Category: Inter 2nd Year

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 4(b) Immune System Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 4(b) Immune System

Very Short Answer Questions

Question 1.
Define the terms immunity and immune system.
Answer:
Immunity :
It is the ability of the host or individual to fight against the disease-causing organisms that is called immunity.

Immune System :
The network of organs, cells, and proteins that protect the body from harmful, infectious agents such as bacteria, viruses, animal parasites, fungi, etc., is called the immune system.

Question 2.
Define the non-specific lines of defence in the body.
Answer:
Non-specific lines of defence is the first line of defence mechanism and are also called as innate immunity, which is inherited by birth. It does not depend on prior contact with the microorganism. Non-specific lines of defence mechanism executed by four barriers namely;

1. Physical barriers
2. Physiological barriers
3. Cellular barriers
4. Cytokine barriers.

Question 3.
Differentiate between mature B-cells and functional B-cells.
Answer:

Mature B-cells	Functional B-cells
1. B-cells arise from stem cells and develop into mature B-cells.	1. Functional B-cells develop from mature B-cells.
2. The mature B -cells express antibodies on their surface to bind and engulf antigen for processing and presenting.	2. Functional B-cells differentiate into memory and plasma cells. Plasma cells produce antibodies, to eliminate antigen.

Question 4.
Write the names of any four mononuclear phagocytes.
Answer:

1. Histocytes – present in the connective tissue
2. Kupffer cells – in the liver
3. Microglia – in the brain
4. Osteoclasts – in the bone.

Question 5.
What are complement proteins?
Answer:
Complement proteins are a group of inactive plasma proteins and cell surface proteins. They are activated in cascade fashion. When activated, they form a membrane attack complex (MAC) that forms a pore in the plasma membrane, allowing ECF to enter the cell and make it swell and burst.

Question 6.
Colostrum is very much essential for the newborn infants.
Answer:
The colostrum secreted by the mother during the initial days of lactation has abdundant IgA antibodies to protect infant from initial sources of infection. .

Question 7.
Differentiate between perforins and granzymes.
Answer:
Perforins :
Perforins are the enzymes produced during the process of cell mediated immunity from cytotoxic T-lymphocytes. Perforins form pores in the cell membrane of the infected cells.

Granzymes:
Granzymes are the enzymes produced during the process of cell mediated immunity from cytotoxic T-lymphocytes. Granzymes enter th6 infected cells through the perfororations and activate certain proteins which help in distinction of the infected cell i.e., called apoptosis.

Question 8.
Explain the mechanism of Vaccinization (or) Immunization.
Answer:
Vaccinization is based on property of the mempry of the immune system. During the process of vaccinization, inactivated or weakend pathogens or antigenic proteins of pathogen * are introduced into the body of the host and they initiate the production of antibodies and also generate memory B-cells and memory T-cells. On subsequent exposures, the memory cell recognizes that pathogen quickly and overcomes the invader with a rapid and massive production of antibodies.

Question 9.
Mention the various types of immunological disorder.
Answer:
There are various types of immunological disorders.

1. Immuno deficiency disorders
2. Hypersensitivity disorders
3. Antoimmune disorders
4. Graft rejection.

Question 10.
More and more people in metro cities of India are prone to allergies. Justify.
Answer:
The people in metro cities of India suffer from allergies leading to asthmatic attacks due to environmental pollutants.

Question 11.
What are auto-immune disorders? Give Any two examples.
Answer:
Generally our immune system can recognize our own proteins (self antigens) and does not attack our own tissues. Unfortunately, in some cases our immune system fails to recognise some of our own body proteins and treats them as foreign antigens, that results in attacks on our own tissues. This leads to some very serious diseases collectively known a autoimmune disease.
Eg: 1. Graves’ disease 2. Rheumatoid arthritis.

Question 12.
How can the graft rejections be avoided in patients?
Answer:
After organ transplantation our body recognises them as foreign and initiate the graft rejection To avoid this tissue and maching and blood group matching are essential before undertaking graft. Even after this the patient has to take immuno-suppressant ‘drugs throughout the life.

Short Answer Questions

Question 1.
Write short notes on B-cells.
Answer:
The lymphocytes capable of producing antibodies and can capture circulating antigens are called B-cells. They are produced from the stem cells in the bone marrow, liver of foetus and bursa of fabricius in birds. Mature B-cells express or display Ig M and Ig D antibodies on their membrane surfaces. As these antibodies can take antigens, the mature B-cells are also called immuno-competent B-cells.

In secondary lymphoid organs these immune-competent B-cells develop into functional immune cells which later differentiate into long lived memory cells and effector plasma cells. The plasma cells produce antibodies specific to the antigen to which they are exposed. Memory cells store information about the specific antigens and show quick response, when the same type of antigen invades the body later.

Question 2.
Write short notes on Immunoglobulins.
Answer:
Whenever pathogen enters our body, the B-lymphocytes produce an army of proteins called antibodies to fight with them. They are highly specialised for binding with specific antigens. The part of an antibody that recognises an antigen is called the paratope antigen binding site.

Based on their mobility, antibodies are of two types.

1. Circulating or free antibodies :
These are present in the body fluids like serum, lymph etc.

2. Membrane bound antibodies :
These are present on the surface of the mature B-cells as well as the memory cells.

Structure :
Immunoglobulin is a Y’ shaped molecule with four polypeptide chains of which two &ye long identical heavy chains (H) and two are small, identical light chains (L). These two chains are linked by disulfide bonds. One end of the antibody molecule is called Fab end (Fragment- antigen binding) and the other end is called Fc end (Fragment-Crystaline). Based on the structure, the antibodies are of five types namely Ig G, Ig A, Ig M, Ig D and Ig E.

Question 3.
Describe various types of barriers of innate immunity.
Answer:
Innate immunity is a non-specific type of defence mechanism which provides the first line of defence mechanism against infections. This is executed by providing different types of barriers like;

a) Physiological:
Skin and mucus membranes are the main physical barriers. Skin prevents the entry of micro-organism, whereas the mucus membranes help in trapping the microbes entering our body.

b) Phyloigical barriers :
Secretions of the body like HCl in the stomach, saliva in the mouth, tears from the eyes are the main physiological barriers against microbes.

c) Cellular barriers :
Certain types of cells like polymorphonuclear leucocytes, monocytes, and natural killer cells in the blood as well as macrophages in the tissues are the main cellular barriers. They phagocytose and destroy the microbes-.

d) Cytokine barriers :
The cytokines secreted by the immune cells like interleukins and interferons are involved in differentiation of cells of immune system and protect the non-infected cells from further infection.

Question 4.
Explain the mechanism of humoral immunity.
Answer:
The immunity mediated by the antibodies that released into the fluids of the body (humors) such as plasma, lymph etc., is called humoral immunity.

Mechanism of humoral immunity :
Whenever the antigen (exogenous) enters into our body, they reach secondary lymphoid organs, where the free antigens bind to Fab end of the membrane bound antibodies that are present on the surface of mature B-cells. They engulf and process antigen. Then they display the antigenic fragments on their membrane with the help of Class-II MHC molecule. Then appropriate T₄ cells recognise them and interact with the antigen-MHC-II complex and release interleukins, which stimulates the B-cells to proliferate and differentiate into memory cells and plasma cells. The plasma cells release specific antibodies into plasma or extra cellular fluids.

These antibodies help in opsonising and immobi – lizing the bacteria, neutralizing and cross linking of antigens leading to agglutination of insoluble antigens and precipitation of soluble antigens. They also activate the phagocytes and complement system.

Question 5.
Explain the mechanism of cell mediated immunity.
Answer:
The immunity mediated by the activated T-cells, natural killer cells etc., is known as cell mediated immunity. It is effective against both exogenous and endogenous antigens.

Mechanism of cell mediated immunity :
Exogenous antigens are processed by the antigen presenting cells (APC), whereas endogenous antigens are processed by altered self cells (ASCs). Then the processed antigenic fragments are displayed on their surface with the help of class-I and class-II MHC molecules of ASCs and APCs respectively They are recognised by TCR of T-cells. The binding of T-cells to APCs or ASCs cause the production of a activated T-cells and T-memory cells.

The activated T_H cells secrete various types of interleukins which transform activated T_C cells into effector cytotoxic T-lymphocytes. They attach to the infected or altered cells and release enzymes like perforins and granzymes. Perforins form pores in the cell membrane of the infected cells. Then granzymes enter the infected cells through these perforations and activates the proteins which help in the distinction of the infected cell by a process called apoptosis The NK cells are similar in their action to CTL’s.

Question 6.
Explain the mechanism by which HIV multiplies and leads to AIDS.
Answer:
AIDS is a non-congenital, transmissible, lethal, sexually transmitted disease caused by Human Immunodeficiency Virus (HIV). HIV is a retrovirus with an envelope enclosing two ss RNA molecules as the genetic material.

Mechanism :
After getting into the body of a person, HIV enters the T_H cells, macrophages, or dendritic cells. In these cells ss RNA of HIV synthesizes a DNA strand complementary to the viral RNA using the enzyme reverse transcriptase. The same enzyme is responsible for the formation of the second DNA strand, complementary to the first strand forming the double-stranded viral DNA. This dsDNA gets incorporated into the DNA of the host’s DNA by a viral enzyme called integrase and it is in the form of a provirus.

Transcription of DNA results in the production of RNA, which can act as the genome for new viruses and can be translated into viral proteins. The various components of the viral particles are assembled and the HIV particles are produced. The infected human cells continue to produce virus particles. New viruses bud off from the host cell and attack other T_H cells. This leads to decrease CD4 receptors containing T_H cells in the infected person leading to immunodeficiency in him, finally causing AIDS.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 4(a) Endocrine System and Chemical Coordination Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 4(a) Endocrine System and Chemical Coordination

Very Short Answer Questions

Question 1.
What is acromegaly? Name the hormone responsible for this disorder.
Answer:
Acromegaly is a hormonal disorder that results when the pituitary gland produces excess growth hormone (GH). This disease is characterised by enlargement of the bones of the jaw, hand and feet, thickended nose, lips, eyelids and wide finger tips and gorilla like appearance of the person affected.

Question 2.
Which hormone is called anti-diuretic hormone? Write the name of the gland that secretes it.
Answer:
Vasopressin is also called as anti-diuretic hormone which is secreted by posterior pituitary.

Question 3.
Name the gland that increases in size during childhood and decreases in size during adulthood. What important role does it play in case of infection?
Answer:
Thymus is small at birth, it increases in size during childhood and reaches maximum size at puberty. During adulthood, it shrinks to its size at birth.

In old person thymus gland is degenerated, resulting in a decreased production of thymosin. Thymosin plays an important role in immune developments. Thus, immune response against infections of old people becomes weak.

Question 4.
Distinguish between diabetes insipidus and diabetes mellitus.
Answer:
Diabetes insipidus :
Deficiency of Vasopressin causes a disease called diabetes insipidus. It does not involve loss of sugar in urine.

Diabetes mellitus :
Under secretion of insulin by the pancreatic gland increases the level Of glucose in blood is called hyperglycemia. Prolonged hyperglycemia leads to a disease called diabetes mellitus, associated with loss of glucose through urine and formation of ketone bodies.

Question 5.
What are Islets of langerhans?
Answer:
The endocrine region of pancreas is called Islets of langerhans where it contain 1 to 2 millions Islets of langerhans. There are two main types of cells α – cells and β – cells.

α – cells produce the hormone glucagon, whereas β – cells produce insulin.

Question 6.
What is insulin shock?
Answer:
Hyper secretion of insulin leads to decreased level of glucose in blood (hypoglycemia) resulting in insulin shock.

Question 7.
Which hormone is commonly known as fight and flight hormone?
Answer:
Epinephrine and norepinephrine hormones are called fight and flight hormones because these hormones are secreted in response to stress and emergency situations.

Question 8.
What are androgens, which cells secrete them?
Answer:
Androgens are male sex hormones usually steroid hormones. E. g: testosterone.
Androgens are produced by the Leydig cells of the testes and to a minor extent by the adrenal glands in both sexes.

Question 9.
What is erythropoietin? What is its function?
Answer:
Erythropoietin is a hormone secreted the juxtaglomerular cells of the kidney. It plays an important role in the erythropoiesis i.e., in formation of RBC. Erythropoietin controls the formation of RBC by regulating the differentiation and proliferation of erythroid progenitor cells in the bone marrow.

Short Answer Questions

Question 1.
List out the names of the endocrine glands present in human beings and mention the hormone they secrete.
Answer:
1) Hypothalamus:
It secretes thyrotropin releasing hormone, corticotropin releasing hormone, gonadotropin releasing hormone, growth hormone releasing hormone, growth hormone release inhibiting hormone, prolactin release inhibiting hormone.

2) Pituitary glands :
Anatomically, it is divided into anterior and posterior pituitary.

Anterior Pituitary :
Produces Growth hormone, Prolactin, Thyroid stimulating hormone, Adreno corticotropic hormone, Follicular stimulating hormone, Luteinizing hormone.

Posterior Pituitary:
It releases two hormones namely Oxytocin and Vasopressin/ADH.

3) Pineal gland :
It secretes a hormone called Melatonin.

4) Thyroid gland :
It produces two hormones namely Thyroxine (T₄) and Tri iodothyronine (T₃).

5) Parathyroid gland :
Secretes a hormone called Parathyroid hormone.

6) Thymus gland :
It secretes peptide hormone called Thymosin.

7) Adrenal gland:
a) Adrenal cortex: GluCo corticoids, Mineralo corticoids, Androgens and Estrogens.
b) Adrenal medulla : Produces Epinephrine, norepinephrine.

8) Pancreas:
It secretes Glucagon and Insulin.

9) Testes :
Which secrete Androgens and Testosterone.

10) Ovaries :
Which produce Estrogen and Progesterone.

Question 2.
Describe the role of hypothalamus as a neuroendocrine organ. ‘
Answer:
Hypothalamus is located below the thalamus. It connects the neural and endocrine systems, as it closely tied to the pituitary gland. It responds to the sensory impulses received from different receptors by sending out appropriate neural or endocrine signals.

The hypothalamus is the master control centre of the endocrine system, as it contains several group of neuro secretary cells called nuclei, which produce hormones, called neuro hormones; These hormones directly control the pituitary gland which in turn secrete hormone that regulate the growth and functioning of other endocrine glands.

For example :
The two types of hormones produced by the hypothalamus are :
1) Releasing hormones :
Which stimulates the secretions of pituitary hormones.
Eg: 1) Thyrotropin releasing hormone – acts on anterior pituitary to release thyroid stimulating hormone.
2) Growth hormone releasing hormone – stimulate the release of growth hormone.

2) Inhibiting hormones :
Which inhibits the secretion of pituitary hormones.
Eg: 1) Growth hormone release inhibiting hormone – which inhibit the release of growth hormone from anterior pituitary.
2) Prolactin release inhibiting hormone – Inhibit the release of prolactin from anterior pituitary.

Question 3.
Give an account of the secretions of pituitary gland.
Answer:
The pituitary gland is also called hypophysis. Anatomically pituitary gland is divided into anterior and posterior pituitary.
I. Anterior Pituitary :
It produces six important peptides. They are ;
1) Growth hormone (GH) Somatotropin :
They promote growth of the entire body by increasing protein synthesis, cell division and cell differentiation.

2) Prolactin:
It causes enlargement of the mammary glands of the breast and initiate the maintenance of lactation in mammals. Prolactin also promote the growth of corpus luteum and stimulate the production of progesterone.

3) Thyroid stimulating hormone” (TSH) :
It stimulates the production of thyroid hormones from thyroid gland.

4) Adreno corticotropic hormone (ACTH) :
Controls the production of steroid hormones called gluco corticoids, by the adrenal cortex.

5) Follicle stimulating hormone (FSH) :
It stimulates growth the development of the ovarian follicles in females. In males FSH along with the androgens, regulates spermatogenesis.

6) Luteinizing hormone (LH) :
In males it stimulates production of androgens. In females it stimulates the ovaries to produce estrogens and progesterone and it maintains corpus luteum.

II. Posterior pituitary :
It stores and releases two hormones called oxytocin and vasopressin.

Oxytocin :
In females it stimulates contraction of pregnant uterus during child birth and ejection of milk from the mammary gland.

Vasopressin (ADH) :
Affects the kidney and stimulates reabsorption of water and electrolytes by the DCT and collecting duct.

Question 4.
Compare a pituitary dwarf and a thyroid dwarf in respect of similarities and dismilarities they posses.
Answer:

Pituitary dwarf	Thyroid dwarf
1. Hypersecretion of growth hormone from pituitary during childhood retards growth resulting in pituitary dwarf.	1. Hyposecretion of thyroid hormones during pregnancy, defective development of baby i.e., physical and mental growth get severely stunted, resulting in thyroid dwarf.
2. Human growth hormone deficiency results in abnormally slow growth and short structure with normal proportion.	2. Deficiency of thyroid hormones by birth results in enlarged head, short limbs, puff eyes, a thick and protudding tongue, dry skin, tow. I Qetc.
3. The pituitary dwarf is sexually and intellectually a normal individual.	3. If the condition not treated the child will grow up dwarf, mentally retarded and sexually sterile.
4. Administration of purified HGH has been shown to induce skeletal growth in these patients.	4. Early treatment can result in normal growth and development.

Question 5.
Explain how hypothyroidism and hyperthyroidism can affect the body.
Answer:
Hypothyroidism:
Inadequate supply of iodine or impairment in the function of thyroid glands leads to decrease in production of thyroid hormones (T₃ & T₄) results in hypothyroidism and enlargement of the thyroid gland called Simple goiter.

During pregnancy due to hypothyroidism, defective development of the growing body leads to a disorder called Cretinism. Physical and mental growth gets severely stunted due to untreated congenital hypothyroidism, stunted growth, mental retardation, low IQ, deafness, and mutism are some characteristics features of this disease.

In adult women it may cause irregular menstrual cycles. Hypothyroidism in adult causes Myxoedema characterized by bagginess under the eyes, puffiness of face, dry skin, slowness in physical and mental activities.

Hyperthyroidism:
Over activity of thyroid, cancer of the gland or development of nodule of thyroid lead to hyper thyroidism. In adults it causes an abnormal growth leads to a disease called Exophthalmic goiter with characteristically protruded eyeballs. Hyperthyroidism also affects the physiology of the body i.e., increased metabolic rate, nervousness, rapid heartbeat, sweating, increased appetite etc.

Question 6.
Write a note on Addison’s disease and Cushing’s Syndrome.
Answer:
Addison’s disease: It is caused due to hyposecretion of glucocorticoids by the adrenal cortex. This disease is characterised by loss of weight, muscle weakness, fatigue and reduced blood pressure. Sometimes darkening of the skin in both exposed and non-exposed parts of the body occurs in this disorder.

Cushing’s Syndrome :
It results due to over production of glucocorticoids. This condition is characterised by breakdown of muscle proteins and redistribution of body fat resulting in spindly arms and legs, a round moon-face, buffalo hump on the back and pendulous abdomen is also observed. Wound healing is poor. The elevated level of cortisols causes hyperglycemia, over deposition of glycogen in liver and rapid gain of weight.

Question 7.
Why does sugar appear in the urine of a diabetic?
Answer:
Hyposecretion of insulin of pancreatic gland increases the level of glucose in blood called hyperglycemia. Prolonged hyperglycemia leads to a disease called diabetes mellitus.

In diabetic patients glucose or sugar appears in urine because kidney plays a special role in the homeostasis of blood glucose.’Glucose is continuously filtered by the glomeruli, reabsorbed and returned to the blood. If the level of glucose in blood is above 160 -180 mg/dl. i.e., in hyperglycemia condition glucose in primary urine is not completely reabsorbed, and returned to the blood. Some of which is retained and excreted in urine.

Question 8.
Describe the male and female sex hormones and their actions.
Answer:
The hormones, which are responsible for the development of secondary sexual characters and changes in different stages of life are called sex hormones.

Male sex hormones :
Androgens :
Androgens are produced by the Leydig cells of the testes and to a minor extent by the adrenal glands in both sexes.

Functions :
→ Growth, development and maintenance of male reproductive organs.
→ Sexual differentiation and secondary sexual characteristics.
→ Spermatogenesis.
→ Male pattern of aggressive behaviour.
→ Increases the protein synthesis and increases the glycolysis.

Female sex hormones:
1) Estrogens :
Synthesized by the follicles and corpus luteum of ovary.
Functions :
→ Development and maintenance of female reproductive organs.
→ Maintenance of menstrual cycle.
→ Development of secondary sexual characters.
→ Estrogen promotes the protein synthesis and calcification and bone growth.

2) Progesterone :
It is synthesized and secreted by corpus luteum and placenta. Functions: required for implantation of fertilised ovum and maintenance of pregnancy.

3) Follicle stimulating and Lutenizing hormones :
Both these hormones produced from anterior pituitary gland in both sexes.

Functions:
Both these hormones play an important role in secondary sexual characters in both sexes.

Question 9.
Write a note on the mechanism of action of hormones.
Answer:
Hormones are primary messengers which interacting with receptors and they generate secondary messengers. These secondary messengers regulate cellular metabolism in the target cells.

Mechanism of action of lipid insoluble hydrophillic hormone :
→ The Hormone binds to a stimulatory membrane bound receptor, and stimulate ‘G’protein.
→ ‘G’ protein of the cell membrane binds to GTP and activates adenylate cyclase.
→ Adenylate Cyclase forms cAMP from ATP.
→ cAMP activates the protein kinase, which activates the enzyme phosphorylase.
→ Phosphorylase further phosphorylate the inactive enzyme and convert it to active form and involved in the metabolic process. Eg : Epinephrine.

Mechanism of action of lipid soluble hormone:
Lipid soluble hormones easily diffuse through the cell membrane.
→ It binds to a specific receptor in the cytoplasm forming hormone receptor complex molecule.
→ This complex molecule enters the nucleus and binds to the DNA and stimulate the production of specific m-RNA molecule.
→ The m-RNA passes into the cytoplasm, where it is involved in the translation process and synthesizes a protein. These proteins produced by the cell as a response of hormone and plays an important role in their respective metabolism.
Eg : Aldosterone

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 3(b) Neural Control and Coordination Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 3(b) Neural Control and Coordination

Very Short Answer Questions

Question 1.
Name the cranial meninges covering the brain of a man.
Answer:
The brain is covered by three connective tissue membranes called meninges.

1. Dura mater
2. Arachnoid mater
3. Pia mater.

Question 2.
What is Corpus callosum?
Answer:
Two cerebral hemispheres are internally connected by a transverse, wide and flat bundle of myelinated fibres beneath the cortex is called corpus callosum.

Question 3.
What do you know about arbor vitae?
Answer:
The white matter of cerebellum is branched, tree like appearance. Hence it is called arbor vitae and is surrounded by a sheath of grey matter.

Question 4.
Why the sympathetic division is called thoraco-lumbar division?
Answer:
The preganglionic sympathetic neurons have their cell bodies in the grey matter of thoracic and lumber regions of the spinal cord. So, sympathetic division is called thoracolumbar division.

Question 5.
Why the para sympathetic division is called cranio sacral division?
Answer:
The cell bodies of the paraganglionic neurons of the parasympathetic division are located in the brain and in the sacral region of the spinal cord. Hence, the parasympathetic is also known as the cranio sacral division.

Question 6.
Distinguish between the absolute and relative refractory periods.
Answer:
Absolute refractory period:
During the absolute, refractory period, even a very strong stimulus cannot initiate a second action potential. This period coincides with the period of depolarization and repolarization.

Relative refractory period:
It is the time during which a second action potential can be initiated by a larger than normal stimulus. It coincides with the period of hyperpolarization.

Question 7.
What is all-or-none principle?
Answer:
The action potential occurs in response to a threshold stimulus (or) supra threshold stimulus but does not occur at subthreshold stimuli. It means the nerve impulse is either conducted totally (or) not conducted at all and this called all-or-none principle.

Question 8.
How do rods and cones of human eye differ from each other chemically and functionally?
Answer:
Rods:
Rods contain a purplish red protein called rhodopsin, which contains a derivative of vitamin A. Rods are concerned with dim light.

Cones :
Cones contain a visual pigment called iodopsin, made of a protein called photopsin and they are important in daylight vision and colour vision.

Question 9.
Distinguish between the blind spot and the yellow spot.
Answer:
Blind Spot :
The region of the retina where the optic nerve exists the eyeball and devoid of rods and cones is called blind spot.

Yellow spot:
The centre of the posterior portion of the retina is called yellow spot.

Question 10.
What is organ of corti?
Answer:
The hearing apparatus that is present in the middle canal of the cochlea is called organ of corti. The organ of corti contains hair cells that act as auditory receptors.

Short Answer Questions

Question 1.
Draw a labelled diagram of the T.S. of the spinal cord of man.
Answer:

Question 2.
Distinguish between somatic and autonomic neural systems.
Answer:

Somatic neural system	Myosin
1. The senso’ry neurons conduct sensory impulses from the different somatic receptors to the CNS.	1. The autonomic neurons are associ-ated with interoceptors.
2. All these sensations are consciously perceived.	2. These sensory signals are generally not continuously perceived.
3. Somatic motor neurons innervate the skeletal muscles and produce voluntary movements.	3. Autonomic motor neurons regulate the involutionary activities of the cardiac muscle, smooth muscle and glands.
4. Acetyl choline is the neurotrans-mitter.	4. Acetyl choline (or) norepinephrine is neurotransmitter.

Question 3.
Give an account of the retina of human eye.
Answer:
Retina is the inner coat of the eye. It consist of a pigmented epithelium and a neural portion. The pigmented epithelium is a sheet of melanin containing epithelial cells. The neural portion has three layers namely photoreceptor Iaydr, bipolar cell layer and ganglion cell layer.

Photoreceptor layer consist of rods and cones. Rods contain a protein called rhodopsin. Rods are concerned with dim light. Cones contain a visual pigment called iodopsin and they are important in daylight vision and colour vision. There are three types of cones and are response to red, green and blue colours.

The centre of the posterior portion of the retina is called yellow spot. A depression present in the yellow spot is called ‘Forea’ contractile and it contains only cones. Forea is responsible for sharp vision. The region of retina which is devoid of rods and cones is known as blind spot (or) optic disc, which form the optic nerve called 2nd cranial nerve.

Question 4.
Give an account of synaptic transmission.
Answer:
A nerve impulse is transmitted from one neuron to another through junction called synapses.

There are two types of synapses. 1) Electrical synapses 2) Chemical synapses.

Electrical synapses :
These synapses are electrically conductive links between two neurons and are also called “gap junctions”. Impulses transmission across an electrical synapses is always faster than that across a chemical synapses.

Chemical synapses :
Chemicals called neuro transmitters are involved in the transmission of impulses at those synapses. When an impulse arrives at the axon terminal, it depolarizes the membrane opening voltage gated calcium channels. Calcium ions stimulate the release of neurotransmitters in the cleft by exocytosis. The released neurotransmitters bind to their specific receptors, present on the post synaptic membrane.

The post synaptic membrane has ligand gated channels. They are ion channels which respond to chemical signals, rather than to changes in the membrane potential. The entry of ions can generate a new potential in the post synaptic neuron. The new potential developed may be either excitatory (or) inhibitory.

Excitatory post synaptic potentials cause depolarisation, where as inhibitory post synaptic potentials cause hyper polarisation of post synaptic membrane.

Question 5.
List out the differences between Sympathetic and Parasympathetic neural system in man.
Answer:

Sympathetic neural system	Parasympathetic neural system
1. SNS originates in the thoracic and lumbar regions of the spinal cord.	1. PNS originates in the cranial region of the brain and the sacral region of the spinal cord.
2. Its ganglia are linked up to form a chain.	2. Its ganglia remain isolated.
3. Preganglionic fibres are short and the postganglionic fibres are long.	3. Preganglionic fibres are long and the postganglionic fibres are short.
4. Norepinephrine is produced at the terminal ends of the post-ganglionic fibres at the synapses on the effectors organ. Hence the system is called adrenergic’ usually.	4. Acetycholine is produced at the ter-minal ends of the postganglionic fi-bres at the effector organ. Hence the system is called cholinergic’ usually.
5. Active during stressful conditions preparing the body to face them.	5. Active during relaxing times, restor-ing normal activity after stress.
6. The-overall effect is excitatory and stimulating.	6. The overall effect is inhibitory.

Long Answer Questions

Question 1.
Give a brief account of the structure and functions of the brain of man.
Answer:
Brain is the site of information, processing and control. It is protected in the cranial cavity and covered by three cranial meninges namely duramater (outer layer), arachnoid mater (thin middle layer) and piamater (inner layer).
The brain can be divided into three major parts called

1. Fore brain
2. Mid brain
3. Hind brain.

1) Fore brain :
The fore brain consists of i) Olfactory bulb ii) Cerebrum and iii) Dience-phalon.

i) Olfactory bulb :
Which receives impulses pertaining to smell from the Olfactory epithe-lium.

ii) Cerebrum :
Cerebrum forms the major part of the human brain. A deep cleft divides the cerebrum longitudinally into two halves, which are termed as the left and right cerebral hemispheres. The hemispheres are connected by a transverse, wide and flat bundle of myelinated fibres beneath the cortex, called corpus callosum. It brings the coordination between the left and right sides of the hemispheres. The surface of the cerebral cortex shows many folds and grooves. The folds are called gyri, the deepest and shallower grooves between folds are called fissures and sulci respectively.

The cerebral cortex contain three functional areas called
a) Sensory areas : receive and interpret the sensory impulses.
b) Motor areas : which control volutntary muscular movements.
c) Association areas : which are neither clearly sensory nor motor in function, they deal integrative functions, such as memory and communications.

The cerebral medulla consist of mostly myelinated axons. Each cerebral hemisphere of the cerebrum is divided into four lobes namely frontal, parietal, temporal and occipital lobes.

iii) Diencephalon :
It contains three main parts namely, a) Epithalamus, b) Thalamus and c) Hypothalamus.
a) Epithalamus :
It is the roof of the diencephalon. It is axon nervous part which is fused with the pia matter to form the anterior choriod plexus. The epithelium of the epithalamus forms a pineal stalk, which ends in a rounded structure called pineal body.

b) Thalamus :
It lies superior to the mid brain. It is the major coordination centre for sensory and motor signalling.

c) Hypothalamus :
It lies at the base of the thalamus. The hypothalamus forms a funnel-shaped downward extension called infundibulum, connecting the hypothalamus with the pituitary gland. It also contains a group of neuro-secretory cells, which secrete hormones called hypothalamic hormones.

Hypothalamus controls and integrates the activities of the autonomous nervous system and it has osmoregulatory, thermoregulatory, thirst, feeding the satiety centres.

Limbic system :
The inner part of cerebral hemisphere and group of associated structures forms limbic system. Limbic system along with hypothalamus is involved in the regulation of sexual behaviour and expression of emotional reactions.

2) Mid brain :
Mid brain is located between the thalamus of the fore brain and pons varolii of hind brain. The ventral portion of mid brain consists of a pair of longitudinal bands of nervous tissues called cerebral peduncles. The dorsal portion of the mid brain consists of four lobes called corpora quadrigmina. The two larger anterior lobes are called superior colliculi, which are concerned with visual function. The smaller posterior lobes are called inferior colliculi and are concerned with auditory functions.

3) Hind brain :
The hind brain comprises of cerebellum, pons varolii and medulla oblongata.
i) Cerebellum :
It is the second largest part of the brain. It consists of two cerebellar hemispheres and a central vermis. Each cerebellar hemisphere consists of three lobes namely anterior, posterior and floccular lobes. It has a branching tree like core of white matter called arbor vitae.

ii) Pons Varolii :
It consists of nerve fibres which form a bridge between the two cerebellar hemispheres. It is a relay station between the cerebellum, spinal cord and the rest of the brain. Pons has the pneumotaxic centre as it regulates the amount of air a person can take in each time.

iii) Medulla Oblongata :
It is the posterior part of brain. It extends from the Pons Varolii above and continuous with the spinal cord below. Medulla includes cardiovasicular, and respiratory centers, the centers for swallowing, vomiting, coughing, sneezing and hiccupping. The mid brain, pons and medulla Oblongata are collectirel referred to as brain stem.

Question 2.
Explain the transmission of nerve impulse through a nerve fibre with the help of suitable diagrams.
Answer:
Nerve impulse is the combination of mechanical, chemical (or) electrical disturbances occur in neuron because of stimulus. The propagation of a impulse along nerve fibre is called transmission. In this process both physical and chemical changes are involved. The entire process is divided into stimulation, excitation, conduction and response.

Resting membrane potential :
The resting membrane potential exists because of a small buildup of negative ions in the axoplasm along the inside of the membrane and an equal buildup of positive ions in the extra cellular fluid along the outer surface of the membrane. Such a Separation of positive and negative electrical chafges is a form of potential energy. In neurons, the resting membrane potential ranges from -40 to -90 mV. A typical value is-70 mV.

At resting phase, the axolemma is polarized. If the inner side becomes less negative, it is said to be depolarized. If the inner side becomes more negative, it is said to be hyperpolarized. During the resting phase the activation gates of sodium are closed, the inactivation gates of sodium are open and the activation gates of potassium are closed.

Sodium-potassium pump : Sodium and potassium ions diffuse inwards and outwards, respectively, down their concentration gradients through leakage channels. Such a movement of ions, if unchecked, would eventually disturb the resting membrane potential. These flows of ions are offset by sodium-potassium pumps (Na⁺/K⁺ ATPases) present in the axonal walls. These pumps expel three Na⁺ ions for each two K⁺ ions imported. As these pumps remove more positive charges from the axoplasm than they bring into it, they contribute to the negativity of the resting membrane potential i.e.,-70mv.

Depolarization (Rising phase):
When a nerve fibre is stimulated, the plasma membrane becomes more permeable to Na⁺ ions than to K⁺ ions as the activation and inactivation voltage gates of sodium open and activation voltage gates of potassium close. As a result the rate of flow of Na+ into the axoplasm exceeds the rate of flow of K⁺ to the ECF. Hence, the axolemma is positively charged inside and negatively charged outside. This reversal of electrical charge is called “depolarization”.

Outer face of the point which is adjacent to the site of depolarization remains positively charged. The electrical potential difference between these two areas is called “action potential”. An action potential occurs in the membrane of the axon of a neuron when depolarization reaches a certain level called ‘threshold potential1 (-55 mV). The particular stimulus which is able to bring the membrane potential to threshold is called ‘threshold stimulus’.

Repolarization (Falling phase) :
As the wave of depolarization passes away from its site of origin to the adjacent point, the activation gates of sodium remain open, inactivation gates of sodium close and activation gates of potassium open at the site of origin of depolarization. As a result the influx of Na⁺ ions into the axoplasm from the ECF is checked and ‘efflux’ of K⁺ ions occurs, which leads to the returning of axolemma to the resting state (exit of potassium ions causes a reversal of membrane potential to negative inside). This is called ‘repolarization’.

Hyperpolarization (Undershoot):
The repolarization typically goes more negative than the resting potential to about -90 mV This is called ‘hyperpolarization’. This occurs because of the increased K⁺ permeability that exists while voltage gated K⁺ channels are open activation and inactivation gates of Na⁺ channels remain closed. The membrane potential returns to its original resting state as the K⁺ channels close completely. As the voltage falls below the -70 mV level of the resting state, it is called ‘undershoot’.

The refractory periods :
The period of time after an action potential begins during which the neuron cannot generate another action potential in response to a normal threshold stimulus is called the ‘refractory period’. There are two kinds of refractory periods, namely the absolute refractory period and the relative refractory period. During the absolute refractory period, even a very strong stimulus cannot initiate a second action potential. The relative refractory period is the time during which a second action potential can be initiated by a larger than normal stimulus.

Conduction speed:
The conduction speed of a nerve impulse depends on the diameter of the axon: the greater the axon’s diameter, the faster the conduction. In a myelinated axon, the voltage-gated Na⁺ and K⁺ channels are concentrated at the nodes of Ranvier. As a result, the impulse ‘jumps’ from one Ranvier’s node to the next, rather than travelling the entire length of the nerve fibre. This mechanism of conduction is called Saltatory conduction. Saltatory conduction is faster (in myelinated fibres) than continuous conduction (in nonmyelinated fibres).

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 3(a) Musculo-Skeletal System Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 3(a) Musculo-Skeletal System

AP Inter 2nd Year Zoology The Musculo Questions and Answers

Very Short Answer Questions

Question 1.
What is a ‘motor unit’ with reference to muscle and nerve?
Answer:
Motor unit is made up of a motor neuron and set of muscle fibres innervated by all the telodendrites.

Question 2.
What is triad system?
Answer:
In a skeletal muscle each transverse tubule (T-Tubule) is flanked on either side by several cfsternae of the sarcoplasmic reticulum. T-tubule and the two terminal cistemae at its sides form the triad system.

Question 3.
Write the difference between actin and myosin.
Answer:

Actin	Myosin
1. Actin is a thin contractile protein.	1. Myosin is a thick contractile protein.
2. It is present in light bands and is called an isotropic band.	2. It is present in dark bands and is called an anisotropic band.
3. Each actin filament is made of two ‘F’ actin molecules helically wound around each other, tropomyosin and a complex, protein called troponin.	3. Each mydsin is made up of monomeric protein called meromyosins. Each meromyosin has’ two parts namely head, and arm (or) neck.

Question 4.
Distinguish between red muscle fibers and white muscle fibers. Ans.
Answer:

Red muscle fiber	White muscle fiber
1. Red muscle fibers are thin and smaller in size.	1. White muscle fibers are thick and larger in size.
2. They are red in colour as they contain large amount of myoglobin.	2. They are white in colour as they contain small amount of myoglobin.
3. They contain numerous mitochondria.	3. They contain less number of mito-chondria.
4. They carry out slow and sustained Contractions for a long period.	4. They cany out fast work for short duration.

Short Answer Questions

Question 1.
Write a short note on sliding filament theory of muscle contraction.
Answer:
The sliding filament theory explains the process of muscle contraction. It was proposed by Jean Hanson and Hugh Huxley. It states that contraction of a muscle fiber takes place by the sliding of the thin filaments over the thick filament, which shorfens the myofibril.

Each muscle fiber contains a special contractile proteins called actin and myosin. Actin is the thin contractile protein present in the light band and is known as the T band, where as myosin is thick contractile protein present in dark band aind is known as ‘A’ band. There is an elastic fiber called ‘Z’ line, that bisets each T band. The central part of the thick filament that pot overlapped by the thin filament is known as the ‘if zone.

During the muscle contraction, the myosin heads bind to the exposed active sites on the actin molecules and form across bridge. As a result the thin filaments are pulled towards the centre of the A band. The ‘Z’ line attached to the actin filaments is also pulled leading to the shortening of the sarcomere i.e., contraction.

During the shortening of the muscle the T bands get reduced in length, whereas the A’ bands retain their length and ‘H’ zone disappears.

Question 2.
Describe the important steps in muscle contraction.
Answer:
During skeletal muscle contraction, the thin filament slides over the thick filament by repeated binding and releases myosin along the filament.
Important steps in muscle contraction :

Step 1:
Muscle contraction is initiated by signals that travel along the axon and reach the neuro muscular junction. As a result, acetyl choline is released into the synaptic cleft by generating an action potential in sarcolemma.

Step 2:
The generation of this action potential releases calcium ions from sarcoplasmic reticulum in the sarcoplasm.

Step 3:
The increased calcium ions in the sarcoplasm leads to the activation of actin sites, then active actin sites are exposed and this allows myosin heads to attach to this site and forms cross bridges by utilising energy from ATP hydrolysis.

Step 4:
The actin filaments are pulled. As a result, the ‘H’ zone reduces. It is at this stage that the contraction of the muscle occurs.

Step 5:
After muscle contraction, the myosin head pulls the actin filament and releases ADP along with phosphate. ATP molecules bind and detach myosin and the cross bridges are broken and decreases the calcium ions contraction. As a result masking the actin filaments and leading to muscle relaxation.

Question 3.
Describe the structure of a skeletal muscle.
Answer:
1) Skeletal muscle is made up of number of muscle bundles (or) fascicles. The fascicles are held together by a common collagenous connective tissue layer called fascia.

2) Each fascicle contains a number of cylindrical muscle fibers. Each muscle fiber is lined by the plasma membrane called sarcolemma enclosing the sarcoplasm.

3) Skeletal muscle fiber is a syncytium as each fiber is formed by fusion of embryonic, mononucleate myoblasts. Hence, the skeletal muscle cells are multinucleate, with characteristically peripheral nuclei.

Question 4.
Write short notes on contractile proteins.
Answer:
Actin and myosins are contractile proteins.
Actin :

1. Each actin filament is made of two ‘F (filamentous) actin molecules helically wound around each other.
2. Each actin is a polymer of monomeric ‘G’ (globular) actin molecules. Two filaments of another protein, called tropomyosin also run close to the ‘F’ actin molecules, throughout their length.
3. A complex protein called troponin is distributed at regular intervals on the tropomyosin.
4. Troponin is made of three polypeptides namely Tn-T, Tn-I and Tn-C. Tn-T binds to tropomyosin, Tn-I inhibits the myosin binding site on the actin, Tn-C can bind to Ca²⁺ when Ca²⁺ ions are not bound to troponic, which block the active site of actin. When calcium ions attaches to Tn-C, the tropomyosin moves away from the active sites, allowing the myosin heads to bind to the active sites of actin.
5. Troponin and tropomyosin are often called regulatory proteins, because of their role in masking and unmasking the active sites.

Myosin:

1. Myosin, is a motor protein that is able to convert chemical energy in the ATP molecules into mechanical energy.
2. Each myosin filament is a polymerized protein, consist of monomeric proteins called meromyosins.
3. Each Meromyosin has two important parts, a globular head with a short arm and tail.
4. The globular head with arm is composed of heavy meromyosin and the tail is made of light meromyosin.
5. The short arm / neck serves as a flexible link between the head and tail regions.
6. There are about 200-300 molecules of myosin per thick filament.
7. The head and short arm project outwards at regular distance and angels from each other from the surface of a polymerized myosin filament and is known as cross arm.
8. Each head has two binding sites, one for ATP and other for an active site on the actine molecule.

Question 5.
Draw a neat labelled diagram of the ultra structure of muscle fibre.
Answer:

Question 6.
Draw the diagram of a sarcomere of skeletal muscle showing different regions.
Answer:

Question 7.
What is Cori cycle? Explain the process.
Answer:
Lactate produced by anaerobic glycolysis in the muscle, moves to the liver and i converted to glucose, which then return to the muscles and is converted back to lactate This two way traffic between skeletal muscle and liver is called the Cori cycle.

Cori cycle :
The lactate produced during rapid contraction of skeletal muscles under low availability of oxygen is partly oxidized and a major part of it is carried to the liver by the blood, where it is converted into pyruvate and then to glucose through gluconeogenesis. The glucose can enter the blood and be carried to muscles and is immediately converted back to lactate. If by this time the muscles have stopped contraction, the glucose can be used to rebuild reserve of glycogen through glycogenesis.

Long Answer Questions

Question 1.
Explain the mechanism of muscle contraction.
Answer:
Mechanism of muscle contraction is best explained by the sliding filament theory. It states that contraction of muscle fiber takes place by the sliding of the thin filament over the thick filaments.

Mechanism of muscle contraction :
1. Excitation of muscle :
a) Muscle contraction is initiated by the signal sent by central nervous system via a motor neuron.
b) A neural signal reaching the neuromuscular junction releases acetyl choline, which generates an action potential in the sarcolemma.
c) When the action potential spreads to the triad system through T-tubules, the cistemae of the sarcoplasmic reticulum release calcium ions into the sarcoplasm.

2. Formation of cross bridge :
a) Increase in the Ca²⁺ level leads to the binding of calcium ions to the subunit Tn-C of the troponin of the actin filament (thin). This makes troponin and tropomyosin complex to move away from the active sites of actin molecules.
b) In this stage the myosin head attaches to the exposed active site of actin and forms cross bridges by utilising energy from ATP hydrolysis.

3. Power stroke :
a) The cross bridge pulls the attached actin filaments, towards the centre of the ‘A’ band.
b) The ‘Z’ lines attached to these actin filaments are also pulled in wards from both sides, there by causing shortening of the sarcomere i.e., contraction.
c) During the shortening of the muscle, the I bands get reduced in length, whereas the ‘A’ bands retain their length.
d) As the thin filaments are pulled deep into the A bands making the H bands narrow, the muscle shows the effect contraction.

4. Recovery stroke :
a) The myosin goes back to its relaxed state and releases ADP.
b) A new ATP molecule binds to the head of myosin and the cross bridge is broken.

5. Relaxation of muscle :
a) When motor impulses stop, the calcium ions are pumped back into the sarcoplasmic cistem&e it results in the marking of active sites of the actin filaments.
b) The myosin heads fail to bind with the active sites of actin.
c) These changes Cause the return of ‘Z’ lines back to their original position i.e., relaxation.

Question 2.
List in sequence, the events .that take place during muscle contraction.
Answer:
During skeletal muscle contraction, the thin filament slides over the thick filament by repeated binding and releases myosin along the filament.

The following events take place during muscle contraction:
1. Muscle contraction is initiated by signals that travel along the axon and reach the neuro muscular junction (or) motor end plate. As a result, acetyl choline is released into the synaptic left by generating an action potential in sarcolemma.

2. The action potential spreads to the triad system through the T-tubules, the cistemae of the sarcoplasmic reticulum release calcium ions into the sarcoplasm.

3. Increase in the calcium ions level leads to the binding of calcium ions to the sub unit Tn-C of the troponin of the thin filament: This makes troponin and tropomyosin complex to remove away from the active sites of actin molecules.

4. In this stage, the myosin head attaches to the exposed site of actin and forms cross
bridge by utilising energy from ATP hydrolysis.

5. The cross bridge pulls the attached actin filaments towards the centre of the ‘A’ band. The ‘Z’ lines attached to these actin filaments are also pulled inwards from both the sides, thereby causing contraction. During the contraction the ‘I’ bands get reduced in length, where as ‘A’ bands retain their size.

6. As the thin filaments are pulled deep into the ’A” bands making the ‘H’ bands narrow, the muscle shows the effect contraction.

Contraction is turned off by the following sequence of events :
7. Acetyl choline at the neuromqscular junction is broken down by acetyl cholinesterase and this terminates the stream of action potentials along the muscle fibre surface.

8. The sarcoplasmic reticulum ceases to release calcium ions and immediately calcium ions are pumped back into the sarcoplasmic cistemae.

9. In the absence of calcium ions a change in the configuration of troponin and . tropomyosin i.e., masking of the active sites of the actin filaments.

10. The myosin heads fail to bind with active sites of actin. These changes cause the return of ‘Z’ lines back to their original position i.e., relaxation.

AP Inter 2nd Year Zoology The Skeleton Questions and Answers

Very Short Answer Questions

Question 1.
Name two cranial sutures and their locations.
Answer:

1. Coronal suture – between the frontal and parietal bones.
2. Lambdoid suture – between the parietal and occipital bones.

Question 2.
Name the keystone bone of the cranium. Where is it located?
Answer:
Sphenoid bone is the keystone bone of the cranium, because it articulates with all the other cranial bones. It is present at the middle part of the base of the skull.

Question 3.
Human skull is described as dicondylic skull. Give the reason.
Answer:
Human skull is described as discondytic skull because, two occipital condyles are present one on each side of the foramen magnum. ‘

Question 4.
Name the ear ossicles and their evolutionary origin in human beings.
Answer:
Each middle ear contains three tiny bones called ear ossicles. They are ;
Malleus – modification of articular
Incus – modified quadrate
Stapes – modified hyomandibula.

Question 5.
Name the type of joint between a) atlas / axis b) carpal / metacarpal of the human thumb.
Answer:
a) Joint between atlas / axis – Pivot joint

b) Joint between carpal / meta carpal of the human thumb – Saddle joints.

Question 6.
Name the type of joint between a) Atlanto – axial joint b) Femur – acetabulum joint.
Answer:
a) Joint between atlanto – axial joint – Pivot joint
b) Joint between Femur – acetabulum joint – Ball and Socket joint.

Question 7.
Name the typen of joint between a) Cranial bones b) Inter-tarsal joint.
Answer:
a) Joint between Cranial bones-Sutures (Fibrous joint) E.g.: Cororial suture, lambdoid suture. . , .
b) Inter-tarsal joint – Gliding joint.

Short Answer Questions

Question 1.
List out the bones of the human cranium.
Answer:
Cranium, the brain box is formed by eight cranial flattened bones. They are ;
i) Frontal bone (1) :
It forms the forehead, anterior part of the cranial floor and roof of the orbit.

ii) Parietal bones (2) :
They form the major portion of the sides and roof of the cranial cavity.

iii) Temporal bones (2):
They form lateral walls of the cranium as well as housing the external ear.

iv) Occipital bone (1):
It forms the posterior part and most of the base of the cranium.

v) Sphenoid bone (1):
It is present at the middle part of the base of the skull. It is also called keystone bone of the cranium.

vi) Ethmoid bone (1) :
It is present on the midline of the anterior part of the cranial floor.

Question 2.
Write short notes on the ribs of human being.
Answer:
The ribs are thin, flat, curved bones that form a protective cage around the organs present in the human chest. They are comprised of 24 bones arranged in 12 pairs. These bones are divided into three categories :

1) True Ribs :
The first seven pairs of ribs are called true ribs. Dorsally, they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilages.

2) False Ribs:
The remaining ribs are called false ribs. The 8th, 9th and 10th pairs of ribs do not atriculate directly with the sternum, but joint the cartilaginous parts of the seventh rib. These are called vertebrochondral (or) false rib.

3) Floating Ribs :
Last two pairs (11th and 12th) of the ribs are not connected ventrally either to sternum or the anterior ribs, hence called floating ribs.
The thoracic vertebrae, ribs and sternum together form the rib cage.

Question 3.
List the bones of the human fore limb.
Answer:
Each fore limb of human is made of 30 bones. They are ;

Humerus :
Long bone in the fore limb that runs from shoulder to elbow.

Radius and Ulna :
These bones form forearm. It is the region betweeen elbow and the wrist.

Carpals :
These are the bones of wrist, eight in number.

Metacarpals :
The metacarpals form the skeleton of the palm. They are five in number.

Phalanges :
These are finger bones, fourteen in number, three for each finger and two for the thumb.

Question 4.
List the bones of the human leg.
Answer:
Each hind limb of human is made of 30 bones. They are ;

Femur :
Femur is the only bone in the thigh. It is the longest, heaviest and strongest bone in human body.

Tibia and fibula :
Both of these bones form lower leg i.e., the region from knee to ankle.

Tarsals :
These are ankle bones, seven in number.

Meta tarsals :
These are five short tubular bones, distal to the tarsals and proximal to phalanges.

Phalanges :
Foot has 14 phalanges, each toe has three phalanges, except for the first toe.

Patella :
It is a cup-shaped bone, covers the kneejoint vertically.

Question 5.
Draw a neat labelled diagram of the skeleton of the fore limb of man.
Answer:

Question 6.
Draw a neat labelled diagram of pelvic girdle.
Answer:

Question 7.
Describe the structure of synovial joint with the help of a neat labelled diagram.
Answer:
Synovial joints are characterised by the presence of a fluid filled synovial cavity between the articulating surfaces of the two bones.

Structure of synovial joint :
Synovial joint is covered by a double layered synovial capsule. The outer layer consist of dense fibrous irregular connective tissue with more collagen fibers. This layer is continuous with the periosteum and resists stretching and prevents the dislocation of joints. Some fibres of these membranes are arranged in bundles called ligaments.

The inner layer of synovial capsule is formed of areolar tissue and elastic fibers. It secretes a viscous synovial fluid which contains hyaluronic acid, phagocytes etc., and acts as a lubricant for the free movement of the joints.

Long Answer Questions

Question 1.
Describe the structure of human skull.
Answer:
The skull is the bony framework of the head. It is consist of the eight cranial and fourteen facial bones.

The cranial bones make up the protective frame of the bone around the brain called cranium.

The cranial bones are :
i) Frontal bone (1) :
It forms the forehead, anterior part of the cranial floor, and the roof of the orbits.

ii) Parietal bones (2) :
They form the major portion of the sides (left and right) and roof of the cranial cavity. They are joined to the frontal bone by a coronal suture and posteriorly to the occipital bone by lambdoid suture.

iii) Temporal bones (2) :
The left and right temporal bones form the lateral walls of the cranium as well as housing the external ear.

iv) Occipital bone (1) :
It forms the posterior part and the most of the base of cranium. It has large opening called foramen magnum. Medulla oblongata passes out through this foramen and joins the spinalcord.

v) Sphenoid bone (1) :
It is present at the middle part of the base of the skull. It is the keystone bone of the cranium, because it atriculates with all other cranial bones.

vi) Ethmoid bone (1) :
It is present on the midline of the anterior part of the cranial floor.

Facial region is made up of fourteen facial bones which form upper and lower jaw and other facial structures.

The facial.bones are :
i) Nasal bones (2):
These are paired bones that form the bridge of the nose.

ii) Maxillae (2) :
Two maxillae join together and form the upper jaw. Maxillae bears sockets for lodging the maxillary teeth.

iii) Zygomatic bones (2) :
These are known as cheek bones.

iv) Lacrimal bones (2) :
These are smallest bones of the face.

v) Palatine bones (2) :
They form the posterior portion of the hard palate.

vi) Nasal conchae (2) :
These are scroll like bones that form a part of lateral wall of the nasal cavity.

vii) Vomers (1) :
It is a triangular bone present on the floor of nasal cavity.

viii) Mandible (1) :
It is the lower jow bone. It is “U” shaped and is the longest and strongest of all the facial bones. It is the only movable skull bone.

Skeletal structures associated with sense organs :
i) Nasal cavity:
It is divided into left and right cavities by vertical partition called the nasal septum.

ii) Orbits:
These are bony depressions, which accommodate the eyeballs and associated structures.

iii) Ear ossicles :
Each middle ear contains three tiny bones, namely malleus, incus, stapes, collectively called ear ossicles.

iv) Hyoid bone :
It is a single U shaped bone present at the base of the buccal cavity between the larynx and the mandible. The hyoid bone keeps the larynx open.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 2(b) Excretory Products and their Elimination Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 2(b) Excretory Products and their Elimination

Very Short Answer Questions

Question 1.
Name the blood vessels that enter and exit the kidney.
Answer:
Renal artery enters kidney and renal vein comes out of the kidney.

Question 2.
What are renal pyramids and renal papillae?
Answer:
The conical-shaped medullary regions of kidney are the renal pyramids. Tips of the renal pyramids which open into pelvis are renal papillae.

Question 3.
What are the columns of Bertin?
Answer:
Columns of Bertin are the medullary extensions of the renal cortex in between the renal pyramids.

Question 4.
Name the structural and functional unit of kidney. What are the two main types of structural units in it?
Answer:
The structural and functional unit of kidney is ‘Nephrons’. The two main parts are
1) Malpighian body (renal corpuscle),
2) Convoluted tube.

Question 5.
Distinguish between cortical and juxta medullary nephrons.
Answer:

1. Cortical nephrons have renal corpuscle in the superficial renal cortex. They have short loop of Henle but without vasa recta.
2. Juxta medullary nephrons re located near the renal medulla. They have loop of Henle and vasa recta.

Question 6.
Define glomerular Alteration.
Answer:
The process of Alteration of blood, which occur between glomerulus and lumen of the Bowman’s capsule due to difference in net pressure is called glomerular Alteration. The filtered fluid which entered the Bowman’s capsule is primary urine or glomerular filtrate which is hypotonic.

Question 7.
Define glomerular Alteration rate (GFR)?
Answer:
The amount of filterate formed by both the kidneys, per minute is called glomerular Alteration rate (GFR). GFR in a healthy individual is approximately 125 ml 11 minute.

Question 8.
What is meant by mandatory reabsorption? In which parts of nephron does it occur?
Answer:
In a healthy individual the GFR is 125 ml/1 minute or 180 ltr per day. About 85% of the filterate formed is reabsorbed in a constant, unregulated fashion by the proximal convoluted tubule and descending limb of Henle’s loop, called mandatory reabsorption.

Question 9.
Distinguish between juxtaglomerular cells and macula densa.
Answer:

1. The cells of the distal convoluted tubule are crowded in the region where distal convoluted tubule makes contact with afferent arteriole. These cells are known as Macula densa.
2. Along side of macula densa, the wall of the afferent arteriole contains the modified smooth muscle fibers called juxtaglomerular cells.

Question 10.
Whait is Juxtaglomerular apparatus?
Answer:
Macula densa along with juxtaglomerular cells form juxtaglomerular apparatus which releases an enzyme Called renin.

Question 11.
Distinguish between the enzymes reniri and rennin.
Answer:
Renin :
Renin is an enzyme produced by the JG cells. This enzyme catalyses the conversion of angiotensinogen into angiotensin -I.

Rennin :
Rennin is also an enzyme found in the gastric juice of infants. It acts on the milk protein casein in the presence of calcium ions and convert it into calcium para caseinate and proteoses.

Question 12.
What is meant by the term osmoregulation?
Answer:
The process of maintaining the quantity of water and-dissolved solutes in balance is referred to as osmoregulation.

Question 13.
What is the role of atrial-natriuretic peptide in the regulation of urine formation?
Answer:
A large increase in blood volume promotes the release of atrial natriuretic peptide from the heart. Atrial natriuretic peptide decreases the absorption of water, Na⁺ from proximal convoluted tubule.

Short Answer Questions

Question 1.
Terrestrial animals are generally either ureotelic or uricotelic and not ammonotelic. Why?
Answer:
Ammonia is highly toxic and readily soluble in water, hence it should be eliminated from the body quickly in a very dilute solution.

Aquatic animals are surrounded by water, so water conservation is not a problem for them. In this manner, they are continuously eliminating ammonia.

On the other hand, terrestrial animals have to conserve water. They cannot waste water. So ammonia in diluted form can’t be eliminated continuously. Since ammonia is highly toxic, it has to be converted to less toxic form, like urea or uric acid.

Urea is 1,00,000 times less toxic than ammonia and requires less water for their excretion. Uric acid is less toxic than urea and being insoluble in water can be excreted as semi solid waste or pellets with very little water. This is the great advantage for animals with little access to water.

Question 2.
Differentiate vertebrates on the basis of the nitrogenous waste products they excrete, giving example.
Answer:
Vertebrates are divided into three categories based on nitrogenous waste excretory products. They are :
1) Ammonotelic animals:
The animals which excrete ammonia as nitrogenous waste products are called ammonotelic animals. These are aquatic organisms.
Ex : Some bony fishes.

2) Ureotelic animals:
The animals which excrete urea as their chief nitrogenous waste are called ureotelic animals.
Ex : Earth worms, cartilaginous fishes, most of the amphibians and mammals.

3) Uricotelic animals:
These animals excrete their nitrogenous waste products in the form of uric acid.
Ex : Reptiles, birds.

Question 3.
Draw labelled diagram of the V.S of kidney.
Answer:

Question 4.
Describe the internal structure of kidney of man.
Answer:

1. Kidney is bean shaped structure, the outer surface of kidney is convex and inner surface is concave where it has a deep notch called hilum.
2. A longitudinal sections of the human kidney shows two distinct regions namely the outer cortex and the inner medulla.
3. Medulla is divided into multiple cone shaped masses of tissue called renal pyramids. The renal pyramids are separated by the projections of the cortex called columns of Berlin.
4. The tips of the pyramids are renal papilla.
5. Renal papilla projects into cup like calyces, formed by the funnel shaped pelvis, which continues out as the ureter. Ureter carries urine into urinary bladder.
6. In cortex and medulla, nearly one million nephrons are present. They are structural and functional units of kidney. They are embedded in the loose connective tissue of cortex and medulla.
7. In addition, kidney contains a network of blood capillaries, lymph sinuses and intestitial fluid in intra cellular spaces.
8. The kidney gets blood supply through renal artery and blood from kidney is carried out by renal vein.

Question 5.
Explain micturition.
Answer:
The process of passing out of urine is called micro nutrition and the neural mechanism involved is called micturition reflex.

Urine is formed by the nephrons is ultimately carried to the urinary bladder where it is stored till a voluntary signal is given by the central nervous system (CNS). This sighal is initiated by the stretching of the urinary bladder as it gets filled with urine. In response, the stretch receptors on the walls of the bladder send signals to the CNS. The CNS passes on motor messages to initiate the contraction of smodth muscles of the bladder and simultaneous relaxation of the urethral sphincter, causing the release of urine.

Question 6.
What is the significance of juxta glomerular apparatus (JGA) in kidney function?
Answer:
Macula densa together with JG cells form juxtaglomerular apparatus (JAG). JAG plays a complex regulating role. A fall in glomerular blood flow or glomerular blood pressure or GFR can activate JG cells to release an enzyme called renin into the blood. This catalyses the conversion of angiotensinogen into angotensin-I which is further converted into angiotensin-II by angiotensin converting enzyme. Angiotensin-II, being a powerful vasoconstrictor increase the glomerular blood pressure and there by GFR.

Angiotensin-II also activates the adrenal cortex to release aldosterone. Aldosterone causes reabsorption of Na⁺ and water from distal convoluted tubule and collecting duct. To reduce loss through urine, and also promote secretion of K⁺ ions into distal convoluted tubule and collecting duct. It leads to increase in the blood pressure and GFR. This complex mechanism is generally known as renin – angiotensin- aldosterone system (RAAS).

Question 7.
Give a brief account of the counter current mechanism.
Answer:
Mammals have the ability to produce concentrated urine. The Henle’s loop and vasa recta plays an important role in this. The flow of the renal filterate in the two limbs of Henle’s loop is in opposite directions and thus form counter current. The flow of blood through vasa recta is also in counter current pattern. The proximity between the Henle’s loop and vasa recta, as well as the counter currents of renal fluid and blood in them help in maintaining an increasing osmolarity towards the inner medullary interstitium.

This gradient is mainly caused by NaCl and urea. NaCl passes out the ascending limb of Henle’s loop, and it enters the blood of the descending limb of vasa recta. NaCl is returned to the intestitium from the ascending portion of the vasa recta. Similarly small amounts of urea enter the thin segment of ascending limb of Henle’s loop which is transported back to the interstitium, from the collecting duct. Transport of these substances facilitated by the special arrangement of Henle’s loop and vasa recta.is called the counter current mechanism.

This mechanism helps to maintain a concentration gradient .in the medullary interstitium. Presehce of such interstitium gradient help easy passage of water from the collecting duct, there by concentrating the urine.

Question 8.
Explain the auto regulatory mechanism of GFR.
Answer:
Auto Regulation of GFR: *The kidneys have built in mechanisms for the regulation of glomerular filtration rate. One such efficient mechanism is carried out by juxta glomerular apparatus. Juxta glomerular apparatus is a special region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact.

A fall in GFR can activate the juxta glomerular cells to release an enzyme called renin, which catalyses the conversion of angiotensinogen into angiotensin-I and further converted to angiotensin-II by action of an enzyme angiotensin converting enzyme. Angiotensin-II’ stimulate the adrenal cortex to secrete aldosterone. Aldosterone causes reabsorption of Na+ and water from DCT and collecting duct to reduce loss through urine and also promotes the secretion of K+ ions into the DCT and CD (collecting duct). It leads.an increase in the blood pressure and GFR.

Question 9.
Describe the role of liver, lungs and skin in excretion.
Answer:
In addition to the kidneys, liver, lungs and skin also play an important role in the elimination of excretory wastes.

a) Liver :
Liver is the largest gland in our body, secretes bile, containing substances like bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins and drags. Most of these substances ultimately pass out along with digestive wastes.

b) Lungs :
Lungs remove large amounts of C02 (18 litres 1 day), various, volatile materials and significant quantities of water.

c) Skin :
Human skin possesses two types of glands namely sweat and sebaceous glands for the elimination of certain substances through their secretions.

• Sweat produced by the sweat glands is a watery fluid containing NaCl, small amount of urea, lactic acid etc.,
• Sebaceous glhnds eliminate certain substances like sterols, hydrocarbons and waxes through sebum. This secretion provides a protective oily covering for the skin.

Question 10.
Name the following.
Answer:
a) A chordate animal having protonephridial type excretatory structures Cephalo chordate.
b) Cortical portions projecting between the medullary pyramids in the human kidney. Columns of Bertini
c) Capillary network paralleing the loop of Henle. Vasa recta.
d) A non chordate animals having green glands as excretory structures. Crustaceans.

Long Answer Questions

Question 1.
Describe the excretory system of man, giving the structure of a nephron.
Answer:
In humans, the excretory system consists of a pair of kidney, a pair of ureters, a urinary bladder and urethra.

Kidney :
Kidneys are reddish brown, bean shaped structures, situated on either side of the vertebral column between the levels of last thoracic and third lumbar vertebrae in a retroperitoneal position. The right kidney is slightly lower than the left one due to the presence of liver.

The outer surface of the kidney is convex and the inner surface is concave, where it has a deep notch called hilum, the point at which the renal artery and nerves enter and renal vein and ureter leave. Each kidney is surrounded by a tough, fibrous tissue, called renal capsule.

Ureter :
These are slender whitish tubes, which emerges from the pelvis of the kidney. The ureter rundown and open into the urinary bladder.

Urinary bladder :
Urinary bladder is a pear shaped like muscular organ. It tempirarily stores the urine, situated in the lower abdominal cavity. The neck of the bladder leads into the urethra. Urethra opens near the vaginal orifice in the female and through the penis in the male.

Structure of nephron:
Each kidney has nearly one million nephrons. These are structural and functional units of kidney, embedded in the loose connective tissue of cortex and medulla. Nephron consist of malpighian body and renal tubule.

I) Malphigian body :
It begins in the cortex of the kidney. It contains Bowman’s capsule and glomerulus.

a) Bowman’s capsule :
It is a thin walled, double layered cup. The inner wall of the Bowman’s capsule has certain unique cells called podocytes.

b) Glomerulus :
It is a dense network of capillaries in the cup of Bowman’s capsule. Afferent arteriole of renal artery enter the cavity of Bowman’s capsule and split into five branches.

They unite and come out of the Bowman’s capsule as an afferent arteriole.

The podocytes of inner wall of Bowman’s capsule wrap around each capillary. The podocytes are arranged in an intricate manner so as to leave some minute spaces called filteration slits. The endothelium cells of capillaries have numerous pores called fenestrations.

II) Renal tubule:
It is narrow, delicate tubule arises from the posterior part of Bowman’s capsule known as neck. It opens into along narrow convoluted tubule with three parts like proximal convoluted tubule, Loop of Henle and Distal convoluted tubule.

a) Proximal convoluted tubule :
It is a lined by simple cuboidal epithelium with brush border to increase area of absorption.

b) Loop of Henle :
It is a hairpin like tubule present in medulla region. It consist of a descending limb and an ascending limb. The proximal part of the ascending limb is thin and the distal part is thick. The thick ascending limb continuous into the distal convoluted tubule.

c) Distal convoluted tubule (DCT) :
It is present in cortex. It is lined by simple cuboidal epithelium. The DCT continuous as the initial collecting duct in the cortex.

Collecting system :
Some initial collecting ducts unite to form straight collecting duct, which passes through the medullary pyramid. In the medulla, the tubes of each pyramid join and form duct of Bellini, which finally opens into tip of the renal papilla.

Capillary network of nephron :
The efferent arteriole emerging from the glomerulus forms a fine capillary network called the peritubular capillaries, around the renal tubule. The portion of the peritubular capillaries that surrounds the loop of Henle is called the vasa recta. The vasa recta is absent or highly reduced in the cortical nephrons. The juxta medullary nephrons possess well developed yasa recta.

Question 2.
Explain the physiology of urine formation.
Answer:
The formation of urine involves three main processes namely

1. Glomerular Alteration
2. Selective reabsorption
3. Tubular secretion.

1) Glomerular Alteration :
It is Arst step in urine formation. The process of Alteration of blood, which occurs between glomerulus and lumen of the Bowman’s capsule due to difference in netpressure is called glomerular Alteration.

The hydrostatic pressure of blood while Aowing in the glomerulus is 60 mm Hg. It is opposed by glomerular colloidal osmotic pressure of 32 mm Hg and Bowman’s capsule hydrostatic pressure of 18 mm Hg.

The net filterate pressure is 10 mm Hg ( 60 – 32 + 18 = 10). This causes the Alteration of blood through the 3 layered filterate membrane formed by endothelium cells of glomerular capillary together with the basement membrane and podocytes of the Bowman’s cup. By the result of glomerular Alteration primary urine or renal Auid is collected in lumen of the Bowman’s capsule.

The primary urine contains almost all the constituents of plasma, except the proteins. The primary urine is hypotonic to the cortical Auid, it passes into the next part of renal tubule.

2) Selective reabsorption:
During the process of glomerular Alteration 125 ml/minute df primary urine is formed. Nearly 99% of which and essential substances are reabsorbed* by renal tubules called selective reabsorption. About 85% of filterate formed (primary urine) is reabsorbed in a constant unregulated manner called obligatory reabsorption.

3) Tubular Secretion :
During the formation of urine, the tubular cells secrete substances such as H⁺, K⁺ and NH₃⁺ into the filterate. Tubular secretion is also an important step in the formation of urine as it helps in maintenance of ionic and acid base balance of the body fluids.

Mechanism of selective reabsorption and secretion takes place is different parts of nephrons.

a) In the proximal convoluted tubule :
Nearly all the essential nutrients and 70-80% of electrolytes and water are reabsorbed by this segment. Na⁺, glucose, amino acids, Cl^– and other essential substances are reabsorbed into blood.

PCT also helps to maintain the pH and ionic balance of body fluids by selective secretion of H⁺ and NH₃ into the filterate and by the absorption of HCO₃^– from it.

b) In the Henle’s loop :
Reabsorption in this segment is minmium.

• The descending loop of Henle is permeable to water and almost impermeable to electrolytes results the filterate concentration gradually increases.
• The ascending limb has two specialized regions, a proximal thin segment in which NaCl diffuses out into interstitial fluid passively, and distal thick segment, in which NaCl is actively pumped out.

The ascending limb is impermeable to water. Thus the filterate becomes progressively more dilute as it moves up to the cortex i.e., towards the DCT.

In the Distal convoluted tubule (DCT) :
It is permeable to water and ions. The reabsorption of water is variable depending on several conditions and is regulated by ADH. DCT is also capable of reabsorption of HCO₃^– and selective secretion of H⁺ and K⁺ ions and NH₃⁺ into DCT from peritubular network, to maintain the pH and sodium – potassium balance in tHe blood.

In the collecting duct (CD) :
Considerable’amount of water could be reabsorbed from this region to produce concentrated urine. This segment allows passage of small amount of urea to medullary interstitium to keep up its osmolarity. It also plays a role in the maintenance of pH and ionic balance of blood by the selective secretion of H⁺ and K⁺ ions.

The renal fluid after the process of facultative reabsorption in the CD, influenced by ADH, constitute the urine, that is sent out. Urine in the CD is hypertonic to the plasma of blood.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 2(a) Body Fluids and Circulation Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 2(a) Body Fluids and Circulation

Very Short Answer Questions

Question 1.
Write the differences between open and closed systems of circulation.
Answer:

Open circulation system	Closed circulation system
1. In this type, blood flows from the heart into the arteries and then into large spaces called sinuses.	1. In this type blood flows through a series of blood vessels.
2. Organs located in the space are bathed by blood.	2. Each organ has blood vessels that carry blood to it.
3. Blood flows slowly because there is no blood pressure after the blood leaves the blood vessels.	3. Blood flows at a high speed because there is high blood pressure after the blood leaves the heart.
4. It is found in Leeches, arthropods, and mollusks.	4. It is found in annelids and chordates.

Question 2.
The sino-atrial node is called the pacemaker of our heart. Why?
Answer:
A sino-atrial node consists of specialized cardiomyocytes. It has the ability to generate action potentials without any external stimuli hence called pacemaker.

Question 3.
What is the significance of the atrioventricular node and atrioventricular bundle in the functioning of the heart?
Answer:
Atrio-ventricular node and atrioventricular bundle plays an important role in the contraction of the ventricles.

Aricular contraction initiated by the wave of excitation from sino-atrial node (SAN) stimulate the atrio-ventricular node from where they are conducted through the bundle of His (atrio-ventricular bundle), its branches and Purkinje fibers to the entire ventricular musculature. This causes the stimulation ventricular systole. It lasts about 0.3 sec.

Question 4.
Name the valves that guard the left and right atrio-ventricular apertures in man.
Answer:
Bicuspid valve (or) Mitral valve – Left atrio-ventricular aperture.
Tricuspid valve – Right atrio-venticular aperture.

Question 5.
Where is the valve of Thebesius in the heart of man.
Answer:
Opening of coronary sinus into left precaval vein is bound by a crescentic fold known as valve of Thebesius.

Question 6.
Name the aortic arches arising from the ventricles of the heart of man.
Answer:

1. Pulmonary arch – arises from right ventricle.
2. Left systemic arch – arises from left ventricle.

Question 7.
Name the heart sounds when they are produced.
Answer:
The lub-dup sounds are produced by heart. The first sound ‘lub1 is caused by closure of the1 AV valves at the beginning of ventricular systole and preventing the back flow of blood. The second heart sound ‘dup’ results from the closure of the semilunar valves at the beginning of ventricle diastole and prevents the back flow of blood.

Question 8.
Define cardiac cycle and cardiac output.
Answer:
Cardiac cycle :
Cardiac events that occur from the beginning of one heart beat to the beginning of the next is called cardiac cycle.

Cardiac output:
The volume of blood pumped out by the heart from each ventricle per minute is termed cardiac output. It is approximately 5 litres.

Question 9.
What is meant by double circulation? What is its significance?
Answer:
The double circulation system of blood flow refers to the separate systems of pulmonary circulation and the systemic circulation. All animals with lungs have a double circulatory system.

In pulmonary circulation deoxygenated blood is pumped away from the heart, via pulmonary artery to the lungs and returns oxygenated blood to the heart via pulmonary vein.

In systemic circulation oxygenated blood away from heart to the rest of the body and returns deoxygenated blood back to the heart.

Question 10.
Why the arteries are more elastic than the vein?
Answer:
Arteries are more elastic than vein because they are structurally designed to withstand tremendous blood pressures.

Veins on the other hand, contain blood at relatively low blood pressure.

Short Answer Questions

Question 1.
Describe the evolutionary change in the structural pattern of the heart among the vertebrates.
Answer:
1) Fishes have the 2-chambered heart with an atrium and a ventricle. Blood passes through the heart only once in a complete circuit hence called single circulation. This means there is no separate circulation for oxygenated and deoxygenated blood.

2) Amphibians have a 3 – chambered heart with two atria and one ventricle, which further evolved in, reptiles, have two atria and an incompletely divided ventricle in which left atrium receives oxygenated blood from the gills / lungs / skin and right atriupi receives blood from the other parts of the body. The two types of, blood get’ mixed in the single ventricle, which pumps out mixed type of blood. Thus these animals show complete double circulation.

3) Birds and mammals possess 4-chambered heart with two atria and two ventricles. In these animals the oxygenated and the deoxygenated types of blood received by left and right atria, passes on to the left and right ventricles, respectively. The ventricles pump the blood out without any mixing of the oxygenated and deoxygenated types of blood. Hence these animals are said to be showing double circulation namely systemic arrd pulmonary circulations.

Question 2.
Describe atria of the. heart of man.
Answer:
Atria are thin walled receiving chambers, form the anterior part of the heart. The right one is larger than the left, they are separated by inter-atrial septum. It has small pore in embryonic stage known as Foramen Ovale. Later it is closed and appears as a depression in the septum known as Fos&a ovalis. If the foramen ovale does not close properly, it is called a patent foramen ovale.

The right atrium receives deoxygenated blood from different parts of the body (except the lungs) through three caval veins like two precaval veins and one post caval vein. The right atrium also receives blood from the walls of the heart through the coronary sinus, whose opening into the right atrium is guarded by a crescentric fold, the valve of Thebesius. Opening of the post caval vein is guarded by the valve of inferior vena cavae or Eustachian valve. It directs the blood to the left atrium through the foramen ovale, in the fetal stage, but in the adults it becomes non functional.

The openings of the precaval veins into the right atrium have no valves. The left atrium receives oxygenated blood from lungs through a pair of pulmonary veins, which opens into the left atrium through a common pore. Atrio-ventricular septum separates atria and ventricles. It has right and left atrio-ventricular apertures.

Tricuspid valve guards the right atrio-ventricular aperture and bicuspid valve (mitral valve) guards the left atrio-ventricular aperture.

Question 3.
Describe the ventricles of the heart of man.
Answer:
Two ventricles right and left form the posterior part of the heart. These are the thick walled blood pumping chambers, separated by inter-ventricular septum. The wall of the left ventricle is thicker than that of the right ventricle. The inner surface of ventricles is raised into muscular ridges or columns known as columnae carneae projecting from the inner walls of the ventricles. Some of them are large and conical and known as papillary muscles. Collagenous cords are known as chordae tendineae are present between atrio-ventricular valves and papillary muscles. They prevent the cusps of the antrio-ventricular valves from bulging too far into atria during ventricular systole.

Question 4.
Draw a labelled diagram of the L.S of the heart of man.
Answer:

Question 5.
Describe the events in a cardiac cycle, briefly.
Answer:
The cardiac events that occur from the beginning of one heart beat to the beginning of the next, is called cardiac cycle. Cardiac cycle consists of three phases namely atrial systole, ventricular systole and cardiac diastole.

i) Atrial systole: It lasts about 0.1 seconds.
→ The SAN generate an action potential which stimulate contraction of atria, which helps in the flow of blood into ventricles by about 30%. The remaining blood flows into the ventricles before the atrial systole.

ii) Ventricular systole : It lasts about 0.3 seconds
→ Ventricles contract and atria relax during this phase.
→ Contraction of ventricles raises the pressure in ventricles due to which AV valves are closed. It causes the first heart sound “Lub”.
→ When pressure in ventricles exceeds the pressure in aortic arches, semilunar valves open. It results the flow of blood from ventricles into aortic arches.

iii) Cardial diastole : It lasts about 0.4 seconds.
→ The ventricles now relax, atria are also in diastolic condition.
→ When pressure in ventricles falls below that in aortic arches, semilunar valves are closed.
→ It causes the second heart sound “dup”.

When pressure in ventricles falls below atrial pressure, AV valves open and ventricular filling begins. The total cycle takes about 0.8 seconds. This gives a heart rate of about 75 beats per minute.

Question 6.
Explain the mechanism of clotting of blood.
Answer:
When a blood vessel is injured a number of physiological mechanisms Eire activated that promote hemostasis, and stops bleeding. Blood clots within 3-6 minutes after damage of a bloodvessel.

Mechanism of blood clotting: Blood clotting takes place in three essential steps, i) Formation of prothrombin activator : It is formed by two pathways.

a) Intrinsic pathway:
It occurs when the blood is exposed to collagen of injured wall of blood vessel. This activates factor XII, and in turn it activates another clotting factor, which activates yet another reaction, which results in the formation of prothrombin activator.

b) Extrinsic pathway:
It occurs when the damaged vascular wall or extra vascular tissue comes into contact with blood. This activates the release of tissue thromboplastin, from the damaged tissue. It activates the factor VII. As a result of these cascade reactions, the final product formed is the prothrombin activator.

ii) Activation of prothrombin:
The prothrombin activator, in the presence of sufficient amount of Ca²⁺, causes the convertion of inactive prothrombin to active thrombin.

iii) Convertion of soluble fibrinogen into fibrin:
Thrombin converts the soluble protein fibrinogen into soluble, fibrin monomers, which are held together by weak hydrogen bonds. The factor XIII replaces hydrogen bonds with covalent bonds and cross links the fibers to form a meshwork and prevent the blood bleeding.

Question 7.
Distinguish between SAN and AVN.
Answer:
Sino-atrial node (SAN) :
It is present in the right upper comer of the right atrium. It is called pacemaker because it generates impulses for beating of heart. The action potential from SAN, stimulate, both atria which causes them to contract. Simultaneously causing the atrial systole. It lasts for 0.1 second.

Atrio ventricular node (AVN) :
It is seen in the lower left corner of the right atrium. AV node is a relay point that relays the action potential received from the SA node to the ventricular musculature through the bundle of His, its branches and Purkinje fibers. This causes the simultaneous ventricular systole. It lasts for about 0.3 seconds.

Question 8.
Distinguish between arteries and veins.
Answer:

Arteries	Veins
1. Arteries carry oxygenated blood, away from the heart except pulmonary artery.	1. Veins carry deoxygenated blood towards the heart except the pulmonary veins.
2. These are bright red in colour.	2. These are dark red in colour.
3. These are mostly deeply seated in the body.	3. Veins are generally superficial.
4. Arteries are thick walled,with elastin and highly muscular.	4. Veins are thin walled and slightly muscular.
5. These possess narrow lumen.	5. These possess wide lumen.
6. Valves are absent.	6. Valves are present whiqh provide undirectional flow of blood.
7. Blood in the arteries flow with more pressure and by jerks.	7. Blood in the veins flow steadily with relatively low pressure.
8. Arteries end in capillaries.	8. Veins start with capillaries.
9. Arteries empty up at the time of death.	9. Veins get filled tip at the time of death.

Long Answer Questions

Question 1.
Describe the structure of the heart of man with the help of neat labelled diagram.
Answer:
Human heart is a hallow muscular, cone shaped, and pulsating organ situated between lungs. It is about the size of a closed fist.

The heart is covered by double walled pericardium, which consists of outer fibrous pericardium and inner serous pericardium. The serous pericardium is double layered, outer parietal layer and inner visceral layer. These two layers are separated by pericardial space, which is filled with pericardial fluid. This fluid reduces friction between the two membranes and allow free movement of the heart.

Human heart has four chambers with two smaller upper chambers called atria and two larger lower chambers called ventricles. Atria and ventricles are separated by a deep transverse groove called coronary sulcus.

i) Atria :
→ Atria are thin walled receiving chambers. The right one is larger than the left.

→ The two atria are separated by thin inter-atrial septum. It has a small pore known as Foramen Ovale in fetal stage Later it is closed and appears as depression (oval patch) known as ‘Fossa ovale’. If the foramen ovale does not close properly it is called a patent foramen ovale.

→ The right atrium receives deoxygenated blood from different parts of the body, through three caval veins like two precaval veins and one post caval vein.

→ The right atrium also receives blood from wall of the heart through coronary sinus, whose opening into the right‘atrium is guarded by the valve of Thebesius.

→ Opening of the post caval vein is guarded by the Eustachian valve. It is functional in fetal stage and directs the blood from post caval vein into left atrium thrdugh foramen ovale. But it is non-functional in adult.

→ The openings of the precaval veins into the right atrium have no valves.

→ Left atrium receives oxygenated blood from lungs through a pair of pulmonary veins, which open into the left atrium through a common pore.

→ Atrio-ventricular septum separates atria and ventricles. It has right and left atrio- venticular aperture’s.

→ Tricuspid valve guards the right atrio-ventricular aperture. Bicuspid valve guards the left atrio-ventricular aperture.

ii) Ventricles :
→ These are the thick walled blood pumping chambers, separated by an interventricular septum. The wall of the left ventricle is thicker than that of the right ventricle as the left ventricle must force the blood to all the parts of the body.

→ The inner surface of the ventricles is raised into muscular ridges called columnae cameae. Some of them are large and conical and known, as papillary muscles. Collagenous cords are known as chordae tendinae are present between atrio-ventricular valves and papillary muscles. They prevent the cusps of valves from bulging too far into atria during ventricular systole.

Nodal tissue :
A specialized cardiac musculature called the nodal tissue is also distributed in the heart.

1. Sino-artrial node (SAN) – Present in the right upper corner of right atrium.
2. Atrio-ventricular node (AVN) – Present in the lower left comer of right atrium.

iii) Aortic arches :
Human heart has two aortic arches.
1) Pulmonary arch :
Arises from the left anterior angle of the right ventricle. It carries deoxygenated blood to lungsf. It’s opening from right ventricle is guarded by pulmonary Valve made with 3 semiluminar valves.

2) Left systemic arch :
Arises from the left ventricle to distribute oxygenated blood tovarious pahs in the body. Its opening is also guarded by aortic valve made with a set of 3 semilunar valves.

A fibrous strand, known as ligamenturri arteriosm is present at the point of contact of the systemic and pulmonary arches. It is the remnant of the ductus arteriosus, which connects the systemic and pulmonary arches in the embryonic stage.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 1(b) Breathing and Exchange of Gases Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 1(b) Breathing and Exchange of Gases

Very Short Answer Questions

Question 1.
Define vital capacity. What is its significance?
Answer:
Vital capacity :
The maximum volume of air a person can breathe in after forced expiration. This includes ERV (Expiratory Reserve Volume), TV (Tidal Volume), and IRV (Inspiratory Reserve Volume) (or) the maximum volume of air a person can breathe out after forced inspiration (VC = TV + IRV + ERV).

Question 2.
What is the volume of air remaining in the lungs after a normal expiration?
Answer:
The volume of air that remains in the lungs after a normal expiration is called ‘Functional Residual Capacity (FRC)’.
FRC = ERV + RV
ERV = 1000 to 1100 ml
RV = 1100 to 1200 ml. So
FRC = 2100 to 2300 ml.

Question 3.
Diffusion of oxygen occurs in the alveolar region only and not in other parts of respiratory system. How do you justify the statement?
Answer:
Alveoli are primary sites of exchange of gas by simple diffusion. Aleveolar region is having enough pressure gradient to facilitate diffusion of gases. Other regions of the respiratory system doesn’t have the required pressure gradient.

High pO₂, low pCO₂. lesser H⁺ concentration, low temperature conditions in alveoli favourable for diffusion of O₂ ahd formation of oxyhaemoglobin. Solubility of gases as well as thickness of the membrane are also some of the important factors that can effect the ratio of diffusion.

Question 4.
What is the effect of pCO₂ on oxygen transport?
Answer:
pCO₂ plays an important role in the transport of oxygen. At the alveolus, the low pCO₂ and high pO₂ favours the formation of oxyhaemoglobin. At the tissues, the high pCO₂ and low pO₂ favours the dissociation of oxygen from oxyhaemoglobin. Hence, the affinity of haemoglobin for oxygen is enhanced by the decrease of pCO₂ in blood. Therefore, oxygen is transported in blood as oxyhaemoglobin and oxygen dissociates from it at the tissues.

Question 5.
What happens to the respiratory process in man going up a hill?
Answer:
When a man is going up a hill or doing some strenous exercise then there is more consumption of oxygen and resulting in more demand of oxygen. As a result there is an increased breathing rate to fill the gap.

Question 6.
What is tidal volume? Find out the tidal volume in a healthy human, in an hour?
Answer:
Tidal Volume (TV) :
Volume of air inspired (or) expired during normal inspiration (or) expiration. It is approximately 500 ml i.e., a healthy man can inhale (or) exhale approximately 6000 to 8000 ml of air per minute (or) 3,60,000 to 4,80,000 ml per hour.

Question 7.
Define oxyhaemoglobin dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Answer:
The oxyhaemoglobin dissociation curve is a graph showing the percentage of oxyhaemoglobin at various partial pressures of oxygen.

Reasons for Sigmoidal pattern :
In alveoli, where there is a high pO₂, low pCO₂ lesser H⁺ and low temperature, the factors are all favourable for formation of oxyhaemoglobin. In the tissues where low pO₂, high pCO₂, high H⁺ concentration and higher temperature exist, the conditions are favourable for dissociation of oxygen from oxyhaemoglobin under these conditions. Oxygen dissociation curve shift away from the Y-axis and form sigmoid curve.

Question 8.
What are conchae?
Answer:
These are curved bones that make up the upper portion of the nasal cavity. There are different conchae in the nose, such as interior concha, medial concha and superior concha. The nasal concha bones are also referred’to as turbinate pones.

Question 9.
What is meant by chloride shift?
Answer:
Chloride shift:
It refers to the exchange of chloride and bicarbonate ions between erythrocytes and plasma. It is also called Hamburger’s phenomenon.

Question 10.
Mention any two occupational respiratory disorders and their causes in human beings?
Answer:
Occupational respiratory disorders ate caused by exposure of the body to the harmful substances.
E.g.:
1) Asbestosis:
It occurs due to chronic exposure to asbestos dust in the people Working in asbestos factory.

2) Silicosis :
It occurs because of long term exposure to ‘silica dust’ in the people working in mining industries, quarries etc.,

Question 11.
Name the muscles that help in normal breathing movements?
Answer:
Muscles of diaphragm and external inter-costal muscles help in the process of normal breathing movements.

Question 12.
Draw a diagram of oxyhaemoglobin dissociation curve?
Answer:

Short Answer Questions

Question 1.
Explain the process of inspiration and expiration under normal conditions.
Answer:
Inspiration : Intake of atmospheric air into the lungs is called inspiration. It is an active process, as it takes place by the contraction of the muscles of the diaphragm and the external inter-costal muscles which extend in between the ribs. The contraction of diaphragm increases the volume of thoracic chamber in the anterio posterior axis. The contraction of external inter costal muscles lifts up the ribs and sternum causing an increase in the dorso- ventral axis.

The overall increase in the thoracic volume causes a similar increase in the pulmonary volume. An increase in the pulmonary volume decreases the intra-pulmonary pressure to less than that of the atmosphere, which forces the air from the outside to move into the lungs, that is inspiration.

Expiration :
Release of alveolar air to the exterior is called expiration. It is a passive process. Relaxation of the diaphragm and external inter-costal muscles returns the diaphragm and sternum to their normal positions, and reduces the thoracic volume and thereby the pulmonary volume. This leads to an increase in the intra-pulmonary pressure to slightly above that of the atmospheric pressure, causing the expulsion of air from the lungs, that is called expiration.

Question 2.
What are the major transport mechanisms for CO₂? Explain.
Answer:
Carbondioxide is transported in three ways.
1. In dissolved state :
7% of CO₂ is transported in dissolved state through plasma.
CO₂ + H₂O → H₂CO₃.

2. As Carbamino compounds:
About 20-25% of CO₂ combine directly with free amino group of haemoglobin and forms Carbamino haemoglobin in a reversible manner.
Hb – NH₂ + CO₂ → Hb – NHCOO^– + H⁺.
pCO₂ and pO₂ could affect the binding of CO₂ to haemoglobin.
— when pCO₂ is high and pO₂ is low as in the tissues, binding of more CO₂ occurs.
— when pCO₂ is low and p02 is high as in the alveoli, dissociation of CO₂ carbamino
– haemoglobin takes place, (i.e., CO₂ which is bound to haemoglobin from the tissues is delivered at the alveoli)

3. As Bicarbonates :
About 70% of CO₂ is transported as bicarbonate. RBCs contain a very high concentration of the enzyme, carbonic anhydrase and a minute quantity of the same is present in plasma too. This enzyme facilitates the following reaction in both the directions.

At the tissues where partial pressure of CO₂ is high due to catabolism, CO₂ diffuses into the blood and forms carbonic acid which dissociates into HCO^–₃ + H⁺

At the alveolar site where pCO² is low, the reaction proceeds in the opposite direction leading to the formation of CO² and water. Thus CO² is mostly trapped as bicarbonate at the tissues and transported to the alveoli where it is dissociated out as CO².

Every 100 ml of deoxygenated blood delivers approximately 4 ml of CO₂ to the alveolar air.

Question 3.
How is respiratory movements regulated in man?
Answer:
In human beings the respiratory movements are regulated by neural system.
1. A special centre present in the medulla region of brain, called ‘respiratory rhythm centre’ is primarily responsible for this regulation.

2. Another centre present in the pons of the brain stem called ‘pneumotaxic centre’ can moderate the functions of the respiratory rhythm centre. Neural signal from and this centre can reduce the duration of inspiration and thereby alter the respiration rate.

3. A chemo-sensitive area is situated adjacent to the respiratory rhythm centre which is highly sensitive to CO₂ and H⁺. Increase in these substances can activate this centre, which inturn can send signals to the respiratory rhythm centre to make necessary adjustments in the respiratory process by which these substances can be eliminated.

4. Receptors associated with aortic arch and carotid artery also recognize changes in CO₂ and H⁺ concentration and send necessary signals to the respiratory rhythm centre for necessary actions.

The role of oxygen in the regulation of the respiratory rhythm is quite insignificant.

Question 4.
Distinguish between a) IRV and ERV b) Inspiratory capacity and Expiratory capacity c) Vital capacity and Total lung capacity.
Answer:
a) IRV and ERV:
IRV (Inspiratory Reserve Volume) :
The maximum volume of air that can be inhaled during forced breathing, in addition to the tidal volume. This is about 2500 ml to 3000 ml.

ERV (Expiratory Reserve Volume) :
The maximum volume of air that can be exhaled during forced breathing in addition to the ‘tidal volume’. This is about 1000 ml to 1100 ml.

b) Inspiratory capacity and Expiratory capacity :
Inspiratory capacity (IC) :
The total volume of air, a person can inhale after normal expiration’. This includes tidal volume and inspiratory reserve volume.
IV = TV + IRV
It is about 3000 ml to 3500 ml.

Expiratory capacity (EC) :
The total volume of air, a person can expire after a ‘normal inspiration’. This includes tidal volume and expiratory reserve volume.
EC = TV + ERV

c) Vital capacity and Total lung capacity:
Vital capacity (VC) :
The maximum volume of air a person can breathe in after ‘forced expiration’. This includes ERV TV and IRV (or) the maximum volume of air, a person can > breathe out after forced inspiration.
VC = TV + IRV + ERV

Total lung capacity (TLC) :
The total volume of air accommodated in the lungs at the end of forced inspiration.
This includes RV ERV, TV and IRV
TLC = ERV + IRV + TV + RV (or)

Question 5.
Describe disorders of respiratory system.
Answer:
Disorders of respiratory system.

1) Asthma:
Asthma is a difficulty in breathing caused due to inflammation of bronchi and bronchioles. Symptoms include coughing, difficulty in breathing and wheezing.

2) Emphysema:
It is a chronic disorder in which alveolar walls are damaged and their walls coalesce due to which respiratory surface area of exchange of gases is decreased. One of the major causes of this

3) Bronchitis :
Bronchitis is the inflammation of the bronchi, resulting in the swelling of mucus lining of bronchi, increased mucus production and decrease in the diameter of bronchi. Symptoms include chronic cough with thick sputum.

4) Pneumonia :
The infection of lungs caused by Streptococcus pneumoniae and also by certain Virus, Fungi, Protozoans and Mycoplasmas. Symptoms include inflammation of lungs, accumulation of mucus in alveoli and impaired exchange of gases, leading to death if untreated.

Occupational dissorders :
These are caused by exposure of the body to the harmful substances.
E.g.:
i) Asbestosis:
It occurs due to chronic exposure to asbestos dust in the people working in asbestos industry.

ii) Silicosis :
It occurs because of long term exposure to silica dust.

iii) Siderosis :
It occurs due to deposition of iron particles in tissues.

iv) Black lung disease :
It develops from inhalation of coal dust.

Long Answer Questions

Question 1.
Describe the respiratory system in man.
Answer:
The human respiratory system composed of external nostrils, nasal chambers, nasopharynx, larynx, trachea, bronchi, bronchioles and lungs. It is responsible for the process of respiration that is vital to the survival of living beings.

1) External nostrils :
A pair of external nostrils opens out above the upper lip. They lead into nasal chambers through the nasal passages.

2) Nasal chambers:
They lie above the palate and are separated from each other by a nasal septum. Each nasal chamber can be differentiated into three parts gamely; i) Vestibular part – which has hair and sabaceous gland’s to prevent the entry of dust particles, ii) Respiratory part – involved in the conditioning the temperature, iii) Olfactory part – is fined by an Olfactory epithelium.

3) Naso-pharynx :
Nasal chambers lead into nasopharynx through a pair of internal nostrils. Nasopharynx is a portion of pharynx, the common chamber for the passage of food and air. Nasopharynx leads into oropharynx, and opens through glottis of larynx into the trachea.

4) Larynx :
This is also called voice box or Adam’s apple, connects the pharynx with the trachea. Larynx is the organ of voice as well as an air passage extending from the root of the tongue to the trachea. It is well developed in man. It consist of a) Vocal cord b) Glottis c) Epiglottis.
a) Vocal cord : These are muscular folds that projects from lateral walls.
b) Glottis : Narrow passage between the true and false vocal cords of the larynx.
c) Epiglottis : It is a thin leaf like elastic cartilaginous flap attached to the thyroid cartilage to prevent the entry of food into the larynx through the glottis.

5) Trachea :
Trachea is also called windpipe. It is a straight tube extending upto the mid-thoracic cavity. The wall of the trachea is supported by 16-20 ‘C’ shaped rings of hyaline cartilage. These rings are incomplete dorsally and keep the trachea always open preventing collapse. Internally the trachea is lined by pseudostratified ciliated epithelium.

6) Bronchi and Bronchioles :
On entering the mid thoracic cavity, trachea divides into right and left primary bronchi. Each primary bronchus enters the corresponding lung and divides into secondary bronchi that further divides into tertiary bronchi. Each tertiary bronchus divides and redivides to form primary, secondary, tertiary, terminal and respiratory bronchioles. Each respiratory bronchiole terminates in a cluster of alveolar ducts which ends in alveolar sacs.

7) Lungs :
These are paired, situated in the thoracic chamber which is anatomically an air tight chamber. Lungs are covered by a doubled layered pleura with pleural fluid between them. It reduces friction on the lung surface. The outer pleural membrane is in close contact with the thoracic lining where as the inner pleural membrane is in contact with lung’s surface. The part starting with external nostrils upto the terminal bronchioles constitute the conducting part, whereas the alveoli and their ducts form the respiratory or exchange part of respiratory system. The conducting part transports the atmospheric air to the alveoli, clears it from foreign particles, humidifies and also bring the inhaled air to the body temperature. Exchange part is the site of actual diffusion of and between blood and atmospheric air.

Question 2.
Write an essay on the transport of oxygen and carbondioxide by blood.
Answer:
Blood is the medium for the transport of oxygen and carbondioxide.

Transport of oxygen :
Oxygen is transported from the lungs to the tissues through the plasma and RBC of the blood. 100 ml of oxygenated blood can deliver 5 ml of O₂ to the tissues under norpial condtions.

i) Transport of oxygen through plasma:
About 3% of O₂ is carried through the blood plasma in dissolved state.

ii) Transport of oxygen by RBC :
about 97% of oxygen is transported by the . haemoglobin of RBC in the blood. Haemoglobin is a red coloured iron containing pigment present in the RBCs. Each haemoglogin molecule can carry a maximum of four molecules of oxygen. Binding of oxygen with haemoglobin is primarily related to the partial pressure of O₂. At lungs, where the partial pressure of O₂ is high, oxygen binds to haemoglobin in a reversible manner to form oxyhaemoglobin. This is called oxygenation of haemoglobin.
Hb + 4O₂ -» Hb (O₂)₄.

At the tissues, where the partial pressure of O₂ is low oxyhaemoglobin dissociates into haemoglobin and oxygen. The other factors such as partial-pressure of CO₂, H⁺ concentration (pH), and the temperature influence the binding of oxygen with haemoglobin. For example in alveoli high pO₂, low pCO₂ high H⁺ concentration lower temperature are favourable for formation of oxyhaemoglobin. In tissues low pO₂, high pCO₂ high H⁺ concentration and high temperature conditions are favourable for. dissociation, of oxygen from oxyhaemoglobin.

Transport of Carbondioxide:
Carbondioxide is transported in three ways,
1. In dissolved state :
7% of CO₂ is transported in dissolved state through plasma.
CO₂ + H₂O → H₂CO₂.

2. As Carbamino compounds:
About 20-25% of CO₂ combine directly with free amino group of haemoglobin and forms Carbmino haemoglobin in a reversible, manner.
Hb – NH₂ + CO₂ → Hb – NHCOO^– +H⁺.

pCO₂ and pO₂ could affect the binding of CO₂ to haemoglobin.

— when pC02 is high and pO₂ is low as in the tissues, binding of more CO₂ occurs.
— when pCO₂ is low and pO₂ is high as in the alveoli, dissociation of CO₂ carbamino – haemoglobin takes place, (i.e., CO₂ which is bound to haemoglobin from the tissues is delivered at the alveoli)

3. As Bicarbonates :
About 70% of CO₂ is transported as bicarbonate. RBCs contain a very high concentration of the enzyme carbonic anhydrase and a minute quantity of the same is present in plasma too. This enzyme facilitates the following reaction in both the directions.

At the tissues where partial pressure of CO₂ is high due to catabolism, CO₂ diffuses into the blood and forais carbonic acid which dissociates into HCO^–₃ + H⁺

At the alveolar site where pCO₂ is low, the reaction proceeds in the opposite direction leading to the formation of CO₂ and water. Thus CO₂ is mostly trapped as bicarbonate at the tissues’and transported to the alveoli where it is dissociated out as CO₂.

Every 100 ml of deoxygenated blood delivers approximately 4 ml of CO₂ to the alveolar air.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 1(a) Digestion and Absorption Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 1(a) Digestion and Absorption

Very Short Answer Questions

Question 1.
Give the dental formula of an adult human being.
Answer:
The arrangement of different types of teeth in each half of both the jaws in the order I, C, PM, M is represented by the dental formula.
In adult it is = \(\frac{2123}{2123}\) = 32.

Question 2.
Bile juice contains no digestive enzymes, yet it is important for digestion. How?
Answer:
Bile juice doesn’t contain enzymes, but it contains bile salts such as sodium/potassium glycocholates and taurocholates, which help in the digestion and absorption of lipids. Bile salts emulsify fats and also render them water-soluble. Bile salts activate the lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.

Question 3.
Describe the role of chymotrypsin. Name two other digestive enzymes of the same category secreted by the same gland.
Answer:
Chymotrypsin plays an important role in digestion of proteins, proteoses and peptones and convert them into tripeptides and dipeptides. Chymotrypsin, trypsin and carboxy peptidase are ehdopeptidases, produced by the pancreas arid involved in the digestion of proteins.

Question 4.
What would happen if, HCl were not secreted in the stomach?
Answer:
HCl is secreted by the glands present on the stomach walls. It provides acidic pH which is optimal for the action of pepsin. HCl activates the pepsinogen into pepsin. Pepsin plays an important role in digestion of proteins. Therefore, if HCl were not secreted in the stomach, then pepsin would not be activated. This would affect on protein digestion.

Question 5.
Explain the terms thecodont and diphyodont dentitions.
Answer:
Thecodont:
Teeth of human beings are embedded in the sockets of the jaw bones is called thecodont.

Diphyodont:
Majority of mammals including human beings form two sets of teeth during their life time, a set of temporary / milk teeth replaced by a set of permanent teeth. This type of dentition is called diphyodont dentition.

Question 6.
What is Autocatalysis? Give two examples.
Answer:
Autocatalysis is the catalysis of a reaction in which the catalyst is one of the product of the reaction (or) catalysis caused by a catalytic agent formed during a chemical reaction.

Question 7.
What is chyme?
Answer:
Semi fluid mass of partly digested acidic food formed in the stomach is called chyme.

Question 8.
Name the different types of Salivary glands of man, and their locations in the human body?
Answer:
There are three pairs of Salivary glands in man.
1. Parotid glands — present below the pinna / inner surface of the cheeks.
2. Sub maxillary (or) Sub mandibular glands —.located at the angles of lower jaw.
3. Sublingual glands — present below the tongue.

Question 9.
Name different types of papillae present on the tongue of man.
Answer:
The upper surface of the tongue has small projections called papillae. In humans the tongue bears 3 (three) types of papillae namely 1) fungi form 2) filiform 3) Circumvallate papillae.

Question 10.
What is the hardest substance in the human body? What is its origin?
Answer:
Enamel of tooth is the hardest substance in the human body, which is secreted by ameloblasts of ectodermal origin.

Question 11.
Name the structure of gut which is vestigial in human beings, but well developed in herbivores. And mention the type of tissue with which it is mostly formed.
Answer:
Appendix is vestigial part in human beings. It is a narrow finger like tubular projection, arises from the caecum. In herbivores it is a functional part and useful in the digestion of cellulose materials. The appendix contain a high concentration of lymphoid tissues. These are highly specialized structures which are a part of the immune system.

Question 12.
Distinguish between deglutition and mastication.
Answer:
Deglutition :
Deglutition is the swallowing of food and involves a complex and coordinated process. It is divided into three phases.

Phase one :
The collection and swallowing of masticated food.

Phase two :
Passage of food through the pharynx into the beginning of the esophagus.

Phase three :
The passage of food into the stomach.

Mastication :
The mastication process includes the biting and tearing of food into manageable pieces. This usually involves using the incisors and canines teeth. The grinding of food is usually performed by the molars and premolars. During the mastication process, food is moistened and mixed with saliva.

Question 13.
Distinguish between diarrhoea and constipation.
Answer:
Diarrhoea :
The abnormal frequency of bowel movement and increased liquidity of the faecal discharge is known as diarrhoea. It reduces the absorption of fqod arid results in loss of water.

Constipation :
A condition ip which the faeces are retained within the rectum as it is hard due to low content of water and the movement of bowel occurs irregularly.

Question 14.
Name two hormones secreted by the duodenal mucosa.
Answer:
The epithelium of duodenum secretes the hormones namely secretin and cholecystokinin (cck).

Question 15.
Distinguish between absorption and assimilation.
Answer:
Absorption :
Absorption is the process by which the end products of digestion pass through the intestinal mucosa into blood (or) lymph. It is carried out by passive, active (or) facilitated transport mechanisms.

Assimilation:
The absorbed substances finally reach the tissues, where food materials become integral components of the living protoplasm and used for the production of energy, growth and repair. This process is called assimilation.

Short Answer Questions

Question 1.
Draw a neat labelled diagram of L.S of a tooth. Ans.
Answer:

Question 2.
Describe the process of digestion of proteins in the stomach.
Answer:
Protein digestion begins in the stomach. The food entered into the stomach is mixed thoroughly with the gastric juice of the stomach by the churning movements of its muscular wall and the product is called chyme. The main components of gastric juice are protein digestive enzymes, hydrochloric acid and mucus.

HCl provides the acidic pH (1.8) which is optimal for the action of pepsin. The proenzymes of gastric juice, the pepsinogen and prorennin, on exposure to hydrochloric acid are convened into the active enzymes, pepsin and rennin respectively. Pepsin converts proteins into proteoses and peptones. Rennin is a proteolytic enzyme found in the gastric juice of infants. It acts on the milk protein, the casein in the presence of calcium ions and converts it into calcium paracaseinate and proteoses. Pepsin acts on calcium paracaseinate and converts it into peptones. The entire process of protein digestion in the stomach takes about 4 hours.

Question 3.
Explain the role ofpancreatic juice in the digestion of proteins.
Answer:
Pancreatic juice is secreted by the pancreas and it plays an important role in protein ‘digestion. Pancreatic juice contains protein hydrolysing enzymes like trypsinogen, chymotrypsinogen and pro carboxy peptidases, but they are inactive enzymes.

Trypsinogen is activated by the enzyme enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. Trypsin itself can similarly activate trypsinogen into trypsin.

Chymotrypsin, Trypsin and Carboxy Peptidase of pancreatic juice act upon proteins, proteoses and peptones in the chyme, result in the formation of tri and dipeptides. Which in turn hydrolysed into aminoacids by the action Of tri aruj di peptidases.

Question 4.
How are polysaccharides and disaccharides digested?
Answer:
Dietary carbohydrates principally consist of polysaccharides :
Starch and glycogen. It contains disaccharides and small amounts of monosaccharides.

Digestion in mouth:
Digestion of carbohydrates starts at the mouth, where they come in contact with saliva during mastication. Saliva contains carbohydrate-splitting enzyme called Salivay amylase (ptyalin). This enzyme hydrolyses the starch into disaccharides (maltose).

Digestion in stomach:
Ptyalin action stops in stomach when pH falls to 3.0. No carbohydrate splitting enzymes are available in gastric juice. Some dietary sucrose may be hydrolysed by HCl.

Digestion in small intestine :
Chyme reaches the duodenum from stomach where it meets pancreatic juice. Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the disaccharides and convert them into monosaccharides.

Question 5.
If you take butter in your food how does it get digested and absorbed in the body? Explain.
Answer:
Butter contains fat. Fats remain mostly undigested in stomach.

Digestion of fat in the small intestine :
The major site of fat digestion is the small intestine. This is due to the presence of a powerful lipase/(steapsin) in the pancreatic juice and bile juice. Bile juice contains bile salts such as Sodium/Potassium glycocholates and taurocholates, which helps in the emulsification of fat i.e., break down of fats into very small micelles. Bile also activates lipases of pancreatic juice (steapsin) and intestinal lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.

Absorption :
Fatty acids and glycerol being insoluble in water cannot be absorbed into the blood directly. They are first modified into small droplets called micelles, which move into intestinal mucosal cells. They are reformed into very small protein coated fat globules called chylomicrons, which are transported into the lymph capillaries in the villi by exocytosis. Then they are ultimately released into blood stream through left subclavian vein via the thoracic duct. These chylomicrons are broken down to fatty acids and glycerol by the action of an enzyme lipoprotein lipase and they diffuse into the adipocytes of the adipose tissue and liver for storage.

Question 6.
What are the functions of liver?
Answer:
Liver performs a variety of functions such as synthesis, storage and secretion of various substances.

1. Liver secretes bile juice, it contains bile salts such as sodium / potassium glycocholates and taurocholates, which helps in digestion and absorption of lipids.
2. Liver plays the key role in carbohydrate metabolism.
   a) Glycogenesis : formation of glycogen from glucose.
   b) Glycogenolysis : breakdown of glycogen into glucose.
   c) Gluconeogenesis : Synthesis of glucose from certain amino acids, lactate (or) glycerol.
3. Liver also plays an important role in synthesis of cholesterol and production of triglycerides.
4. Deamination of proteins occurs in the liver.
5. Liver is the chief organ of detoxification of toxic substances that enter the gut along with food.
6. Liver acts as thermoregulatory organ.
7. Liver acts as a haemopoietic organ in the foetus and erythroclastic organ in the adult.
8. The liver synthesizes the plasma proteins such as albumin, globulins, blood clotting factors such as fibrinogen / prothrombin, etc., and the anticoagulant called heparin.
9. The lactic acid formed during anaerobic muscle contraction is converted into glycogen (gluconeogenesis) in the liver by Cori cycle.
10. Kupffer cells are the largest phagocytic cells which remove unwanted substances and microbes that attack the liver by phagocytosis.

Long Answer Questions

Question 1.
Describe the physiology of digestion of various types of food in the human digestive system
Answer:
Digestion is the process of convertion of complex non-diffusible food substances into simple diffusible forms. The process of digestion is accomplished by mechanical and biochemical process.

I. Digestion in the buccal cavity :
Buccal cavity performs two major functions, mastication of food and facilitation of swallowing. Teeth and tongue with the help of saliva masticate and mix up the food thoroughly. Mucus in saliva helps in lubricating and adhering the masticated food particles into bolus. The saliva secreted into oral cavity contains electrolytes such as Na⁺, K⁺, Cl^–, HCO₃^– arid enzymes like salivary amylase (ptyalin) and lysozyme. Carbohydrates digestion starts in the buccal cavity, about 30% of starch is hydrolysed here into a disaccharide called maltose by the enzyme amylase (ptyaline). Lysozyme acts as antibacterial agent that prevents infections.

II. Digestion in the stomach :
As the bolus enters into stomach starch digestion stops and protein digestion begins. The food entered into stomach is mixed thoroughly with gastric juice of the stomach by the churning movement of its muscular wall and the product is called chyme. The mucus and bicarbonates present in the gastric juice act as lubricant and protect the mucosal epithelium from HCl. HCl in the stomach provides the acidic pH (1.8) which is optimal for the action of pepsin.

The proenzymes of gastric juice, the pepsinogen and prorennin on exposure to HCl are converted into the active enzymes, pepsin and rennin respectively. Pepsin converts proteins into proteoses and peptones. Rennin found in gastric juice of infants. It acts on the milk protein, the casein in the presence of calcium ions convert into calcium paracaseinate and proteoses. Pepsin acts on paracaseinate and convert it into peptones. The entire process of protein digestion in stomach takes about 4 hours.

III. Digestion in the small intestine :
Various types of movements are generated by the muscular external layer of small intestine. These movements help in thorough mixing of the food with bile, pancreatic juice and intestinal juice in the intestine and thereby facilitate digestion. The duodenal cells of the proximal part produces large amount of bicarbonates to completely neutralize any gastric acid that passes further down into the digestive tract.

i) Digestion of proteins :
Pancreatic juice contains protein hydrolysing enzymes like trypsinogen, chymotrypsin and procarboxy peptidases, but they are inactive enzymes.

Trypsinogen is activated by the enzyme enterokinase secreted by the intestinal mucosa into active trypsin which intum activate the other enzymes in the pancreatic juice. Trypsin itself can similarly activate trypsinogen into trypsin.

Chymotrypsin, trypsin and carboxy peptidase of pancreatic juice act upon proteins, proteoses and peptones in the chyme, result in the formation of tri and dipeptides which inturn hydrolysed into amino acids by the action of tri and dipeptidases.

ii) Digestion of fats:
Bile salts of bile help in the emulsification of fat i.e., breakdown of fats into very small micelles. Bile also activates lipases of pancreatic juice (steapsin) and intestinal lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.

iii) Digestion of Carbohydrates :
Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the disaccharides and convert them into monosacharides.

iv) Digestion of nucleic acids :
Nucleases of the pancreatic juice act on the nucleic acids to form nucleotides and nucleosides. Nucleotidases and nucleosidases of the intestinal juice convert the nucleotides and nucleosides into pentose sugar and nitrogen bases.

The end products of digestion pass through the intestinal mucosa into blood (or) lymph is carriedout by passive, active (or) facilitated transport mechanisms.

Question 2.
Explain the digestive system of man with neat labelled diagram.
Answer:

The digestive system is a group of organs and tissues involve in the breaking down of ingested food in the alimentary canal into a form that can be absorbed ai a assimilated by the tissues of the body.

Human digestive system consists of the alimentary canal and the associated glands. Alimentary canal / Digestive tract:

The alimentary canal of man begins with the anterior opening, the mouth and ends with the posterior opening, the anus.

Parts of the alimentary canal / digestive tract:

1. Mouth and Buccal (oral) cavity
2. Pharynx
3. Oesophagus
4. Stomach
5. Small intestine
6. Large intestine

1. Mouth and Buccal (oral) cavity:
Mouth is the first part of the alimentary canal. It is formed by the cheek on either side and boardered by the movable upper and lower lips, leads into the buccal (or) oral cavity. The palate separate the ventral buccal cavity from the dorsal nasal chamber and facilitates chewing and breathing simultaneously. The jaw bones bear teeth and tongue occurs at the base of the buccal cavity.

i) Teeth :
These are ecto-mesodermal in origin. An adult human has 32 permanent teeth, which are of four different types namely, incisors (I), canines (C), premolars (PM), and molars (M). These are useful in cutting, tearing and grinding of food. The arrangement’ of teeth is represented by dental formula. In adult human
it is = \(\frac{2123}{2123}\) = 32

ii) Tongue :
It is a freeely movable muscular sense organ, attached to the floor of the oral cavity by a fold of tissue called frenulum. The upper surface of the tongue has small projections called papillae, some of which bear taste buds. The tongue acts as universal toothbrush and helps in mixing saliva with food, taste detection, deglutition and speaking.

2. Pharynx:
It is a muscular tube connecting the oral cavity and oesophagus and trachea. It is a common passage for food and air. It is divided into nasopharynx, oropharynx and laryngo pharynx. Oesophagus and trachea open into the laryngopharynx. The trachea open into the laryngopharynx through the glottis. A cartilaginous flap called epiglottis prevents the entry of food into glottis during swallowing.

3. Oesophagus:
It is a thin long muscular tube (9 to 12 inches). The semisolid digested food from pharynx enters the oesophagus. Oesophagus is separated by the Cardiac sphincter from stomach. When the food reaches lower end of Oesophagus the cardiac sphincter opens allowing the food to enter the stomach.

4. Stomach :
It is a wide ‘J’ shaped muscular sac, located iii the upper left portion of the abdominal cavity just below the diaphragm. It has three major parts, an anterior cardiac portion into which the oesophagus opens, a middle large fundic region and a posterior pyloric portion which opens into the first part of the small intestine through the pyloric aperture which is guarded by the pyloric sphincter.

5. Small intestine :
The small intestine is the longest part of alimentary canal. It has three regions namely proximal duodenum middle long coiled jejunum and distal highly coiled ileum. Duodenum receives the hepato-pancreatic duct. Ileum opens into the large intestine.

6. Large intestine :
It consist of caecum, colon and rectum. Caecum is a small blind sac. A narrow finger like vestigial tubular organ arises from caecum called appendix. The caecum opens into colon which is‘divided into an ascending, a transverse, a descending parts and a sigmoid colon that continues behind into rectum. Rectum is a small dilated sac which leads into anal canal that opens out through the anus.

Digestive glands :
1. Salivary glands : There are three pairs of glands in man.
i) Parotid glands
ii) Sub-maxillary glands
iii) Sub-lingual glands
They secrete saliva, which mainly contains salivary amylase and lysozyme.

2. Gastric glands :
These are located in the wall of the stomach beneath the surface
epithelium, Gastric glands are of three types namely
i) Cardiac glands – secrete mucus
ii) Pyloric glands – secrete mucus and hormone gastrin
iii) Fundic / Oxyntic glands – secrete mucus, proenzymes like pepsinogen and prorennin, HCl, intrensic factor and some amount of gastric lipase.

3. Intestinal glands :
They are of two types
i) Brunner’s glands
ii) Crypts of lieberkuhn
which secrete intestinal juice contains peptidases, disaccharidases, enterokinase and lysozyme.

4. Liver :
Liver is the largest gland in human. Liver secretes bile juice, contains bile salts, which play a very important role in lipid digestion.

5. Pancreas :
The pancreas is the second largest gland in human. Exocrine part of pancreas secretes pancreas juice contains sodium bicarbonates, trypsinogen, chymotrypsinogen, carboxy peptidase, steapsin, -pancreatic amylase and nucleases such as DNAase and RNAase.

Students get through AP Inter 2nd Year Physics Important Questions 16th Lesson Communication Systems which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 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.

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 ? [A.P. 17; A.P., T.S. Mar. 16, T.S. Mar. 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. [T.S. Mar. 15, 17, A.P. Mar. 16]
Answer:
The basic methods of modulation are :

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

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.

Textual Examples

Question 1.
A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna Is 50 m. What is the maximum distance between them for satisfactory communication in LOS mode? Given radius of earth 6.4 × 10⁶ m.
Solution:
d_m = \(\sqrt{2 \mathrm{Rh}_{\mathrm{T}}}+\sqrt{2 \mathrm{Rh}_{\mathrm{R}}}\)
d_m = \(\sqrt{2 \times 64 \times 10^5 \times 32}+\sqrt{2 \times 64 \times 10^5 \times 50 \mathrm{~m}}\)
=64 × 10² × \(\sqrt{10}\) +8 × 10³ × \(\sqrt{10}\)m = 144 × 10² × \(\sqrt{10}\)m
= 45.5 km.

Question 2.
A message signal of frequency 10 kHz and peak voltage of 10 volts Is used to modulate a carrier of frequency 1 MHz and peak voltage of 20 volts. Determine
(a) modulation index
(b) the side bands produced.
Solution:
a) Modulation index = \(\frac{10}{2}\) = 0.5
b) The side bands are at (1000 + 10 kHz) = 1010 kHz and (1000 – 10 kHz) = 990 kHz.

Students get through AP Inter 2nd Year Physics Important Questions 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 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 a pure tetravalent semiconductor, it is called an n-type semiconductor.
• In an n-type semiconductor majority, of charge carriers are electrons and the 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.

Question 3.
What is a p-type semiconductor ? What are the majority and minority charge carriers in it ? [A.P. & T.S. Mar. 17]
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.
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.

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 i) forward and ii) reverse bias ?
Answer:
i) 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.

ii) 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.

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 rectifiers is 81.2%.

Question 7.
What is Zener voltage (V_z) 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]
Answer:

Question 11.
Define 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 = \(\frac{\mathrm{V}_0}{\mathrm{~V}_{\mathrm{i}}}\)

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 are called universal gates ? [T.S. Mar. 15]
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:

Short Answer Questions

Question 1.
What is a rectifier ? Explain the working of half wave and full Wave rectifiers with diagrams. [A.P. Mar. 17]
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:

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 R_L.
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 r_f = Forward resistance of a diode; R_L = 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.

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 R_L.
4. During positive half cycles of ac, the diode D₁ is forward biased and current flows through the load resistance R_L. 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 R_L. 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 2.
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 suitablyjoined to n-type semiconductor, a p-n junction diode is formed. ’
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 combihe 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.

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”.

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 apd electric current flows in the circuit.
Hence in forward bias resistance of diode is low. This position is called switch on position.
Reverse bias:

“When the negative terminal of the battery is connected to p-side and positive terminal of the battery is connected to 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 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.

Zener diode can be used as a voltage regulator. In- general zener diode is connected in reverse bias in the circuits.

1. The zener diode is connected to a battery, through a resistance R. The battery reverse biases the zener diode.
2. The load resistance R_L 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 R_L 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 R_L varies. Thus, zener diode works as voltage regulator.
8. If I is the input current, I_Z and I_L are zener and load currents.
   I = I_Z + I_L; V_in = IR + V_Z
   But V_0ut = V_Z
   ∴ V_out = V_in – IR

Question 4.
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 not 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 with majority carriers near the junction.

On recombination of electrons and hole, the 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 5.
Define NAND and NOR gates. Give their truth tables. [T.S. Mar. 17]
Answer:
NAND gate : NAND gate is a combination of AND gate and NOT gate.

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
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.

1. If both inputs are low, output is high.
   A = 0, B = 0, X = 1
2. If any input is high, the output is low.
   A = 0, B = 1, X = 0
   A = 1, B = 0, X = 0
3. If both inputs are high, the output is low.
   A = 1, B = 1, X = 0
   NOR gate is also a universal gate.
   

Question 6.
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.
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.

Working : When light (E = hv) falls at the junction, electron – hole pairs are generated near the 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.

I-V characteristics : I – V characteristics of solar cell is drawn in the fourth quadrant of the coordinate axes. Because it does not draw current.

Uses : They are used in calculators, wrist watches, artificial satellites etc.

Question 7.
Explain the operation of a NOT gate and give its truth table. [IPE 15, T.S.]
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.

1. If input is low, output is high.
   A = 0, X = \(\overline{0}\) = 1
2. If input is high, output is low.
   A = 1, X = \(\overline{1}\) = 0
   

Question 8.
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.

In truth table logic function is written as A or B ‘OR’ logic function is represented by the symbol ‘plus’.
Q = A + B
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 OV represents the logical value 0.

When A = 0, B = 0 both the diodes are reverse biased and there is no current through the resistance. So, the potential at Q is zero i.e., Q = 0. When A = 0 or B = 0 and the other equal to a potential of 5 V the diode whose anode is at a potential of 5 V is forward – biased and that diode behaves like a closed switch. The output potential then becomes 5 V i.e., Q = 1. When both A and B are 1, both the diodes are forward-biased and the potential at Q is same as that at A and B which is 5 V i.e., Q = 1. The output is same as that of the OR gate.

Question 9.
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 output terminal.

• If both the inputs are low (or) one of the inputs is low.
  • The output 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
  
  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.

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 (0). 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 = T 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 semi-conductor. In this excess holes in addition to those formed due to thermal energy are available for conduction in the valence band and the conductivity of semiconductor increases.

Question 10.
Explain how transistor can be used as a switch ?
Answer:
To understand the operation of transistor as a switch.

1. As long as V_i is low and unable to forward bias the transistor, V₀ is high (at V_cc).
2. If V_i is high enough to drive the transistor into saturation then V₀ is low, very near to zero.
   
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.
   

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 (r_f) = 20Ω
R_L = 2kΩ = 2000 Ω
η = \(\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}=\frac{0.406 \times 2000}{20+2000} \times 100=\frac{812 \times 100}{2020}\)
η = 40.2%.

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 R_L = 1300 Ω
r_f = 9Ω
η = \(\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}=\frac{0.812 \times 1300}{9+1300} \times 100 ; \eta=\frac{8120 \times 13}{1309}\)
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 (∆I_C) = 1mA = 10^-3 A
Change in base current (∆I_B) = 20 μA = 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 R_L = 2k ohm and the input resistance R_i = 1 k ohm. If the current gain is 50, calculate voltage gain of the amplifier.
Solution:
R_L = 2kΩ = 2 × 10³ Ω
R_i = 1kΩ = 1 × 10³ Ω
β = 50.
Voltage gain (A_V) = β × \(\frac{\mathrm{R}_{\mathrm{L}}}{\mathrm{R}_{\mathrm{i}}}=\frac{50 \times 2 \times 10^3}{1 \times 10^3}\)

A_V = 100.

Textual Examples

Question 1.
C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors ?
Solution:
The 4 bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out and electron from these atoms(i.e., ionisation energy E_g) will be least for Ge, following by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C.

Question 2.
Suppose a pure Si crystal has 5 × 10²⁸ atoms m^-3. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that n. = 1.5 × 10¹⁶ m^-3.
Solution:
Note that thermally generated electrons (n_i ~ 10¹⁶ m^-3) are negligibly small as compared to those produced by doping.
Therefore, n_i ≈ N_D.
Since n_en_h = n_i², The number of holes, n_h = (1.5 × 10¹⁶)² / 5 × 10²⁸ × 16^-6
n_h = (2.25 × 10³²)/(5 × 10²²) ~ 4.5 × 10⁹m^-3

Question 3.
Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction ?
Solution:
No ! Any slab, howsoever flat, will have roughness much large than the inter-atomic crystal spacing(~2 to 3 A°) and hence continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers.

Question 4.
The V-I characteristics of a silicon diode are shown in the Fig. Calculate the resistance of the diode at (a) I_D = 15 mA and (b) V_D = -10 V.

Solution:
Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law.
a) From the curve, at I = 20 mA, V = 0.8 V, I = 10 mA, V = 0.7 V
r_fb = ∆V/∆I = 0.1V/10mA = 10Ω
b) From the curve at V = -10 V, I = -1 µA.
Therefore r_rb = 10V/1µA = 1.0 × 10⁷ Ω

Question 5.
In a Zener regulated power supply a Zener diode’with V_z = 6.0 V is used for regulation. The load current is to be 4.0 mA and the unregulated input is 10,0 V. What should be the value of series resistor R_s?
Solution:
The value of Rs should be such that the current through the Zener diode is much larger than the load current. This is to have good load regulation. Choose Zener current as five times the load current, i.e., I_z = 20 mA. The total current through R_S is , therefore, 24 mA. The voltage drop across R_S is 10.0 – 6.0 = 4.0 V This gives R_S = 4.0V/(24 × 10^-3) A = 167Ω. The nearest value of carbon resistor is 150 Ω. So, a series resistor of 150 Ω is appropriate. Note that slight variation in the value of the resistor does not matter, what is important is that the current I_Z should be sufficiently larger than I_L.

Question 6.
The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~µA). What is the reason then to operate the photodiodes in reverse bias ?
Solution:
Consider the case of an n-type semiconductor. Obviously, the majority carrier density (n) is considerably larger than the minority hole density p (i.e., n> > p). On illumination, let the excess eletrons and holes generated be ∆n and ∆p, respectively,
n’ = n + ∆n
p’ = p + ∆p
Here n’ and p’ are the electron and hole concentrations at any particular illumination and n and p are carriers concentration when there is no illumination. Remember ∆n = ∆p and n > > p. Hence, the fractional change in the majority carriers (i.e., ∆n/n) would be much less than that in the minority carriers (i.e,, ∆p/p). In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes-are preferably used in the reverse bias condition for measuring light intensity.

Question 7.
Why are Si and GaAs are preferred materials for solar cells ?
Solution:
The solar radiation spectrum received by us is shown in figure.

The maxima is near 1.5 eV. For photo-excitation, hv > E_g. Hence, semiconductor with band gap ~1.5 eV or lower is likely to give better solar conversion efficiency. Silicon has E_g ~ 1.1 eV while for GaAS it is ~ 1.53 eV. In fact, GaAs is better On spite of its higher band gap) than Si because of its relatively higher absorption coefficient. If we choose materials like CdS or CdSe(E_g ~ 2.4 eV), we can use only the high energy component of the solar energy for photo-conversion and a significant part of energy will be of no use.

The question arises: why we do not use material like pbS(E_g ~ 0.4 eV) which satisfy the condition hv > E_g for v maxima corresponding to the solar radiation spectra ? if we do so, most of the solar radiation will be absorbed on the top-layer of solar cell mid will not reach in or near the depletion region. For effective electron-hole separation, due to the junction field, we want the photo-generation to occur in the junction region only.

Question 8.
From the output charactristics shown in fig, calculate the values of β_ac and β_dc of the transistor when V_CE is 10 V and I_C = 4.0 mA.

Solution:
β_ac = \(\left(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\right)_{\mathrm{V}_{\mathrm{CE}}}\) ; β_dc = \(\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}\)
For determining β_ac and β_dc at the stated values of V_CE and I_C one can proceed as follows. Consider any two characteristics for two values of I_B which lie above and below the given value of I_C. Here I_C = 4.0 mA. (Choose characteristics for I_B = 30 and 20μA.) At V _CE = 10V
we read the two values of Ic from the graph. Then
∆I_B = (30 – 20)μA = 10μA. ∆I_C
= (4.5 – 3.0) mA = 1.5 mA
Therefore, β_ac =1.5 mA/ 10μA = 150
For determining β_dc either estimate the value of I_B corresponding to I_C = 4.0 mA at V_CE = 10V or calculate the two values of β_dc for the two characteristics chosen and find their mean.
Therefore, for I_C = 4.5 mA and I_B = 30 μA
β_dc = 4.5 mA/30 μA = 150 and for I_C = 3.0 mA/ and I_B = 20 μA
β_dc = 3.0 mA / 20 μA= 150
Hence, β_dc = (150 + 150)/ 2 = 150

Question 9.
In Fig. the V_BB supply can be varied from OV to 5.0 V. The Si transistor has β_dc = 250 and R_B = 100 kΩ, R_C = 1 KΩ, V_CC = 5.0V. Assume that when the transistor is saturated, V_CE = 0V and V_BE = 0.8V. Calculate
(a) the minimum base current, for which the transistor will reach saturation. Hence,
(b) determine V₁ for when the transistor is ‘switched on’,
(c) find the ranges of V₁ for which the transistor is ‘switched of and ‘switched on’.

Solution:
Given at stauration
V_CE = OV, V_BE = 0.8V
V_CE = V_CC – I_CR_C
I_C = V_CC / R_C = 5.0V/1/0 kΩ = 5.0mA
Therefore, I_B = I_C/β
= 5.0 mA/250 = 20μA
The input voltage at which the transistor will go into saturation is given by
V_IH = V_BB = I_BR_B + V_BE
= 20μA × 100 kΩ + 0.8V = 2.8V
The value of input voltage below which the transistor remains cutoff is given by
V_IL = 0.6V, V_IH = 2.8V
Between 0.0V and 0.6V the transistor will be in the ‘switched off-state. Between 2.8V and 5.0V, it will be in ‘switched on’ state.
Note that the transistor is in active state when I_B varies from 0.0mA to 20mA. In this range, I_C = βI_B is valid. In the saturation range I_C ≤ βI_B.

Question 10.
For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2.0 kΩ is 2.0 V. Suppose the current amplification factor of the transistor is 100, what should be the value of R_B in series with V_BB supply of 2.0 V if the dc base current has to be 10 times the signal current. Also calculate the dc drop across the collector resistance. (Refer to Fig)

Solution:
The output ac voltage is 2.0 V. So, the ac collector current i_C= 2.0/2000 = 1.0 mA. The signal current through the base is, therefore given by i_B = i_C/β = 1.0 mA/100 = 0.010 mA. The dc base current has to be 10 × 0.010 = 0.10 mA
From V_BB = V_BE+ I_B R_B R_B = (V_BB – V_BE)/I_B. Assuming V_BE = 0.6V
R_B = (2.0 – 0.6)/0.10 = 14kΩ
The dc collector current I_C = 100 × 0.10 = 10 mA.

Question 11.
Justify the output waveform (Y) of the OR gate for the following inputs A and B given in fig.

Solution:
Note the following :

• At t ≤ t₁; A = 0, B = 0; Hence Y = 0
• For t₁ to t₂; A = 1, B = 0; Hence Y = 1
• For t₂ to t₃; A = 1, B = 1; Hence Y = 1
• For t₃ to t₄; A = 0, B = 1; Hence Y = 1
• For t₄ to t₅; A = 0, B = 0; Hence Y = 0
• For t₅ to t₆; A = 1, B = 0; Hence Y = 1
• For t > t₆; A = 0,B = 1; Hence Y = 1

Therefore the waveform Y will be as shown in the Fig.

Question 12.
Take A and B input waveforms similar to that in Example 11. Sketch the output waveform obtained from AND gate.
Solution:

• For t ≤ t₁; A = 0, B = 0; Hence Y = 0
• For t₁ to t₂; A = 1, B = 0; Hence Y = 0
• For t₂ to t₃; A = 1, B = 1; Hence Y = 1
• For t₃ to t₄; A = 0, B = 1; Hence Y = 0
• For t₄ to t₅; A = 0, B = 0; Hence Y = 0
• For t₅ to t₆; A = 1, B = 0; Hence Y = 0
• For t > t₆; A = 0,B = 1; Hence Y = 0

Based on the above, the output waveform for AND gate can be drawn as given below.

Question 13.
Sketch the output Y from a NAND gate having inputs A and B given below :
Solution:

• For t ≤ t₁; A = 1, B = 1; Hence Y = 0
• For t₁ to t₂; A = 0, B = 0; Hence Y = 1
• For t₂ to t₃; A = 0, B = 1; Hence Y = 1
• For t₃ to t₄; A = 1, B = 0; Hence Y = 1
• For t₄ to t₅; A = 1, B = 1; Hence Y = 0
• For t₅ to t₆; A = 0, B = 0; Hence Y = 1
• For t > t₆; A = 0,B = 1; Hence Y = 1
  

Students get through AP Inter 2nd Year Physics Important Questions 14th Lesson Nuclei which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 14th Lesson Nuclei

Very Short Answer Questions

Question 1.
The half life of ⁵⁸Co is 72 days. Calculate Its average life. [Board model Paper]
Answer:
T_half = 0.693 × T_Mean ⇒ T_Mean = \(\frac{T_{\text {half }}}{0.693}=\frac{72}{0.693}\) = 103.8 days.

Question 2.
Why do all electrons emitted during p-decay not have the same energy?
Answer:
When a neutron is converted into a proton, an electron and neutron are emitted along with it.
\({ }_1^1 \mathrm{n} \longrightarrow{ }_1^1 \mathrm{H}+{ }_{-1}^0 \mathrm{e}+\mathrm{v}\)
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, ail electrons do not have same energy.

Question 3.
Neutrons are the best projectiles to produce nuclear reactions. Why ?
Answer:
Neutrons are uncharged particles. So they do not get deflected by the electric and magnetic fields. Hence Neutrons are considered as best projectiles in nuclear reaction.

Question 4.
Neutrons cannot produce ionization. Why ?
Answer:
Because neutrons are uncharged particles and cannot produce ionization.

Question 5.
What are delayed neutrons ?
Answer:
Neutrons are emitted in the fission products after, sometime are called delayed neutrons.

Question 6.
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 7.
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 8.
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 9.
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 10.
Define Becquerel and Curie.
Answer:
Becquerel: 1 disintegration or decay per second is called Becquerel. It is SI unit of activity.
1 disintegration or decay
i.e., Becquerel = \(\frac{1 \text { disintegration or decay }}{\text { second }}\)
Curie : 3.7 × 10¹⁰ decays per second is called Curie.
1 Curie : 1 Ci = \(\frac{3.7 \times 10^{10} \text { decays }}{\text { second }}\) = 3.7 × 10¹⁰Bq.

Question 11.
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 12.
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, Beryllium.

Question 13.
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.
Write a short note on the discovery of neutron.
Answer:

1. Bothe and Becker found that when beryllium is bombarded with a – particles of energy 5 MeY which emitted a highly penetrating radiation.
2. The equation for above process can be written as
   \({ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{13} \mathrm{C}\) + γ – (radiation energy)
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.
   \({ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{12} \mathrm{C}+{ }_0^1 \mathrm{n}+\mathrm{Q}\)
   

Question 2.
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 – N, 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 n-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 3.
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 = \(\frac{-\left(\frac{d N}{d t}\right)}{N}\)

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 A 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 = N₀ e^-λt
4. \(\frac{\mathrm{N}_0}{2}\) = N₀ e^-λT
   e^λT = 2
   λT= ln 2
   T = \(\frac{\ln 2}{\lambda}=\frac{2.303 \log _{10}^2}{\lambda}\)
   ∴ T = \(\frac{0.693}{\lambda}\)

Question 4.
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 \({ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n} \rightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_0^1 \mathrm{n}+\mathrm{Q}\)
Where Q is the energy released.
Q = (Total mass of reactants – Total mass of product) C²
= [(Mass of \({ }_{92}^{235} \mathrm{U}\) + Mass of \({ }_0^1 \mathrm{n}\)) – (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

Question 5.
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 6.
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 outside 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 can 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 107 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.

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 (∆M) : 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, Em, of its constituents.
Mass defect, (∆M) = [Zm_p + (A – Z)m_n – M]

2. Binding energy: The energy required to break the nucleus into its constituent nucleons is called the binding energy.
Binding Energy, (E_b) = ∆MC² = [Zm_p + (A – Z)m_n – M] 931.5 MeV
Nuclear binding energy is an indication of the stability of the nucleus.
Nuclear binding energy per nucleon E_bn = \(\frac{E_{\mathrm{b}}}{\mathrm{A}}\)

3. The following graph represents how the binding energy per nucleon varies with the mass number A.

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.

Question 2.
What is radioactivity ? State the law of radioactive decay. Show that radioactive decay is exponential in nature. [T.S. Mar. 16]
Answer:
1. Radioactivity : The nuclei of certain elements disintegrate spontaneously by emitting alpha (α), beta (β) and gamma (δ) rays. This phenomenon is called Radioactivity or Natural radioactivity.

2. Law of radioactivity decay : “The rate of radioactive decay \(\left(\frac{d N}{d t}\right)\) (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 = – λ Ndt …………………….. (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
\(\int \frac{\mathrm{dN}}{\mathrm{N}}=-\lambda \int \mathrm{dt}\)
ln N = – λt + C …………….. (3)
Where C = Integration constant.

6. At t = O; N = N₀. Substituting in eq. (3), we get ln N₀ = C
∴ ln N = -λt + ln N₀
ln N – ln N₀ = – λt
ln (\(\frac{\mathrm{N}}{\mathrm{N}_0}\)) = – λt
∴ N = N₀ e^-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.

Sample Problems

Question 1.
The half life of radium is 1600 years. How much time does lg of radium take to reduce to 0.125g. [IPE 2016 (TS)]
Answer:
Half – life of radium = 1600 years;
\(\frac{N}{N_0}=\frac{1}{2^n} \Rightarrow \frac{0.125}{1}=\frac{1}{2^n} \Rightarrow \frac{125}{1000}=\frac{1}{2^n} \Rightarrow \frac{1}{8}=\frac{1}{2^n} \Rightarrow \frac{1}{2^3}=\frac{1}{2^n} \Rightarrow n=3\)
∴ Time taken = Half life × no. of Half lives = 1600 × 3 = 4800 years

Question 2.
Plutonium decays with a half-life of 24,000 years. If plutonium is stored for 72,000 years, what fraction of it remains ?
Answer:
Half life of plutonium, T = 24000 years; Stored time of plutonium, t = 72000 years
no. of half lives, n = \(\frac{t}{T}=\frac{72000}{24000}\) = 3; Fraction of plutonium remains = \(\frac{N}{N_0}=\frac{1}{2^n}=\frac{1}{2^3}=\frac{1}{8}\)

Question 3.
Explain the principle and working of a nuclear reactor with the help of a labelled diagram. [A.P. & T.S. Mar. 17, A.P. Mar. 16 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 nuclear reactor consists of (1) Fuel (2) Moderator (3) Control rods (4) Radiation shielding (5) Coolant.

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 is used 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 cadnium 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 is 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.

Question 4.
Explain the source of stellar energy. Explain the carbon – nitrogen cycle, proton – proton cycle occuring 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 cylce.

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.

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 latex]{ }_2^4 \mathrm{He}[/latex] and two protons releasing a total amount of energy 25.71 MeV The nuclear fusion reactions are given below.

Problems

Question 1.
Compare the radii of the nuclei of mass numbers 27 and 64.
Solution:
A₁ = 27; Asub>2 = 64
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left[\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right]^{1 / 3}\) [∵ R = R₀A^1/3]
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left[\frac{27}{64}\right]^{\frac{1}{3}}=\frac{3}{4}\)
∴ R₁ : R₂ = 3 : 4

Question 2.
Find the energy required to split \({ }_8^{16} \mathrm{O}\) nucleus into four α-particles. The mass of an a-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 X 4.002603) – 15.994915] u × c²
= [16.010412 – 15.994915] u × c²
= (0.015497) 931.5 MeV = 14.43 MeV

Question 3.
A certain substance decays to 1/232 of its initial activity in 25 days. Calculate its half-life.
Solution:
Fraction of substance decays
= \(\frac{\text { Quantity remains }}{\text { Initial quantity }}\)
= \(\frac{1}{2^n}=\frac{1}{32}=\frac{1}{2^5}\)
∴ n = 5
Duration of time = 25 days
We know (n) = \(\frac{\text { Duration of time }}{\text { Half life time }}\)
∴ Half life time = \(\frac{\text { Duration of time }}{\mathrm{n}}\)
\(\frac{25}{5}\) = 5 days

Question 4.
One gram of radium is reduced by 2 milli- gram 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 = \(\frac{\mathrm{N}}{\mathrm{N}_0}\) ⇒ e^λt = \(\frac{\mathrm{N}_0}{\mathrm{~N}}\) ⇒ λt = log_e[latex]\frac{\mathrm{N}_0}{\mathrm{~N}}[/latex]
λt = 2.303 log [latex]\frac{\mathrm{N}_0}{\mathrm{~N}}[/latex]
λt = 2.303 log [latex]\frac{1}{0.998}[/latex]
= 2.303 log (1.002)
= 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

Question 5.
If one microgram of \({ }_92^{235} \mathrm{U}\) is completely destroyed in an atomhomb, how much energy will be released ?
Solution:
m = 1 μg = 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 6.
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 × 10⁶ W
P = \(\frac{\mathrm{nE}}{\mathrm{t}} \Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=\frac{\mathrm{P}}{\mathrm{E}}=\frac{10^6}{200 \times 10^6 \times 1.6 \times 10^{-19}}\)
= \(\frac{1}{32}\) × 10¹⁸
∴ P = 0.03125 × 10¹⁸
= 3.125 × 10⁶

Textual Examples

Question 1.
Given the mass of iron nucleus as 55.85u and A = 56, find the nuclear density ?
Solution:
m_Fe = 55.85, u = 9.27 × 10^-26 kg
Nuclear density = \(\frac{\text { mass }}{\text { volume }}\)
= \(\frac{9.27 \times 10^{-26}}{(4 \pi / 3)\left(1.2 \times 10^{-15}\right)^3} \times \frac{1}{56}\)
= 2.29 × 10¹⁷ kg m^-3
The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus.

Question 2.
Calculate the energy equivalent of 1 g of substance.
Solution:
Energy, E = 10^-3 × (3 × 10⁸)² J
E = 10^-3 × 9 × 10¹⁶ = 9 × 10¹³ J
Thus, if one gram of matter is converted to energy, there is a release of an enormous amount of energy.

Question 3.
Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of \({ }_8^{16} \mathrm{O}\) in MeV/c².
Solution:
1 u = 1.6605 × 10^-27 kg
To convert it into energy units, we multiply it by c² and find, that energy. equivalent
= 1.6605 × 10^-27 × 2.9979 × 10⁸ kg m²/s²
= 1.4924 × 10^-10 J
= \(\frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}}\) eV
= 0.9315 × 10⁹ eV = 931.5 MeV
or, 1 u = 931.5 MeV/c².
For \({ }_8^{16} \mathrm{O}\), ∆M = 0.13691 u = 0.13691 × 931.5 MeV/c² = 127.5 MeV/c²
The energy needed to separate \({ }_8^{16} \mathrm{O}\) into hs constituents is thus 127.5 MeV/c².

Question 4.
The half-life of \({ }_{92}^{238} \mathrm{U}\) undergoing a – decay is 4.5 × 10⁹ years. What is the activity of 1 g sample of \({ }_{92}^{238} \mathrm{U}\) ?
Solution:
T_1/2 = 4.5 × 10⁹ y
= 4.5 × 10⁹ y × 3.16 × 10⁷ s/y.
= 1.42 × 10¹⁷ s
One k mol of any isotope contains Avogadro’s number of atoms, and so 1 g of \({ }_{92}^{238} \mathrm{U}\) contains
\(\frac{1}{238 \times 10^{-3}}\) k mol × 6.025 × 10²⁶ atoms/kmol
= 25.3 × 10²⁰ atoms.
The decay rate R is
R = λN
= \(\frac{0.693}{\mathrm{~T}_{1 / 2}}\) N = \(\frac{0.693 \times 25.3 \times 10^{20}}{1.42 \times 10^{17}}\) S^-1
= 1.23 × 10⁴ S^-1
= 1.23 × 10⁴ Bq

Question 5.
Tritium has a half-life of 12.5 y undergoing beta decay. What fraction of sample of pure tritium will remain undecayed after 25 y.
Solution:
By definition of half-life, half of the initial i sample will remain undecayed after 12.5 y. In the next 12.5 y, one-half of these nuclei would have decayed. Hence, one fourth of the sample of the initial pure s tritium will remain undecayed.

Question 6.
We are given the following atomic masses:
\({ }_{92}^{238} \mathrm{U}\) = 238.05079 u
\({ }_{2}^{4} \mathrm{He}\) = 4.00260 u
\({ }_{90}^{234} \mathrm{Th}\) = 234.04363 u
\({ }_{1}^{1} \mathrm{H}\) = 1.00783 u
\({ }_{91}^{237} \mathrm{Pa}\) = 237.05121 u
Here the symbol Pa is for the element protactinium (Z = 91).
a) Calculate the energy released during the alpha decay of \({ }_{92}^{238} \mathrm{U}\).
b) Show that \({ }_{92}^{238} \mathrm{U}\) cannot spontaneously emit a proton.
Solution:
a) The alpha decay of \({ }_{92}^{238} \mathrm{U}\) is given by

The energy released in this process is given by
Q = (M_U – M_Th – M_He) c²
Substituting the atomic masses as given in the data, we find
Q = (238.05079 – 234.04363 – 4.00260)u × c²
= (0.00456 u) c²
= (0.00456 u) (931.5 MeV/u)
= 4.25 MeV

b) If \({ }_{92}^{238} \mathrm{U}\) spontaneously emits a proton, the decay process would be
\({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{91}^{237} \mathrm{~Pa}+{ }_1^1 \mathrm{H}\)
The Q for this process to happen is
= (M_U – M_Pa – M_H)c²
(238.05079 – 237.05121 – 1.00783) u × c²
=(- 0.00825 u) c²
= – (0.00825 u) (931.5 MeV/u)
= -7.68 MeV
Thus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply an energy of 7.68 MeV to a \({ }_{92}^{238} \mathrm{U}\) nucleus to make it emit a proton.

Students get through AP Inter 2nd Year Physics Important Questions 13th Lesson Atoms which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 13th Lesson Atoms

Very Short Answer Questions

Question 1.
What is the physical meaning of negative energy of an electron’ ?
Answer:
The ‘negative energy of an electron’ indicates that the electron is bound to the nucleus due to force of attraction.

Question 2.
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 3.
Name a physical quantity whose dimensions are the same as those of angular momentum.
Answer:
Planck’s constant.

Question 4.
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 5.
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 6.
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 7.
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 8.
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 will agree with the values of wavelengths observed experimentally by Lyman.

Question 9.
Write down a table giving longest and shortest wavelengths of different spectral series.
Answer:
Wavelength limits of some spectral series of hydrogen.

Question 10.
Give two drawbacks of Rutherford’s atomic model.
Answer:
Drawbacks of Rutherford’s atomic model:

1. As the revolving electron loses energy continuously, it must spiral inwards and eventually fall into the nucleus. But matter .is stable, we cannot expect the atom collapse.
2. The atoms should emit continuous spectrum, but what we observe is only a line spectrum.

Question 11.
If the kinetic energy of revolving electron in an orbit is K, what is its potential energy and total energy ?
Answer:
For an electron revolving round the nucleus, total energy is always negative and it is numerically equal to kinetic energy.
∴ Total energy = -Kinetic energy = -K
Potential energy is always negative and PE = 2 × TE = -2K

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 asymtotic direction in which it receeds.
3. The relation between b and θ is b = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Ze}^2}{\mathrm{E}}\) cot \(\frac{\theta}{2}\) where E = K.E of α – particle = \(\frac{1}{2}\) mυ².

Question 2.
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 cannot 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.
   
   The closest distance of approach,
   d = \(\frac{1}{4 \pi \varepsilon_0} \times \frac{Z e^2}{\left(\frac{1}{2} m v^2\right)}\)
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 3.
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^-10 m.
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 is provided by the electrostatic force of attraction between the electrons and the nucleus.

Drawbacks : 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 4.
What are the limitations of Bohr’s theory of hydrogen atom ? [A.P. Mar. 17; 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 elliptical 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 5.
Write a short note on Debroglie’s explanation of Bohr’s second postulate of quantization. [T.S. Mar. 17]
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 = \(\frac{\mathrm{nh}}{2 \pi}\) where n = 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 n^th circular orbit of radius r_n, total distance covered = circumference of the orbit = 2πr_n
   ∴ For permissible orbit, 2πr_n = nλ
7. According to Debrogile, λ = \(\frac{h}{m v_n}\)
   Where υ_n is speed of electron revolving in n^th orbit
   ∴ 2πr_n = \(\frac{\mathrm{nh}}{\mathrm{m} v_{\mathrm{n}}}\)
   mυ_nr_n = \(\frac{\mathrm{nh}}{2 \pi}=\mathrm{n}\left(\frac{\mathrm{h}}{2 \pi}\right)\)
   i.e., angular momentum of electron revolving in n^th orbit must be an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\), which is the quantum condition proposed by Bohr in second postulate.

Question 6.
Explain the different types of spectral series in hydrogen atom. [T.S. Mar. 16, A.P. Mar. 15]
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

Long Answer Questions

Question 1.
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. [A.P. Mar. 16]
Answer:
a) Basic postulates of Bohr’s theory are

1. The electron revolves round a nucleus in an atom in various orbits known as stationary orbits. The electrons cannot 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 \(\frac{\mathrm{h}}{2 \pi}\)
   i.e., mυ_nr_n = \(\frac{\mathrm{nh}}{2 \pi}\) …………… (1)
   where n = 1, 2, 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 circular orbit of radius r_n.
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.

3) Radius of the oribit: Substituting the value of (6) in (2),
\(\frac{\mathrm{m}}{\mathrm{r}_{\mathrm{n}}}\left(\frac{\mathrm{n}^2 \mathrm{~h}^2}{4 \pi^2 \mathrm{r}_{\mathrm{n}}^2 \mathrm{~m}^2}\right)=\frac{\mathrm{Ke}^2}{\mathrm{r}_{\mathrm{n}}^2}\)
r_n = \(\frac{n^2 h^2}{4 \pi^2 m K e^2}\) ……………. (1)
∴ r_n = 0.53 n²

4) Total energy (E_n) : Revolving electron possess K.E. as well as P.E.

Problems

Question 1.
The radius of the first electron orbit of a hydrogen atom is 5.3 x HTum. What is the radius of the second orbit ?
Solution:
∴ r_n ∝ n²
\(\) ⇒ r₂ = 4r₁
∴ r₂ = 4 × 5.3 × 10^-11 = 2.12 × 10^-10m.

Question 2.
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 1^st orbit, E = -3.4 eV
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 = \(\frac{\mathrm{U}}{2}\) – u = \(\frac{\mathrm{-U}}{2}\)
U = -2E
∴ U = -2 × -3.4 = 6.8 eV.

Question 3.
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 . . ,

Question 4.
Prove that the ionisation energy of hydrogen atom is 13.6 eV.
Solution:
n = 1 corresponds to ground state.
E = \(\frac{-13.6}{\mathrm{n}^2}\) eV
E = \(\frac{-13.6}{\mathrm{1}^2}\) 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 5.
Calculate the ionization energy for a lithium atom.
Solution:
For 3^Li7 atom, Z = 3, n = 2 [∵Li = 1s² 2s¹]
E_n = \(\frac{13.6 \mathrm{Z}^2}{\mathrm{n}^2}\) eV
= \(\frac{13.6 \times(3)^2}{4}\) = 30.6 eV
∴ Ionization energy of Lithium = 30.6eV

Question 6.
The wavelength of the first member of Lyman series is 1216 A. Calculate the wavelength of second member of Balmer series.
Solution:

Textual Examples

Question 1.
In the Rutherford’s nuclear model of the atom, the nucleus (radius about 10^-15m) is analogous to the sun about which the electron move in orbit (radius ≈ 10^-10m) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the sun than actually it is ? The radius of earth’s orbit is about 1.5 × 10¹¹m. The radius of sun is taken as 7 × 10⁸ m.
Solution:
The ratio of the radius of electron’s orbit to the radius of nucleus is (10^-10 m)/(10^-15 m) = 105, that is, the radius of the electron’s orbit is 10⁵ times larger than the radius of nucleus. If the radius of the earth’s orbit around the sun were 10⁵ times larger than the radius of the sun, the radius of the earth’s orbit would be 10⁵ × 7 × 10⁸ m = 7 × 10¹³ m. This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the sun.

It implies that an atom contains a much greater fraction of empty space than our solar system does.

Question 2.
In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV α-particle before it comes momentarily to rest and reverses its direction?
Solution:
The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an α-particle and a gold nucleus is conserved. The system’s initial mechanical energy is E_i; before the particle and nucleus interact, and it is equal to its mechanical energy E_f when the α-particle momentarily stops. The initial energy E_t is just the kinetic energy K of the incoming α-particle. The final energy E_f is just the electric potential energy U of the system. The potential energy U can be calculated from Equation.
F = \(\frac{1}{4 \pi \varepsilon_0} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{r}^2}\)
Let d be the centre-to-centre distance between the α-particle and the gold nucleus when the α-particle is at its stopping point. Then we can write the conservation of energy
E_i = E_f as
K = \(\frac{1}{4 \pi \varepsilon_0} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{d}}=\frac{2 \mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{~d}}\)
Thus the distance of closest approach d is given by
d = \(\frac{2 \mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{~K}}\)
The maximum kinetic energy found in α-particles of natural origin is 7.7 MeV or 1.2 × 10^-12 J. Since 1/4πε₀ = 9.0 × 10⁹N m²/C². Therefore with e = 1.6 × 10^-19C, we have,
d = \(\frac{(2)\left(9.0 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^2 \mathrm{Z}}{1.2 \times 10^{-12} \mathrm{~J}}\)
= 3.84 × 10^-16 Zm
The atomic number of foil material gold is Z = 79, so that
d (Au) = 3.0 × 10^-14m = 30 fm. (1 fm (i.e. fermi) = 10^-15m.)
The radius of gold nucleus is, therefore, less than 3.0 × 10^-14 m. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the a-particle. Thus, the α-particle reverses its motion without ever actually touching the gold nucleus.

Question 3.
It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.
Solution:
Total energy of the electron in hydrogen atom is – 13.6 eV = -13.6 × 1.6 × 10^-19J
= -2.2 × 10^-18 J.
Thus from equation, E = – \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) we have
– \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) = 2.2 10^-18 J
This gives the orbital radius
r = – \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) = \(\frac{\left(9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^2}{(2)\left(-2.2 \times 10^{-18} \mathrm{~J}\right)}\)
= 5.3 × 10^-11 m.
The velocity of the revolving electron can be computed from Equation r = –\(\frac{e^2}{4 \pi \varepsilon_0 r m v^2}\)
with
m = 9.1 × 10^-31kg,
υ = \(\frac{\mathrm{e}}{\sqrt{4 \pi \varepsilon_0 \mathrm{mr}}}\) = 2.2 × 10⁶ m/s

Question 4.
According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom.
Solution:
From Examjple 3 we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10^-11 m is 2.2 × 10^-6 m/s. Thus, the frequency of the electron moving around the proton is
v = \(\frac{v}{2 \pi \mathrm{r}}=\frac{2.2 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}}{2 \pi\left(5.3 \times 10^{-11} \mathrm{~m}\right)}\)
≈ 6.6 × 10¹⁵ Hz.
According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is 6.6 × 10¹⁵Hz.

Question 5.
A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite.
Solution:
From equation, we have
mυ_nr_n = nh/2π
Here m = 10 kg and r_n = 8 × 10⁶ m. We have the time period T of the circling satellite as 2h. That is T = 7200 s.
Thus the velocity υ_n = 2π r_n/T
The quantum number of the orbit of satellite
n = (2π r_n)² × m(T × h)
Substituting the values,
n = (2π × 8 × 10⁶m)² × 10/(7200 s × 6.64 × 10^-34 J s)
= 5.3 × 10⁴⁵
Note that the quantum number for the satellite motion is extremely large. In fact for such large quantum numbers the results of quantisation conditions tend to those of classical physics.

Question 6.
Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.
Solution:
The Rydberg formula is
hc/λ_if = \(\left(\frac{1}{n_{\mathrm{f}}^2}-\frac{1}{\mathrm{n}_{\mathrm{i}}^2}\right)\)
The wavelengths of the first four lines in the Lyman series correspond to transitions from n_i = 2, 3, 4, 5 to n_f = 1. We know that
\(\frac{\mathrm{me}^4}{8 \varepsilon_0^2 \mathrm{~h}^2}\) = 13.6 eV = 21.76 × 10^-19 J
Therefore,

Substituting, n_i = 2, 3, 4, 5 we get λ₂₁ = 1218 Å, λ₃₁ = 1028 Å, λ₄₁ = 974.3 Å, and λ₅₁ = 951.4 Å.