 = 200 × \(\frac{6 \times 10^{23}}{235}\) × 1000 MeV
= 5.106 × 1026 MeV = 5.106 × 1026 × 1.6 × 10-13J
= 8.17 × 103 J
Time for which reactor operates \(\frac{80}{100}\) × 5
years = 4 years.
Total energy generated in 5 years.
= 1000 × 106 × 60 × 60 × 24 × 365 × 4J
∴ Amount of U consumed in 5 years
= \(\frac{1000 \times 10^6 \times 60 \times 60 \times 24 \times 365 \times 4}{8.17 \times 10^{13}} \mathrm{~kg}\)
= 1544 kg
∴ Initial amount of 92U235 = 2 × 1544 kg
= 3088 kg

Question 19.
How long can an electric lamp. of 100W be kept glowing by fusion of 20 kg of deuterium ? Take the fusion reaction as \({ }_1^2 \mathrm{H}\) + \({ }_1^2 \mathrm{H}\) → \({ }_2^3 \mathrm{He}\) + n + 3.27 MeV
Solution:
Number of deuterium atoms in 2.0 kg
\(\frac{6.023 \times 10^{23} \times 2000}{2}\) = 6.023 × 1026
Energy released when 2 atoms füse = 3.27 MeV
∴ Total energy released
= \(\frac{3.27}{2}\) × 6.023 × 1026 MeV
= 1.635 × 6.023 × 1026 × 1.6 × 10-13 j
= 15.75 × 103 J
Enery consumed by the bulb/sec = 100 J
∴ Time for which bulb will glow
= \(\frac{15.75 \times 10^{13}}{100} \mathrm{~S}\)
= \(\frac{15.75 \times 10^{11}}{60 \times 60 \times 24 \times 365}\) = 4.99 × 107 years

Question 20.
Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint : The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.
Solution:
For head on collision distance between centres of two deuterons = r = 2 × radius
r = 4 fm = 4 × 10-15 m
Charge of each deuteron e = 1.6 × 10-10 C
Potential energy
\(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\) = \(\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}{4 \times 10^{-15}}\) Joule
= \(\frac{9 \times 1.6 \times 1.6 \times 10^{-14}}{4 \times 1.6 \times 10^{-16}}\)KeV
PE = 360 KeV
P.E = 2 × K.E of each deuteron = 360 KeV
K.E of each deuteron = \(\frac{360}{2}\) = 180 KeV
This is a measure of height of Coulomb barrier

Question 21.
From the relation R = R0A1/3, where R0, is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e., independent of A.
Solution:
Density of nuclear matter
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 29
ρ = \(\frac{\mathrm{mA}}{\frac{4}{3} \pi \mathrm{R}^3}\), where m is average mass of a nucleon
Using R = R0A1/3 we get
ρ = \(\frac{3 \mathrm{~mA}}{4 \pi\left(\mathrm{R}_0 \mathrm{~A}^{1 / 3}\right)^3}\) = \(\frac{3 \mathrm{~m}}{4 \pi \mathrm{R}_0^3}\)
As R0 is constant, therefore ρ is constant.

Question 22.
For the β+ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K – shell, is captured by the nucleus and a neutrino is emitted).
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 30
Show that if β+ emission is energetically allowed, electron capture is necessarily allowed but not vice-versa.
Solution:
The β+ emission from a nucleus ZXA may be represented as
zXA = z-1YA + 1e0 + v + Q1 —– (i)
The other competing process of electron capture may be represented as
-1e0 + ZXA = Z-1yA + v + Q2 —– (ii)
The energy released in Q1 in (1. is given by
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 31
Note that mN here denotes mass of nucleus and m denotes the mass of atom similarly from (ii)
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 32
Ir Q1 > 0 then Q2 > 0
i.e., If positron emission is energetically allowed electron capture is necessarily allowed. But Q2 > 0 does not necessarily mean Q1 > 0. Hence the reverse is not true.

Additional Exercises

Question 1.
In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are \({ }_{12}^{24} \mathrm{Mg}\) (23.98504u), \({ }_{12}^{25} \mathrm{Mg}\) (24.98584u) and \({ }_{12}^{26} \mathrm{Mg}\) (25.98259u). The natural abundance of \({ }_{12}^{24} \mathrm{Mg}\) is 78.99% by mass. Calculate the abundances of other two isotopes.
Solution:
Let the abundance of \({ }_{12} \mathrm{Mg}^{25}\) by mass be x% therefore, abundance of 12Mg26 by mass
= (100 – 78.99 – x%)
= (21.01 – x%)
Now average atomic mass of magnesium is
24.312 =
\(\frac{23.98504 \times 78.99+24.98584+25.98529(21.01-\mathrm{x})}{100}\)
on solving we get x = 9.303% for 12Mg25 and for 12Mg26 (21.01 – x) = 11.71%

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 2.
The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei \({ }_{20}^{41} \mathrm{Ca}\) and \({ }_{13}^{27} \mathbf{Al}\) from the following data: .
m(\({ }_{20}^{40} \mathrm{Ca}\)) = 39.962591 u
m(\({ }_{20}^{41} \mathrm{Ca}\)) = 40.962278 u
m({ }_{13}^{26} \mathrm{Ca}) = 25.986895 u
m({ }_{13}^{27} \mathrm{Ca}) = 26.981541 u
Solution:
When a neutron is separated from 20Ca41 we are left with
20Ca40 i.e. 20Ca41 → 20Ca40 + 0n1
Now mass defect
ΔM = m(20Ca40) + mn – m (20Ca41)
= 39.962591 + 1.008665 – 40.962278
= 0.008978 a.m.u
∴ Neutron seperation energy
= 0.008978 × 931MeV
= 8.362 MeV
similarly 13Al27 → 13Al26 + 0n1
∴ Mass defect, ΔM = m (13Al26) + mn – m(13Al27)
= 25.986895 + 1.008665 – 26.981541
= 0.013845 u
∴ Neutron seperation energy. = 0.0138454 × 931MeV
= 12.89 MeV

Question 3.
A source contains two phosphorous radio nuclides \({ }_{15}^{32} \mathbf{P}\) (T1/2 = 14.3d) and \({ }_{15}^{33} P\) (T1/2 = 25.3d). Initially, 10% of the decays come from \({ }_{15}^{33} \mathrm{P}\). How long one must wait until 90% do so ?
Solution:
Suppose initially the source has 90% 15pt32 and 10% \({ }_{15} \mathrm{P}_{\mathrm{t}}^{32}\), say 9x gram P2 and x gram of P1.

After t days, suppose the source has 90% \({ }_{15} \mathbf{P}_2^{33}\) and 10% \({ }_{15} \mathrm{P}_{\mathrm{t}}^{32}\) i.e., y gram of P2 and 9y gram of P1
we have to calculate :
from \(\frac{\mathrm{N}}{\mathrm{N}_0}\) = \(\left(\frac{1}{2}\right)^n\) = \(\left(\frac{1}{2}\right)^{t / T}\) = 2-i/T
N = N02-t/T
y = 9×2-t/14.3 for P2 and 9y = x 2-t/25.3 for P1
Dividing we get
\(\frac{1}{9}\) = 9 × 2(t/25.3 – t/14.3.)
or \(\frac{1}{81}\) = 2-11t/25.3 × 14.3
log 1 – log 81 = \(\frac{-11 \mathrm{t}}{25.3 \times 14.3}\) log 2
0 – 1 – 9085 = \(\frac{-11 \mathrm{t}}{25.3 \times 14.3}\) × 0.3010.
t = \(\frac{25.3 \times 14.3 \times 1.9085}{11 \times 0.3010}\) = 208.5 days

Question 4.
Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α – particle. Consider the following decay processes :
\({ }_{88}^{223} \mathrm{Ra}\) → \({ }_{82}^{209} \mathrm{~Pb}\) + \({ }_6^{14} \mathrm{C}\)
\({ }_{88}^{223} \mathrm{Ra}\) → \({ }_{86}^{219} \mathrm{~Pb}\) + \({ }_2^4 \mathrm{He}\)
Calculate the Q-values for these decays and determine that both are energetically allowed.
Solution:
i) For the decay process
88Ra 223 → 82pb209 + 6C14 + Q
mass defect, ΔM = mass of Ra223 – (mass of pb209 + mass of C14)
= 223.01850 – (208.98107 + 14.00324)
= 0.03419u
Q = 0.03419 × 931 MeV = 31.83 MeV

ii) For the decay process
88Ra23 → 86Rn219 + 2He4 + Q mass defect, ΔM = mass of Ra223 – (mass of Rn219 + mass of He4) = 223.01850 – (219.00948 + 4.00260)
= 0.00642 u
∴ Q = 0.00642 × 931 MeV = 5.98 MeV
As Q values are positive in both the cases, therefore both the decays are energetically possible.

Question 5.
Consider the fission of \({ }_{92}^{238} \mathbf{U}\) by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are \({ }_{58}^{140} \mathrm{Ce}\) and \({ }_{44}^{99} \mathrm{Ru}\). Calculate Q for this fission process. The relevant atomic and particle masses are
m(\({ }_{92}^{238} \mathrm{U}\)) = 238.05079 u
m(\({ }_{58}^{140} \mathrm{Ce}\)) = 139.90543 u
m(\({ }_{44}^{99} \mathrm{Ru}\))= 98.90594 u
Solution:
For this fission reaction,
92U238 + on1 → 58Ce140 + 44Ru99 + Q
mass defect ΔM = mass of U238 + mass of n – (mass of Ce140 + mass of Ru99.
= 238.05079 + 1.00867 – (139.90543 + 98.90594)
= 0.24809U
∴ Q = 0.24809 × 931 MeV = 230.97 Mev

AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei

Question 6.
Consider the D-T reaction (deuterium- tritium fusion)
\({ }_1^2 \mathrm{H}\) + \({ }_1^3 \mathrm{H}\) → \({ }_2^4 \mathrm{He}\) + n
a) Calculate the energy released in MeV in this reaction from the data:
m\(\left({ }_1^2 \mathrm{H}\right)\) = 2.014102 u
m\(\left({ }_1^3 \mathrm{H}\right)\) = 3.016049 u
b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the Coulomb repulsion between the two nuclei ? To what temperature must the gas heated to
initiate the reaction ?
(Hint : Kinetic energy required for one fusion event = average thermal kinetic energy available with the interacting particles = 2(3kt/2); k = Boltzman’s constant, T = absolute temperature..
Solution:
a) For the process 1H2 + 1H3 + 2He4 + n + Q
Q = [m(1H2) + m (1H3) + m(2He4) – mn] × 931 MeV
= (2.014102 + 3.016049 – 4.002603
1.00867) × 931 MeV
= 0.018878 × 931 = 17.58 MeV

b) Repulsive potential energy of two nuclei when they almost touch each other is
= \(\frac{q^2}{4 \pi \varepsilon_0(2 r)}\) = \(\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)^7}{2 \times 2 \times 10^{-15}}\) Joule
= 5.76 × 10-14 Joule

Classically KE atleast equal to this amount is required to overcome Coulomb repulsion. Using the relation
K.E. = 2 × \(\frac{3}{2}\) KT
T = \(\frac{\mathrm{K} \cdot \mathrm{E}}{3 \mathrm{k}}\) = \(\frac{5.76 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}}\) = 1.39 × 109K
In actual practise the temperature required for trigerring the reaction is somewhat less.

Question 7.
Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. You are given that
m(198Au) = 197.968233 u
m(198Hg) = 197.966760 u.
AP Inter 2nd Year Physics Study Material Chapter 14 Nuclei 33
Solution:
Energy corresponding to r1
E1 = 1.088 – 0 = 1.088 MeV
= 1.088 × 1.6 × 10-13 Joule
Frequency v1 = \(\frac{E_1}{h}\)
= \(\frac{1.088 \times 1.6 \times 10^{-13}}{6.6 \times 10^{-34}}\)
= 2.63 × 1020 HZ
similarly v2 = \(\frac{\mathrm{E}_2}{\mathrm{~h}}\)
= \(\frac{0.412 \times 1.6 \times 10^{-12}}{6.6 \times 10^{10}}\)
= 9.98 × 1013 Hz
and v3 = \(\frac{E_3}{h}\)
= \(\frac{(1.088-0.412) \times 1.6 \times 10^{-13}}{6.6 \times 10^{20} \mathrm{~Hz}}\)
Maximum K.E. of β1 particle
Kmax (β1)= [m(79Au198 – mass of Second excited state of 80Hg198] × 931 MeV
= [m(79Au198) – m(82Hg198) – \(\frac{1.088}{931}\)] × 931 MeV
= 931 [197.968233 – 197.966760] – 1.088 MeV
= 1.371 – 1.088 = 0.283 MeV
similarly kmax (β2) – 0.957 MeV

Question 8.
Calculate and compare the energy released by a. fusion of 1.0 kg of hydrogen deep within Sun and b. the fission of 1.0 kg of 235U in a fission reactor.
Solution:
In sun, four hydrogen nuclei fuse to form a helium nucleus with the release of 26 MeV energy.
∴ Energy released by fusion of 1 kg of hydrogen = \(\frac{6 \times 10^{23} \times 26}{4}\) × 103 MeV
As energy released in fission of one atom of 92U236 = 200 MeV
Energy released in fission of 1 kg of 92U238
= \(\frac{6 \times 10^{23} \times 1000}{235}\) × 200 MeV
E2 = 5.1 × 1026 MeV
\(\frac{\mathrm{E}_1}{\mathrm{E}_2}\) = \(\frac{39 \times 10^{26}}{5.1 \times 10^{26}}\) = 7.65
i.e., Energy released in fusion is 7.65 times the energy released in fission.

Question 9.
Suppose India had a target of producing by 2020 AD, 2,00,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy of thermal energy produced in a reactór was 25%.
How much amount of fissionable uranium would our country need per year by 2020 ? Take the heat energy per fission of 235U to be about 200 MeV.
Solution:
Total targeted power = 2 × 105 MW
Total Nuclear power = 10% of 2 × 105 MW
= 2 × 104 MW
Energy produced in fission = 200 MeV
Effeciency of power plant =25%
∴ Energy converted into electrical energy per fission = \(\frac{25}{100}\) × 200 = 50 MeV
= 50 × 1.6 × 10-13 Joule.
Total electrical energy to be produced :
= 2 × 104 MW = 2 × 104 × 106 Watt
= 2 × 1010 Joule/Sec
= 2 × 1010 × 60 × 60 × 24 × 365 Joule / year
No. of fissions in one year
= \(\frac{2 \times 10^{10} \times 60 \times 60 \times 24 \times 365}{50 \times 1.6 \times 10^{-13}}\)
= 2 × \(\frac{36 \times 24 \times 365}{8}\) × 1024
Mass of 6.023 × 1023 atoms of U235 = 235 gm = 235 × 10-3 kg
Mass of \(\frac{2 \times 36 \times 24 \times 365}{8}\) × 1024 atoms
= \(\frac{235 \times 10^{-3}}{6.023 \times 10^{23}}\) × \(\frac{2 \times 36 \times 24 \times 365 \times 20^{24}}{8}\)
= 3.08 × 104 Kg
Hence mass of Uranium needed per year = 3.08 × 104 Kg.

AP Inter 2nd Year Physics Study Material Chapter 13 Atoms

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 13th Lesson Atoms Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 13th Lesson Atoms

Very Short Answer Questions

Question 1.
What is the angular momentum of electron in the second orbit of Bohr’s model of hydrogen atom ?
Answer:
Angular momentum of electron in second orbit of Hydrogen atom
L = \(\frac{2 \mathrm{~h}}{2 \pi}\) = \(\frac{h}{\pi}\) (∵ L = \(\frac{h h}{2 \pi}\))

Question 2.
What is the expression for fine structure constant and what is its value?
Answer:
Formula for fine structure constant
α = \(\frac{\mathrm{e}^2}{2 \varepsilon_0 \mathrm{ch}}\); value of α = \(\frac{1}{137}\)

AP Inter 2nd Year Physics Study Material Chapter 13 Atoms

Question 3.
What is the physical meaning of ‘negative energy of an electron’ ?
Answer:
The ‘negative energy of an electron’ indicates that the electron bound to the nucleus due to force of attraction.

Question 4.
Sharp lines are present in the spectrum of a gas. What does this indicate ?
Answer:
Sharp lines in the spectrum of gas indicates bright lines against dark background.

Question 5.
Name a physical quantity whose dimensions are the same as those of angular momentum.
Answer:
Planck’s constant.

Question 6.
What is the difference between α – particle and helium atom ?
Answer:
Alpha particle

  1. It is a + 2e charged Helium nucleus.
  2. It contains 2 protons and 2 neutrons.

Helium atom

  1. It has no charge.
  2. It contains 2 protons, 2 electrons and 2 neutrons.

Question 7.
How is impact parameter related to angle of scattering ?
Answer:
The impact parameter related to angle of scattering is given by b = \(\frac{1}{4 \pi \varepsilon_0} \times \frac{\mathrm{Ze}^2}{\left(\frac{1}{2} m v^2\right)} \cot \theta_2\)

Question 8.
Among alpha, beta and gamma radiations, which get affected by the electric field ?
Answer:
Alpha and Beta radiations are get affected by the electric field.

Question 9.
What do you understand by the ‘phrase ground state atom’ ?
Answer:
If the electron is present in the ground state, it is called ground state atom.

Question 10.
Why does the mass of the nucleus not have any significance in scattering in Rutherford’s experiment ?
Answer:
The size of the atom is 10-10 m and size of the nucleus is 10-15 m. Hence atom has large empty space. So the mass of nucleus has no significance in Rutherford’s scattering experiment.

Question 11.
The Lyman series of hydrogen spectrum lies in the ultraviolet region. Why ? (A.P. Mar. ’15)
Answer:
The calculated values of wavelengths lie in the ultraviolet region of the spectrum well agree with the values of wavelengths observed experimentally by Lyman.

Question 12.
Write down a table giving longest and shortest wavelengths of different spectral series.
Answer:
Wavelength limits of some spectral series of hydrogen.
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 1

Question 13.
The wavelengths of some of the spectral lines obtained in hydrogen spectrum are 1216 A, 6463 A and 9546A. Which one of these wavelengths belongs to the Paschen series ?
Answer:
The wavelength of spectral line 9546A belongs to the Paschen series.

AP Inter 2nd Year Physics Study Material Chapter 13 Atoms

Question 14.
Give two drawbacks of Rutherford’s atomic model.
Answer:
Drawbacks of Rutherford’s atom model:

  1. As the revolving electron loses energy continuously, it must spiral inwards and eventually fall into the nucleus. But matter is stable, we can not expect the atom collapse.
  2. The atoms should emit continuous spectrum, but what we observe is only a line spectrum.

Short Answer Questions

Question 1.
What is impact parameter and angle of scattering ? How are they related to each other ?
Answer:

  1. Impact parameter (b) : Impact parameter is defined as the perpendicular distance of the initial velocity vector of the alpha particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom.
  2. Scattering angle (θ): The scattering angle (θ) is the angle between the asymtotic direction of approach of the α – particle and the asymptotic direction in which it receeds.
  3. The relation between b and θ is b = \(\frac{1}{4 \pi \varepsilon_0} \frac{Z \mathrm{e}^2}{E} \cot \frac{\theta}{2}\) where E = K.E. of α – particle = \(\frac{1}{2} \mathrm{mv}^2\)

Question 2.
Derive an expression for potential and kinetic energy of an electron in any orbit of a hydrogen atom according to Bohr’s atomic model. How does P.E change with increasing n. (T.S. Mar. ’15)
Answer:

  1. According fo Bohr electrostatic force of attraction, Fe between the revolving electrons and nucleus provides the necessary centripetal force Fc to keep them in their orbits.
  2. Thus for dynamically state orbit in a hydrogen atom.
    Fc = Fe ⇒ \(\frac{m v^2}{r}\) = \(\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r_2}\)
  3. The relation between the orbit radius and the electron velocity is r = 2
    4e0 (m )
  4. The kinetic energy (K) and electrostatic potential energy (υ) of the electron in hydrogen atom are
    K = \(\frac{1}{2} \mathrm{mv}^2\) = \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 r}\) and υ = \(\frac{-\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\)
  5. The total energy E of the electron in a hydrogen atom is
    E = K + U = \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) – \(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\)
    ∴ E = \(\frac{-\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\)
  6. With increase in ‘nr potential energy (U) also increases.

Question 3.
What are the limitations of Bohr’s theory of hydrogen atom ? (Mar. ’14)
Answer:
Limitations of Bohr’s theory of Hydrogen atom :

  1. This theory is applicable only to simplest atom like hydrogen, with z = 1. The theory fails in case of atoms- of other elements for which z > 1.
  2. The theory does not explain why orbits of electrons are taken as circular, while elliptlical orbits are also possible.
  3. Bohr’s theory does not say anything about the relative intensities of spectral lines.
  4. Bohr’s theory does not take into account the wave properties of electrons.

Question 4.
Explain the distance of closest approach and impact parameter.
Answer:
Distance of closest approach :

  1. Suppose an α-particle with initial kinetic energy (K.E) is directed towards the centre of the nucleus of an atom.
  2. On account of Coulomb’s repulsive force between nucleus and alpha particle, kinetic energy of alpha particle goes on decreasing and in turn, electric potential energy of the particle goes on increasing.
  3. At certain distance ‘d’ from the nucleus, K. E of α-particle reduces to zero. The particle stops and it can not go closer to the nucleus. It is repelled by the nucleus and therefore it retraces its path, turning through 180°.
  4. Therefore, the distance d is known as the distance of closest of approach.
    AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 2
    The closest distance of approach,
    d = \(\frac{1}{4 \pi \varepsilon_0} \times \frac{Z e^2}{\left(\frac{1}{2} m v^2\right)}\)
    AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 3
  5. Impact parameter (b) : Impact parameter is defined as the ⊥r distance of the initial velocity vector of the α – particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom.

Question 5.
Give a brief account of Thomson model of atom. What are its limitations ?
Answer:
Thomson’s model of atom :

  1. According to Thomson’s model, every atom consists of a positively charged sphere of radius of the order of 10-10 m in which entire mass and positive charge of the atom are uniformly distributed.
    AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 4
  2. Inside this sphere, the electrons are embedded like seeds in a watermelon or like plums in a pudding.
  3. The negative charge of electrons is equal to the positive charge of the atom. Thus atom is electrically neutral.

Limitations :

  1. It could not explain the origin of spectral series of hydrogen and other atoms, observed experimentally.
  2. It could not explain large angle scattering of a-particles from thin metal foils, as observed by Rutherford.

Question 6.
Describe Rutherford atom model. What are the draw backs of this model.
Answer:
Rutherford atom model: The essential features of Rutherford’s nuclear model of the atom or planetary model of the atom are as follows :

  1. Every atom consists of tiny central core, called the atomic nucleus, in which the entire positive charge and almost entire mass of the atom are concentrated.
  2. The size of nucleus is of the order of 10-15m, which is very small as compared to the size of the atom which is of the order of 10-10m.
  3. The atomic nucleus is surrounded by certain number of electrons. As atom on the whole is electrically neutral, the total negative charge of electrons surrounding the nucleus is equal to total positive charge on the nucleus.
  4. These electrons revolve around the nucleus in various circular orbits as do the planets around the sun. The centripetal force required by electron for revolution in provided by the electrostatic force of attraction between the electrons and the nucleus.

Draw backs : According to classical E.M. theory.

  1. The revolving electron loses energy continuously, it must spiral inwards and eventually fall into the nucleus. As matter is stable, we cannot expect the atoms to collapse.
  2. Since the frequency of radiation emitted is the same as the frequency of revolution, the atom should radiate a continuous spectrum, but what we observe is only a line spectrum.

Question 7.
Distinguish between excitation potential and ionization potential.
Answer:
Excitation Potential:

1) When the electron jumps from lower orbit to higher orbit by absorbing energy is called excited electron and the process is known as excitation. The minimum accelerating potential which provides an electron energy sufficient to jump from the inner most orbit (ground state) to one of the outer orbits is called excitation potential or resonance potential.

2) a) For example, in case of hydrogen atom,
E1 = -13.6 eV. E2 = -3.4 eV E3 = -1.51eV and soon, E∞ = 0
∴ Energy required to raise an electron from ground state (n = 1) to first excited state
(n = 2) is E = E2 – E1 = -3.4 – (-13.6) = 10.2 eV.
The corresponding excitation potential = 10.2 Volt,

b) Similarly, energy required to raise an electron from ground state (n = 1) to second excited , state (n = 3) is
E = E3 – E1 = -1.51 – (-13.6) = -1.51 + 13.6 = 12.09 eV
The corresponding excitation potential = 12.09 Volt and so on.

3) The excitation potential of an atom is not one. It can have many values, depending on the state to which the atom is excited.

Ionisation potential:

  1. The energy supplied is so large that it can remove an electron from the outer most orbit of an atom, the process is called Ionisation. Thus ionisation is the phenomenon of removal of an electron from the outer most orbit of an atom.
  2. The minimum accelerating potential which would provide an electron energy sufficient just to remove it from the atom is called Ionisation potential.
  3. For example, total energy of electron in ground state of hydrogen atom, + 13.6 eV energy is required.
    ∴ Ionisation energy of hydrogen atom = 13.6 eV.
    Ionisation potential of hydrogen atom = 13.6 Volts.
  4. The general expression for ionisation potential of an atom is V = \(\frac{13.6 \mathrm{Z}^2}{\mathrm{n}^2}\) volt, Where Z is the charge number of the atom and n is number of orbit from which electron is to be removed.
  5. For a given element, ionisation potential is fixed, but for different elements, ionisation potentials are different.

AP Inter 2nd Year Physics Study Material Chapter 13 Atoms

Question 8.
Explain the different types of spectral series in hydrogen atom. (A.P. Mar. ’19, ’15; T.S. Mar. ’16)
Answer:
The atomic hydrogen emits a line spectrum consisting of five series.

  1. Lyman series : v = Rc \(\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\) where n = 2, 3, 4, ……
  2. Balmer series : v = Rc\(\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\) where n = 3, 4, 5, ………
  3. Paschen series : v = Rc\(\left(\frac{1}{3^2}-\frac{1}{n^2}\right)\) where n = 4, 5, 6, …….
  4. Brackett series : v = Rc\(\left(\frac{1}{4^2}-\frac{1}{n^2}\right)\) where n = 5, 6, 7, ……
  5. Pfund series : v = Rc\(\left(\frac{1}{5^2}-\frac{1}{n^2}\right)\) where n = 6, 7, 8,……..

Question 9.
Write a short note on Debroglie’s explanation of Bohr’s second postulate of quantization.
Answer:
Debroglie’s explanation of Bohr’s second postulate of quantization :

  1. The second postulate of Bohr atom model says that angular momentum of electron orbiting around the nucleus is quantized i.e., mυr = \(\frac{\text { nh }}{2 \pi}\) where m = 1, 2, 3,….
  2. According to Debroglie, the electron in its circular orbit, as proposed by Bohr, must be seen as a particle wave.
  3. When a string fixed at two ends is plucked, a large number of wavelengths are excited and standing wave is formed.
  4. It means that in a string, standing waves form when total distance travelled by a wave down the string and back is an integral number of wavelengths.
  5. According to Debroglie, a stationary orbit is that which contains an integral number of Debrogile waves associated with the revolving electron.
  6. For an electron revolving in nth circular orbit of radius rn, total distance covered = circumference òf the orbit = 2πrn
    ∴ For permissible orbit, 2πrn = nλ
  7. According to Debrogile, λ = \(\frac{h}{m v_n}\) Where υn is speed of electron revolving in nth orbit
    ∴ mυnrn = \(\frac{\mathrm{nh}}{2 \pi}\) = \(\mathrm{n}\left(\frac{\mathrm{h}}{2 \pi}\right)\)
    i.e., angular momentum of electron revolrmg in nth orbit must be an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\), which is the quantum condition proposed by Bohr in second postulate.

Long Answer Questions

Question 1.
Describe Geiger-Marsden Experiment on scattering of α – particles. How is the size of the nucleus estimated in this experiment ?
Answer:

  1. The experimental set up used by Rutherford and his colaborators, Geiger and Marsden is shown in fig.
    AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 5
  2. The α-particles emitted by radio active source contained in a lead cavity are collimated into a narrow beam with the help of a lead slit (collimator).
  3. The collimated beam is allowed to fall on a thin gold foil of thickness of the order of 2.1 × 10-7m.
  4. The α-particles scattered in different directions are observed through a rotatable detector consisting of zinc sulphide screen and a microscope.
  5. The α-particles produce bright flashes or scintillations on the ZnS screen.
  6. These are observed in the microscope and counted at different angles from the direction of incidence of the beam.
  7. The angle θ of deviation of an α-particle from its original direction is called its scattering angle θ.

Observations : We find that

  1. Most of the alpha particles pass straight through the gold foil. It means they do not suffer any collision with gold atoms.
  2. Only about 0.14% of incident α-particles scatter by more than 1°.
  3. About one α-particle in every 8000 α-particles deflect by more than 90°.

Estimation of size of the nucleus :

  1. This led to Rutherford postulate, that the entire positive charge of the atom must be concentration in a tiny central core of the atom. This tiny central core of each atom was called atomic nucleus.
  2. The electrons would be moving in orbits about the nucleus just as the planets do around the sun.
  3. Rutherford’s experiments suggested the size of the nucleus to be about 10-15m to 10-14m. From kinetic theory, the size of an atom was known to be 10-10m, about 10,000 to 1,00,000 times larger than the size of the nucleus.

Question 2.
Discuss Bohr’s theory of the spectrum of hydrogen atom.
Answer:

  1. According to Bohr’s model an electron continuous to revolve round the nucleus in fixed, stationary orbits. This is called groupd state of the atom. In ground state there is no emission of radiation.
  2. But when some energy is given to an atom the electron absorbs this energy. This is called excited state of the atom. In this state the electron jumps to the next higher orbit. But it can remain 10-8 sec and it immediatly returns back to its ground state and the balance of the energy is emitted out as a spectral line.
  3. According to Bohr’s third postulate, the emitted energy is given by E = hv = E2 – E1
    AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 6

Spectral series of Hydrogen atom:

AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 7

Hydrogen atom has five series of spectral lines: They are

1. Lyman series: When an electron jumps from the outer orbits to the first orbit, the spectral lines are in the ultra – violet region. Here n1 = 1, n2 = 2, 3, 4, 5….
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\mathrm{n}_2^2}\right]\) = \(\mathrm{R}\left[1-\frac{1}{\mathrm{n}_2^2}\right]\)

2) Balmer Series : When an electron jumps from the outer orbits to the second orbit, the
spectral Lines are in the visible region. Here n1 = 2, n2 = 3, 4, 5…
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{\mathrm{n}_2^2}\right]\)

3) Paschen series : When an electron jumps from the outer orbits to the third orbit, the spectral lines are in the near infrared region. Here n1 = 3, n2 = 4, 5, 6 ….
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left[\frac{1}{3^2}-\frac{1}{\mathrm{n}_2{ }^2}\right]\)

4) Brackett series : When an electron jumps from outer orbits to the forth orbit, the spectral lines are in the infrared region. Here n1 = 4, n2 = 5, 6, 7…….
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left[\frac{1}{4^2}-\frac{1}{\mathrm{n}_2^2}\right]\)

5) Pfund series : When an electron jumps from outer orbits to the fifth orbit, the spectral lines are in the far infrared region. Here n1 = 5, n2 = 6, 7, 8, ………
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left[\frac{1}{5^2}-\frac{1}{\mathrm{n}_2^2}\right]\)

AP Inter 2nd Year Physics Study Material Chapter 13 Atoms

Question 3.
State the basic postulates of Bohr’s theory of atomic spectra. Hence obtain an expression for the radius of orbit and the energy of orbital electron in a hydrogen atom.
Answer:
a) Basic postulates of Bohr’s theory are
1) The electron revolves round a nucleus is an atom in various orbits known as stationary orbits. The electrons can not emit radiation when moving in their own stationary levels.

2) The electron can revolve round the nucleus only in allowed, orbits whose angular momentum is the integral multiple of \(momentum is the integral multiple of
i.e., mυnrn = [latex]\frac{\mathrm{nh}}{2 \pi}\) ———> (1)
where n = 1, 2, 3…..

3) If an electron jumps from higher energy (E2) orbit to the lower energy (E1) orbit, the difference of energy is radiated in the form of radiation.
i.e., E = hv = E2 – E1 ⇒ v = \(\frac{E_2-E_1}{h}\) ——> (2)

b) Energy of emitted radiation : In hydrogen atom, a single electron of charge — e, revolves around the nucleus of charge e in a ciccular orbit of radius rn.

1) K.E. of electron : For the electron to be in circular orbit, centripetal force = The electrostatic force of attraction between the electron and nucleus,
From Coulomb’s law, \(\frac{\mathrm{m} \dot{v}_n^2}{r_n}\) = \(\frac{\mathrm{Ke}^2}{\mathrm{r}_{\mathrm{n}}^2}\) ——> (3)
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 8
where K = \(\frac{1}{4 \pi \varepsilon_0}\) —–> (4)
\(m v_n^2\) = \(\frac{\mathrm{Ke}^2}{\mathrm{r}_{\mathrm{n}}}\) —–> (5)
\(m v^2 r_n\) = Ke2 ——-> (6)
Dividing (5) by (1), υn = Ke2 × \(\frac{2 \pi}{\mathrm{nh}}\)
From (3), kinetic energy K = \(\frac{1}{2} m v_n^2\) = \(\frac{\mathrm{Ke}^2}{2 r_{\mathrm{n}}}\)

2) Potential energy of electron:
P.E. of electron, U = \(\frac{\mathrm{Ke}}{\mathrm{r}_{\mathrm{n}}} \times-\mathrm{e}\) [∵ W = \(\frac{I^{\prime}}{4 \pi \varepsilon_0} \frac{Q}{d}\) × -Q]
∴ U = \(\frac{-\mathrm{Ke}^2}{\mathrm{r}_{\mathrm{n}}}\)

3) Radius of the orbit: Substituting the value of (6) in (2),
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 9

4) Total energy (En) : Revolving electron posses K.E. as well as P.E.
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 10

Textual Exercises

Question 1.
The radius of the first electron orbit of a hydrogen atom is 5.3 × 10-11m. What is the radius of the second orbit ?
Solution:
rn ∝ n2
\(\frac{\mathrm{r}_2}{\mathrm{r}_1}\) = \(\frac{2^2}{1^2}\) = \(\frac{4}{1}\) ⇒ r2 = 4r1

Question 2.
Determine the radius of the first orbit of the hydrogen atom. What would be the velocity and frequency of the electron in the first orbit ?
Solution:
Given: h = 6.62 × 10-34 J-s,
m = 9.1 × 10-31kg,
e = 1.6 × 10-19 C,
k = 9 × 109Nm2C-2, n = 1

i)
r1 = \(\frac{n^2 h^2}{4 \pi^2 \mathrm{mke}^2}\)
= \(\frac{(1)^2 \times\left(6.62 \times 10^{-34}\right)^2}{4 \times(3.14)^2 \times 9.1 \times 10^{-31} \times 9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}\)
∴ r1= 0.529 A \(\simeq\) 0.53 A

ii)
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 11

iii)
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 12

Question 3.
The total energy of an electron in the first excited state of the hydrogen atom is -3.4eV. What is the potential energy of the electron in this state ?
Solution:
In 1st orbit, E = -3.4eV
Total energy E = \(\frac{\mathrm{KZe}^2}{2 \mathrm{r}}\) – \(\frac{\mathrm{KZe}^2}{\mathrm{r}}\)
\(\frac{\mathrm{KZe}^2}{\mathrm{r}}\) = U(say)
E = \(\frac{\mathrm{U}}{2}-\mathrm{U}\) = \(\frac{-\mathrm{U}}{2}\)
U = -2E
∴ U = -2 × -3.4 = 6.8 eV.

Question 4.
The total energy of an electron in the first excited state of hydrogen atom is -3.4eV. What is the kinetic energy of the electron in this state ?
Solution:
In Hydrogen like atom, we know that
K = – Total energy E
Here E = – 3.4eV
∴ K = -(-3.4) = 3.4 eV

AP Inter 2nd Year Physics Study Material Chapter 13 Atoms

Question 5.
Find the radius of the hydrogen atom in its ground state. Also calculate the velocity of the electron in n = 1 orbit. Given
h = 6.63 × 10-34 J s, m = 9.1 × 10-31 kg, e = 1.6 × 10-19 C, K = 9 × 109N m2C-2
Solution:
n = 1, h = 6.63 × 10-34 J-s,
m = 9.1 × 10-31 kg
e = 1.6 × 10-19C,
K = 9 × 109 Nm2C-2
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 13

Question 6.
Prove that the ionisation energy of hydrogen atom is 13.6 eV.
Solution:
n = 1 corresponds to ground state.
E = \(\frac{-13.6}{n^2} e V\)
E = \(\frac{-13.6}{1^2} \mathrm{eV}\)
E = -13.6 eV
∴ The minimum energy required to free the electron from the ground state of hydrogen atom
= 13.6 eV.
∴ Ionisation energy of hydrogen atom = 13.6 eV

Question 7.
Calculate the ionization energy for a lithium atom.
Solution:
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 14
∴ Ionization energy of Lithium = 30.6eV.

Question 8.
The wavelength of the first member of Lyman series is 1216 A. Calculate the wavelength of second member of Balmer series.
Solution:
\(\frac{1}{\lambda}\) = \(R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)
For 1st member of Lyman series, λ = 1216; n1 = 1, n2 = 2
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 15
For 2nd member of Balmer senes,
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 16

Question 9.
The wavelength of first member of Balmer series is 6563 A. Calculate the wavelength of second member of Lyman series.
Solution:
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
For 1st member of Balmer senes,
\(\frac{1}{6563}\) = \(\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\)
\(\frac{1}{6563}\) = \(\frac{5 R}{36}\) —–> (1)
For 2nd member of Lyman senes,
\(\frac{1}{\lambda^1}\) = \(\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{3^2}\right)\)
\(\frac{1}{\lambda^1}\) = \(\frac{8 \mathrm{R}}{9}\) —–> (2)
\(\frac{(1)}{(2)}\) ⇒ \(\frac{\lambda^1}{6563}\) = \(\frac{5 \mathrm{R}}{36} \times \frac{9}{8 \mathrm{R}}\)
λ’ = \(\frac{5}{32} \times 6563\)
∴ λ’ = 1025.5A

Question 10.
The second member of Lyman series in hydrogen spectrum has wavelength 5400 A. Find the wavelength of first member.
Solution:
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
For second member of Lyman senes,
\(\frac{1}{5400}\) = \(R\left(\frac{1}{1^2}-\frac{1}{3^2}\right)\) ⇒ \(\frac{1}{5400}\) = \(\frac{8 \mathrm{R}}{9}\) —-> (1)
For first member of Lyman series,
\(\frac{1}{\lambda^1}\) = \(\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\)
\(\frac{1}{\lambda^1}\) = \(\frac{3 R}{4}\) —–> (2)
\(\frac{(1)}{(2)}\) ⇒ \(\frac{\lambda^1}{5400}\) = \(\frac{8 R}{9} \times \frac{4}{3 R}\)
∴ λ’ = \(\frac{32}{27}\) × 5400 = 6400A.

Question 11.
Calculate the shortest wavelength of Balmer series. Or Calculate the wavelength of the Balmer senes limit. Given : R = 10970000m-1.
Solution:
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
R = 10970000 = 1.097 × 107 ms-1
For Balmer senes limit n1 = 2 and n2 = ∞
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{\infty}\right)\) ⇒ \(\frac{1}{\lambda}\) = \(\frac{R}{4}\)
λ = \(\frac{4}{\mathrm{R}}\) = \(\frac{4}{1.097 \times 10^7}\) = 3646.3A

AP Inter 2nd Year Physics Study Material Chapter 13 Atoms

Question 12.
Using the Rydberg formula, calcûlate the wavelength of the first four spectral lines in the Balmer series of the hydrogen spectrum.
Solution:
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 17

Additional Exercises

Question 1.
Choose the correct alternative from the clues given at the end of the each statement:
a) The size of the atom in Thomson’s model is ……… the atomic size in Rutherford’s model. (much greater than / no different from / much less than.)
b) In the ground state of ……. electrons are in stable equilibrium, while in …… electrons always experience a net force. (Thomson’s model / Rutherford’s model.)
c) A classical atom based on …… is doomed to collapse. (Thomson’s model / Rutherford’s model).
d) An atom has a nearly continuous mass distribution in a ……. but has a highly non-uniform mass distribution in ………. (Thomson’s model / Rutherford’s model.)
e) The positively charged part of the atom possesses most of the mass in …….. (Rutherford’s model / both the models.)
Answer:
a) No different from
b) Thomson’s model, Rutherford’s model
c) Rutherford’s model
d) Thomson’s model, Rutherford’s model
e) Both the models.

Question 2.
Suppose you are given a chance to repeat the alpha – particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect ?
Answer:
The basic purpose of scattering experiment is defeated because solid hydrogen will be much lighter target compared to the alpha particle acting as projectile. According to theory of elastic the collisions, the target hydrogen will move much faster compared to alpha after collision. We cannot determine the size of hydrogen nucleus.

Question 3.
What is the shortest wavelength present in the Paschen series of spectral lines ?
Answer:
From Rydberg’s formula
\(\frac{\mathrm{hc}}{\lambda}\) = 13.6 × 1.6 × 10-19\(\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
For shortest wavelength in Paschen series n2 = ∞ and n1 = 3
\(\frac{\mathrm{hc}}{\lambda}\) = 21.76 × 10-19\(\left[\frac{1}{3^2}-\frac{1}{\infty^2}\right]\)
= 2.42 × 10-19
λ = \(\frac{\mathrm{hc}}{2.42 \times 10^{-19}}\) = \(\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2.42 \times 10^{-19}} \mathrm{~m}\)
= 8.1818 × 10-7m = 818.18nm.

Question 4.
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level ?
Answer:
Here E = 2.3eV = 2.3 × 1.6 × 10-19 J
As E = hv
∴ v = \(\frac{\mathrm{E}}{\mathrm{h}}\) = \(\frac{2.3 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}}\) = 5.6 × 104 Hz

Question 5.
The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state ?
Answer:
Total energy E = -13.6 eV
K.E = -E = 13.6 eV
RE. = -2.K.E = -2 × 13.6
= -27.2eV.

Question 6.
A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
Answer:
For ground state n1 = 1 and n2 = 4
Energy of photon absorbed E = E2 – E1
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 18

Question 7.
a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2 and 3 levels.
b) Calculate the orbital period in each of these levels.
Answer:
a) From v = \(\frac{c}{n} \alpha\), where α = \(\frac{2 \pi \mathrm{Ke}^2}{\mathrm{ch}}\) = 0.0073
v1 = \(\frac{3 \times 10^8}{1}\) × 0.0073 = 2.19 × 106 m/s
v2 = \(\frac{3 \times 10^8}{2}\) × 0.0073 = 1.095 × 106 m/s
v3 = \(\frac{3 \times 10^8}{3}\) × 0.0073 = 7.3 × 105 m/s

b) Orbital period, T = \(\frac{2 \pi r}{V}\), As r1 = 0.53 × 10-10m
T1 = \(\frac{2 \pi \times 0.53 \times 10^{-10}}{2.19 \times 10^6}\) = 1.52 × 10-16S
As r2 = 4r1 and V2 = \(\frac{1}{2} V_1\)
T2 = 8T2 = 8 × 1.52 × 10-6 S = 1.216 × 10-15S
As r3 = 9r1 and V3 = \(\frac{1}{3} \mathrm{~V}_1\)
T3 = 27T1= 27 × 1.52 × 10-16 S = 4.1 × 10-15S

Question 8.
The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10--11m. What are the radii of the n = 2 and n = 3 orbits?
Answer:
As r = n2r
∴ r2 = 4r1 = 4 × 5.3 × 10-11 m = 2.12 × 10-10m ,
and r3 = 9r1 = 9 × 5.3 × 10-11 = 4.77 × 10-10m.

Question 9.
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Answer:
In ground state, energy of gaseous hydrogen at room temparature = -13.6eV, when it is bombarded with 12.5 eV electron beam, the energy becomes 13.6 + 12.5 = -1.1eV.
The electron would jump from n = 1 to n = 3 where E3 = \(\frac{-13.6}{32}\) = -1.5eV
On de — excitation the electron may jump from n = 3 to n = 2 giving rise to Balmer series. It may also jump from n = 3 to n = 1 giving rise to Lýman series.

AP Inter 2nd Year Physics Study Material Chapter 13 Atoms

Question 10.
In accordance with the Bohr’s model, find the quantum number that characterises the earths revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104m/s. (Mass of earth = 6.0 × 1024 kg.)
Answer:
Here r = 1.5 × 1011m, V = 3 × 104m/s, m = 6.0 × 1024kg
According to Bohrs model mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
n = \(\frac{2 \pi \mathrm{mvr}}{\mathrm{h}}\) = 2 × \(\frac{22}{7}\) × \(\frac{6.0 \times 10^{24} \times 3 \times 10^4 \times 1.5 \times 10^{11}}{6.6 \times 10^{-34}}\)
= 2.57 × 1074, which is too large.

Question 11.
Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.
a) Is the average angle of deflection of α — particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
b) Is the probability of backward scattering (i.e., scattering of α – particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model ?
c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α – particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide ?
d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of a – particles by a thin foil ?
Answer:
a) About the same this is because we are talking of average angle of deflection.
b) Much less, because in Thomson’s model there is no such massive central core called the nucleus as in Rutherford’s model.
c) This suggests that scattering is predominantly due to a single collision increases with the number of target atoms which increases linearly with the thickness of the foil.
d) In Thomson model, positive charge is uniformly distributed in the spherical atom. Therefore a single collision causes very little deflection. Therefore average scattering angle can be explained only by considering multiple scattering may be ignored as a first approximation.

Question 12.
The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10-40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
Answer:
The radius of the first Bohr orbit of a hydrogen atom is
r0 = \(\frac{4 \pi \varepsilon_0(h / 2 \pi)^2}{m_e \mathrm{e}^2}\)
If we consider the atom bound by the gravitational force
= \(\left(\frac{\mathrm{Gm}_{\mathrm{p}} \mathrm{m}_{\mathrm{e}}}{\mathrm{r}^2}\right)\). We should replace \(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0}\) by (Gmpme). In that case radius of first Bohr orbit of hydrogen atom would be given by r0 = \(\frac{(\mathrm{h} / 2 \pi)^2}{\mathrm{Gm}_p \mathrm{~m}^2 \mathrm{e}}\)
Putting the standard values we get
r0 = \(\frac{\left(6.6 \times 10^{-34} / 2 \pi\right)^2}{6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times\left(9.1 \times 10^{-31}\right)^2}\)
= 1.2 × 1029 metre.
This is much greater than the estimated size of the whole universe!

Question 13.
Obtain an expression for the frequency of radiation emitted when a hydrogen atom de- excites from level n to level (n – 1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.
Answer:
The frequency v of the emitted radiation when a hydrogen atom de-excites from level n to level (n – 1) is
E = hv = E2 – E1
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 19
In Bohr’s Atomic model, velocity of electron in nth orbit is v = \(\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}\)
and radius of nth orbital is v = \(\frac{n^2 h^2}{4 \pi^2 m K e^2}\)
Frequency of revolution of electron v = \(\frac{\mathrm{V}}{2 \pi \mathrm{r}}\) = \(\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}\)
and radius of nth orbital is r = \(\frac{n^2 h^2}{4 \pi^2 m K e^2}\)
Frequency of revolution of electron
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 20
which is the same as (i) .
Hence for large values of n1 classical frequency of revolution of electron in nth orbit is the same as the frequency of radiation emitted when hydrogen atom de-excites from level (n) to level (n – 1)

AP Inter 2nd Year Physics Study Material Chapter 13 Atoms

Question 14.
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size ? Why is an atom not, say, thousand times bigger than its typical size ? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 1010m).
a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.
b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non- relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the ‘Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.
Answer:
a) Using fundamental constants e, me and c, we construct a quantity which has the dimensions of length. This quantity is \(\left(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{~m}_{\mathrm{e}} \mathrm{c}^2}\right)\)
Now \(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{~m}_{\mathrm{e}} \mathrm{c}^2}\) = \(\frac{\left(1.6 \times 10^{-19}\right)^2 \times 9 \times 10^9}{9.1 \times 10^{-31}\left(3 \times 10^8\right)^2}\) = 2.82 × 10-15m
This is of the order of atom sizes.

b) However when we drop c and use hc, me and e to construct a quantity which has dimensions of length the quantity we obtain is
\(\frac{4 \pi \varepsilon_0(\mathrm{~h} / 2 \pi)}{\mathrm{m}_{\mathrm{e}} \mathrm{e}^2}\)
AP Inter 2nd Year Physics Study Material Chapter 13 Atoms 21
= 0.53 × 10-10m
This is of the order of atom sizes.

Question 15.
The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV
a) What is the kinetic energy of the electron in this state ?
b) What is the potential energy of the electron in this state ?
c) Which of the answers above would change if the choice of the zero of potential energy is changed ?
Answer:
We know kinetic energy of electron = \(\frac{K Z e^2}{2 r}\)
and P.E of electron = \(\frac{-\mathrm{KZe}^2}{\mathrm{r}}\)
P.E. = -2 (kinetic energy).
In this calculation electric potential and hence potential energy is zero at infinity.
Total energy = PE + KE = -2KE + KE = -KE
a) In the first excited state total energy = -3.4eV
∴ K.E = -(-3.4eV) = + 3.4 eV
b) P. E of electron in this first excited state = -2KE = -2 × 3.4 = -6.8eV.
c) If zero of potential energy is changed, KE does not change and continues to be +3.4 eV However, the P.E. and total energy of the state would change with the choice of zero of potential energy.

Question 16.
If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun ?
Answer:
Bohr’s quantisation postulate is in terms of Plank’s constant (h), But angular momenta associated with planetary motion are = 1070 h (for earth). In terms of Bohr’s quantisation posulate this will correspond to n = 107. For such large values of n the differences in successive energies and angular momenta of the quantised levels are so small, that the levels can be considered as continuous and not discrete.

Question 17.
Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (µ–) of mass about 207me orbits around a proton].
Answer:
A muonic hydrogen is the atom in which a negatively charged muon of mass about 207 me revolves around a proton.
In Bohr’s atom model as, r ∝ \(\frac{1}{\mathrm{~m}}\)
\(\frac{\mathrm{r}_\mu}{\mathrm{r}_{\mathrm{e}}}\) = \(\frac{\mathrm{m}_{\mathrm{e}}}{\mathrm{m}_\mu}\) = \(\frac{\mathrm{m}_{\mathrm{e}}}{207 \mathrm{~m}_{\mathrm{e}}}\) = \(\frac{1}{207}\)
Here re is radius of first orbit of electron in hydrogen atom = 0.53A = 0.53 × 10-10m.
rm = \(\frac{\mathrm{r}_{\mathrm{e}}}{207}\) = \(\frac{0.53 \times 10^{-10}}{207}\) = 2.56 × 10-13m
Again in Bohr’s atomic model
E ∝ m
∴ \(\frac{\mathrm{E}_\mu}{\mathrm{E}_{\mathrm{e}}}\) = \(\frac{\mathrm{m}_\mu}{\mathrm{m}_{\mathrm{e}}}\) = \(\frac{207 \mathrm{~m}_{\mathrm{e}}}{\mathrm{m}_{\mathrm{e}}}\), Eμ = 207Ee
As ground state energy of electron in hydrogen atom is Ee = -13.6 eV
Eμ = 207(-13.6)eV = -2815.2eV
= -2.8152KeV.

AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 12th Lesson Dual Nature of Radiation and Matter Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 12th Lesson Dual Nature of Radiation and Matter

Very Short Answer Questions

Question 1.
What are “cathode rays”?
Answer:
Cathode rays are streams of fast-moving electrons or negatively charged particles.

Question 2.
What important fact did Millikan’s experiment establish?
Answer:
Millikan’s experiment established that electric charge is quantized. That means the charge on anybody (oil drop) is always an integral multiple of the charge of an electron, i.e., Q = ne.

AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter

Question 3.
What is “work function” ? (A.P. Mar. ’19 & T.S. Mar. ’15)
Answer:
The minimum energy required to liberate an electron from photometal surface is called work function, ϕ0.

Question 4.
What is “photoelectric effect” ?
Answer:
When light of sufficient energy is incident on the photometal surface electrons are emitted. This phenomenon is called photoelectric effect.

Question 5.
Give examples of “photosensitive substances”. Why are they called so ?
Answer:
Examples of photosensitive substances are Li, Na, K, Rb and Cs etc.
The work function of alkali metals is very low. Even the ordinary visible light, alkali metals can produce photoelectric emission. Hence they are called photosensitive substances.

Question 6.
Write down Einstein’s photoelectric equation. (A.P. Mar. ’15)
Answer:
Einstein’s photoelectric equation is given by Kmax = \(\frac{1}{2} \mathrm{mv}_{\max }^2\) = hυ – ϕ0.

Question 7.
Write down de-Broglie’s relation and explain the terms there in. (A.P. & T.S. Mar. ’16)
Answer:
The de-Broglie wave length (λ) associated with a moving particle is related to its momentum (p) is λ = \(\frac{h}{p}\) = \(\frac{\mathrm{h}}{\mathrm{mv}}\), where h is planck’s constant.

AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter

Question 8.
State Heisenberg’s Uncertainly Principle. (A.P. Mar. ’19) (Mar. ’14)
Answer:
Uncertainity principle states that “it is impossible to measure both position (Δx) and momentum of an electron (Δp) [or any other particle] at the same time exactly”, i.e., Δx . Δp ≈ h where Δx is uncertainty in the specification of position and Δp is uncertainty in the specification of momentum.

Short Answer Questions

Question 1.
What is the effect of
(i) intensity of light
(ii) potential on photoelectric current ?
Answer:
(i) Effect of intensity of light on photoelectric current:

1) When the intensity (I) of incident light, with frequency greater than the threshold frequency (υ > υ0) is increased then the number of photoelectrons emitted increases i.e. the value of photoelectric current (i) increases, ie. i ∝ I.
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 10

ii) The effect of potential on photoelectric current:

  1. On increasing the positive potential on collecting electrode, the photoelectric current increases. At a particular positive potential, the photocurrent becomes maximum which is known as saturated current.
    AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 11
  2. On increasing the value of negative potential on collecting electrode, the photoelectric current gradually goes on decreasing. At a particular negative potential the value of photoelectric current becomes zero. This is known as stopping potential.
  3. Stopping potential does not depend on the intensity of incident light. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.

Question 2.
Describe an experiment to study the effect of frequency of incident radiation on ‘stopping potential’.
Answer:
Experimental study of the effect of frequency of incident radiation on stopping potential:

  1. The experimental set up is shown in fig.
  2. Monochromatic light of sufficient energy (E = hv) from source ‘s’ is incident on photosensitive plate ‘C’ (emitter), electrons are emitted by it.
  3. The electrons are collected by the plate A (collector), by the electric field created by the battery.
  4. The polarity of the plates C and A can be reversed by a commutator.
    AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 12
  5. For a particular frequency of incident radiation, the minimum negative (retarding) potential V0 given to the plate A for which the photo current stops or becomes zero is called stopping potential.
  6. The experiment is repeated with different frequencies, and their different stopping potential are measured with voltmeter.
  7. From graph, we note that
    1. The values of stopping potentials are different for different frequencies.
    2. The value of stopping potential is more negative for radiation of higher incident frequency.
    3. The value of saturation current depends on the intensity of incident radiation but it is independent of the frequency of the incident radiation.
      AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 13

Question 3.
Summarise the photon picture of electromagnetic radiation.
Answer:
We can summarise the photon picture of electromagnetic radiation as follows.

  1. In interaction of radiation with matter, radiation behaves as if it is made up of particles called photons.
  2. Each photon has energy E\(\left[\begin{array}{l}
    =\mathrm{hv} \\
    =\frac{\mathrm{hc}}{\lambda}
    \end{array}\right]\) and momentum P \(\left[\begin{array}{l}
    =\frac{h v}{c} \\
    =\frac{h}{\lambda}
    \end{array}\right]\) and speed c, the speed of light.
  3. By increasing the intensity of light of given wave length, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation.
  4. Photons are not deflected by electric and magnetic field. This shows that photons are electrically neutral.
  5. In a photon-particle collision (such as photo-electron collision), the energy and momentum
    are conserved. However the number of photons may not be conserved in a collision. One photon may be absorbed or a new photon may be created.
  6. The rest mass of photon is zero. According to theory of relativity, the mass of moving particle is given by m = \(\frac{\mathrm{m}_0}{\sqrt{1-\frac{v^2}{c^2}}}\) where v is velocity of particle and c is velocity of light.

AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter

Question 4.
What is the deBroglie wavelength of a ball of mass 0.12 Kg moving with a speed of 20 ms-1 ? What can we infer from this result ?
Answer:
Given, m = 0.12 kg; υ = 20 m/s; h = 6.63 × 10-34 J-s;
λ = \(\frac{h}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{0.12 \times 20}\) = \(\frac{6.63 \times 10^{-34}}{2.4}\) ∴ λ = 2.762 × 10-34 m = 2762 × 10-21 A.

Long Answer Questions

Question 1.
How did Einstein’s photoelectric equation explain the effect of intensity and potential on photoelectric current ? How did this equation account for the effect of frequency of incident light on stopping potential ? (T.S. Mar. ’19)
Answer:

  1. Einstein postulated that a beam of light consists of small energy packets called photons or quanta.
  2. The energy of photon is E = hv. Where ‘h’ is Planck’s constant; v is frequency of incident light (or radiation).
  3. If the absorbed energy of photon is greater than the work function (ϕ0 = hυ0), the electron is emitted with maximum kinetic energy i.e., kmax = \(\frac{1}{2} m_{\max }^2\) = eV0 = hv – ϕ0. This equation is known as Einstein’s photoelectric equation.
  4. Effect of intensity of light on photoelectric current:
    When the intensity (I) of incident light, with frequency greater thanthe threshold frequency (υ > υ0) is increased then the number of photoelectrons emitted decreases i.e. the value of photoelectric current (i) increases, le. i ∝ I.
    AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 14
  5. The effect of potential on photoelectric current:
    1. On increasing the positive potential on collecting electrode, the photoelectric current increases. At a particular positive potential, the photocurrent becomes maximum which is known as saturated current.
    2. On increasing the value of negative potential on collecting electrode, the photoelectric current gradually goes on decreasing. At a particular negative potential the value of photoelectric current becomes zero. This is known as stopping potential (v0).
      AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 15
    3. Stopping potential does not depend on the intensity of incident light. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.
  6. The effect of frequency of incident radiation on stopping potential:
    On increasing the frequency of incident light, the value of stopping potential goes on increasing gradually as shown in fig. That means kmax increases eV0 also increases.
    AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 16
  7. From the graph, we note that
    1. For a given photosensitive metal, the cut off potential (v0) varies linearly with the frequency of the incident radiation.
    2. For a given photosensitive metal, there is a certain minimum cut off frequency v0 (called threshold frequency) for which the stopping potential is zero.
      AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 17
  8. From the graph we note that
    1. The value of cut-off potential is different for radiation of different frequency.
    2. The value of stopping potential is more negative for radiation of higher incident frequency.
  9. From above experiments, it is found that, if the incident radiation is of higher frequency than that of threshold frequency, the photoelectric emission is possible.

Question 2.
Describe the Davisson and Germer experiment. What did this experiment conclusively prove?
Answer:
Davisson and Germer experiment:

  1. The experimental arrangement is schematically shown in fig.
  2. Electrons from a filament F are rendered into a fine beam by applying a positive potential to the cylinder A.
  3. A fine narrow beam of electrons is incident on the nickel crystal. The electrons are scattered in all directions by the atoms of the crystal.
  4. The intensity of the electron beam scattered in a given direction, is measured by the electron detector (collector). The detector can be moved on a circular scale and is connected to a sensitive galvanometer, which records the current.
  5. The deflection of the galvanometer is proportional to the intensity of the electron beam entering collector.
  6. The apparatus is enclosed in an evacuated chamber.
  7. The experiment was performed by varying the accelerating voltage from 44 V to 68 V. It is found that the intensity is maximum at 50° for a critical energy of 54 V
    AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 18
  8. For θ = 50°, the glancing angle, ϕ (angle between the scattered beam of electron with the plane of atoms of the crystal) for electron beam will be given by
    ϕ + θ + ϕ = 180°
    ϕ = \(\frac{1}{2}\left[180^{\circ}-50^{\circ}\right]\) = 65°
    AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 19
  9. According to Bragg’s law for first order diffraction maxima (n = 1), we have 2 d sin ϕ = 1 × λ ⇒ λ = 2 × 0.91 × sin 65° = 1.65A = 0.165 nm. (experimentally).
    [∵ for Nickel crystal interatomic separation d = 0.91 A]
  10. According to de-Broglie hypothesis, the wavelength of the wave associated with electron is given by λ = AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 37
    = 1.67A = 0.167 nm, (Theoritically).
  11. The experimentally measured wavelength was found to be in confirmity with proving the existence of de-Broglie waves.

Textual Exercises

Question 1.
Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Solution:
Given voltage V = 30 kV = 30 × 103 V; e = 1.6 × 10-19 C; h = 6.63 × 10-34 j-s C = 3 × 108 m/s
a) Maximum frequency, v = \(\frac{\mathrm{eV}}{\mathrm{h}}\) = \(\frac{1.6 \times 10^{-19} \times 30 \times 10^3}{6.63 \times 10^{-34}}\) = 7.24 × 1018 Hz

b) Minimum wavelength of X-ray, λ = \(\frac{\mathrm{C}}{\mathrm{v}}\) = \(\frac{3 \times 10^8}{7.24 \times 10^{18}}\) = 0.414 × 10-10 Hz
∴ λ = 0.0414 × 10-9m = 0.0414 nm.

AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter

Question 2.
The work function of caesium metal is 2.14 eV. When light of frequency 6 × 1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
(b) stopping potential and
(c) maximum speed of the emitted photoelectrons ?
Solution:
Given ϕ0 = 2.14 eV; v = 6 × 1014 Hz
a) KEmax = hv – ϕ0 = \(\frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}\) – 2.14 ∴ KEmax = 0.35 eV

b) KEmax = eV0 ⇒ 0.35 eV = eV0 ∴ V0 = 0.35 V
c) KEmax = \(\frac{1}{2} m v_{\max }^2\) ⇒ \(v_{\max }^2\) = \(\frac{2 K_{\max }}{m}\) = \(\frac{2 \times 0.35 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}\) (∴ e = 1.6 × 10-19 C)
\(v_{\max }^2\) = 0.123 × 1012 ⇒ υmax = \(\sqrt{1230 \times 10^8}\) = 35.071 × 104 m/s ∴ υmax = 350.71 km/s.

Question 3.
The photoelectric cut-off voltage in certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ?
Solution:
Given, V0 = 1.5 V; e = 1.6 × 10-19 C, KEmax = eV0 = 1.6 × 10-19 × 1.5 = 2.4 × 10-19 J.

Question 4.
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam ? (Assume the beam to have uniform cross-section which is less than the target area), and,
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon ?
Solution:
Given, λ = 632.8 nm = 632.8 × 10-9m; p = 9.42 mW = 9.42 × 10-3W
h = 6.63 × 10-34 J-s; c = 3 × 10-3 m/s

a) E = \(\frac{h c}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10}{632.8 \times 10^{-9}}\) = 3.14 × 10-19 J.
Momentum of each photon, p = \(\frac{\mathrm{h}}{\lambda}\) = \(\frac{6.63 \times 10^{-34}}{632.8 \times 10^{-9}}\) = 1.05 × 10-27kg \(\frac{\mathrm{m}}{\mathrm{s}}\)

b) No. of photons per second,
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 20
∴ N = 3 × 1016 photons/s
c) Since, PHydrogen = Pphoton
⇒ mυ = p ⇒ υ = \(\frac{\mathrm{p}}{\mathrm{m}}\) = \(\frac{1.05 \times 10^{-27}}{1.66 \times 10^{-27}}\) [∴ mH = 1.66 × 10-27 kg] ∴ υ = 0.63 m/s.

Question 5.
The energy flux of sunlight reaching the surface of the earth is 1.388 × 103 W/m2. How many photons (nearly) per square metre are incident on the Earth per second ? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Solution:
Given, P = 1.388 × 103 W/m2; λ = 550 nm = 550 × 10-9 m
h = 6.63 × 10-34 J-s; e = 3 × 108 m/s
Energy of each photon E = \(\frac{\mathrm{hc}}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{550 \times 10^{-9}}\) = 3.616 × 10-19 J
No. of photons incident on the earths surface, N = \(\frac{\mathrm{P}}{\mathrm{E}}\) = \(\frac{1.388 \times 10^3}{3.66 \times 10^{-19}}\)
∴ N = 3.838 × 1021 photons/m2 – s.

AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter

Question 6.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10-15 V s. Calculate the value of Planck’s constant.
Solution:
Given, slope of graph tan θ = 4.12 × 10-15 V — s; .
e = 1.6 × 10-19 c.
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 21
For slope of graph, tan θ = \(\frac{\mathrm{V}}{\mathrm{v}}\)
We know that hv = eV
\(\frac{\mathrm{V}}{\mathrm{v}}\) = \(\frac{h}{e}\) ⇒ \(\frac{\mathrm{h}}{\mathrm{e}}\) = 4.12 × 10-15; h = 4.12 × 10-15 × 1.6 × 10-19 = 6.592 × 10-34 J-s

Question 7.
A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light ?
(b) At what rate are the photons delivered to the sphere ?
Solution:
Given, P = 100 W; λ = 589 nm = 589 × 10,sup>-9 m; h = 6.63 × 10-34 J – S; c = 3 × 108 m/s
a) E = \(\frac{\text { hc }}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{589 \times 10^{-9}}\) = 3.38 × 10-19J = \(\frac{3.38 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV = 2.11 eV.
b) No. of photons delivered per second, N = \(\frac{\mathrm{P}}{\mathrm{E}}\) = \(\frac{100}{3.38 \times 10^{-19}}\) = 3 × 1020 photons/s

Question 8.
The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Solution:
Given, v0 = 3.3 × 1014 Hz; v = 8.2 × 1014 Hz; e = 1.6 × 10-19 c; KE = eV0 = hv – hv0
V0 = \(\frac{h\left(v-v_0\right)}{e}\) = \(\frac{6.63 \times 10^{-34} \times(8.2-3.3) \times 10^{14}}{1.6 \times 10^{-19}}\) = \(\frac{6.63 \times 10^{-34} \times 10^{14} \times 4.9}{1.6 \times 10^{-19}}\) ∴ V0 = 2.03 V.

Question 9.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm ?
Solution:
Given, ϕ0 = 4.2 eV = 4.2 × 1.6 × 10-19 J = 6.72 × 10~1S J
λ = 330 nm = 330 × 10-9 m; h = 6.63 × 10-34 J – s ⇒ c = 3 × 108 m/s
E = \(\frac{\text { hc }}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{330 \times 10^{-9}}\) ∴ E = 6.027 × 10-19J
As E < ϕ0, no photoelectric emission takes place.

Question 10.
Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons ?
Solution:
Given, v = 7.21 × 1014 Hz; m = 9.1 × 10-31 kg; υmax = 6 × 105 m/s
KEmax = \(\frac{1}{2} \mathrm{mv}_{\max }^2\) = hv – hv0 = h(v – v0)
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 22

Question 11.
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made. .
Solution:
Given, λ = 488 nm = 488 × 10-9 m; V0 = 0.38 V; e = 1.6 × 10-19 c; h = 6.63 × 10-34 J – s
c = 3 × 108 m/s ⇒ KE = eV0 = \(\frac{\mathrm{hc}}{\lambda}\) – ϕ ⇒ 1.6 × 10-19 × 0.38 = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}\) – ϕ0
6.08 × 10-20 = 40.75 × 10-20 – ϕ0 ⇒ (40.75 – 6.08) × 10-20 = 34.67 × 10-20 J
= \(\frac{34.67 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}\) ∴ ϕ0 = 2.17 eV.

Question 12.
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V
Solution:
Given, V = 56 V; e = 1.6 × 10-19 c; m = 9 × 10-31 kg
a) As KE = \(\frac{\mathrm{P}^2}{2 \mathrm{~m}}\) ⇒ 2m (KE) = P2 ⇒ P = \(\sqrt{2 \mathrm{~m}(\mathrm{KE})}\) = \(\sqrt{2 \mathrm{~m} \mathrm{eV}}\) [∵ KE = eV]
∴ P = \(\sqrt{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-31} \times 56}\) = 4.02 × 10-24 kg – m/s
b) λ = \(\frac{12.27}{\sqrt{V}}\) A = \(\frac{12.27}{\sqrt{56}}\) A = 0.164 × 10-9m ∴ λ = 0.164 nm.

Question 13.
What is the
(a) momentum,
(b) speed, and
(c) de Brogue wavelength of an electron with kinetic energy of 120 eV.
Solution:
Given, KE = 120 eV; m = 9.1 × 10-3 kg; e = 1.6 × 10-19 c
a) P = \(\sqrt{2 m(K E)}\) = \(\sqrt{2 \times 9.1 \times 10^{-31} \times\left(120 \times 1.6 \times 10^{-19}\right)}\) ∴ P = 5.91 × 10-24 kg – m/s
b) υ = \(\frac{\mathrm{p}}{\mathrm{m}}\) = \(\frac{5.91 \times 10^{-24}}{9.1 \times 10^{-31}}\) = 6.5 × 106 m/s .
c) λ = \(\frac{12.27}{\sqrt{\mathrm{V}}}\) A = \(\frac{12.27}{\sqrt{120}}\) A = 0.112 × 10-9 m ∴ λ = 0.112 nm.

AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter

Question 14.
The wavelength of light from the spectral emission line of sodium is 589 nm Find the kinetic energy at which (a) an electron, and (b) a neutron, and would have the same de Brogue wavelength.
Solution:
Given, λ = 589 mn = 589 × 10-9 m; me = 9.1 × 10-31 kg.
mn = 1.67 × 10-27 kg; h = 6.62 × 10-34 J – s.
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 23

Question 15.
What is the de Brogue wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 × 10-9 kg drifting with a speed of 2.2 m/s?
Solution:
a) Given, for bullet m = 0.040 kg and o = 1000 m/s = 103 m/s
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{0.040 \times 10^3}\) = 1.66 × 10-35m
b)Given, for ball m = 0.060 kg and υ = 1 m/s ⇒ λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{0.060 \times 1}\) = 1.1 × 10-32 m
c) Given, for a dust particle m = 1 × 10-9 kg and υ = 2.2 m/s
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{1 \times 10^{-9} \times 2.2}\) = 3.0 × 10-25 m.

Question 16.
An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Solution:
Given, λ = 1 mm = 10-9m; h = 6.63 × 10-34 J-S; c = 3 × 108 m/S
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 24

Question 17.
(a) For what kinetic energy of a neutron will the associated de Brogue wavelength be 1.40 × 10-10 m?
(b) Also find the de Brogue wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.
Solution:
(a) Given, for neutron, λ = 1.40 × 10-10 m and m = 1.675 × 10-27 kg
KE = \(\frac{\mathrm{P}^2}{2 \mathrm{~m}}\) = \(\frac{h^2}{2 \mathrm{~m} \lambda^2}\) = \(\frac{\left(6.63 \times 10^{-34}\right)^2}{2 \times\left(1.40 \times 10^{-10}\right)^2 \times 1.675 \times 10^{-27}}\) ∴ KE = 6.686 × 10-21J

b) Given, T = 300 k and K = 1.38 × 10-23 J/K
KE = \(\frac{3}{2}\) KT = \(\frac{3}{2}\) × 1.38 × 10-21 × 300 = 6.21 × 10-21 J
λ = \(\frac{h}{\sqrt{2 m(K E)}}\) = \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}}}\) ∴ λ = 1.45 × 10-10m = 1.45 A

Question 18.
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Solution:
The momentum of a photon of frequency v, wavelength λ is given by p = \(\frac{\mathrm{hv}}{\mathrm{c}}\) = \(\frac{\mathrm{h}}{\lambda}\)
λ = \(\frac{\mathrm{h}}{\mathrm{p}}\) ⇒ de-Broglie wavelength of photon, λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{h}{p}\) = \(\frac{\frac{\mathrm{h}}{\mathrm{hv}}}{\mathrm{c}}\) = \(\frac{\mathrm{c}}{\mathrm{v}}\)
Thus, the wavelength of electromagnetic radiation is equal to the de-Broglie wavelength.

Question 19.
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K ? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Solution:
Given, T = 300 k; K = 1.38 × 10-23 J/k; m = 28.0152u = 28.0152 × 1.67 × 10-27 kg;
h = 6.63 × 10-34 Js; Mean KE of molecules \(\frac{1}{2}\) mυ2 = \(\frac{3}{2}\) KT
υ = \(\sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}}\) = \(\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{28.0152 \times 1.66 \times 10^{-27}}}\)
∴ υ = 516.78 m/s
de-Broglie wavelength, λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{28.0152 \times 1.66 \times 10^{-27} \times 516.78}\) = 2.75 × 10-11 m
∴ λ = 0.0275 × 10-19 m = 0.028 nm.

Additional Exercises

Question 1.
(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 1011 C kg-1.
(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong ? In what way is the formula to be modified ?
Solution:
a) Given, V = 500 V, \(\frac{\mathrm{e}}{\mathrm{m}}\) = 1.76 × 1011 C/kg; KE = \(\frac{1}{2} \mathrm{mv}^2\) = eV

b) V = 10 MV = 107 V; υ = \(\sqrt{\frac{\mathrm{e}}{\mathrm{m}} \times 2 \mathrm{~V}}\) = \(\sqrt{1.76 \times 10^{11} \times 2 \times 10^7}\) ∴ υ = 1.8762 × 109 m/s
This speed is greater than speed of light, which is not possible. As o approaches to c, then mass m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)

AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter

Question 2.
(a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s-1 is subject to
a magnetic field of 1.30 × 10-4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011 C kg-1.
(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam ? If not, in what way is it modified ?
[Note : Exercises 20(b) and 21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]
Solution:
a) Given, υ = 5.20 × 106 m/s; B = 1.30 × 10-4 T; \(\frac{\mathrm{e}}{\mathrm{m}}\) = 1.76 × 1011 C/kg
Since centripetal force is balanced by Force due to magnetic field, \(\frac{\mathrm{m} v^2}{\mathrm{r}}\) = Bυ
[∵ (\(\vec{v} \times \vec{B}\)) = e υ B sin 90° = Beυ]
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 25

b) Given, E = 20 MeV = 20 × 1.6 × 10-13J; me = 9.1 × 10-31 kg
E = \(\frac{1}{2} \mathrm{mv}^2\)
⇒ v = \(\sqrt{\frac{2 E}{m}}\) = \(\sqrt{\frac{2 \times 20 \times 1.6 \times 10^{-13}}{9.1 \times 10^{-32}}}\) ∴ v = 2.67 × 109 m/s

As υ > C, the formula used in (a) r = \(\frac{\mathrm{mv}}{\mathrm{eB}}\) is not valid for calculating the radius of path of 20 MeV electron beam because electron with such a high energy has velocity in relatistic domain i.e., comparable with the velocity of light and the mass varies with the increase in velocity but we have taken it a constant.
∴ m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\) ⇒ Thus, the modified formula will be r = \(\frac{\mathrm{mv}}{\mathrm{eB}}\) = \(\left[\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\right] \frac{v}{e B}\)

Question 3.
An electron gun with its collector at a potential of 100V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~ 10-2 mm of Hg). A magnetic field of 2.83 × 10-4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons and emitting light by electron capture; this method is known as the fine beam tube’ method.) Determine e/m from the data.
Solution:
Given, V = 100 V; B = 2.83 × 10-4 T; m = 9.1 × 10-31 kg; e = 1.6 × 10-19 C;
r = 12 cm = 0.12m; KE = \(\frac{1}{2} \mathrm{mv}^2\) = eV ⇒ \(\frac{1}{2}\) × 9.1 × 10-31 × υ2 = 1.6 × 10-19 × 100
υ2 = \(\frac{2 \times 1.6 \times 10^{-17}}{9.1 \times 10^{-3.1}}\) = 3.516 × 1013 ∴ υ = \(\sqrt{3.516 \times 10^{13}}\) = 5.93 × 106 m/s
Specific charge of electron, \(\frac{\mathrm{e}}{\mathrm{m}}\) = \(\frac{v}{r B}\) [∵ \(\frac{\mathrm{mv}^2}{\mathrm{r}}\) = Beυ] = \(\frac{5.93 \times 10^6}{2.83 \times 10^{-4} \times 0.12}\)
∴ \(\frac{\mathrm{e}}{\mathrm{m}}\) = 1.74 × 1011 C/kg.

Question 4.
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 A. What is the maximum energy of a photon in the radiation ?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube ?
Solution:
a) Given, λ = 0.45 A = 0.45 × 10-10 m; E = \(\frac{\mathrm{hc}}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{0.45 \times 10^{-10} \times 1.6 \times 10^{-19}}\) eV
∴ E = 27.6 × 103 eV = 27.6 KeV

b) In X-ray tube, accelerating voltage provides the energy to the electrons which produce X-rays. For getting X-rays, photons of 27.51 KeV is required that the incident electrons must posess kinetic energy of 27.61 KeV.
Energy = eV = E; eV = 27.6 KeV; V = 27.6 KV .
So, the order of accelerating voltage is 30 KV.

Question 5.
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray ? (1 BeV = 109 eV)
Solution:
Given, energy of 2 γ-rays, 2E = 10.2 BeV
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 27
⇒ 2\(\frac{\mathrm{hc}}{\lambda}\) = 10.2 BeV [∵ E = \(\frac{\mathrm{hc}}{\lambda}\)] ⇒ λ = \(\frac{2 \mathrm{hc}}{10.2 \mathrm{BeV}}\)
Here h = 6.63 × 10-34 J-S; c = 3 × 108m/s, 1 BeV = 109 eV = 109 × 1.6 × 10-19J
⇒ λ = \(\frac{2 \times 6.63 \times 10^{-34} \times 3 \times 10^8}{10.2 \times 10^9 \times 1.6 \times 10^{-19}}\) ∴ λ = 2.436 × 10-16 m

Question 6.
Estimating the following two numbers should bé interesting. The first number will tell you why radio engineers do not need to worry much about photons ! The second number tells you why our eye can never count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~ 10-10 W m-2). Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about 6 × 1014 Hz.
Solution:
a) Given, P = 10kW = 10 × 103 W; λ = 500m; h = 6.63 × 10-34 J – s; C = 3 × 108
The no. of photons emitted per second, N = \(\frac{\mathrm{P}}{\mathrm{E}}\) = \(\frac{\mathrm{P}}{\frac{\mathrm{hc}}{\lambda}}\) = \(\frac{\mathrm{p} \lambda}{\mathrm{hc}}\) = \(\frac{10 \times 10^3 \times 500}{6.63 \times 10^{-34} \times 3 \times 10^8}\)
∴ N = 2.51 × 1031 photons/s

b) Given, v = 6 × 10-4 Hz; I = \(\frac{E}{A-t}\) = 10-10 W/m2; Area of pupil, A = 0.4 cm2 = 0.4 × 10-4 m2.
Total energy falling on pupil in unit time, E’ = IA = 10-10 × 0.4 × 10-4 ∴ E’ = 4 × 10-155 J/s
Energy of each photon, E” = hv = 6.63 × 10-34 × 6 × 1014 = 3.978 × 10-19 J
No. of photons per second, N = \(\frac{E^{\prime}}{E^{\prime \prime}}\) = \(\frac{4 \times 10^{-15}}{3.978 \times 10^{-19}}\) = 1.206 × 104 photons/s
As this number is not so large a: in part (a), so it is large enough for us never to sense the individual photons by our eye.

AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter

Question 7.
Ultraviolet light of wavelength 2271 A from a 1oo W mercury source irradiates a photo cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (-105 W m-2) red light of wavelength 6328 A produced by a He-Ne laser?
Solution:
Given, for UV light, λ = 2271A = 2271 × 10-10 m
V0 = 1.3 V; P = 100W; h = 6.63 × 10-34 J-s; c = 3 × 108 m/s
From Einstein’s equation E = KE + ϕ0, hυ = eV0 + ϕ0
ϕ0 = \(\frac{\mathrm{hc}}{\lambda}\) – eV0 = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2271 \times 10^{-10}}\) – 1.6 × 10-19 × 1.3 = 8.758 × 10-19 – 2.08 × 10-19
ϕ0 = \(\frac{6.678 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV = 4.17 eV ∴ ϕ0 = 4.2 eV
Given, for red light, λ = 6328Å = 6328 × 10-10m
E = \(\frac{\mathrm{hc}}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6328 \times 10^{-10}}\) = \(\frac{3.143 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV ∴ E = 1.96 eV
Here, E < ϕ0, So, the photocell will not respond to this red light. (It is independent of intensity).

Question 8.
Monochromatic radiation of wavelength 640.2 nm (1 nm = 10-9 m) from a neon, lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Solution:
Given, for Neon X = 640.2 nm = 640.2 × 10-9 m ; V0 = 0.54 V
h = 6.63 × 10-34 J-s; c = 3 × 108 m/s; e = 1.6 × 10-19 C
ϕ = \(\frac{\mathrm{hc}}{\lambda}\) – eV0 = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{640.2 \times 10^{-9}}\) – 1.6 × 10-19 × 0.54
= 3.1 × 10-19 – 0.864 × 10-19 = 2.236 × 10-19J = \(\frac{2.236 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV ∴ ϕ = 1.4 eV
For iron; given ϕ0 = 1.4eV; λ = 427.2 nm = 427.2 × 10-9 m
Let V0 be the new stopping potential, eV0 = \(\frac{\mathrm{hc}}{\lambda}\) – ϕ0
eV0’ = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{427.2 \times 10^{-9} \times 1.6 \times 10^{-19}}\) – 1.4 = 1.51 eV. Required stopping potential V0‘ = 1.51 V.

Question 9.
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:
λ1 = 3650Å, λ2 = 4047Å, λ3 = 4358Å, λ4 = 5461 Å, λ5 = 6907Å,
The stopping voltages, respectively, were measured to be:
V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V, V05 = 0V.
Determine the value of Plancks constant h, the threshold frequency and work function
for the material.
[Note : You will notice that, to get h from the data, you will need to know e(which you can take to be 1.6 × 10-19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]
Solution:
Given λ1 = 3650 A = 3650 × 10-10 m
λ2 = 4047 A = 4047 × 10-10 m
λ3 = 4358 A = 4358 × 10-10 m
λ4 = 5461 A = 5461 × 10-10 m
λ5 = 6907 A = 6907 × 10-10 m
V01 = 1.28V, V02 = 0.95, V03 = 0.74 V; V05 = 0

a) v1 = \(\frac{\mathrm{c}}{\lambda_1}\) = \(\frac{3 \times 10^8}{3650 \times 10^{-10}}\) = 8.219 × 1014 Hz
v2 = \(\frac{\mathrm{c}}{\lambda_2}\) = \(\frac{3 \times 10^8}{4047 \times 10^{-10}}\) = 7.412 × 1014 Hz
v3 = \(\frac{\mathrm{c}}{\lambda_3}\) = \(\frac{3 \times 10^8}{4358 \times 10^{-10}}\) = 6.884 × 1014 Hz
v4 = \(\frac{\mathrm{c}}{\lambda_4}\) = \(\frac{3 \times 10^8}{5461 \times 10^{-10}}\) = 5.493 × 1014 Hz
v5 = \(\frac{\mathrm{c}}{\lambda_5}\) = \(\frac{3 \times 10^8}{6907 \times 10^{-10}}\) = 4.343 × 1014 Hz
As the graph between V0 and frequency v is a straight line.
The slope of this graph gives the values of \(\frac{\mathrm{h}}{\mathrm{e}}\)
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 28
∴ \(\frac{\mathrm{h}}{\mathrm{e}}\) = \(\frac{V_{01}-V_{04}}{v_1-v_4}\) = \(\frac{1.28-0.16}{(8.219-5.493) \times 10^{14}}\)
h = \(\frac{1.12 \times 1.6 \times 10^{-19}}{2.726 \times 10^{14}}\) = 6.674 × 10-34 J . s

b) ϕ0 = hv0 = 6.574 × 10-34 × 5 × 1014
= 32.870 × 10-20 J = \(\frac{32.870 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}\)
= 2.05 eV

Question 10.
The work function for the following metals is given:
Na : 2.75 eV; K: 2.30 eV; Mo : 4.17 eV; Ni : 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 A from a He-Cd laser placed 1 m away from the photocell ? What happens if the laser is brought nearer and placed 50 cm away?
Solution:
Given λ = 3300 A = 3300 × 10-10 m
Energy of incident photon, E = \(\frac{\mathrm{hc}}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3300 \times 10^{-10} \times 1.6 \times 10^{-19}}\) ∴ E = 3.75 eV
Here Na, K has lesser work function than 3.75 eV. So, they produce photoelectric effect. If the laser is brought nearer then only the intensity change or the number of photoelctrons change.

Question 11.
Light of intensity 10-5 W m-2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Solution:
Given, I = 10-5 W/m2; A = 2 cm2 = 2 × 10-4 m2; ϕ0 = 2eV
Let t be the time.
The effective atomic area of Na = 10-20 m2 and it contains one conduction electron per
atom.
No. of conduction electrons m five layers
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 29
We know that sodium has one free electron (or conduction electron) per atom.
Incident power on the surface area of photocell
= Incident intensity × Area on the surface area of photo cell
= 10-5 × 2 × 10-4 = 2 × 10-9 W.
The electron present in all the 5 layers of sodium will share the incident energy equally.
Energy absorbed per second per electron, E = AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 30
= \(\frac{2 \times 10^{-9}}{10^{17}}\) = 2 × 10-26 W.
Time required for emission by each electron,
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 31 which is about 0.5 yr.
The answer obtained implies that the time of emission of electron is very large and is not agreement with the observed time of emission. There is no time lag between the incidence of light and the emission of photoelectron.
Thus, it is implied that the wave theory cannot be applied in this experiment.

Question 12.
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative. comparison. take the wavelength of the probe equal to 1 A, which is of the order of interatomic spacing in the lattice) (me = 9.11 × 10-31 kg).
Solution:
Given λ = 1 A = 10-10 m ; me = 9.11 × 10-31 kg; h = 6.63 × 10-34 J – s; c = 3 × 108 m/s
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 32
Thus, for the same wavelength a X-ray photon has much KE than an electron.

Question 13.
(a) Obtain the de Brogue wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable ? Explain. (mn = 1.675 × 10-27 kg)
(b) Obtain the de Brogue wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffráction experiments.
Solution:
a) Given, KE = 150 eV; m = 1.675 × 10-27 kg
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 33
The interatomic spacing is 10-10 m, which is greater than this wavelength. So, neutron beam of 150 eV is not suitable for diffraction experiment.

b) T = t + 273 = 27 + 273 = 300 K; K = 1.38 × 10-23 J/mol/K
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 34
This wavelength is order of interatomic spacing. So, the neutron beam first thermalised and then used for diffraction.

Question 14.
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
Solution:
Given, V = 50 KV s 50000 V
λ =
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 38
= 0.055 A ⇒ λ = 5.5 × 10-12 m; For yellow light (λ) = 5.9 × 10-7m
As resolving power (RP) ∝ \(\frac{1}{\lambda}\)
⇒
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 35

AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter

Question 15.
The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10-15 m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.5 11 MeV.)
Solution:
Given λ = 10-15 m; E = 0.5 11 MeV; P = \(\frac{\mathrm{h}}{\lambda}\) = \(\frac{6.63 \times 10^{-34}}{10^{-15}}\) = 6.63 × 10-19 kgm/s
Rest mass energy; E0 = m0c2 = 0.511 MeV = 0.511 × 1.6 × 10-13 T.
From relativistic theory, E2 = p2c2 + \(m_0^2 c^4\)
= (3 × 108 × 6.63 × 10-19)2 + (0.511 × 10-13 × 1.6)2 = 9 × (6.63)2 × 10-22.
As the rest mass energy is negligible ∴ Energy E = \(\sqrt{p^2 c^2}\) = pc = 6.63 × 10-19 × 3 × 108
= \(\frac{1.989 \times 10^{-10}}{1.6 \times 10^{-19}}\)eV = 1.24 × 109 eV = 1.24 BeV
Thus, to energies the electron beam, the energy should be of the order of BeV.

Question 16.
Find the typical de Brogue wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure; and compare It with the mean separation between two atoms under these conditions.
Solution:
Given T = 27 + 273 = 300 K; K = 1.38 × 10-23 J/mol/K; p = 1 atm = 1.01 × 105 Pa
AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter 36
We can see that the wave length with mean separation r, it can be observed (r >> λ) that separation is larger than wave length.

Question 17.
Compute the typical de Broglie wavelength of an electron in a metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10-10 m.
[Note : Exercise 35 and 36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.]
Solution:
Given, T = 27 + 273 = 300 K; r = 2 × 10-10m
Momentum, P = \(\sqrt{3 \mathrm{mKT}}\) = \(\sqrt{3 \times 9.11 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}\) = 1.06 × 10-25 kg-m/s
λ = \(\frac{\mathrm{h}}{\mathrm{p}}\) = \(\frac{6.63 \times 10^{-34}}{1.06 \times 10^{-25}}\) = 62.6 × 10-10m; Mean separation, r = 2 × 10-10 m
\(\frac{\lambda}{r}\) = \(\frac{62.6 \times 10^{-10}}{2 \times 10^{-10}}\) = 31.3
We can see that de-Broglie wavelength is much greater than the electron separation.

Question 18.
Answer the following questions :
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e; (-1/3)e]. Why do they not show up in Millikan’s oil-drop experiment ?
Solution:
The quarks have fractional charges. These quarks are bound by forces. These forces become stronger when the quarks are tried to be pulled apart. That is why, the quarks always remain’ together. It is due to this reason that tough fractional charges exists in nature but the observable charges are always integral multiple of charge of electron.

(b) What is so special about the combination e/m ? Why do we not simply talk of e and m separately ?
Solution:
The motion.of electron in electric and magnetic fields are governed by these two equations.
\(\frac{1}{2} \mathrm{mv}^2\) = eV or Beυ = \(\frac{m v^2}{\mathrm{r}}\)
In these equations, e and m both are together i.e. there is no equation in which e or m are alone. So, we always take e/m.

(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures ?
Solution:
At ordinary pressure, only very few positive ions and electrons are produced by the ionisation of gas molecules. They are not able to reach the respective electrodes and becomes insulators. At low pressure, density decreases and the mean free path becomes large. So, at high voltage, they acquire sufficient amount of energy and they collide with molecules for further ionisation. Due to this, the number of ions in a gas increases and it becomes a conductor.

(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic ? Why is there an energy distribution of photoelectrons ?
Solution:
Because all the electrons in the metal do not belong to same level but they occupy a continuous band of levels, therefore for the given incident radiation, electrons come out from different levels with different energies.

(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations :
E = h v, p = \(\frac{\mathbf{h}}{\lambda}\)
But while the value of λ is physically significant, the value of v (and therefore, the value of the phase speed v λ) has no physical significance. Why ?
Solution:
As λ = \(\frac{\mathrm{h}}{\mathrm{p}}\) = p = \(\frac{h}{\lambda}\) ⇒ E = hv = \(\frac{\mathrm{hc}}{\lambda}\)
Energy of moving particle E’ = \(\frac{p^2}{2 m}\) = \(\frac{1}{2} \frac{\left(\frac{h}{\lambda}\right)^2}{m}\) = \(\frac{1}{2} \frac{h^2}{\lambda^2 \mathrm{~m}}\). For the relation of E and p, we note that there is a physical significance of λ but not for frequency v.