Inter 2nd Year

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material 8th Lesson Applied Biology Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material 8th Lesson Applied Biology

Very Short Answer Questions

Question 1.
What are the factors constitute dairying?
Answer:

1. Selection of good breeds having high yielding potential, combined with disease resistance ones.
2. Proper housing with adequate water, feed, ventilation suitable temperature, etc.

Question 2.
Mention any two advantages of inbreeding.
Answer:

1. Inbreeding increases homozygosity. Thus inbreeding is necessary if we want to evolve a pure line animal.
2. It helps in the accumulation of superior genes and the elimination of less desirable genes.

Question 3.
Distinguish between out-cross and cross-breed.
Answer:
Out cross :
The offspring formed by mating of animals within the same breed, but having no ancestors on either side of pedigree for 4-6 generations.

A single out cross helps to overcome inbreeding depression.

Cross breed :
The offspring formed by a mating between superior males of one breed and superior females another breed.

Cross breed shows desirable qualities of two different breeds to be combined.

Question 4.
Define the terms layer and broiler.
Answer:
Layer :
The birds which are raised exclusively for the production of eggs are called layers.

Boiler :
The birds which are raised only for their meat are called broilers.

Question 5.
What is apiculture?
Answer:
Apiculture is the maintenance of hives of honeybees for the production of honey and wax.

Apiculture is an age-old cottage industry.

Question 6.
Distinguish between a drone and worker in honey bee colony.
Answer:

Drones	worker bees
1) These are fertile males.	1) These are sterile female.
2) These are developed from unfertilized ova by male parthenogenesis.	2) These are developed from fertilized eggs.
3) These are short lived.	3) These live for two and three months.

Question 7.
Define the term Fishery.
Answer:
Fishery is an industry devoted to the catching, processing for storage in freezers and selling of fish, shellfish or any other aquatic animals for human consumption.

Question 8.
Differentiate aquaculture and pisciculture.
Answer:

Aquaculture	Pisciculture
Culturing of fishes and other aquatic organisms under regulated conditions to achieve better production.	Culturing of exclusively fin fishes under regulated conditions to achieve better production.

Question 9.
Explain the term hypophysation.
Answer:
Making the fishds to breed artificially to meet the demand of carpseed as called hypophysation.

Question 10.
List out any two Indian carps and two exotic carps.
Answer:
Indian carps :

1. Catla catla (catla)
2. Cirrhinus mrigala (mrigal)

Exotic carps :

1. Grass carp
2. Silver carp

Question 11.
Mention any four fish by-products.
Answer:

1. Shark and cod liver oils
2. Fish guano
3. Shagreen
4. Isinglass.

Question 12.
How many aminoacids and polypeptide chains are present in insulin?
Answer:
Human insulin is made up of 51 aminoacids arranged in two polypeptides.
– polypeptide chain A with 21 aminoacids
– Polypeptide chain B with 30 aminoacids.

Which are held together by disulphide linkages.

Question 13.
Define the term vaccine.
Answer:
Vaccine is biological preparation that improves immunity to a particular disease. A vaccine typically contains live attenuated an inactivated disease causing organism. The toxins or one of the surface proteins of pathogens are also used in the preparation of vaccines.

Question 14.
Mention any two features of PCR.
Answer:

• Very low concentration of bacteria or viruses can be detected by amplification of their nucleic acids by PCR.
• PCR helps to detect very low amounts of DNA by amplification of the small DNA fragments.

PCR is now routinely used for detection of HIV in suspected cases, detection of mutations and genetic disorders.

Question 15.
What does ADA strand for? Deficiency of ADA causes which disease?
Answer:
ADA stands for adenosine deaminase. Deficiency of adenosine deaminase (ADA) causes severe combined immuno deficiency (SCID).

Question 16.
Define the term transgenic animal.
Answer:
Animals that have their own genome and had their DNA manipulated to possess and express an extra or foreign gene is known as transgenic animals.

Question 17.
What is popularly called “Guardian anger of Cell Genome?
Answer:
The protein p53 is a tumor suppressor protein, which plays an important role with reference to the ”G1 check point”. In the regulation of cell division cycle. It guards the integrity of the DNA. So it is also called guardian angel of cell’s genome.

Question 18.
List out any four features of cancer cells.
Answer:

• Loss of contact inhibition
• Reduced intra cellular adhesion
• Immortalization
• Loss of anchorage dependence

Question 19.
How do we obtain radiographs?
Answer:
A beam of X-rays is produced by an X-ray generator and is projected on the body parts. X-rays that pass through the body parts are recorded on a photographic film. Photographs developed using X-rays are known as radiographs.

Question 20.
What is tomogram?
Answer:
Tomogram is a recorded image formed by computed tomography which shows the 3-D cross sectional pictures of the part of the body and displays the picture on the screen.

Question 21.
MRI scan is harmless. Justify.
Answer:
MRI does not use ionizing radiation, as involved in X-rays, and is generally safe and harmless procedure.

Question 22.
What is electrocardiography and what are the normal components of ECG?
Answer:
Electrocardiography is a commonly used, non invasive procedure for recording electrical changes in the heart.

Normal components of ECG:
(i) Waves (ii) Intervals (iii) Segments (iv) Complexes.

Question 23.
What does prolonged F-R interval indicate?
Answer:
Prolonged P-R interval indicates delay in conduction of impulses from S-A node to the A-V node.

P-R interval is prolonged in bradycardia.

Question 24.
Differentiate between primary and secondary antibodies.
Answer:

Primary antibodies	Secondary antibodies
1) These antibodies are formed against the specific antigen.	1) These antibodies are formed against the foreign primary antibody.
2) These antibodies reacts with the antigens of interest.	2) These antibodies react with the primary antibodies.

Question 25.
Which substances in a sample are detected by direct and indirect ELISA respectively.
Answer:

1. Direct ELISA – used to detect antigens present in the sample.
2. Indirect ELISA – used to detect antibodies present in the sample.

Short Answer Questions

Question 1.
What are the various methods employed in animal breeding to improve livestock?
Answer:
Animal breeding is the method of mating closely related individuals.
There are broadly two methods in animal breeding. (1) In breeding (2) Out breeding
1) In breeding:
When crossing is done between animals of the same breed it is called in breeding. In breeding is of two types (a) Close breeding (b) Line breeding.
a) Close breeding:
Close breeding is mating between male parent and female offspring and/or female with male offspring.

b) Line breeding :
Line breeding is the selective breeding of animals for a desired feature by mating them within a closely related line. It leads to upgrading of a desired commercial character.

2) Out breeding:
Out breeding is the breeding of the unrelated animals. Out breeding is of three types (a) Out-crossing (b) Cross-breeding (c) Interspecific hybridisation.

a) Out-crossing :
Mating of animals within the same breed, but having no common ancestors on either side of pedigree for 4-6 generations. The off spring of such mating is known as an out-cross.

b) Cross-breeding :
In this method, superior males of one breed are mated with superior females of another breed. The offspring of such a mating is said to be a cross breed.

c) Interspecific hybridisation :
In this method, male and female animals of two different related species are mated. The progeny may combine desirable features of both the parents and is different from both the parents.

Question 2.
Define the term breed. What are the objectives of animal breeding ?
Answr:
Breed:
A breed is a group of animals related by descent and similar in most characters such as general appearance, size, configuration and features with other members of the same species.

Jersery and Brown Swiss are example of foreign breeds of cattle. These two varieties of cattle have the ability to produce abundant quantities of milk. This milk is very nutritious with high protein content.

Objects of animal breeding :

1. To produce disease resistant animals.
2. Increase in the quality and quantity of milk, meat, wool etc.,
3. Fast growth rate.
4. Enhanced productive life by improving the genetic merit of livestock.
5. Early maturity
6. Economy of feed

Question 3.
Explain the role of animal husbandry in human welfare.
Answer:
Animal husbandary deals with the scientific management of livestock. It includes various aspects such as feeding, breeding and control diseases to raise the population of livestock. Animal husbandary usually includes buffaloes, cows, pigs, horses, cattle, sheep, camels, goats, poultry, fish etc which are useful for humans in various ways.

These animals are managed for production of commercially important products such as milk, meat, wool, egg, honey, silk etc. The increase in human population has increased the demand of these products. Hence it is necessary to improve the management of livestock scientifically. ,

Question 4.
List out the various steps involved in MOET.
Answer:
The following are the steps involved in Multiple Ovulation and Embryo Transfer /MOET):

• A cow is administrated hormones, with FSH like activity.
• This induces follicular maturation and super ovulation.
• In Super ovulation instead of one egg, which they produce per cycle, they produce 6 – 8 eggs.
• The cow is either mated with elite bull or artificially inseminated.
• The embryos are at 8-32 called stages are recovered non-aurgically and transferred to surrogate mother, when the embryo develops into complete animal.

Now the genetic mother is ready for another round of super ovulation. This technology is in use for cattle, sheep, rabbits, buffaloes etc. to produce high yielding ones.

Question 5.
Write short notes on controlled breeding experiments.
Answer:
Controlled breeding experiments are carried out using artificial insemination and multiple ovulation and embryo transfer technology.

• In this technique the semen is collected from superior bulls. This semen can be used immediately or can be frozen and used later period. It can be transported in a frozen form to place where a female is housed.
• Meanwhile a cow or animal is administered hormones, with FSH like activity.
• These hormones induces follicular maturation and super ovulation.
• Now the cow is artificially inseminated for fertilisation.
• The embryos are at 8-32 celled stages are recovered non-surgically and transferred to surrogate mother uterus for further development.

This technology is use for cattle, sheep, rabbits, buffaloes etc. By using this method we can produce high milk and meat yielding animals and also control the venereal diseases.

Question 6.
Explain the important components of poultry management.
Answer:
Important components of poultry management:

Selection of disease free and suitable breeds:
The selected’breeds should be disease free and get acclimatised to a wide range of climatic conditions. Eg: In India Hybrid layers-BV 300, Hyline, Poona – Pearls etc., Broiler strains – Hubbard, Vencobb etc.

Feed management:
Balanced diet is must to maximise the yield. Brooder, chick mash, grower mash, prelayer mash and layer mash are fed to layers at different stages. Likewise pre starter mash, starter mash and finish mash are the feed given to broilers. Safewater should be supplied through waterers at all times.

Health care :
Vaccination against viral diseases and using antibodies for both bacterial and fungal diseases.

In addition to the above hygiene, proper and safe farm conditions ensure better produce.

Question 7.
Discuss in brief about ‘AvianFlu’.
Answer:
AvianFlu or birdFlu is an important disease affecting poultry birds and man.

Causative organism :
AvianFlu or birdFlu is caused by an “avianFlu virus” the H5NI. The virus that causes the bird infection infects humans too. It is a pandemic disease.

Mode of infection:
Infection may be spread simply by touching contaminated surfaces. Birds infected by this type of influenza, continue to release the virus as in their faeces and saliva for as long as 10 days.

Symptoms:
In humans it causes typical-flu-like symptoms, include cough, diarrhoea difficulty in breathing, fever, headache, malaise, muscle aches and sore throat.

Prevention :

• Avoiding consumption of under cooked chicken.
• People who work for poultry birds should use protective clothing and special breathing masks.
• Complete culling of infected flock by burying or burning them.

Question 8.
Explain in brief about queen bee.
Answer:

• Queen bee is the largest individual in the colony.
• It is a fertile diploid female, one per bee hive and the egg layer of the colony.
• She lives for about five years and her only function is to lay eggs.
• The queen bee during its nuptial flight receives sperms from a drone and stores in the spermathecae and lays two types of eggs, the fertilised and unfertilised.
• All fertilised eggs develop into females.
• All the larvae developing from the fertilised eggs are fed with the royal jelly for first four days only. Afterwards royal jelly is fed only to the bee that is bound to develop into next queen, whereas the other larvae fed on bee bread become workers.

Question 9.
Honey bees are economically important – justify.
Answer:

• Honeybees are economically important insects in the world. Because honeybee products like Honey, wax, propolis and beevenom have more economic importance.
• Honey – It is a rich source of fructose, glucose, water minerals and vitamins.
• Bee’s wax – It is used in the preparation of cosmetics, polishes of various kinds and candles.
• Propolis – Propolis is used in the treatment of inflammation and superficial bums.
• Bee’s Venom – Which extracted .from the string of worker bees is used in the treatment of rheumatoid arthritis.
• Pollination – Bees are the pollinators of our crop plants such as sunflower, Brassica, Apple and Pear.

Question 10.
What are the various factors required for Bee keeping?
Answer:
Bee keeping or apiculture is the maintenance of hives of honeybees for the production of honey and wax.

Factors required for successful Bee keeping :

1. Knowledge of nature and habits of honeybees.
2. Selection of suitable location for keeping the beehives. ‘
3. Raising a hive with the help of a queen and small group of worker bees.
4. Management of beehives during different seasons.
5. Knowledge of handling procedures and collection of honey and bee wax.

Question 11.
Fisheries have carved a niche in Indian economy. Explain.
Answer:
Fisheries have carved a niche in Indian economy, as fisheries have more economic importance.

As food :
Fish meat, in general is a good source of proteins, vitamins, minerals and rich in iodine. Tunas, shrimps and crabs are not only edible but also have export value.

Byproducts :

1. Shark and Cod liver oils – are good source of vitamins A and D.
   Oils from Sardine and Salmon- are good source of Omega 3 – fatty acids.
2. Fish guano from Scarp fish – used as fertilizer.
3. Shagree and I$inglass – used in clarification of wines.

Question 12.
Explain in brief structure of Insulin.
Answer:

Insulin is a poly peptide hormone produced by the β – cells of islets of langerhans of pancreas. It is the first protein produced by recombinant DNA technology.

Structure of Insulin :
Human insulin is made up of 51 aminoacids arranged in two polypeptide chains. The chain A has 21 aminoacids while chain B has 30 aminoacids. Both are held together by two interchain disulfide bridges, connecting A₇ to B₇ arid A₂₀ to B₁₉. In addition, there is an intrachain disulfide link in chain A between the aminoacids 6 and 11.

In mammals, including humans, insulin is synthesized as a pro-hormone, which contains an extra stretch called the ‘c’ peptide. This ‘c’ peptide is not present in the mature insulin and is removed during maturation into insulin. .

Question 13.
Define Vaccine and discuss about types of Vaccines.
Answer:
A Vaccine is a biological preparation that improves immunity to a particular diseases. A Vaccine typically contains inactive or attenuated disease causing microorganisms. The toxin or one of the surface proteins of the microorganisms are also used in preparing vaccines.

Types of Vaccines :
1) Attenuated whole agent vaccines :
They contain disabled line microorganisms. Mostly they are antiviral. Eg: Vaccines against Yellow fever, measles, rubella and mumps and the bacterial disease such as typhoid.

2) Inactivated whole agent vaccines :
They contain killed microbes. Eg : Vaccines against influenza, cholera, hepatitis A, rabies etc.

3) Toxoids:
They contain toxoids which are inactivated exotoxins of certain microbes.
Eg : The vaccines against diphtheria and tetanus.

Question 14.
Write in brief the types of gene therapy. .
Answer:
Gene therapy is the insertion of genes into an individual’s cells and tissues to treat a
There are two approaches to achieve gene therapy :

1. Somatic line therapy
2. Germ line gene therapy

1) Somatic line therapy:
In this type of therapy, functional genes are introduced into somatic cells of a patient. The approach is to correct a disease phenotype by treating defect in somatic cells in the affected person. The changes effected in this type of gene therapy are nonfinheritable.
Somatic line therapy is of two types :
a) Ex-vivo gene therapy:
In which the cell are collected from patient, modified outside the body and then transplanted back Eg: SCID.

b) In-vivo gene therapy :
In this therapy, the genes are changed in cells, while they are still inside the body Eg : Cystic fibrosis.’

2) Germ line gene therapy:
In this type of therapy, functional genes are introduced into sperms or ova and are thus integrated into their genomes. Therefore the changes or modifications become heritable. Due to various technical and ethical reasons, the germ line gene therapy remained at infant stage.

Question 15.
List out any four salient features of cancer cells.
Answer:
Salient features of Cancer cells :
Loss of contact inhibition :
Normal cells in a culture stop growing when their plasma membranes come into contact with one another. This inhibition of growth after contact is called contact inhibition. Cancer cells lose this property.

Reduced intracellular adhesion :
When normal cells growing in medium, the cells are joined by intracellular adhesion proteins called cadherins. They are missing in Cancer cells.

Immortalisation :
Normal cell culture does not survive indefinitely. They undergo apoptosis. Where as Cancer cells do not undergo apoptosis.

Loss of anchorage dependence :
Most normal cells must be attached to a rigid substratum in order to grow Cancer cells can grow even when they are not attached to the substratum.

Increased growth of blood vessels :
When tumors grow in size diffusion of oxygen and nutrients become restricted and so tumors resort to attracting more blood vessels from their surrounding matrix.

Question 16.
Explain the different types of cancers.
Answer:
Based on the origin Cancers are classified into :
1) Carcinomas :
These are malignant tumor of epithelial cells. They are originating from the epithelial tissues of skin, lining of the respiratory, digestive, urinary and genejal systems or cells from various glands breast and nervous tissue etc. 85% of Cancers are Carcinomas.

2) Sarcomas :
These are malignant tumors of connective tissues or organs that originate from mesoderm. About 2% of tumors are Sarcomas.

3) Leukemias :
These are malignant tumors of stem ceils of hematopoietic tissues, resulting in unrestrained production of WBC. They are liquid tumors. About 4% of Cancers are Leukemias.

4) Lymphomas :
These are malignant tumors of secondary lymphoid organs like spleen, and lymphnodes. About 4% of Cancers are Lymphomas.

Question 17.
Write about the procedure involved in MRI. X jmsin
Answer:
MRI Scan is a diagnostic radiology technique that uses magnetism, radiowaves and a computer to produce images of body components.

Procedure :
MRI Scanner is giant circular magnetic tube.

• The patient is placed on a movable bed that is inserted into the magnet.
• Human body is mainly composed of water which contains two protons.
• The magnet creates a strong magnetic field that makes these proton align with the direction of the magnetic field.
• A second radiofrequency electromagnetic field is then turned on for a brief period. The protons absorb some energy from these radio waves.
• When this second radio frequency emitting field is turned off, the protons release energy at a radio frequency which can be detected by the MRI scanner.
• Different types of tissues emit different quanta of energy. Abnormal tissues such as tumors can be detected because the protons in different types of tissues return to their equilibrium state at different rates.
• Tissues of bone with less water content look different in MRI, and pathological tissues also can be detected.

The information received is processed by computer and generated an image.

Question 18.
Write briefly about different waves and intervals in an ECG. X
Answer:
ECG (electrocardiography) is commonly used, non-invasive procedure fro recording electrical changes in the heart.

The graphic record which is called an electrocardiogram, shows the series of waves that occur during each cardiac cycle.

The normal ECG consists of (i) Waves (ii) Intervals (iii) Segments (iv) Complexes.

i) Waves :

• The waves in a normal record are named P, Q, R, S and T in that order.
• A typical ECG tracing of a normal heartbeat consists of (I) a ’P’ wove (II) a ‘QRS complex of waves’ (III) a T Wave.
• P wave: It represents the atrial systole and shows that the impulse is passing through atria. The duration of P. Wave is 0.1 sec.
• QRS complex of wave : It represents ventricular systole. Q wave is small negative., R-wave is tall positive and S wave is a negative wave. Its duration is 0.08 to 0.1 sec.
• T wave: It represents the ventricular repolarization. It is a positive wave,’its duration is 0.2 sec.

ii) Intervals:
P-R intervals :
P-R intervals is the interval between the onset of p wave and the onset of Q wave. P-R interval is normally. 0.12 – 02 sec.

Q-T intervals :
The interval between the onset of Q wave and the end of the T-wave. It represents the electrical activity in muscle of the ventricles. It lasts for about 0.4 seconds.

R-R intervals:
It signifies the duration of one cardiac cycle and lasts for about 0.8 sec

Segments :
S-T segment is the time period between the end of the ‘S’ wave and the onset of the T-wave. It is an isoelectric or zero voltage period.

Question 19.
Discuss briefly the process of indirect ELISA.
Answer:
Enzyme linked immunosorbent assay is a tool of clinical immunology to detect, antigens or antibodies in a given sample. ELISA is of two types (1) Direct ELISA (2) Indirect EUSA.

Indirect ELISA:
It is used to detect antibodies present in the serum of the patient or given sample.

Protocol

• A known antigen is added to the well, which absorbed on the surface of well.
• Patients antiserum is added to AG coated well.
• Allowed to react antibodies present in the serum with the antigen, coated on the surface of the well.
• Washed the well to remove the any unbound free antibodies present in the well.
• Enzyme linked antihuman serum globulins are added. They bind to the antibody which is already bound to the antigen.
• Washed it to remove excess antibodies present m the well.
• Enzyme substrate is added and the reaction produces a visible colour change which can be measured by a spectro photometer.

If there are no antibodies (i.e., anti HIV antibodyies in the serum sample, there is no binding of primary antibodies to the antigens and so enzyme linked secondary antibodies do not bind to the primary antibodies. There cannot be any enzymatic reaction and so no colour change is observed the test is said to be negative.

Question 20.
Write short note on EEG.
Answer:
Electro encephalography is the process of recording the electrical activity of the brain with help of an EEG machine and some electrodes placed all over the scalp.

The waves recorded by an EEG consist of synchronized waves which are common in normal healthy people and, in certain neurological conditions the waves are desynchronized. The wave pattern can be broadly classified into alpha, beta, delta and theta wave pattern.

Alpha waves :
They are rhythmical 8-13 cycleslsec. This type of Wave pattern is seen in persons who are drowsy or sleepy with closed eyes.

Beta waves:
These waves occur at a high frequency of 13-40 cycleslsec their amplitude is low. There are desynchronized waves recorded in person who are mentally very active and tense.

Delta waves :
Their frequency is quite low i.e., less than 3 cycleslsec. They have high amplitude. They are common in early childhood in awaken condition. In adults, they occur in deep sleep, epilepsy, mental depression etc. .

Theta waves:
Their frequency is between 4 and 7 cycleslsec. These waves are common in children of less than 5 years of age and emotional stress in adults.

Uses :

• EEG is useful tool in diagnosing neurological apd sleep disorders.
• The diagnostic application of EEG is the diagnosis of epilepsy.
• EEG is also useful in the diagnosis of coma and brain death.

Long Answer Questions

Question 1.
Write in detail about outbreeding.
Answer:
Out breeding is the breeding of the unrelated animals, it is the cross between different breeds.
Out breeding is of three types

1. Out crossing
2. Cross breeding
3. Inter specific hybridisation.

1) Out crossing :
It is the practice of mating of animals with in the same breed, but having no common ancestors on either side of the pedigree for 4-6 generations. The offspring of such a mating is known as an outcroas. It is the best breading method for animals that are below average in milk production, growth rate etc.

2) Cross breeding :
In this method, superior males of one breed are mated with superior females of another breed. The offspring of such a mating is said to be a cross breed. Cross breeding allows the desirable qualities of two different breeds to be combined. The progeny is not only used for commercial production but also inbreeding and selection to develop stable breeds which may be superior to existing breeds.
Eg : Hisardale is a new breed of sheep developed by crossing Bikaneri ewes and Marino rams. ‘ . .

3) Inter specific hybridisation :
In this method, male and female animals of two different related species are mated. The progeny may combine desirable features of both the parents and is different from both the parents.
Eg: 1) When a male donkey is crossed with a female horse, it leads to the production of “mule” (sterile/
2) When a male horse is crossed with a female donkey “hinny” (sterile) is produced. Mules have considerable economic value.

Question 2.
Explain in detail clinical inferences from ECG. –
Answer:
ECG is commonly used, non-invassive procedure for recording electrical changes in the heart The graphic record is called an electrocardiogram, shows the series of waves that occur during each cardiac cycle.

Normal ECG consist of waves, intervals, segments and complexes.

Waves :
A typical ECG tracing normal heart beat consist of a ‘P’ wave a QRS complex of waves, a T wave.

P wave :
It represents the atrial systole and shows that the impulse is passing through atria. The duration of P wave is 0./ sec.

QRS complex of wave:
It represents ventricular systole Q wave is small negative, R-Wave is tall positive and S-yvave is a negative wave. It’s donation is 0.08 to 0.1 sec.

T wave :
It represents the ventricular repolarization. It is a positive wave, its duration is 0.2 sec.

Intervals:
P – R intervals :
It is the interval between 9nset of P wave and onset of Q wave. P-R interval is normally 0.12-0.2 sec.

Q – T intervals :
The interval between the onset of Q wave and the end of the • T-wave. It represents the electrical activity in muscle of the ventricles. It lasts for about 0.4 sec.

R – R intervals :
It signifies the duration of one cardiac cycle and lasts for about 0.8 sec . .

Segments :
S-T segment is the time-period between the end of the ‘S’ wave and the onset of the T-wave. it is an isoelectric or zero voltage period.

Clinical inferences from ECG :

1. Enlarged P wave – indicates enlarged atria
2. Variation in the duration, amplitude and morphology of the QRS complex – indicates disorders such as block of conduction of impulses through the branches of the bundle of His.
3. Prolonged P-R interval duration – indicates delay in conduction of impulses from S-A node to the A – V node.
   P-R interval is prolonged in bradycardia and shortened in tachycardia.
4. Prolonged Q-T interval – indicates myocardial infraction and hypothyroidism.
5. Shortened Q.T interval – indicates hyper calcemio.
6. Elevated S – T segment – indicates myocardial infarction.
7. Tall T wave – indicates hyperkalemia.
8. Small, flat or inverted T wave – indicates hypokalemia.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material 6th Lesson Genetics Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material 6th Lesson Genetics

Very Short Answer Questions

Question 1.
What is Pleiotropy?
Answer:
The ability of a gene to have multiple phenotypic effects because it influences a number of characters simultaneously is known as Pleiotropy.
Eg: Phenylketonuria.

Question 2.
What are the antigens causing ‘ABO’ blood grouping? Where are they present?
Answer:
Isoagglutinogen A (antigen A) and Isoagglutinogen B (antigen B) are the antigens responsible for ABO blood grouping. These antigens are present on the surface of red blood cells.

Question 3.
What are the antibodies of ABO blood grouping? Where are they present?
Answer:
Isoagglutinin A (anti A) and Isoagglutinin B (anti B) are the antibodies of ABO blood grouping. These antibodies are present in the blood plasma.

Question 4.
What are multiple alleles?
Answer:
If a gene has more than two alleles then they are said to be multiple alleles.
Eg : In humans ABO blood groups are the best example for multiple allelism.

Question 5.
What is erythroblastosis foetalis?
Answer:
Erythroblastosis foetalsis is an alloimmune condition that develops in an Rh positive foetus whose father is Rh positive and mother is Rh negative.

In this disorder the antibodies developed against the Rh antigen in mother, cross, placenta and destroy the RBC cells of the Rh^+ve foetus during second pregnancy.

Question 6.
A child has blood group ‘O’. If the father has blood group A and mother blood group B, work out the genotypes of the parents and possible genotypes of the other off spring.
Answer:
Child blood group is ‘O’, and ‘O’ has the genotype I°I°. Hence, if father has blood group A and mother has blood group B, then the possible genotype of the parents will be I^AI° and I^BI° respectively.

Genotypes of the off springs
I^AI^B – AB blood group
I^AI° – A blood group
I^BI° – B blood group
I°I° – O blood group

Question 7.
What is the genetic basis of blood types in ABO system in man?
Answer:
Three alleles of gene I are responsible for ABO blood grouping. They are I^A, I^B and I°.
I^AI^A / I^AI° – for A blood group
I^BI^B / I^B I° – for B blood group
I^AI^B – for AB blood group
I° I° – for O blood group

Question 8.
What is polygenic inheritance?
Answer:
Polygenic inheritance is a cumulative effect of two or more genes on a single phenotypic character. Eg: Skin colour in humans.

Question 9.
Compare the importance of Y-chromosome in human being and Drosophila.
Answer:
In human beings Y-chromosomes are responsible for the development of maleness.

In Drosophila Y-chromosome, lacks male determing factors, but contains only genetic information essential to male fertility.

Question 10.
Distinguish between heterogametic and homogametic sex determination systems.
Answer:

Heterogametic	Homogametic
1. It is the condition in which two types of gametes are formed. Eg: XY-in humans.	1. It is the condition, in which similar type of gametes are formed. Eg: XX in females.
2. They play a very important role in deciding the sex of the off spring	2. It self, it can’t decide the sex of the progeny.

Question 11.
What is haplo-diploidy?
Answer:
It is a mechanism of sex determination. In this system the sex of the offspring is determined by the number of sets of chromosomes. Eg : Honeybees.

In honeybees fertilised eggs developed into female and unfertilised eggs developed into male. This means male have half the number of chromosomes ie., haploid and the females have double the number i.e., diploid hence the name haplo.-diploidy.

Question 12.
What are barr bodies?
Answer:
Barr bodies are condensed heterochromatin in one of the ‘X’ chromosome found in the somatic cells of diploid females. These were observed by Murray. L.Barr in female cats and Moor? and Barr in female human beings.

Question 13.
What is Klinefelter’s syndrome?
Answer:
Klinefelter’s syndrome is caused by trisome in 23rd pair. A klinefelter’s male possesses an additional X chromosome along with the normal XY (i.e.,47 chromosomes).

Symptoms :
Hypogonadism, sterility, enlargement of breast, high pitched voice etc., Somatic cells of Klinefelter male exhibits barr bodies in their nuclei.

Question 14.
What is Turner’s syndrome?
Answer:
It is an allosomal disorder. The Karyotype is 45, it is due to monosomy in 23rd pair. These females have 42 autosomes and one X-chromosome.

Symptoms:
Short structure, gonadal dysgenesis, webbed neck, broad shield like chest and widely spaced nipples etc. Turners female does not show barr bodies.

Question 15.
What is Down syndrome?
Answer:
Down Syndrome is a genetic condition that causes delay in physical and intellectual development. The cause of this genetic disorder is the presence of an additional copy of the chromosome numbered 21.

Symptoms :
The affected individual is short, with small round head, furrowed tongue and partially open mouth. Physical and mental development is retarded.

Question 16.
What is Lyonisation?
Answer:
Lyonisation is a process by which one of two copies of X-ehromosome present in the body cells of female mammals is inactivated. The inactive X-chromosome is transcriptionally inactive called heterochromatic body.

Question 17.
What is sex-linked inheritance?
Answer:
The inheritance of a trait that is determined by a gene located on one of the sex chromosome is called sex linked inheritance.
Eg: Colour blindness, Haemophilia etc.

Question 18.
Define hemizygous condition?
Answer:
The condition in which thd genes are present on non-homologous portion of either X- chromosome (or) Y-chromosomes. For these genes, related alleles are absent on corresponding paired chromosomes.
Eg : X-linked genes and Y-linked genes in males.

Question 19.
What is crisscross inheritance?
Answer:
Crisscross inheritance is also called as skip generation inheritance. The X-linked recessive characters do not occur in one generation. They skip it off’in that generation and are expressed in the next generation. Eg: X-linked recessive characters – Colour blindness.

Colour blindness is transmitted from grandfather to his grandson through a carrier daughter.

Question 20.
Why are sex – linked recessive characters more in male human beings?
Answer:
Sex linked recessive characters are more in males because these genes located in the X-chromosome. Male possess only one X-chromosome and female possess two ‘X’ chromosomes. So male needs only one copy of the mutant allele to express the phenotype.

Question 21.
Why are sex – linked dominant characters more in female human beings?
Answer:
In sex-linked dominant inheritance, the gene responsible for genetic disorder is located on the X-chromosome, and only one copy of the allele is sufficient to cause the disorder. Females are more likely to be affected by sex-linked dominant characters as the females have 2X-chromosomes, they have double chance to inherit the character.

Question 22.
What are sex limited characters?
Answer:
The genes for sex limited characters are autosomal genes present in both males and- females. Their phenotypic expression is limited to only one sex due to internal hormonal. environment. Eg: Development of breast in women, beard in man.

Question 23.
What are sex influenced characters?
Answer:
The genes for sex influenced characters are autosomal genes present in both males and females. In sex influenced inheritance, the genes’ behave differently in the two sexes. Probably because sex hormones,provide different cellular environment in males and females. Thus heterozygous phenotype may exhibit one phenotype in males and the contrasing one in females. E.g.: Baldness in humans.

Question 24.
How many base pairs are observed in human genome? What is the average number of base pairs in a human gene?
Answer:

1. 3164.7 million nucleotide base pairs were observed in a human genome.
2. The average number of base pairs in human gene is 3000.

Question 25.
What is junk DNA?
Answer:
The entire DNAin the nucleus does not code for proteins. Some DNA codes for specific proteins and Some DNA involve in the regulation of expression of genes, codes for proteins. The remaining non-functional DNA is called junk DNA.

Question 26.
What are VNTRs?
Answer:
These are repetitive DNA composed of a number of copies of short sequence. The VNTR of two persons generally shows variations, they differ in the number of tandem repeats or the sequence of bases. They are useful as genetic markers.

Question 27.
List out any two applications.of DNA fingerprinting technology.
Answer:

1. Medico-legal cases – Establishing paternity and (or) maternity more accurately.
2. Forensic analysis – Positive identification of a suspect in a crime.

Short Answer Questions

Question 1.
Briefly mention the contribution of T.H. Morgan to genetics.
Answer:
1) T.H. Morgan worked on Drosophila melanogaster for experimental verification of the chromosomal theory of inheritance to discover the basis for the variatiori that sexual reproduction produced.

2) He also Carried out dihybrid crosses in Drosophila to study the independent inheritance of two pairs of characters. He formulated the chromosomal theory of linkage. He defined linkage as co-existence of two or more genes in the same chromosome. His experiments have also proved that tightly linked genes show very low recombination while loosely linked genes show higher recombination.

3) T.H. Morgan worked on Drosophila melanogaster to analyse the behaviour of the two alleles of a fruit fly’s eye – colour gene and he discovered sex lihked inheritance.

4) Morgan discovery that transmission of X-chromosome in Drosophila correlates with the inheritance of an eye colour trait was the first solid evidence indicating that a specific gene is associated with a specific chromosome.

Question 2.
What is pedigree analysis? Suggest how such an analysis, can be useful?
Answer:
Pedigree analysis is a record of inheritance of certain traits over two or more ancestral generations of a person in the form of a diagram of family tree.

Uses:
→ Pedigree analysis is useful to study the inheritance of a specific trait, abnormality or disease etc.,
→ It also helps to work out the possible genotypes from the knowledge of the respective phenotypes.
→ It is useful to study the pattern of inheritance of a dominant or a recessive trait.
→ The possible genetic make up of a person for a trait can also be known with the help of pedigree chart.

Genetic counselors use pedigree chart for analysis of various traits and diseases in family and predict their inheritance patterns. It is useful in preventing hemophilia, sickle cell anemia and other genetic disorders in the future generations.

Question 3.
How is sex determined in human beings?
Answer:
The sex determination in humans is XX-XY type. In human beings both females and males have same number of chromosomes i.e., 23 pairs. Out of 23 pairs, 22 pairs are exactly same in males and females. These are called autosomes. In addition to these (autosomes) female possesses two ‘X’ chromosomes while male possess one ‘X’ and one ‘Y’ chromosome. During spermatogenesis among males, two types of gametes are produced. 50% of the total sperm produced carry the X-chromosome and the rest 50% has Y-chromosomes besides the autosomes. Females however, produce only one type of ovum with an X’ chromosome.

There is equal probability of fertilisation of ovum with sperm carrying either X or Y chromosome. In case the ovum is fertilised with sperm carrying X-chromosome, the zygote develops into a female and the fertilisation of ovum with Y-chromosome carrying sperm results into male offspring. Thus, the sex of the child depends on the type of sperm involved in the fertilisation.

Question 4.
Describe erythroblastosis foetalis.
Answer:
Erythroblastosis foetalis develops in an Rh positive foetus, whose father is Rh positive and mother is Rh negative. In Rh positive person rhesus antigens are present on the surface of blood cells where as in Rh negative person Rhesus antigens are absent.

During the process of delivery, the foetal blood cells may pass through the ruptured placenta into the Rh negative maternal blood. The mother’s, immune system recognises the Rh antigens and gets sensitized and produces Rh antibodies. These antibodies are Ig G type they can pass through placenta. Generally first child is not effected because child is delivered by the time of the mother gets sensitized and produce antibodies.

During second pregnancy, if the second child is Rh positive, these antibodies cross the placenta, enter the foetal blood circulation and destroy the Rh positive blood cell of foetus (haemolysis), leads to haemolytic anemia and jaundice. To compensate the haemolysis of blood cells there is a.rapid production of RBC’s from the bone marrow, and but also from liver and spleen. Now many large and immature blood cells in erythrobtast stage are released into circulation. Because of this disease is called erythroblastosis foetalis.

Question 5.
Mention any two autosomal genetic disorders with their symptoms.
Answer:
1) Sickle-cell anaemia :
It is an autosomal recessive genetic disorder, characterised * by rigid, sickle-shaped red blood cells in hypoxia conditions.

Sickle cell anaemia is due to point mutation in the DNA that codes for p – globin polypeptide chain of haemoglobin molecule, causing the replacement of the glutamic acid in the sixth position by valine.

Symptoms:
The sickled erythrocytes are fragile and their continuous breakdown leads to anaemia called sickle-cell anaemia.

The sickled cells block the capillaries resulting in poor blood supply to tissue. This leads to physical weakness, pain, organ danjage and even paralysis.

2) Phenylketonuria :
It is an autosomal recessive, metabolic genetic disorder caused by a mutation in a gene code for phenylalanine hydroxylase, located in chromosome 12 The affected individual lacks the phenylalanine hydroxylase enzyme, that converts the aminoacid phenylalanine into tyrosine, results in accumulation of phenylalanine in tissues later it is converted to phenylpyruvate and their derivatives. All these metabolities are excreted in urine.

Symptoms :
Accumulation of these substances in the brain causes mental retardation, failure to walk or talk, failure of growth etc.

Question 6.
Describe the genetic basis of ABO blood grouping.
Answer:
Bernstein proposed the genetic basis of ABO blood grouping. The genetic basis of ABO blood grouping is mainly dependent on the three alleles I^A, I^B and I° of the gene I, located in chromosome 9. The alleles I^A and I^B are responsible for the production of the respective antigens A and B on the surface of RBC. The allele I° does not produce any antigen on the surface of RBC. The alleles I^A and I^B are dominant to the allele lp, but codominant to each other (I^A = I^B >I°).

A child receives one of the three alleles from each parent, giving rise to six possible genotypes and four possible blood types. The genotypes are I^AI^A, I^AI°, I^BI^B, I^BI°, I^AI^B, I°I°.

The phenotypic expression of I^AI^A and I^AI° are A-type blood,

The phenotypic expression of I^BI^B and I^BI° are B-type blood, and that of I^AI^B is AB blood type. The phenotype I°I° is ‘O’ – type blood.

Question 7.
Describe male heterogamety.
Answer:
In this mechanism, the female sex has two ‘X’ chromosomes, while the male sex has only a single X chromosome. The heterogametic .male may be of the following two.types.

1) XX – XO type :
In certain insects belonging to orders Hemiptera (true bugs), Orthoptera (grass hoppers) and Dictyoptera (cockroaches), female has two X chromosomes (XX) and are, thus homogametic, while male has only single X’ chromosome (XO). The male being heterogametic sex produces two types of sperms, half with X chromosome and half without X chromosome in equal proportions. The sex of the offspring depends upon the sperm that fertilises the egg, each of which carries a single X chromosome.

Thus fertilisation between male and female gametes always produced zygotes with one X Chromosome from the female, but only 50% of the zygotes have an additional X Chromosome from the male. In this way, XO’ and ‘XX’ types would be formed in equal proportions, the former being males and the latter being females.

2) XX – XY type :
In man, other mammals, certain insects including Drosophila, the females possess two X chromosomes (XX) and are thus homogametic, produce one kind of eggs, each one with one X chromosome. While the males possess one X and one Y chromosome (XY) and are hence, heterogametic. They produce two kinds of sperms, half with X chromosome and half with Y chromosome. The sex of embryo depends on the kind of sperm. An egg fertilised by a X bearing sperm, produces a female, but if fertilised by a Y- bearing sperm, a male is produced.

Question 8.
Describe female hetergamety.
Answer:
In this method of sex determination the male produces similar type of gametes, while female produces dissimilar gametes. The heterogametic females may be of following two types.

i) ZO – ZZ :
This mechanism is found in certain moths and butterflies. In this case, female possesses one single ‘Z’ chromosome and hence is heterogametic, producing two kinds of eggs half with Z chromosome and another half without any Z chromosome. Male possesses two Z chromosomes and thus homogametic, producing single type of sperms, each carries single Z chromosome. The sex of the offspring depends on the kind of egg.

ii) ZW – ZZ:
This system is found in certain insects (gypsy moth) and vertebrates such as fishes, reptiles and birds. In this system, the female is heterogametic and produces two types of gametes, one with ‘Z’ chromosome and the other with ‘W chromosome. On the other hand, male is homogametic and produces all sperms of same type carrying one ‘Z’ chromosome. The sex of the offspring depends on the kind of egg being fertilised. The ‘Z’ chromosome bearing eggs produce males, but the W chromosome bearing eggs produces females.

Question 9.
Describe the Genic Balance Theory of sex determination.
Answer:
Genic balance mechanism of determination of sex was first observed and studied by C B.Bridges in 1921 in Drosophila. According to this mechanism, the sex of an individual in Drosophila melanogaster is determined by a balance between the genes for femaleness located in the X-chromosome and those for maleness located in autosomes. Hence, the sex of an individual is determined by the ratio of number of its X chromosomes and that of its autosomal sets, the Y chromosome being unimportant.

Individuals with sex index of 0.5 develop into normal males and those with sex index of 1 into normal females. If the sex index is between 0.5 and 1, the resulting individuals is called inter sex. Such individuals are sterile. Some flies have sex index of > 1, such flies are called super females or metafemales. Super male flies have a sex index value of < 0.5 and are also weak, sterile and non-viable.

Sex index = X/A	Phenotypes
0.5	Male
1.0	Female
Between 0.5 and 1	Inter sex
Below 0.5	Metamale
Above 1.0	Metafemale

Bridges drew, crossed a triploid females (3A + XXX) with normal diploid males (2A + XY). From such a cross he obtained normal diploid females, males, triploid females, intersexes, metamales and metafemales.

Question 10.
Explain in the inheritance of sex linked recessive character in human being.
Answer:
The sex linked recessive characters in human beings are : Colour blindness, Hemophilia etc.,

Colour Blindness :
Colour blindness if particular trait in human beings renders them unable to differentiate between the red colour and green colour. The gene for this colour blindness is located on X-chromosome. Colour blindness is recessive to normal vision so that if colour blind man marries a normal (homozygous) vision woman, all the sons and daughters are normal but daughters are heterozygous, which means that these daughters would be carriers to this trait. If such a carrier woman marries a man with normal vision, all the daughters and half of the sons have normal vision and half of sons are colour blind.

If carrier married to normal male

Hemophilia :
Hemophilia is the most notorious disease which is more common in men than women. This is also known as bleeder’s disease. It is the recessive character and is, therefore, masked in the heterozygous condition. Individuals suffering with this disease lack a factor responsible for clotting of blood. Consequently even a minor cut on the or in body surface may cause prolonged bleeding leading to death. Since it is caused by recessive X-lined gene, a lady may carry the disease and would transmit it to 50% of her sons, even if the father is normal.

Question 11.
Describe the experiment conducted by Morgan to explain sex linkage.
Answer:
Morgan worked on Drosophila melanogaster to analyse the behaviour of the two alleles of a fruitfly eye colour gene. From this work he discovered sex linkage.

Morgan’s experiment: When he crossed a white eyed (mutant) male to a normal (wild) red eyed female, in the F₁ generation all the males and females were red eyed.

When F₁ generation red eyed female was crossed to a red eyed male, in the F₂ generation all the females and 50% of males were red eyed and remaining 50% males were white eyed. This type of inheritance of a character from grand father to grand son is called criss cross inheritance.

In reciprocal cross, in which a white eyed female was crossed to a red eyed male, the F₁ resultant male offsprings had white eyes while the female offspring had red eyed. This proves that the allele responsible for the white eye is sex linked and recessive.

Question 12.
Explain the inheritance of sex influenced characters in human beings. ~
Answer:
Sex influenced genes are the autosomal genes, present in both males and females. In sex influenced inheritance, the genes behave differently in the two sexes, because the sex hormones provide different cellular environment in males and females Eg .’Baldness in humans.

The allele for baldness behave dominant (B) in males but recessive (b) in females. Pattern of baldness in man

Genotype	Male	Female
BB	Baldness	Baldness (less affect)
Bb	Baldness	Non-bald
bb	Non-bald	Non-bald

If a heterozygous non-bald woman (Bb) married a heterozygous bald man (Bb), in the offspring the ratio of bald to non bald ill the male progeny is 3:1 while in females it is 1:3.

Question 13.
A man and woman of normal vision have one son and one daughter. Son is colour blind and his son is with normal vision. Daughter is with normal vision, but one of her sons is colour blind and the other is normal. What are the genotypes of the father, mother, son and daughter?
Answer:
Man and woman of normal vision having colour blind son and normal vision daughter. So the genotype of women is carrier i.e., “X⁺X^–” and man is normal i.e., “X⁺Y”.

In the above cross colour blind son marries a normal woman his son will be normal.

Daughter with normal vision are of her son is colour blind means she must be carrier i.e., X⁺X^–“.

From the above reasons the genotype, of
Father is – X⁺Y Normal
Mother is – X⁺X^– Normal (carrier)
Son is – X^–Y Colour blind
Daughter is – X⁺X^– Normal (carrier)

Question 14.
A colour blind man married a woman who is the daughter of a colour blind father and mother homozygous normal vision. What is the probability of their daughters being colour blind?
Answer:
A colourblind man married a woman, who is daughter of a colourblind father and mother homozygous normal vision that means the woman is carrier i.e., the genotype is ‘X⁺X^–‘.

Here all women (daughters) are carriers, i.e., X⁺X^–
A cross between colour blin$ man a woman from the above result

From the above cross the probability of their daughter being colour blind is 50% or 1/2 among the daughters or 1/4 among their child’s

Question 15.
A heterozygous bald man who is non-haemophilic, married a woman who is homozygous for the non-bald trait and is haemophilic. What is the probability of her male children become bald and haemophilic?
Answer:
Man is heterozygous bald and non-haemophilic = Bb X⁺Y
Woman is homozygous non-bald and haemophilic = bb X^–X^–

Thus the probability of bald and haemophilic male is 1/2 i.e, 50% among males produced.

Question 16.
A woman’s father shows ‘IF but her mother and husband are normally pigmented. What will be the phenotypic ratio of her children?
Answer:
In continentia pigment is an uncommon disorder, inherited on an X-Iinked dominant manner. In this condition, a random loss of melanin from skin leads to mosaic appearance of skin. It is occur much more often in females than in males.

A woman’s father shows IP but her mother is normally pigmented, that means the woman also shows IP.

Cross between women with IP and normal male (husbend)

The phenotypic of children is 1 : 1

Question 17.
Write the salient of features of HGP.
Answer:
Salient features of HGP:

• The human genome comprised of 3164.7 million nucleotide bases.
• Human genome contains 30,000 genes.
• Each gene consist of ah average 3000 bases. ‘
• Functions of 50% of genes discovered are unknown.
• All proteins are coded by less than 2% of the genome.
• Majority of the genome consisted by repeated sequences.
• Chromosome one has highest number of genes i.e., 2,968 genes and Y chromosome has the fewest genes i.e., 231 genes.
• It is also identified that 1.4 millions locations, where Single base DNA difference (SNPs) occurs in humans. This information promises to revolutionise the process of finding chromosomal locations for disease associated sequences and tracing human history.

Question 18.
Describe the steps involved in DNA finger printing technology.
Answer:
DNA finger printing is a method for indentifying an individual by particular structure of their DNA.

Steps involved in DNA finger printing :
1. Obtaining DNA:
The DNA sample is collected from blood, Saliva, hair root, semen etc.,

2. Fragmentating DNA:
The DNA is treated with restriction enzymes to cut DNA at specific sites and form smaller fragments, .

3. Separation of DNA fragments:
The DNA fragments are separated by electrophoresis based on their charge and molecular weight.

4. Denaturaling of DNA:
The DNA on the gel is denatured by using alkaline chemicals.

5. Blotting :
Through a blotting technique the DNA fragments on the gel is transferred to nylon membrane. •

6. Using probes to identify specific DNA :
A radioactive probe is added to the DNA bands, which is complementary to the DNA bands, which is complementary to those of interested gene fragment.

7. Hybridization with probe :
After the probe hybridizes, excess probe washed off by washing. A photographic film is placed on the membrane containing DNA hybrids.

8. Exposure on film to make a DNA finger print:
The radio active label exposes the film to form bands corresponding to specific DNA bands.

Those bands form a pattern of bare which constitute a DNA finger print.

Long Answer Questions

Question 1.
What are multiple alleles? Describe multiple alleles with the help of ABO blood groups in man.
Answer:
Generally a gene has two alternative forms called allele. Sometimes a gene may have more than two alleles. These are referred to as multiple alleles. When more than two alleles exist fn a population of a specific organism, the phenomenon is called multiple allelism. Multiple’alleles cannot be observed in the genotype of a diploid individual, but can be observed in a population.

The number of genotypes that can occur for multiple alleles is given by the expression where ‘n’ = number of alleles.

ABO blood groups are the best example for multiple allelism in human beings.

The ABO blood group system was proposed by Karl Landsteiner. The blood groups A, B, AB and O types are characterised by the presence or absence of antigens on the surface of RBC. Blood type ‘A’ person have antigen A on their RBCs and anti-B antibodies in the plasma. Blood type ‘B’ person have antigen B. On their RBCs and anti-A. antibodies in the plasma. Blood type ‘AB’ person have antigens A

Blood group	Antigen on RBC	Antibodles in Plasma
A	A	b
B	B	a
AB	AB	—
O	—	A, b

Bernstein discovered that these phenotypes were’inherited by the interaction of three autosomal allies’ of the gene named T, located on chromosome 9. I_A, I_B and I° are the three alleles of the gene I. The alleles I_A and I_B are responsible for the production of the respective antigens A and B. The allele I° does not produce any antigen. The alleles I_A and I_B are dominant to the allele I° but co-dominant to each other (I_A = I_B > I°).

A child receives one of the three alleles from each parent, giving rise to six possible genotypes and four possible blood types. The genotypes are I_AI_A, I_AI°, I_BI_B, I_BI°, I_AI_B, I° I°. The phenotypic expressions of I_A I_A and I_A I°, are A-type blood, the phenotypic expression of I_B I_B and I_BI° are B-type blood, and that of I_A I_B, is AB blood type. The phenotype I° I° (ii) is ’O-type’ blood.

Question 2.
Describe chromosomal theory of sex determination.
Answer:
Chromosomal sex determination :
The chromosomes, which’ determine the somatic characters of an individual are known as autosomes. These chromosomes do not differ in morphology and number in male and female sex. Those chromosomes, which differ in morphology and number in male and female sex and contain genes responsible for the determination of sex are known as allosomes or sex chromosomes. There are two types of sex chromosomal mechanisms :
a) Heterogametic male and
b) Heterogametic female

a) Heterogametic male :
In this mechanism, the female sex has two ‘X’ chromosomes, While the male sex has only a single X chromosome.
The heterogametic male may be of the following two types :
(i) XX – XO (ii) XX – XY

i) XX – XO type :
In certain insects belonging to orders Hemiptera (true bugs), Orthoptera (grass hoppers) andDictyoptera (cockroaches) female has two X” chromosomes (XX) and are, thus homogametic, while male has only siftgle X” chromosome (XO). The, male being heterogametic sex produces two types of spgrms, half with X chromosome and . half without X chromosome in equal proportions. The sex of the offspring depends upon the sperm that fertilises the egg, each of which carries a singfe X chromosome. Thus fertilisation between male and female gametes always produced zygotes with oiie X chromosome from the female, but only 50% of the zygotes have an additional X Chromosome from the male. In this way, XO’ and ‘XX’ types Would be formed in equal proportions, the former being males and the latter being females.

ii) XX – XY type :
In man, other mammals, certain, insects including Drosphila, the females possess two X chromosomes (XX) and are thus homogametic, produce one kind of eggs, each one with one X chromosome. While the males possess one X and one Y chrbmosoihe (XY) and are hence, heterogametic. They produce two kinds of sperms, half with X chromosome and half with Y chromosome. The sex of embryo depends on the kind of sperm. An egg fertilised by a X bearing sperm, produces a female, but if fertilised by a Y bearing sperm, a male is produced.

b) Heterogametic female :
In this method of sex determination, the maid produces similar type of gametes, while female produce dissimilar gametes. The heterogametic fehiales may be of following two types, (i) ZO – ZZ (ii) ZW – ZZ.

i) ZO – ZZ :
This mechanism is found in certain moths and butterflies. In this case, female possesses one single ‘Z’ chromosome and hence is heterogametic, producing two kinds of eggs half witji Z chromosome and another half without any Z chromosome. Male possesses two Z chromosomes and thus homogametic, producing single type of sperms, each carries single Z chromosome. The sex of the offspring depends on the kind of egg.

ii) ZW – ZZ:
This system is found in certain insects (gypsy moth) and vertebrates such as fishes, reptiles and birds. In this system, the female is heterogametic and pi duces two types of gametes, onfe with ‘Z’ chromosome and the other with W chromosome. On the other hand, male is homogametic and produces all sperms of same type carrying one ‘Z’ chromosome. The sex of the offspring depends on the kind of egg being fertilised. The ‘Z’ chromosome bearing eggs produce males, but the W chromosome bearing eggs produces females.

Question 3.
What is crisscross inheritance ? Explain the inheritance of one sex linked recessive characters in human beings.
Answer:
The X-linked genes are represented twice in female (because female has two ‘X’- chromosomes) and once in males, (because male has one X-chromosome). In male single. X-linked recessive gene express it phenotypically, in contrast to female in which two ‘X’ linked recessive genes are necessary for the determination of a single phenotypic trait related to sex.

The recessive X-linked genes have chracteristic crisscrossinheritance.

Crisscross inheritance :
The inheritance of X-linked recessive trait (genes) to his grandson (F₂) through his daughter (carrier) is called crisscross inheritance. Crisscross inheritance can be explained in humans by sex-linked recessive disorder, colour blindness.

Colourblindness :
Colour blindness is a particular trait in human beings render them unable to .differentiate between red and green colour. The gene for this colour blindness is- located on X-chromosome. Colour blindness is recessive to normal vision so that if colour blind man marries a normal vision (homozygous) woman, all the sons and daughters are normal but daughter are heterozygous, which means that these daughters would be carrier for this trait. If such carrier woman marries a man with normal vision all the. daughters and half of the sons have normal vision and half of sons are colour blind.

Colour blind trait is inhereted from a male parent to his grandson through carrier daughter i.e., this trait shows crisscross inheritance

If carrier female is married to normal male

Characteristics of X-linked recessive traits :

• They never passed from father to son.
• Males are much more likely to be affected because they need only one copy of the mutant allele to express the phenotype.
• Affected males get the disease from their carrier mother only.
• Sons of heterozygous female (i.e., carrier female) have 50% chance of receiving mutant alleles. These disorders are typically passed from an affected grandfather to 50% of his grandsons.
• The X-linked recessive traits shows Crisscross pattern of inhertance.
  Eg : Colourblindness, Hemophilia, Muscular dystrophy etc.,

Question 4.
Write an essay on common genetic disorders.
Answer:
A number of disorders in human beings have been found to be associated with the inheritance of changed or altered genes of chromosomes.

Genetic disorders broadly grouped into two categories :
(1) Mendalian disorders, (2) Chromosomal disorders

1) Mendelian disorders :
These are genetic disorders showing Mendelian pattern of inheritance, caused by a single mutation in structure of DNA.

Most common and prevalent Mendelian disorders are: Haemophilia, Cystic fibrosis, sickle cell anaemia, colour blindness, phenyl ketonuria, thalassemia etc.,

I. Haemophilia : It is also called as bleeder’s disease.
(a) Haemophilia-A:
This is sex linked recessive disorder, transmitted by females and affecting males. Haemophilia-A is the rhost common clotting abnormality and is due to the deficiency of clotting factor VIII.

Symptoms :
The affected individuals have prolonged clotting time and suffer from internal bleeding. .

(b) Haemophilia-B :
This is due to the deficiency of clotting factor IX.

symptoms :
Symptoms are similar to that found in haemophilia-A.

II. Sickle-cell anaemia :
It is an autosomal recessive genetic disorder, characterised by rigid, sickle-shaped red blood cells in hypoxia condition. It is due to point mutation in the P-globin gene causing replacement of glutamic acid in the sixth position by valine.

Symptoms :
Haemolysis leads to sickle-cell anaemia sickle cells block. The capillaries resulting in poor blood supply to tissue leads to’ physical weakness, pain, organdamage, paralysis etc.,

III. Phenylketonuria:
This is an autosomal recessive metabolic genetic disorder caused by a mutation in the gene codes for phenylalanine hydroxylase. This enzyme catalyses the convertion of phenylalanine into tyrosine. Defect of this enzyme leads to accumulation of phenylalanine derivatives like pheriylpyruvate, phenylacetate etc.,

Symptoms :
Mental retardation, failure to walk or talk, failure of growth etc.,

IV. Colour blindness :
It is a sex linked disorder. It is the inability to differentiate between some colours. This phenotypic trait is dumb mutation in certain genes located in X-chromosome.

Symptoms : Protanopia – red colour blindness
Deuteranopia – green colour blindness
Tritanopia – blue colour blindness

V. Thalassemia :
Thalassemia is an autosome linked recessive blood disorder. Thalassemias are characterised by a defect in the a or 13 Globin chain, resulting in production of abnormal haemoglobin molecules leads to anaemia.

Symptoms : Anaemia .

VI. Cystic fibrosis :
It is an autosomal recessive genetic disorder. It is the result of mutation in the gene that influences salt and water movement across epithelial cell membrane.

Symptoms :
The mucus builds up in organs such as lungs, pancreas, GI tracts etc., If they are not treated it may lead to death.

2. Chromosomal disorders:
Chromosomal disorders are caused by errors in the number or structure of chromosomes.

Allosomal disorders :
I. Klinefelter’s syndrome :
This genetic disorder due to the presence of additional X-‘ chromosome along with the normal XY.
Symptoms : The resulting young sterile male shows feeble breast, small testicles, rounded hips etc., .

II. Turner’s syndrome:
A female with 44 autosomes with one X-chromosome, such females are sterile.
Symptoms : Short structure, webbed neck, broad shield chest with widely spaced nipples, poorly developed ovaries etc.,

Autosomal disorders :
I. Down syndrome (Trisomy 21):
The cause of this genetic disorder is the presence of an additional copy of chromosome numbered 21.
Symptoms : Small rounded head, furrowed tongue and partially open mouth mental retardent etc., –

II. Edwards syndrome (Trisomy 18):
This is due to presence of an extra copy of genetic material on the 18^th chromosome, either in whole or a part.
Symptoms : Majority of people with the syndrome die during the foetal stage due to defect in heart and kidney. .

III. Patau syndrome (Trisomy 13):
Patau syndrome is due to presence of an addition copy of chromosome number 13.
Symptoms : Kidney and heart defects, intellectual disability etc.,

Question 5.
Why is the human genome project c,ailed a mega project?
Answer:
Human genome project was an international effort formally begun in October, 1990. The HGP was a 13-year project coordinated by the U.S. Department of Energy and National Institute of Health. During early years of the HGP, the Wellcome trust became major partner and additional contributions came from Japan, France, Germany, China and others.

The total expenditure of this project is 3 billion dollars. This proeject almost completed in 2003.

Goals of HGP:

• Identify all the genes (20,000-25,000) in human DNA.
• Determine the sequence of entire human DNA.
• Improve tools for data analysis.
• Address the ethical, legal and social issues that may arise from the project.

Genome sequencing:
DNA sequencing is the process of determine the exact order of the 3 billion paired chemical building blocks that make up the DNA of the 22 autosomes X and Y chromosomes.

→ For sequencing the total DNA from a cell is isolated and converted into random fragments of relatively smaller size by using restriction enzymes and cloned in suitable most using specialised vectors.

→ The cloning results in the amplification of DNA fragments which are used for sequencing the bases.

→ Bacteria, yeast are most commonly used hosts and vectors are called ‘BAG’ and YAC’.

→ The fragments were sequenced using automated DNA sequencers that worked based on the principle of Sangers dideoxy method.

→ To allign these sequence a specialised computer based programs were developed, because it is humanly not possible.

→ These sequences were subsequently annotated and were assigned to each chromosome.

Salient features of human genome :
The human genome comprised of 3164.7 million nucleotide bases.

• Human genome contains 30,000 genes.
• Each gene consist of an average 30,000 bases.
• Functions are unknown for over 50% of the genes discovered.
• Lessthan 2% (nearly 1.5%) of the genome codes for proteins.
• Majority of the genome consisted of repeated sequences.
• Chromosome one has the highest number of genes i.e., 2,968 genes*and Y- chromosome has the fewest genes i.e., (231) genes.
• It is also identified that 1.4 million locations where single base DNA differences (SNPs) occur in humans. This information promises to revolutionise the process of finding chromosomal locations for disease associated sequences and tracing human history.

Advantages of HGP:

1. Identification and mapping of the genes responsible for diseases helps in diagnosis, treatement and prevention of these diseases.
2. It is useful to know the gene expression of different species, cellular growth, differentiation and evolutionary biology.
3. To improve gene therapy for genetic disorders.
   Becuase on the above views the human genome project was called a mega project.

Question 6.
What is DNA finger printing ? Mention its applications.
Answer:
DNA finger printing is a method for identifying individuals by the particular structure of their DNA.

Human DNA consists of 3 billion nucleoticdes, 90% Of which are identical among all individuals. No two people have exactly the same sequence of base in their DNA. Restricion fragment length polymorphism are characteristic to every person’s DNA. They are called Variable Number Tandem Repeats (VNTRs) and are useful as “Genetic markers”. The VNTRs of two persons generally show variations. DNA finger printing involves in dentifying differences in some specific regions in DNA sequence called repetitive DNA. These sequences show high degree of polymorphism and*form the basis of DNA finger printing.

Protocol of DNA finger printing :
1. Obtaining DNA:
DNA sample is collected from the blood, saliva, hair roots, semen etc. If needed many copies of the DNA is amplified by using PCR.

2. Fragmenting DNA (or) Restriction Digestion :
DNA sample is treated with restriction enzymes to cut the DNA at a specific sites and form smaller fragments.

3. Separation of DNA fragments by electrophoresis :
By using agarose gel electrophresis the DNA fragments are separated based on their charge and molecular weight.

4. Denaturing DNA:
The. DNA on the gel is denatured to form single stranded DNA strands using alkaline chemicals.

5. Blotting :
A thin’nylon membrane is placed over the size fractioned DNA strands and covered by paper towels. As the towels draw moisture the DNA strands are transferred on to-the nylon membrane by capillary action. This process is called blotting.

6. Using probes to identify specific DNA:
A radio active probe is added to the DNA bands. The probe is a single stranded DNA molecule that is complementary to the gene of interest in the sample under study. The probe attaches by base pairing to those restriction fragments that are complementary to its sequence.

7. Hybridization with probe :
After the probe hybridizes, the excess probe is washed off by washing. A photographic film is placed on the membrane containing DNA hybrids.

8. Exposure on film to make a DNA finger print:
The radioactive label exposes the film to form an image in the form of bands corresponding to specific DNA bands. These bands form a pattern of bars which constitute a DNA finger print.

Applications of DNA finger printing :

1. Conservation of wild life : Protection of endangered species, by maintaining their records for identification of tissues of the dead endangered organisms.
2. Taxonomical applications : Study of Phytogeny.
3. Pedigree analysis : Inheritance pattern of gene through generations.
4. Anthropological studies : Charting of origiij and migration of human population.
5. Medico-legal cases : Establishing paternity and or maternity more accurately.
6. Forensic analysis : Positive identification of a suspect in a crime.

The Process of DNA finger printing :
1. The process begins with a blood or cell sample from which the DNA is extracted.

2. The DNA is out into fragments using a restriction enzyme. The fragments are then separated into bands by electrophoresis through an agarose gel.

3. The DNA band pattern is transferred to a nylon membrane.

4. A radio active DNA probe is introduced. The DNA probe binds to specific DNA sequences on the nylon membrane.

5. The excess probe material is washed away leaving the unique DNA band pattern.

6. The radioactive DNA pattern is transferred to X-ray film by direct exposure. When developed, the resultant visble pattern is the DNA finger print.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 5(b) Reproductive Health Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 5(b) Reproductive Health

Very Short Answer Questions

Question 1.
What are the measures one has to take to prevent contracting STDs?
Answer:
The measures one has to be taken to prevent STDs are

1. Avoiding sex with unknown partners / multiple partners.
2. Using condoms compulsorily during coitus.
3. Consulting qualified doctor for early detection of STDs and getting complete treatment in case of infections.

Question 2.
What in your view are the reasons for population explosion, especially in India?
Answer:
The reasons for population explosion in India are

1. Illiteracy among people
2. Decline in death rate
3. Increased health care facilities.

Question 3.
It is true that ’MTP is not meant for population control’. Then why did the Government of India legalize MTP?
Answer:
’Medical Termination Pregnancy’ (MTP) or induced abortion is the procedure to terminate pregnancy with the help of medications. Government of India legalized MTP in 1971 to avoid its misuse, this is necessary to keep a check on indiscriminate and illegal female foeticides.

Question 4.
What is amniocentesis? Name any two disorders that can be detected by amniocentesis.
Answer:
Amniocentesis is a diagnostic procedure to detect genetic defects in the unborn baby, in which amniotic fluid is collected from foetus and diagnosed for abnormalities. Down’s syndrome, Turner’s syndrome and Edward’s syndrome can be detected by amniocentesis.

Question 5.
Mention the advantages of ‘lactational amenorrhea method’?
Answer:
Lactational amenorrhea is the absence of menstruation as long as mother breast feeds her baby.

The advantages of ‘lactational amenorrhea’ are

1. As long as the mother fully breast feeds her child, chances of conception are almost zero.
2. Breast feeding babies will have enhanced immunity, protection against allergies.

Short Answer Questions

Question 1.
Briefly describe the common sexually transmitted diseases in human beings.
Answer:
Sexually transmitted diseases (STDs) : Diseases or infections which are transmitted through sexual contact (intercourse) are’ collectively called sexually transmitted diseases (STDs) or Veneral Diseases (VDs) or Reproductive Tract Infections (RTI).

Most common STDs and their causative organisms are shown in the following table.

Name of the Disease	Causative organism
1. Gonorrhea	Neisseria gonorrhoeae (bacteria)
2. Syphilis	Treponema pallidum (spirochete bactrium)
3. Genital herpes	Herpes simplex virus (HSV)
4. Genital warts, cervical cancer	Human papilloma virus (HPV)
5. Trichomoniasis	Trichomonas vaginalis (a protozoan parasite)
6. Chlamydiasis	Chlamydia trachomatis (bacteria)
7. Hepatitis-B	HBV
8. HIV infection/AIDS	HTV (Human immunodeficiency virus)

Except for Hepatitis-B, genital herpes and HIV infection, all the above diseases are completely curable if they are detected early and treated properly.

The common modes of transmission of STDs are :

1. Sharing injection needles
2. Sharing surgical instrument with infected persons
3. Transfusion of contaminated blood
4. Ffom mother to foetus.

The common symptoms of most of the STDs are :

1. Itching
2. Fluid discharge
3. Slight pain and swelling in genital region
4. Pelvic inflammatory diseases
5. Abortions
6. Still births
7. Ectopic pregnancies
8. Infertility and cancer of reproductive tract persons in the age group of 15-24 years are more vulnerable to contract STDs.

The measures to be taken to prevent STDs are

1. Avoiding sex with unknown / multiple partners.
2. Using condoms compulsorily during coitus.
3. Consulting qualified doctor for early detection of STDs and getting complete treatment in case of infections.

Question 2.
Describe the surgical methods of contraception.
Answer:
Surgical procedure to prevent pregnancy is known as sterilization. There are two surgical methods of contraception. They are
a) Vasectomy b) Tubectomy

a) Vasectomy :
It is carried out in male. A small part of the vas deferens on either side is removed or tied up through a small incision on the scrotum. Thus the sperms are prevented from reaching the seminal vesicle so the semen in vasectomised males does not contain sperms.

b) Tubectomy:
It is the contraceptive method in females. A small part of the fallopian tube on both sides is removed or tied up through a small incision made in the abdomen or through vagina. This will block the entry of ova into the fallopian tubes and thus pregnancy is prevented.

Question 3.
Write short notes on any two of the following.
a) IVF b) ICSI e) IUDs
Answer:
a) IVF :
Fertilization of ovum by sperm outside the body of a woman is called in Vitro Fertilization (IVF). The resultant early embryonic stage is transferred into the mother’s uterus for further development (Embryo Transfer or Intra Uterine Trdnsfer – IUT).

In this method, which is popularly known as Test tube baby procedure, ova from the wife or female donor and sperms from the husband or male donor are collected, mixed and induced to form zygote under simulated conditions in the laboratory. If the mother’s uterus is not medically fit to receive the embryo produced invitro, it can be implanted in the uterus of surrogate mother is who willing to carry this embryo.

b) ICSI:
Intra Cytoplasmic Sperm Injection is another specialised procedure in which a sperm is directly injected into the ovum with the help of microscopic needle to form an embryo in the laboratory. Later the embryo is transferred to the uterus or fallopian tube for further development. This method is employed to assist the couple where there are problems with the sperms such as decreased sperm count.

c) IUDs :
Intra Uterine Devises (IUDs) are used by females in a process of contraception. IUDs are inserted into the uterus by doctors or trained nurses through vagina.

IUDs promote phagocytosis of sperms by white blood corpuscles within the Uterus and the copper ions released suppress the motility, viability and fertilizing capacity of the spermatozoa. The hormone releasing IUDs, makes the uterus unsuitable for implantation and the cervix hostile to sperms. IUDs are ideal contraceptives to females who want to delay or have space between children. This is a widely accepted method of contraception in India.

Type of IUDs	Kxample
1. Non medicated	Lippes loop
2. Copper releasing	CuT, Cu7, multiload 375
3. Hormone releasing	Progestasert, LNG-20

Question 4.
Suggest some methods to assist infertile couples to have children.
Answer:
The infertility may be due to physical, genetic, certain diseases, drugs, immunological or even psychological. Infertility clinics could help in diagnosis and corrective treatment of some of these disorders and enable the couples to have children in natural way.

In the cases where such corrections are not possible, the couple could be assisted to have children through special techniques known as Assisted Reproductive Technology (ART). The following are some important techniques employed in ART.

1) IVF :
In Vitro Fertilization is a process in which fertilization of ovum by sperm done outside the woman’s body. In this method, popularly known as ‘Test Tube Baby Procedure’, ova from wife or female donor and sperms from husband / male donor are collected, mixed and induced to form zygote under simulated conditions in the laboratory. If the mother’s uterus is not medically fit to receive the embryo produced invitro, it can be implanted in the uterus of another woman (surrogate mother).

2) ZIFT :
Zygote Intra Fallopian Transfer is another technique used to overcome infertility. The ovum is extracted and fertilized invitro and the zygote is transferred to the woman’s fallopian tube to complete its further course of development.

3) GIFT:
Gamete Intra Fallopian Transfer is a procedure done for women who cannot produce ova either due to defect or diseases in ovaries, but still, can provide suitable environment for fertilization and further development of the embryo in their uterus. In these cases, ovum is collected from donor is transferred to the fallopian tube of recipient woman for fertilization.

4) ICSI:
Intracytoplasmic Sperm Injection is another specialised procedure in which a sperm is directly injected into ovum with the help of microscopic needle to form an embryo in vitro. Later the embryo is transferred to the uterus or fallopian tube for further development. This method is employed assist the couple where there are problems with the sperms such as decrease in sperm count.

5) AI:
Artificial Insemination is done in a case where male is unable to inseminate the female or due to very low sperm count in the ejaculate. In this technique, semen is collected from the husband or healthy donor and is introduced into the uterus (Intra Uterine Insemination-IUI) for achieving fertilization.

Question 5.
Is sex education necessary in schools? Why?
Answer:
Governmental and non-governmental agencies have taken various steps to educate people on reproduction-related issues using audio-visual and print media. Introduction of sex education in schools will provide right information about the reproductive organs, adolescence , and related changes, safe and hygienic sexual practices, sexually transmitted diseases such as HIV etc, would help people, especially those in adolescent age group lead a reproductively healthy life.

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 are the first line of defence mechanism and are also called 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 non-congential, 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, the 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 responsible for formation of 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 virus and it 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 another T_H cells. This leads to decrease CD4 receptors containing T_H cells in the infected person leading to the immuno deficiency in him, finally causes AIDS.

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(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 molluscs.	4. It is found in annelids and chordates.

Question 2.
Sino-atrial node is called the pacemaker of our heart. Why?
Answer:
Sino-atrial node consists of specialized cardio myocytes. It has the ability to generate action potentials without any external stimuli henc’e called pacemaker.

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

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• Chapter 3 Consignment Notes
• Chapter 4 Not-for-Profit Organizations Notes
• Chapter 5 Partnership Accounts Notes
• Chapter 6 Admission of a Partner Notes
• Chapter 7 Retirement / Death of a Partner Notes
• Chapter 8 Company Accounts Notes
• Chapter 9 Computerised Accounting System Notes
• Chapter 10 Accounts from Incomplete Records Notes

AP Inter 2nd Year Accountancy Notes in Telugu Medium

• Chapter 1 వినిమయ బిల్లులు Notes
• Chapter 2 తరుగుదల Notes
• Chapter 3 కన్సైన్మెంటు Notes
• Chapter 4 వ్యాపారేతర సంస్థల ఖాతాలు Notes
• Chapter 5 భాగస్వామ్య ఖాతాలు Notes
• Chapter 6 భాగస్తుని ప్రవేశం Notes
• Chapter 7 భాగస్తుని విరమణ / మరణము Notes
• Chapter 8 కంపెనీ ఖాతాలు Notes
• Chapter 9 కంప్యూటరైజ్డ్ అకౌంటింగ్ Notes
• Chapter 10 అసంపూర్తి రికార్డుల నుంచి ఖాతాలు (ఒంటి పద్దు విధానం) Notes

These TS AP Intermediate 2nd Year Accountancy Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations. These Inter 2nd Year Accountancy Notes will enable students to study smartly and get a clear idea about each and every concept discussed in their syllabus.

Students can go through Telangana & Andhra Pradesh BIEAP TS AP Inter 2nd Year Commerce Notes Pdf Download in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP Sr Inter 2nd Year Commerce Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics.

Students can also go through AP Inter 2nd Year Commerce Study Material and AP Inter 2nd Year Commerce Important Questions for exam preparation.

AP Intermediate 2nd Year Commerce Notes

• Chapter 1 Entrepreneurship Notes
• Chapter 2 Domestic and International Trade Notes
• Chapter 3 Business Services Notes
• Chapter 4 Financial Markets Notes
• Chapter 5 Consumer Protections Notes

AP Inter 2nd Year Commerce Notes in Telugu Medium

• Chapter 1 ఎంట్రప్రిన్యూర్షిప్ Notes
• Chapter 2 స్వదేశీ, విదేశీ వర్తకం Notes
• Chapter 3 వ్యాపార సేవలు Notes
• Chapter 4 విత్త మార్కెట్లు Notes
• Chapter 5 వినియోగదారుల రక్షణ Notes

These TS AP Intermediate 2nd Year Commerce Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations. These Inter 2nd Year Commerce Notes will enable students to study smartly and get a clear idea about each and every concept discussed in their syllabus.

Students can go through Telangana & Andhra Pradesh BIEAP TS AP Inter 2nd Year Civics Notes Pdf Download in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP Sr Inter 2nd Year Civics Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics.

Students can also go through AP Inter 2nd Year Civics Study Material and AP Inter 2nd Year Civics Important Questions for exam preparation.

AP Intermediate 2nd Year Civics Notes

AP Inter 2nd Year Civics Notes in English Medium

• Chapter 1 The Constitution of India Notes
• Chapter 2 Fundamental Rights and Directive Principles of State Policy Notes
• Chapter 3 Union Executive Notes
• Chapter 4 Union Legislature Notes
• Chapter 5 Union Judiciary Notes
• Chapter 6 State Executive Notes
• Chapter 7 State Legislature Notes
• Chapter 8 State Judiciary Notes
• Chapter 9 Union-State Relations Notes
• Chapter 10 Local Governments in India Notes
• Chapter 11 Elections and Representation Notes
• Chapter 12 Political Parties Notes
• Chapter 13 Recent Developments in Andhra Pradesh and India Notes

AP Inter 2nd Year Civics Notes in Telugu Medium

• Chapter 1 భారత రాజ్యాంగం Notes
• Chapter 2 ప్రాథమిక హక్కులు – ఆదేశక సూత్రాలు Notes
• Chapter 3 కేంద్ర కార్యనిర్వాహక శాఖ Notes
• Chapter 4 కేంద్ర శాసననిర్మాణ శాఖ Notes
• Chapter 5 కేంద్ర న్యాయశాఖ Notes
• Chapter 6 రాష్ట్ర కార్యనిర్వాహక శాఖ Notes
• Chapter 7 రాష్ట్ర శాసననిర్మాణ శాఖ Notes
• Chapter 8 రాష్ట్ర న్యాయశాఖ Notes
• Chapter 9 కేంద్ర – రాష్ట్ర సంబంధాలు Notes
• Chapter 10 భారతదేశంలో స్థానిక ప్రభుత్వాలు Notes
• Chapter 11 ఎన్నికలు – ప్రాతినిధ్యం Notes
• Chapter 12 రాజకీయ పార్టీలు Notes
• Chapter 13 ఆంధ్రప్రదేశ్, ఇండియాలలో ఇటీవలి పరిణామాలు Notes

These TS AP Intermediate 2nd Year Civics Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations. These Inter 2nd Year Civics Notes will enable students to study smartly and get a clear idea about each and every concept discussed in their syllabus.

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Civics Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 2nd Year Civics Blue Print Weightage 2022-2023, Telugu Academy Intermediate 2nd Year Civics Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 2nd Year Study Material Pdf.

Students can also read AP Inter 2nd Year Civics Syllabus & AP Inter 2nd Year Civics Important Questions for exam preparation. Students can also go through AP Inter 2nd Year Civics Notes to understand and remember the concepts easily.

AP Intermediate 2nd Year Civics Study Material Pdf Download | Sr Inter 2nd Year Civics Textbook Solutions

AP Inter 2nd Year Civics Study Material in English Medium

• Chapter 1 The Constitution of India
• Chapter 2 Fundamental Rights and Directive Principles of State Policy
• Chapter 3 Union Executive
• Chapter 4 Union Legislature
• Chapter 5 Union Judiciary
• Chapter 6 State Executive
• Chapter 7 State Legislature
• Chapter 8 State Judiciary
• Chapter 9 Union-State Relations
• Chapter 10 Local Governments in India
• Chapter 11 Elections and Representation
• Chapter 12 Political Parties
• Chapter 13 Recent Developments in Andhra Pradesh and India

AP Inter 2nd Year Civics Study Material in Telugu Medium

• Chapter 1 భారత రాజ్యాంగం
• Chapter 2 ప్రాథమిక హక్కులు – ఆదేశక సూత్రాలు
• Chapter 3 కేంద్ర కార్యనిర్వాహక శాఖ
• Chapter 4 కేంద్ర శాసననిర్మాణ శాఖ
• Chapter 5 కేంద్ర న్యాయశాఖ
• Chapter 6 రాష్ట్ర కార్యనిర్వాహక శాఖ
• Chapter 7 రాష్ట్ర శాసననిర్మాణ శాఖ
• Chapter 8 రాష్ట్ర న్యాయశాఖ
• Chapter 9 కేంద్ర – రాష్ట్ర సంబంధాలు
• Chapter 10 భారతదేశంలో స్థానిక ప్రభుత్వాలు
• Chapter 11 ఎన్నికలు – ప్రాతినిధ్యం
• Chapter 12 రాజకీయ పార్టీలు
• Chapter 13 ఆంధ్రప్రదేశ్, ఇండియాలలో ఇటీవలి పరిణామాలు

TS AP Inter 2nd Year Civics Weightage Blue Print 2022-2023

TS AP Inter 2nd Year Civics Weightage 2022-2023 | TS AP Inter 2nd Year Civics Blue Print 2022

Intermediate 2nd Year Civics Syllabus

TS AP Inter 2nd Year Civics Syllabus

Chapter 1 The Constitution of India
The Constitution, Elements of a Constitution, Indian Constitution — Its Historical background, Making of the Constitution, Sources of the Constitution, Preamble of the Constitution, Salient features of Indian Constitution

Chapter 2 Fundamental Rights and Directive Principles of State Policy
Fundamental Rights, Evolution of Fundamental Rights, Characteristic Features of Fundamental Rights, Analysis of Fundamental Rights, Restrictions on Fundamental Rights, Significance of Fundamental Rights, Directive Principles of State Policy, Evolution of Directive Principles, Characteristic Features of Directive Principles of State Policy, Types of Directive Principles of State Policy, Significance of Directive Principles of State Policy, Differences between Fundamental Rights and Directive Principles of State Policy, Changing Relationship between Fundamental Rights and Directive Principles, Implementation of Directive Principles of State Policy, Fundamental Duties, Relevance of Fundamental Duties, Significance of Fundamental Duties

Chapter 3 Union Executive
Union Executive, The President of India, The Vice-President of India, Prime Minister of India, Union Council of Ministers, Role of Union Cabinet, Collective Responsibility

Chapter 4 Union Legislature
Union Legislature (Parliament), Unique Features of Union Legislature, Lok Sabha, The speaker of the Lok Sabha, Rajya Sabha, Chairman of Rajya Sabha, Powers and Functions of Union Legislature (Indian Parliament), Types of bills in Parliament, Law making procedure in Parliament, Important matters in Parliament, Parliamentary Committees, Amendment Procedure of Indian Constitution, Significance of Union Legislature (Parliament)

Chapter 5 Union Judiciary
The Supreme Court of India, Powers, and Functions of the Supreme Court, Judicial Review, Public Interest Litigation (PIL), Independence of Judiciary, The Attorney General of India

Chapter 6 State Executive
State Executive, The Governor, Powers and Functions, The Chief Minister, Powers and Functions of the Chief Minister, Relationship of the Chief Minister with Governor, Position, and Significance of the Chief Minister, The State Council of Ministers, Powers and Functions of State Council of Ministers, Position of the State Council of Ministers, Relationship between the Governor and the State Council of Ministers

Chapter 7 State Legislature
Legislative Assembly, Powers, and Functions of State Legislative Assembly, State Legislative Council, Powers and Functions of State Legislative Council, Supremacy of Legislative Assembly over the Legislative Council, Position of the State Legislature, Brief history of Andhra Pradesh Legislature, Legislative Committees

Chapter 8 State Judiciary
High Court, Powers, and Functions, District Level Judiciary, State Advocate General

Chapter 9 Union-State Relations
Union – State Relations, Legislative Relations, Administrative Relations, Financial Relations, Finance Commission, Planning Commission or NITI Aayog, National Development Council, National Integration Council, Inter-State Council, Sarkaria Commission, Punchchi Commission, Tension Areas in Union State Relations, Trends In Union-State Relations

Chapter 10 Local Governments in India
Local Governments in India, Historical Background, Rural Local Governments in India, Constitution (73rd Amendment) Act 1992, Types of Rural Local Governments (Panchayat Raj Institutions), Urban Local Governments in India, Constitution (74th Amendment) Act 1992, Types of Urban Local Government, District Collector

Chapter 11 Elections and Representation
Elections and Democracy, Electoral Functions, Election System in India, Features of Indian Electoral System, Methods of Election, ElectIon Process, Corrupt Practices in Elections, Electoral Offences, Breach of Official Duty, Representation, Election Commission of India, Powers and Functions of the Election Commission, Role of the Commission, Electoral Reforms

Chapter 12 Political Parties
Meaning and Definitions, Characteristics of Political Parties, Types of Political Parties, Functions of Political Parties, Party System, Types of Party System, Party System in India, Characteristics of Indian Party System, Major National Political Parties in India, One Party Dominance, Major Regional Political Parties In India, Types of Regional Political Parties, Significance of Regional Parties in Indian Politics

Chapter 13 Recent Developments in Andhra Pradesh and India
Re-Organization of States, The Birth of Andhra State, Emergence of Andhra Pradesh, Political Crisis in 1969 and1972, Bifurcation of Andhra Pradesh, National Human Rights Commission, State Human Rights Commissions, Right to Information Act 2005

We hope that this Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Civics Study Material Textbook Solutions Guide PDF Free Download 2022-2023 in English Medium and Telugu Medium helps the student to come out successful with flying colors in this examination. This Sr Inter 2nd Year Civics Study Material will help students to gain the right knowledge to tackle any type of questions that can be asked during the exams.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 13th Lesson Atoms Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 13th Lesson Atoms

Very Short Answer Questions

Question 1.
What is the angular momentum of electron in the second orbit of Bohr’s model of hydrogen atom ?
Answer:
Angular momentum of electron in second orbit of Hydrogen atom
L = \(\frac{2 \mathrm{~h}}{2 \pi}\) = \(\frac{h}{\pi}\) (∵ L = \(\frac{h h}{2 \pi}\))

Question 2.
What is the expression for fine structure constant and what is its value ?
Answer:
Formula for fine structure constant
α = \(\frac{\mathrm{e}^2}{2 \varepsilon_0 \mathrm{ch}}\); value of α = \(\frac{1}{137}\)

Question 3.
What is the physical meaning of ‘negative energy of an electron’ ?
Answer:
The ‘negative energy of an electron’ indicates that the electron bound to the nucleus due to force of attraction.

Question 4.
Sharp lines are present in the spectrum of a gas. What does this indicate ?
Answer:
Sharp lines in the spectrum of gas indicates bright lines against dark background.

Question 5.
Name a physical quantity whose dimensions are the same as those of angular momentum.
Answer:
Planck’s constant.

Question 6.
What is the difference between α – particle and helium atom ?
Answer:
Alpha particle

1. It is a + 2e charged Helium nucleus.
2. It contains 2 protons and 2 neutrons.

Helium atom

1. It has no charge.
2. It contains 2 protons, 2 electrons and 2 neutrons.

Question 7.
How is impact parameter related to angle of scattering ?
Answer:
The impact parameter related to angle of scattering is given by b = \(\frac{1}{4 \pi \varepsilon_0} \times \frac{\mathrm{Ze}^2}{\left(\frac{1}{2} m v^2\right)} \cot \theta_2\)

Question 8.
Among alpha, beta and gamma radiations, which get affected by the electric field ?
Answer:
Alpha and Beta radiations are get affected by the electric field.

Question 9.
What do you understand by the ‘phrase ground state atom’ ?
Answer:
If the electron is present in the ground state, it is called ground state atom.

Question 10.
Why does the mass of the nucleus not have any significance in scattering in Rutherford’s experiment ?
Answer:
The size of the atom is 10^-10 m and size of the nucleus is 10^-15 m. Hence atom has large empty space. So the mass of nucleus has no significance in Rutherford’s scattering experiment.

Question 11.
The Lyman series of hydrogen spectrum lies in the ultraviolet region. Why ? (A.P. Mar. ’15)
Answer:
The calculated values of wavelengths lie in the ultraviolet region of the spectrum well agree with the values of wavelengths observed experimentally by Lyman.

Question 12.
Write down a table giving longest and shortest wavelengths of different spectral series.
Answer:
Wavelength limits of some spectral series of hydrogen.

Question 13.
The wavelengths of some of the spectral lines obtained in hydrogen spectrum are 1216 A, 6463 A and 9546A. Which one of these wavelengths belongs to the Paschen series ?
Answer:
The wavelength of spectral line 9546A belongs to the Paschen series.

Question 14.
Give two drawbacks of Rutherford’s atomic model.
Answer:
Drawbacks of Rutherford’s atom model:

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

Short Answer Questions

Question 1.
What is impact parameter and angle of scattering ? How are they related to each other ?
Answer:

1. Impact parameter (b) : Impact parameter is defined as the perpendicular distance of the initial velocity vector of the alpha particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom.
2. Scattering angle (θ): The scattering angle (θ) is the angle between the asymtotic direction of approach of the α – particle and the asymptotic direction in which it receeds.
3. The relation between b and θ is b = \(\frac{1}{4 \pi \varepsilon_0} \frac{Z \mathrm{e}^2}{E} \cot \frac{\theta}{2}\) where E = K.E. of α – particle = \(\frac{1}{2} \mathrm{mv}^2\)

Question 2.
Derive an expression for potential and kinetic energy of an electron in any orbit of a hydrogen atom according to Bohr’s atomic model. How does P.E change with increasing n. (T.S. Mar. ’15)
Answer:

1. According fo Bohr electrostatic force of attraction, F_e between the revolving electrons and nucleus provides the necessary centripetal force F_c to keep them in their orbits.
2. Thus for dynamically state orbit in a hydrogen atom.
   F_c = F_e ⇒ \(\frac{m v^2}{r}\) = \(\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r_2}\)
3. The relation between the orbit radius and the electron velocity is r = 2
   4e0 (m )
4. The kinetic energy (K) and electrostatic potential energy (υ) of the electron in hydrogen atom are
   K = \(\frac{1}{2} \mathrm{mv}^2\) = \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 r}\) and υ = \(\frac{-\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\)
5. The total energy E of the electron in a hydrogen atom is
   E = K + U = \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) – \(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\)
   ∴ E = \(\frac{-\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\)
6. With increase in ‘nr potential energy (U) also increases.

Question 3.
What are the limitations of Bohr’s theory of hydrogen atom ? (Mar. ’14)
Answer:
Limitations of Bohr’s theory of Hydrogen atom :

1. This theory is applicable only to simplest atom like hydrogen, with z = 1. The theory fails in case of atoms- of other elements for which z > 1.
2. The theory does not explain why orbits of electrons are taken as circular, while elliptlical orbits are also possible.
3. Bohr’s theory does not say anything about the relative intensities of spectral lines.
4. Bohr’s theory does not take into account the wave properties of electrons.

Question 4.
Explain the distance of closest approach and impact parameter.
Answer:
Distance of closest approach :

1. Suppose an α-particle with initial kinetic energy (K.E) is directed towards the centre of the nucleus of an atom.
2. On account of Coulomb’s repulsive force between nucleus and alpha particle, kinetic energy of alpha particle goes on decreasing and in turn, electric potential energy of the particle goes on increasing.
3. At certain distance ‘d’ from the nucleus, K. E of α-particle reduces to zero. The particle stops and it can not go closer to the nucleus. It is repelled by the nucleus and therefore it retraces its path, turning through 180°.
4. Therefore, the distance d is known as the distance of closest of approach.
   
   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 5.
Give a brief account of Thomson model of atom. What are its limitations ?
Answer:
Thomson’s model of atom :

1. According to Thomson’s model, every atom consists of a positively charged sphere of radius of the order of 10^-10 m in which entire mass and positive charge of the atom are uniformly distributed.
   
2. Inside this sphere, the electrons are embedded like seeds in a watermelon or like plums in a pudding.
3. The negative charge of electrons is equal to the positive charge of the atom. Thus atom is electrically neutral.

Limitations :

1. It could not explain the origin of spectral series of hydrogen and other atoms, observed experimentally.
2. It could not explain large angle scattering of a-particles from thin metal foils, as observed by Rutherford.

Question 6.
Describe Rutherford atom model. What are the draw backs of this model.
Answer:
Rutherford atom model: The essential features of Rutherford’s nuclear model of the atom or planetary model of the atom are as follows :

1. Every atom consists of tiny central core, called the atomic nucleus, in which the entire positive charge and almost entire mass of the atom are concentrated.
2. The size of nucleus is of the order of 10^-15m, which is very small as compared to the size of the atom which is of the order of 10^-10m.
3. The atomic nucleus is surrounded by certain number of electrons. As atom on the whole is electrically neutral, the total negative charge of electrons surrounding the nucleus is equal to total positive charge on the nucleus.
4. These electrons revolve around the nucleus in various circular orbits as do the planets around the sun. The centripetal force required by electron for revolution in provided by the electrostatic force of attraction between the electrons and the nucleus.

Draw backs : According to classical E.M. theory.

1. The revolving electron loses energy continuously, it must spiral inwards and eventually fall into the nucleus. As matter is stable, we cannot expect the atoms to collapse.
2. Since the frequency of radiation emitted is the same as the frequency of revolution, the atom should radiate a continuous spectrum, but what we observe is only a line spectrum.

Question 7.
Distinguish between excitation potential and ionization potential.
Answer:
Excitation Potential:

1) When the electron jumps from lower orbit to higher orbit by absorbing energy is called excited electron and the process is known as excitation. The minimum accelerating potential which provides an electron energy sufficient to jump from the inner most orbit (ground state) to one of the outer orbits is called excitation potential or resonance potential.

2) a) For example, in case of hydrogen atom,
E₁ = -13.6 eV. E₂ = -3.4 eV E₃ = -1.51eV and soon, E_∞ = 0
∴ Energy required to raise an electron from ground state (n = 1) to first excited state
(n = 2) is E = E₂ – E₁ = -3.4 – (-13.6) = 10.2 eV.
The corresponding excitation potential = 10.2 Volt,

b) Similarly, energy required to raise an electron from ground state (n = 1) to second excited , state (n = 3) is
E = E₃ – E₁ = -1.51 – (-13.6) = -1.51 + 13.6 = 12.09 eV
The corresponding excitation potential = 12.09 Volt and so on.

3) The excitation potential of an atom is not one. It can have many values, depending on the state to which the atom is excited.

Ionisation potential:

1. The energy supplied is so large that it can remove an electron from the outer most orbit of an atom, the process is called Ionisation. Thus ionisation is the phenomenon of removal of an electron from the outer most orbit of an atom.
2. The minimum accelerating potential which would provide an electron energy sufficient just to remove it from the atom is called Ionisation potential.
3. For example, total energy of electron in ground state of hydrogen atom, + 13.6 eV energy is required.
   ∴ Ionisation energy of hydrogen atom = 13.6 eV.
   Ionisation potential of hydrogen atom = 13.6 Volts.
4. The general expression for ionisation potential of an atom is V = \(\frac{13.6 \mathrm{Z}^2}{\mathrm{n}^2}\) volt, Where Z is the charge number of the atom and n is number of orbit from which electron is to be removed.
5. For a given element, ionisation potential is fixed, but for different elements, ionisation potentials are different.

Question 8.
Explain the different types of spectral series in hydrogen atom. (A.P. Mar. ’19, ’15; T.S. Mar. ’16)
Answer:
The atomic hydrogen emits a line spectrum consisting of five series.

1. Lyman series : v = Rc \(\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\) where n = 2, 3, 4, ……
2. Balmer series : v = Rc\(\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\) where n = 3, 4, 5, ………
3. Paschen series : v = Rc\(\left(\frac{1}{3^2}-\frac{1}{n^2}\right)\) where n = 4, 5, 6, …….
4. Brackett series : v = Rc\(\left(\frac{1}{4^2}-\frac{1}{n^2}\right)\) where n = 5, 6, 7, ……
5. Pfund series : v = Rc\(\left(\frac{1}{5^2}-\frac{1}{n^2}\right)\) where n = 6, 7, 8,……..

Question 9.
Write a short note on Debroglie’s explanation of Bohr’s second postulate of quantization.
Answer:
Debroglie’s explanation of Bohr’s second postulate of quantization :

1. The second postulate of Bohr atom model says that angular momentum of electron orbiting around the nucleus is quantized i.e., mυr = \(\frac{\text { nh }}{2 \pi}\) where m = 1, 2, 3,….
2. According to Debroglie, the electron in its circular orbit, as proposed by Bohr, must be seen as a particle wave.
3. When a string fixed at two ends is plucked, a large number of wavelengths are excited and standing wave is formed.
4. It means that in a string, standing waves form when total distance travelled by a wave down the string and back is an integral number of wavelengths.
5. According to Debroglie, a stationary orbit is that which contains an integral number of Debrogile waves associated with the revolving electron.
6. For an electron revolving in nth circular orbit of radius r_n, total distance covered = circumference òf 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
   ∴ mυ_nr_n = \(\frac{\mathrm{nh}}{2 \pi}\) = \(\mathrm{n}\left(\frac{\mathrm{h}}{2 \pi}\right)\)
   i.e., angular momentum of electron revolrmg 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.

Long Answer Questions

Question 1.
Describe Geiger-Marsden Experiment on scattering of α – particles. How is the size of the nucleus estimated in this experiment ?
Answer:

1. The experimental set up used by Rutherford and his colaborators, Geiger and Marsden is shown in fig.
   
2. The α-particles emitted by radio active source contained in a lead cavity are collimated into a narrow beam with the help of a lead slit (collimator).
3. The collimated beam is allowed to fall on a thin gold foil of thickness of the order of 2.1 × 10^-7m.
4. The α-particles scattered in different directions are observed through a rotatable detector consisting of zinc sulphide screen and a microscope.
5. The α-particles produce bright flashes or scintillations on the ZnS screen.
6. These are observed in the microscope and counted at different angles from the direction of incidence of the beam.
7. The angle θ of deviation of an α-particle from its original direction is called its scattering angle θ.

Observations : We find that

1. Most of the alpha particles pass straight through the gold foil. It means they do not suffer any collision with gold atoms.
2. Only about 0.14% of incident α-particles scatter by more than 1°.
3. About one α-particle in every 8000 α-particles deflect by more than 90°.

Estimation of size of the nucleus :

1. This led to Rutherford postulate, that the entire positive charge of the atom must be concentration in a tiny central core of the atom. This tiny central core of each atom was called atomic nucleus.
2. The electrons would be moving in orbits about the nucleus just as the planets do around the sun.
3. Rutherford’s experiments suggested the size of the nucleus to be about 10^-15m to 10^-14m. From kinetic theory, the size of an atom was known to be 10^-10m, about 10,000 to 1,00,000 times larger than the size of the nucleus.

Question 2.
Discuss Bohr’s theory of the spectrum of hydrogen atom.
Answer:

1. According to Bohr’s model an electron continuous to revolve round the nucleus in fixed, stationary orbits. This is called groupd state of the atom. In ground state there is no emission of radiation.
2. But when some energy is given to an atom the electron absorbs this energy. This is called excited state of the atom. In this state the electron jumps to the next higher orbit. But it can remain 10^-8 sec and it immediatly returns back to its ground state and the balance of the energy is emitted out as a spectral line.
3. According to Bohr’s third postulate, the emitted energy is given by E = hv = E₂ – E₁
   

Spectral series of Hydrogen atom:

Hydrogen atom has five series of spectral lines: They are

1. Lyman series: When an electron jumps from the outer orbits to the first orbit, the spectral lines are in the ultra – violet region. Here n₁ = 1, n₂ = 2, 3, 4, 5….
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{\mathrm{n}_2^2}\right]\) = \(\mathrm{R}\left[1-\frac{1}{\mathrm{n}_2^2}\right]\)

2) Balmer Series : When an electron jumps from the outer orbits to the second orbit, the
spectral Lines are in the visible region. Here n₁ = 2, n₂ = 3, 4, 5…
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{\mathrm{n}_2^2}\right]\)

3) Paschen series : When an electron jumps from the outer orbits to the third orbit, the spectral lines are in the near infrared region. Here n₁ = 3, n₂ = 4, 5, 6 ….
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left[\frac{1}{3^2}-\frac{1}{\mathrm{n}_2{ }^2}\right]\)

4) Brackett series : When an electron jumps from outer orbits to the forth orbit, the spectral lines are in the infrared region. Here n₁ = 4, n₂ = 5, 6, 7…….
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left[\frac{1}{4^2}-\frac{1}{\mathrm{n}_2^2}\right]\)

5) Pfund series : When an electron jumps from outer orbits to the fifth orbit, the spectral lines are in the far infrared region. Here n₁ = 5, n₂ = 6, 7, 8, ………
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left[\frac{1}{5^2}-\frac{1}{\mathrm{n}_2^2}\right]\)

Question 3.
State the basic postulates of Bohr’s theory of atomic spectra. Hence obtain an expression for the radius of orbit and the energy of orbital electron in a hydrogen atom.
Answer:
a) Basic postulates of Bohr’s theory are
1) The electron revolves round a nucleus is an atom in various orbits known as stationary orbits. The electrons can not emit radiation when moving in their own stationary levels.

2) The electron can revolve round the nucleus only in allowed, orbits whose angular momentum is the integral multiple of \(momentum is the integral multiple of
i.e., mυ_nr_n = [latex]\frac{\mathrm{nh}}{2 \pi}\) ———> (1)
where n = 1, 2, 3…..

3) If an electron jumps from higher energy (E₂) orbit to the lower energy (E₁) orbit, the difference of energy is radiated in the form of radiation.
i.e., E = hv = E₂ – E₁ ⇒ v = \(\frac{E_2-E_1}{h}\) ——> (2)

b) Energy of emitted radiation : In hydrogen atom, a single electron of charge — e, revolves around the nucleus of charge e in a ciccular orbit of radius 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,
From Coulomb’s law, \(\frac{\mathrm{m} \dot{v}_n^2}{r_n}\) = \(\frac{\mathrm{Ke}^2}{\mathrm{r}_{\mathrm{n}}^2}\) ——> (3)

where K = \(\frac{1}{4 \pi \varepsilon_0}\) —–> (4)
\(m v_n^2\) = \(\frac{\mathrm{Ke}^2}{\mathrm{r}_{\mathrm{n}}}\) —–> (5)
\(m v^2 r_n\) = Ke² ——-> (6)
Dividing (5) by (1), υ_n = Ke² × \(\frac{2 \pi}{\mathrm{nh}}\)
From (3), kinetic energy K = \(\frac{1}{2} m v_n^2\) = \(\frac{\mathrm{Ke}^2}{2 r_{\mathrm{n}}}\)

2) Potential energy of electron:
P.E. of electron, U = \(\frac{\mathrm{Ke}}{\mathrm{r}_{\mathrm{n}}} \times-\mathrm{e}\) [∵ W = \(\frac{I^{\prime}}{4 \pi \varepsilon_0} \frac{Q}{d}\) × -Q]
∴ U = \(\frac{-\mathrm{Ke}^2}{\mathrm{r}_{\mathrm{n}}}\)

3) Radius of the orbit: Substituting the value of (6) in (2),

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

Textual Exercises

Question 1.
The radius of the first electron orbit of a hydrogen atom is 5.3 × 10^-11m. What is the radius of the second orbit ?
Solution:
r_n ∝ n²
\(\frac{\mathrm{r}_2}{\mathrm{r}_1}\) = \(\frac{2^2}{1^2}\) = \(\frac{4}{1}\) ⇒ r₂ = 4r₁

Question 2.
Determine the radius of the first orbit of the hydrogen atom. What would be the velocity and frequency of the electron in the first orbit ?
Solution:
Given: h = 6.62 × 10^-34 J-s,
m = 9.1 × 10^-31kg,
e = 1.6 × 10^-19 C,
k = 9 × 10⁹Nm²C^-2, n = 1

i)
r₁ = \(\frac{n^2 h^2}{4 \pi^2 \mathrm{mke}^2}\)
= \(\frac{(1)^2 \times\left(6.62 \times 10^{-34}\right)^2}{4 \times(3.14)^2 \times 9.1 \times 10^{-31} \times 9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}\)
∴ r₁= 0.529 A \(\simeq\) 0.53 A

ii)

iii)

Question 3.
The total energy of an electron in the first excited state of the hydrogen atom is -3.4eV. What is the potential energy of the electron in this state ?
Solution:
In 1^st orbit, E = -3.4eV
Total energy E = \(\frac{\mathrm{KZe}^2}{2 \mathrm{r}}\) – \(\frac{\mathrm{KZe}^2}{\mathrm{r}}\)
\(\frac{\mathrm{KZe}^2}{\mathrm{r}}\) = U(say)
E = \(\frac{\mathrm{U}}{2}-\mathrm{U}\) = \(\frac{-\mathrm{U}}{2}\)
U = -2E
∴ U = -2 × -3.4 = 6.8 eV.

Question 4.
The total energy of an electron in the first excited state of hydrogen atom is -3.4eV. What is the kinetic energy of the electron in this state ?
Solution:
In Hydrogen like atom, we know that
K = – Total energy E
Here E = – 3.4eV
∴ K = -(-3.4) = 3.4 eV

Question 5.
Find the radius of the hydrogen atom in its ground state. Also calculate the velocity of the electron in n = 1 orbit. Given
h = 6.63 × 10^-34 J s, m = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C, K = 9 × 10⁹N m²C^-2
Solution:
n = 1, h = 6.63 × 10^-34 J-s,
m = 9.1 × 10^-31 kg
e = 1.6 × 10^-19C,
K = 9 × 10⁹ Nm²C^-2

Question 6.
Prove that the ionisation energy of hydrogen atom is 13.6 eV.
Solution:
n = 1 corresponds to ground state.
E = \(\frac{-13.6}{n^2} e V\)
E = \(\frac{-13.6}{1^2} \mathrm{eV}\)
E = -13.6 eV
∴ The minimum energy required to free the electron from the ground state of hydrogen atom
= 13.6 eV.
∴ Ionisation energy of hydrogen atom = 13.6 eV

Question 7.
Calculate the ionization energy for a lithium atom.
Solution:

∴ Ionization energy of Lithium = 30.6eV.

Question 8.
The wavelength of the first member of Lyman series is 1216 A. Calculate the wavelength of second member of Balmer series.
Solution:
\(\frac{1}{\lambda}\) = \(R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)
For 1^st member of Lyman series, λ = 1216; n₁ = 1, n₂ = 2

For 2^nd member of Balmer senes,

Question 9.
The wavelength of first member of Balmer series is 6563 A. Calculate the wavelength of second member of Lyman series.
Solution:
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
For 1^st member of Balmer senes,
\(\frac{1}{6563}\) = \(\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\)
\(\frac{1}{6563}\) = \(\frac{5 R}{36}\) —–> (1)
For 2^nd member of Lyman senes,
\(\frac{1}{\lambda^1}\) = \(\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{3^2}\right)\)
\(\frac{1}{\lambda^1}\) = \(\frac{8 \mathrm{R}}{9}\) —–> (2)
\(\frac{(1)}{(2)}\) ⇒ \(\frac{\lambda^1}{6563}\) = \(\frac{5 \mathrm{R}}{36} \times \frac{9}{8 \mathrm{R}}\)
λ’ = \(\frac{5}{32} \times 6563\)
∴ λ’ = 1025.5A

Question 10.
The second member of Lyman series in hydrogen spectrum has wavelength 5400 A. Find the wavelength of first member.
Solution:
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
For second member of Lyman senes,
\(\frac{1}{5400}\) = \(R\left(\frac{1}{1^2}-\frac{1}{3^2}\right)\) ⇒ \(\frac{1}{5400}\) = \(\frac{8 \mathrm{R}}{9}\) —-> (1)
For first member of Lyman series,
\(\frac{1}{\lambda^1}\) = \(\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\)
\(\frac{1}{\lambda^1}\) = \(\frac{3 R}{4}\) —–> (2)
\(\frac{(1)}{(2)}\) ⇒ \(\frac{\lambda^1}{5400}\) = \(\frac{8 R}{9} \times \frac{4}{3 R}\)
∴ λ’ = \(\frac{32}{27}\) × 5400 = 6400A.

Question 11.
Calculate the shortest wavelength of Balmer series. Or Calculate the wavelength of the Balmer senes limit. Given : R = 10970000m^-1.
Solution:
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
R = 10970000 = 1.097 × 10⁷ ms^-1
For Balmer senes limit n₁ = 2 and n₂ = ∞
\(\frac{1}{\lambda}\) = \(\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{\infty}\right)\) ⇒ \(\frac{1}{\lambda}\) = \(\frac{R}{4}\)
λ = \(\frac{4}{\mathrm{R}}\) = \(\frac{4}{1.097 \times 10^7}\) = 3646.3A

Question 12.
Using the Rydberg formula, calcûlate the wavelength of the first four spectral lines in the Balmer series of the hydrogen spectrum.
Solution:

Additional Exercises

Question 1.
Choose the correct alternative from the clues given at the end of the each statement:
a) The size of the atom in Thomson’s model is ……… the atomic size in Rutherford’s model. (much greater than / no different from / much less than.)
b) In the ground state of ……. electrons are in stable equilibrium, while in …… electrons always experience a net force. (Thomson’s model / Rutherford’s model.)
c) A classical atom based on …… is doomed to collapse. (Thomson’s model / Rutherford’s model).
d) An atom has a nearly continuous mass distribution in a ……. but has a highly non-uniform mass distribution in ………. (Thomson’s model / Rutherford’s model.)
e) The positively charged part of the atom possesses most of the mass in …….. (Rutherford’s model / both the models.)
Answer:
a) No different from
b) Thomson’s model, Rutherford’s model
c) Rutherford’s model
d) Thomson’s model, Rutherford’s model
e) Both the models.

Question 2.
Suppose you are given a chance to repeat the alpha – particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect ?
Answer:
The basic purpose of scattering experiment is defeated because solid hydrogen will be much lighter target compared to the alpha particle acting as projectile. According to theory of elastic the collisions, the target hydrogen will move much faster compared to alpha after collision. We cannot determine the size of hydrogen nucleus.

Question 3.
What is the shortest wavelength present in the Paschen series of spectral lines ?
Answer:
From Rydberg’s formula
\(\frac{\mathrm{hc}}{\lambda}\) = 13.6 × 1.6 × 10^-19\(\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
For shortest wavelength in Paschen series n₂ = ∞ and n₁ = 3
\(\frac{\mathrm{hc}}{\lambda}\) = 21.76 × 10^-19\(\left[\frac{1}{3^2}-\frac{1}{\infty^2}\right]\)
= 2.42 × 10^-19
λ = \(\frac{\mathrm{hc}}{2.42 \times 10^{-19}}\) = \(\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2.42 \times 10^{-19}} \mathrm{~m}\)
= 8.1818 × 10^-7m = 818.18nm.

Question 4.
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level ?
Answer:
Here E = 2.3eV = 2.3 × 1.6 × 10^-19 J
As E = hv
∴ v = \(\frac{\mathrm{E}}{\mathrm{h}}\) = \(\frac{2.3 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}}\) = 5.6 × 10⁴ Hz

Question 5.
The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state ?
Answer:
Total energy E = -13.6 eV
K.E = -E = 13.6 eV
RE. = -2.K.E = -2 × 13.6
= -27.2eV.

Question 6.
A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
Answer:
For ground state n₁ = 1 and n₂ = 4
Energy of photon absorbed E = E₂ – E₁

Question 7.
a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2 and 3 levels.
b) Calculate the orbital period in each of these levels.
Answer:
a) From v = \(\frac{c}{n} \alpha\), where α = \(\frac{2 \pi \mathrm{Ke}^2}{\mathrm{ch}}\) = 0.0073
v₁ = \(\frac{3 \times 10^8}{1}\) × 0.0073 = 2.19 × 10⁶ m/s
v₂ = \(\frac{3 \times 10^8}{2}\) × 0.0073 = 1.095 × 10⁶ m/s
v₃ = \(\frac{3 \times 10^8}{3}\) × 0.0073 = 7.3 × 10⁵ m/s

b) Orbital period, T = \(\frac{2 \pi r}{V}\), As r₁ = 0.53 × 10^-10m
T₁ = \(\frac{2 \pi \times 0.53 \times 10^{-10}}{2.19 \times 10^6}\) = 1.52 × 10^-16S
As r₂ = 4r₁ and V₂ = \(\frac{1}{2} V_1\)
T₂ = 8T₂ = 8 × 1.52 × 10^-6 S = 1.216 × 10^-15S
As r₃ = 9r₁ and V₃ = \(\frac{1}{3} \mathrm{~V}_1\)
T₃ = 27T₁= 27 × 1.52 × 10^-16 S = 4.1 × 10^-15S

Question 8.
The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10-^-11m. What are the radii of the n = 2 and n = 3 orbits?
Answer:
As r = n²r
∴ r² = 4r₁ = 4 × 5.3 × 10^-11 m = 2.12 × 10^-10m ,
and r₃ = 9r₁ = 9 × 5.3 × 10^-11 = 4.77 × 10^-10m.

Question 9.
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Answer:
In ground state, energy of gaseous hydrogen at room temparature = -13.6eV, when it is bombarded with 12.5 eV electron beam, the energy becomes 13.6 + 12.5 = -1.1eV.
The electron would jump from n = 1 to n = 3 where E₃ = \(\frac{-13.6}{32}\) = -1.5eV
On de — excitation the electron may jump from n = 3 to n = 2 giving rise to Balmer series. It may also jump from n = 3 to n = 1 giving rise to Lýman series.

Question 10.
In accordance with the Bohr’s model, find the quantum number that characterises the earths revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴m/s. (Mass of earth = 6.0 × 10²⁴ kg.)
Answer:
Here r = 1.5 × 10¹¹m, V = 3 × 10⁴m/s, m = 6.0 × 10²⁴kg
According to Bohrs model mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
n = \(\frac{2 \pi \mathrm{mvr}}{\mathrm{h}}\) = 2 × \(\frac{22}{7}\) × \(\frac{6.0 \times 10^{24} \times 3 \times 10^4 \times 1.5 \times 10^{11}}{6.6 \times 10^{-34}}\)
= 2.57 × 10⁷⁴, which is too large.

Question 11.
Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.
a) Is the average angle of deflection of α — particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
b) Is the probability of backward scattering (i.e., scattering of α – particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model ?
c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α – particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide ?
d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of a – particles by a thin foil ?
Answer:
a) About the same this is because we are talking of average angle of deflection.
b) Much less, because in Thomson’s model there is no such massive central core called the nucleus as in Rutherford’s model.
c) This suggests that scattering is predominantly due to a single collision increases with the number of target atoms which increases linearly with the thickness of the foil.
d) In Thomson model, positive charge is uniformly distributed in the spherical atom. Therefore a single collision causes very little deflection. Therefore average scattering angle can be explained only by considering multiple scattering may be ignored as a first approximation.

Question 12.
The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10^-40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
Answer:
The radius of the first Bohr orbit of a hydrogen atom is
r₀ = \(\frac{4 \pi \varepsilon_0(h / 2 \pi)^2}{m_e \mathrm{e}^2}\)
If we consider the atom bound by the gravitational force
= \(\left(\frac{\mathrm{Gm}_{\mathrm{p}} \mathrm{m}_{\mathrm{e}}}{\mathrm{r}^2}\right)\). We should replace \(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0}\) by (Gm_pm_e). In that case radius of first Bohr orbit of hydrogen atom would be given by r₀ = \(\frac{(\mathrm{h} / 2 \pi)^2}{\mathrm{Gm}_p \mathrm{~m}^2 \mathrm{e}}\)
Putting the standard values we get
r₀ = \(\frac{\left(6.6 \times 10^{-34} / 2 \pi\right)^2}{6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times\left(9.1 \times 10^{-31}\right)^2}\)
= 1.2 × 10²⁹ metre.
This is much greater than the estimated size of the whole universe!

Question 13.
Obtain an expression for the frequency of radiation emitted when a hydrogen atom de- excites from level n to level (n – 1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.
Answer:
The frequency v of the emitted radiation when a hydrogen atom de-excites from level n to level (n – 1) is
E = hv = E₂ – E₁

In Bohr’s Atomic model, velocity of electron in n^th orbit is v = \(\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}\)
and radius of n^th orbital is v = \(\frac{n^2 h^2}{4 \pi^2 m K e^2}\)
Frequency of revolution of electron v = \(\frac{\mathrm{V}}{2 \pi \mathrm{r}}\) = \(\frac{\mathrm{nh}}{2 \pi \mathrm{mr}}\)
and radius of n^th orbital is r = \(\frac{n^2 h^2}{4 \pi^2 m K e^2}\)
Frequency of revolution of electron

which is the same as (i) .
Hence for large values of n₁ classical frequency of revolution of electron in n^th orbit is the same as the frequency of radiation emitted when hydrogen atom de-excites from level (n) to level (n – 1)

Question 14.
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size ? Why is an atom not, say, thousand times bigger than its typical size ? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10¹⁰m).
a) Construct a quantity with the dimensions of length from the fundamental constants e, m_e, and c. Determine its numerical value.
b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non- relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the ‘Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, m_e, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.
Answer:
a) Using fundamental constants e, me and c, we construct a quantity which has the dimensions of length. This quantity is \(\left(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{~m}_{\mathrm{e}} \mathrm{c}^2}\right)\)
Now \(\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{~m}_{\mathrm{e}} \mathrm{c}^2}\) = \(\frac{\left(1.6 \times 10^{-19}\right)^2 \times 9 \times 10^9}{9.1 \times 10^{-31}\left(3 \times 10^8\right)^2}\) = 2.82 × 10^-15m
This is of the order of atom sizes.

b) However when we drop c and use hc, m_e and e to construct a quantity which has dimensions of length the quantity we obtain is
\(\frac{4 \pi \varepsilon_0(\mathrm{~h} / 2 \pi)}{\mathrm{m}_{\mathrm{e}} \mathrm{e}^2}\)

= 0.53 × 10^-10m
This is of the order of atom sizes.

Question 15.
The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV
a) What is the kinetic energy of the electron in this state ?
b) What is the potential energy of the electron in this state ?
c) Which of the answers above would change if the choice of the zero of potential energy is changed ?
Answer:
We know kinetic energy of electron = \(\frac{K Z e^2}{2 r}\)
and P.E of electron = \(\frac{-\mathrm{KZe}^2}{\mathrm{r}}\)
P.E. = -2 (kinetic energy).
In this calculation electric potential and hence potential energy is zero at infinity.
Total energy = PE + KE = -2KE + KE = -KE
a) In the first excited state total energy = -3.4eV
∴ K.E = -(-3.4eV) = + 3.4 eV
b) P. E of electron in this first excited state = -2KE = -2 × 3.4 = -6.8eV.
c) If zero of potential energy is changed, KE does not change and continues to be +3.4 eV However, the P.E. and total energy of the state would change with the choice of zero of potential energy.

Question 16.
If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun ?
Answer:
Bohr’s quantisation postulate is in terms of Plank’s constant (h), But angular momenta associated with planetary motion are = 10⁷⁰ h (for earth). In terms of Bohr’s quantisation posulate this will correspond to n = 10⁷. For such large values of n the differences in successive energies and angular momenta of the quantised levels are so small, that the levels can be considered as continuous and not discrete.

Question 17.
Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (µ^–) of mass about 207m_e orbits around a proton].
Answer:
A muonic hydrogen is the atom in which a negatively charged muon of mass about 207 m_e revolves around a proton.
In Bohr’s atom model as, r ∝ \(\frac{1}{\mathrm{~m}}\)
\(\frac{\mathrm{r}_\mu}{\mathrm{r}_{\mathrm{e}}}\) = \(\frac{\mathrm{m}_{\mathrm{e}}}{\mathrm{m}_\mu}\) = \(\frac{\mathrm{m}_{\mathrm{e}}}{207 \mathrm{~m}_{\mathrm{e}}}\) = \(\frac{1}{207}\)
Here r_e is radius of first orbit of electron in hydrogen atom = 0.53A = 0.53 × 10^-10m.
r_m = \(\frac{\mathrm{r}_{\mathrm{e}}}{207}\) = \(\frac{0.53 \times 10^{-10}}{207}\) = 2.56 × 10^-13m
Again in Bohr’s atomic model
E ∝ m
∴ \(\frac{\mathrm{E}_\mu}{\mathrm{E}_{\mathrm{e}}}\) = \(\frac{\mathrm{m}_\mu}{\mathrm{m}_{\mathrm{e}}}\) = \(\frac{207 \mathrm{~m}_{\mathrm{e}}}{\mathrm{m}_{\mathrm{e}}}\), E_μ = 207E_e
As ground state energy of electron in hydrogen atom is E_e = -13.6 eV
E_μ = 207(-13.6)eV = -2815.2eV
= -2.8152KeV.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 12th Lesson Dual Nature of Radiation and Matter Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 12th Lesson Dual Nature of Radiation and Matter

Very Short Answer Questions

Question 1.
What are “cathode rays”?
Answer:
Cathode rays are streams of fast-moving electrons or negatively charged particles.

Question 2.
What important fact did Millikan’s experiment establish?
Answer:
Millikan’s experiment established that electric charge is quantised. That means the charge on anybody (oil drop) is always an integral multiple of the charge of an electron, i.e., Q = ne.

Question 3.
What is a “work function”? (A.P. Mar. ’19 & T.S. Mar. ’15)
Answer:
The minimum energy required to liberate an electron from a photo metal surface is called the work function, ϕ₀.

Question 4.
What is “photoelectric effect” ?
Answer:
When light of sufficient energy is incident on the photometal surface electrons are emitted. This phenomenon is called photoelectric effect.

Question 5.
Give examples of “photosensitive substances”. Why are they called so ?
Answer:
Examples of photosensitive substances are Li, Na, K, Rb and Cs etc.
The work function of alkali metals is very low. Even the ordinary visible light, alkali metals can produce photoelectric emission. Hence they are called photosensitive substances.

Question 6.
Write down Einstein’s photoelectric equation. (A.P. Mar. ’15)
Answer:
Einstein’s photoelectric equation is given by K_max = \(\frac{1}{2} \mathrm{mv}_{\max }^2\) = hυ – ϕ₀.

Question 7.
Write down de-Broglie’s relation and explain the terms there in. (A.P. & T.S. Mar. ’16)
Answer:
The de-Broglie wave length (λ) associated with a moving particle is related to its momentum (p) is λ = \(\frac{h}{p}\) = \(\frac{\mathrm{h}}{\mathrm{mv}}\), where h is planck’s constant.

Question 8.
State Heisenberg’s Uncertainly Principle. (A.P. Mar. ’19) (Mar. ’14)
Answer:
Uncertainity principle states that “it is impossible to measure both position (Δx) and momentum of an electron (Δp) [or any other particle] at the same time exactly”, i.e., Δx . Δp ≈ h where Δx is uncertainty in the specification of position and Δp is uncertainty in the specification of momentum.

Short Answer Questions

Question 1.
What is the effect of
(i) intensity of light
(ii) potential on photoelectric current ?
Answer:
(i) Effect of intensity of light on photoelectric current:

1) When the intensity (I) of incident light, with frequency greater than the threshold frequency (υ > υ₀) is increased then the number of photoelectrons emitted increases i.e. the value of photoelectric current (i) increases, ie. i ∝ I.

ii) The effect of potential on photoelectric current:

1. On increasing the positive potential on collecting electrode, the photoelectric current increases. At a particular positive potential, the photocurrent becomes maximum which is known as saturated current.
   
2. On increasing the value of negative potential on collecting electrode, the photoelectric current gradually goes on decreasing. At a particular negative potential the value of photoelectric current becomes zero. This is known as stopping potential.
3. Stopping potential does not depend on the intensity of incident light. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.

Question 2.
Describe an experiment to study the effect of frequency of incident radiation on ‘stopping potential’.
Answer:
Experimental study of the effect of frequency of incident radiation on stopping potential:

1. The experimental set up is shown in fig.
2. Monochromatic light of sufficient energy (E = hv) from source ‘s’ is incident on photosensitive plate ‘C’ (emitter), electrons are emitted by it.
3. The electrons are collected by the plate A (collector), by the electric field created by the battery.
4. The polarity of the plates C and A can be reversed by a commutator.
   
5. For a particular frequency of incident radiation, the minimum negative (retarding) potential V₀ given to the plate A for which the photo current stops or becomes zero is called stopping potential.
6. The experiment is repeated with different frequencies, and their different stopping potential are measured with voltmeter.
7. From graph, we note that
   1. The values of stopping potentials are different for different frequencies.
   2. The value of stopping potential is more negative for radiation of higher incident frequency.
   3. The value of saturation current depends on the intensity of incident radiation but it is independent of the frequency of the incident radiation.
      

Question 3.
Summarise the photon picture of electromagnetic radiation.
Answer:
We can summarise the photon picture of electromagnetic radiation as follows.

1. In interaction of radiation with matter, radiation behaves as if it is made up of particles called photons.
2. Each photon has energy E\(\left[\begin{array}{l}
   =\mathrm{hv} \\
   =\frac{\mathrm{hc}}{\lambda}
   \end{array}\right]\) and momentum P \(\left[\begin{array}{l}
   =\frac{h v}{c} \\
   =\frac{h}{\lambda}
   \end{array}\right]\) and speed c, the speed of light.
3. By increasing the intensity of light of given wave length, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation.
4. Photons are not deflected by electric and magnetic field. This shows that photons are electrically neutral.
5. In a photon-particle collision (such as photo-electron collision), the energy and momentum
   are conserved. However the number of photons may not be conserved in a collision. One photon may be absorbed or a new photon may be created.
6. The rest mass of photon is zero. According to theory of relativity, the mass of moving particle is given by m = \(\frac{\mathrm{m}_0}{\sqrt{1-\frac{v^2}{c^2}}}\) where v is velocity of particle and c is velocity of light.

Question 4.
What is the deBroglie wavelength of a ball of mass 0.12 Kg moving with a speed of 20 ms^-1 ? What can we infer from this result ?
Answer:
Given, m = 0.12 kg; υ = 20 m/s; h = 6.63 × 10^-34 J-s;
λ = \(\frac{h}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{0.12 \times 20}\) = \(\frac{6.63 \times 10^{-34}}{2.4}\) ∴ λ = 2.762 × 10^-34 m = 2762 × 10^-21 A.

Long Answer Questions

Question 1.
How did Einstein’s photoelectric equation explain the effect of intensity and potential on photoelectric current ? How did this equation account for the effect of frequency of incident light on stopping potential ? (T.S. Mar. ’19)
Answer:

1. Einstein postulated that a beam of light consists of small energy packets called photons or quanta.
2. The energy of photon is E = hv. Where ‘h’ is Planck’s constant; v is frequency of incident light (or radiation).
3. If the absorbed energy of photon is greater than the work function (ϕ₀ = hυ₀), the electron is emitted with maximum kinetic energy i.e., k_max = \(\frac{1}{2} m_{\max }^2\) = eV₀ = hv – ϕ₀. This equation is known as Einstein’s photoelectric equation.
4. Effect of intensity of light on photoelectric current:
   When the intensity (I) of incident light, with frequency greater thanthe threshold frequency (υ > υ₀) is increased then the number of photoelectrons emitted decreases i.e. the value of photoelectric current (i) increases, le. i ∝ I.
   
5. The effect of potential on photoelectric current:
   1. On increasing the positive potential on collecting electrode, the photoelectric current increases. At a particular positive potential, the photocurrent becomes maximum which is known as saturated current.
   2. On increasing the value of negative potential on collecting electrode, the photoelectric current gradually goes on decreasing. At a particular negative potential the value of photoelectric current becomes zero. This is known as stopping potential (v₀).
      
   3. Stopping potential does not depend on the intensity of incident light. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.
6. The effect of frequency of incident radiation on stopping potential:
   On increasing the frequency of incident light, the value of stopping potential goes on increasing gradually as shown in fig. That means k_max increases eV₀ also increases.
   
7. From the graph, we note that
   1. For a given photosensitive metal, the cut off potential (v₀) varies linearly with the frequency of the incident radiation.
   2. For a given photosensitive metal, there is a certain minimum cut off frequency v₀ (called threshold frequency) for which the stopping potential is zero.
      
8. From the graph we note that
   1. The value of cut-off potential is different for radiation of different frequency.
   2. The value of stopping potential is more negative for radiation of higher incident frequency.
9. From above experiments, it is found that, if the incident radiation is of higher frequency than that of threshold frequency, the photoelectric emission is possible.

Question 2.
Describe the Davisson and Germer experiment. What did this experiment conclusively prove?
Answer:
Davisson and Germer experiment:

1. The experimental arrangement is schematically shown in fig.
2. Electrons from a filament F are rendered into a fine beam by applying a positive potential to the cylinder A.
3. A fine narrow beam of electrons is incident on the nickel crystal. The electrons are scattered in all directions by the atoms of the crystal.
4. The intensity of the electron beam scattered in a given direction, is measured by the electron detector (collector). The detector can be moved on a circular scale and is connected to a sensitive galvanometer, which records the current.
5. The deflection of the galvanometer is proportional to the intensity of the electron beam entering collector.
6. The apparatus is enclosed in an evacuated chamber.
7. The experiment was performed by varying the accelerating voltage from 44 V to 68 V. It is found that the intensity is maximum at 50° for a critical energy of 54 V
   
8. For θ = 50°, the glancing angle, ϕ (angle between the scattered beam of electron with the plane of atoms of the crystal) for electron beam will be given by
   ϕ + θ + ϕ = 180°
   ϕ = \(\frac{1}{2}\left[180^{\circ}-50^{\circ}\right]\) = 65°
   
9. According to Bragg’s law for first order diffraction maxima (n = 1), we have 2 d sin ϕ = 1 × λ ⇒ λ = 2 × 0.91 × sin 65° = 1.65A = 0.165 nm. (experimentally).
   [∵ for Nickel crystal interatomic separation d = 0.91 A]
10. According to de-Broglie hypothesis, the wavelength of the wave associated with electron is given by λ =
    = 1.67A = 0.167 nm, (Theoritically).
11. The experimentally measured wavelength was found to be in confirmity with proving the existence of de-Broglie waves.

Textual Exercises

Question 1.
Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Solution:
Given voltage V = 30 kV = 30 × 10³ V; e = 1.6 × 10^-19 C; h = 6.63 × 10^-34 j-s C = 3 × 10⁸ m/s
a) Maximum frequency, v = \(\frac{\mathrm{eV}}{\mathrm{h}}\) = \(\frac{1.6 \times 10^{-19} \times 30 \times 10^3}{6.63 \times 10^{-34}}\) = 7.24 × 10¹⁸ Hz

b) Minimum wavelength of X-ray, λ = \(\frac{\mathrm{C}}{\mathrm{v}}\) = \(\frac{3 \times 10^8}{7.24 \times 10^{18}}\) = 0.414 × 10^-10 Hz
∴ λ = 0.0414 × 10^-9m = 0.0414 nm.

Question 2.
The work function of caesium metal is 2.14 eV. When light of frequency 6 × 10¹⁴ Hz is incident on the metal surface, photoemission of electrons occurs. What is the
(a) maximum kinetic energy of the emitted electrons,
(b) stopping potential and
(c) maximum speed of the emitted photoelectrons ?
Solution:
Given ϕ₀ = 2.14 eV; v = 6 × 10¹⁴ Hz
a) KE_max = hv – ϕ₀ = \(\frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}\) – 2.14 ∴ KE_max = 0.35 eV

b) KE_max = eV0 ⇒ 0.35 eV = eV₀ ∴ V₀ = 0.35 V
c) KE_max = \(\frac{1}{2} m v_{\max }^2\) ⇒ \(v_{\max }^2\) = \(\frac{2 K_{\max }}{m}\) = \(\frac{2 \times 0.35 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}\) (∴ e = 1.6 × 10^-19 C)
\(v_{\max }^2\) = 0.123 × 10¹² ⇒ υ_max = \(\sqrt{1230 \times 10^8}\) = 35.071 × 10⁴ m/s ∴ υ_max = 350.71 km/s.

Question 3.
The photoelectric cut-off voltage in certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ?
Solution:
Given, V₀ = 1.5 V; e = 1.6 × 10^-19 C, KE_max = eV₀ = 1.6 × 10^-19 × 1.5 = 2.4 × 10^-19 J.

Question 4.
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam ? (Assume the beam to have uniform cross-section which is less than the target area), and,
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon ?
Solution:
Given, λ = 632.8 nm = 632.8 × 10^-9m; p = 9.42 mW = 9.42 × 10^-3W
h = 6.63 × 10^-34 J-s; c = 3 × 10^-3 m/s

a) E = \(\frac{h c}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10}{632.8 \times 10^{-9}}\) = 3.14 × 10^-19 J.
Momentum of each photon, p = \(\frac{\mathrm{h}}{\lambda}\) = \(\frac{6.63 \times 10^{-34}}{632.8 \times 10^{-9}}\) = 1.05 × 10^-27kg \(\frac{\mathrm{m}}{\mathrm{s}}\)

b) No. of photons per second,

∴ N = 3 × 10¹⁶ photons/s
c) Since, P_Hydrogen = P_photon
⇒ mυ = p ⇒ υ = \(\frac{\mathrm{p}}{\mathrm{m}}\) = \(\frac{1.05 \times 10^{-27}}{1.66 \times 10^{-27}}\) [∴ m_H = 1.66 × 10^-27 kg] ∴ υ = 0.63 m/s.

Question 5.
The energy flux of sunlight reaching the surface of the earth is 1.388 × 10³ W/m². How many photons (nearly) per square metre are incident on the Earth per second ? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Solution:
Given, P = 1.388 × 10³ W/m²; λ = 550 nm = 550 × 10^-9 m
h = 6.63 × 10^-34 J-s; e = 3 × 10⁸ m/s
Energy of each photon E = \(\frac{\mathrm{hc}}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{550 \times 10^{-9}}\) = 3.616 × 10^-19 J
No. of photons incident on the earths surface, N = \(\frac{\mathrm{P}}{\mathrm{E}}\) = \(\frac{1.388 \times 10^3}{3.66 \times 10^{-19}}\)
∴ N = 3.838 × 10²¹ photons/m² – s.

Question 6.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10^-15 V s. Calculate the value of Planck’s constant.
Solution:
Given, slope of graph tan θ = 4.12 × 10^-15 V — s; .
e = 1.6 × 10^-19 c.

For slope of graph, tan θ = \(\frac{\mathrm{V}}{\mathrm{v}}\)
We know that hv = eV
\(\frac{\mathrm{V}}{\mathrm{v}}\) = \(\frac{h}{e}\) ⇒ \(\frac{\mathrm{h}}{\mathrm{e}}\) = 4.12 × 10^-15; h = 4.12 × 10^-15 × 1.6 × 10^-19 = 6.592 × 10^-34 J-s

Question 7.
A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light ?
(b) At what rate are the photons delivered to the sphere ?
Solution:
Given, P = 100 W; λ = 589 nm = 589 × 10,sup>-9 m; h = 6.63 × 10^-34 J – S; c = 3 × 10⁸ m/s
a) E = \(\frac{\text { hc }}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{589 \times 10^{-9}}\) = 3.38 × 10^-19J = \(\frac{3.38 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV = 2.11 eV.
b) No. of photons delivered per second, N = \(\frac{\mathrm{P}}{\mathrm{E}}\) = \(\frac{100}{3.38 \times 10^{-19}}\) = 3 × 10²⁰ photons/s

Question 8.
The threshold frequency for a certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Solution:
Given, v₀ = 3.3 × 10¹⁴ Hz; v = 8.2 × 10¹⁴ Hz; e = 1.6 × 10^-19 c; KE = eV₀ = hv – hv₀
V₀ = \(\frac{h\left(v-v_0\right)}{e}\) = \(\frac{6.63 \times 10^{-34} \times(8.2-3.3) \times 10^{14}}{1.6 \times 10^{-19}}\) = \(\frac{6.63 \times 10^{-34} \times 10^{14} \times 4.9}{1.6 \times 10^{-19}}\) ∴ V₀ = 2.03 V.

Question 9.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm ?
Solution:
Given, ϕ₀ = 4.2 eV = 4.2 × 1.6 × 10^-19 J = 6.72 × 10~1S J
λ = 330 nm = 330 × 10^-9 m; h = 6.63 × 10^-34 J – s ⇒ c = 3 × 10⁸ m/s
E = \(\frac{\text { hc }}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{330 \times 10^{-9}}\) ∴ E = 6.027 × 10^-19J
As E < ϕ₀, no photoelectric emission takes place.

Question 10.
Light of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons ?
Solution:
Given, v = 7.21 × 10¹⁴ Hz; m = 9.1 × 10^-31 kg; υ_max = 6 × 10⁵ m/s
KE_max = \(\frac{1}{2} \mathrm{mv}_{\max }^2\) = hv – hv₀ = h(v – v₀)

Question 11.
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made. .
Solution:
Given, λ = 488 nm = 488 × 10^-9 m; V₀ = 0.38 V; e = 1.6 × 10^-19 c; h = 6.63 × 10^-34 J – s
c = 3 × 10⁸ m/s ⇒ KE = eV₀ = \(\frac{\mathrm{hc}}{\lambda}\) – ϕ ⇒ 1.6 × 10^-19 × 0.38 = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}\) – ϕ₀
6.08 × 10^-20 = 40.75 × 10^-20 – ϕ₀ ⇒ (40.75 – 6.08) × 10^-20 = 34.67 × 10^-20 J
= \(\frac{34.67 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}\) ∴ ϕ₀ = 2.17 eV.

Question 12.
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V
Solution:
Given, V = 56 V; e = 1.6 × 10^-19 c; m = 9 × 10^-31 kg
a) As KE = \(\frac{\mathrm{P}^2}{2 \mathrm{~m}}\) ⇒ 2m (KE) = P² ⇒ P = \(\sqrt{2 \mathrm{~m}(\mathrm{KE})}\) = \(\sqrt{2 \mathrm{~m} \mathrm{eV}}\) [∵ KE = eV]
∴ P = \(\sqrt{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-31} \times 56}\) = 4.02 × 10^-24 kg – m/s
b) λ = \(\frac{12.27}{\sqrt{V}}\) A = \(\frac{12.27}{\sqrt{56}}\) A = 0.164 × 10^-9m ∴ λ = 0.164 nm.

Question 13.
What is the
(a) momentum,
(b) speed, and
(c) de Brogue wavelength of an electron with kinetic energy of 120 eV.
Solution:
Given, KE = 120 eV; m = 9.1 × 10^-3 kg; e = 1.6 × 10^-19 c
a) P = \(\sqrt{2 m(K E)}\) = \(\sqrt{2 \times 9.1 \times 10^{-31} \times\left(120 \times 1.6 \times 10^{-19}\right)}\) ∴ P = 5.91 × 10^-24 kg – m/s
b) υ = \(\frac{\mathrm{p}}{\mathrm{m}}\) = \(\frac{5.91 \times 10^{-24}}{9.1 \times 10^{-31}}\) = 6.5 × 10⁶ m/s .
c) λ = \(\frac{12.27}{\sqrt{\mathrm{V}}}\) A = \(\frac{12.27}{\sqrt{120}}\) A = 0.112 × 10^-9 m ∴ λ = 0.112 nm.

Question 14.
The wavelength of light from the spectral emission line of sodium is 589 nm Find the kinetic energy at which (a) an electron, and (b) a neutron, and would have the same de Brogue wavelength.
Solution:
Given, λ = 589 mn = 589 × 10^-9 m; m_e = 9.1 × 10^-31 kg.
m_n = 1.67 × 10^-27 kg; h = 6.62 × 10^-34 J – s.

Question 15.
What is the de Brogue wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 × 10^-9 kg drifting with a speed of 2.2 m/s?
Solution:
a) Given, for bullet m = 0.040 kg and o = 1000 m/s = 10³ m/s
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{0.040 \times 10^3}\) = 1.66 × 10^-35m
b)Given, for ball m = 0.060 kg and υ = 1 m/s ⇒ λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{0.060 \times 1}\) = 1.1 × 10^-32 m
c) Given, for a dust particle m = 1 × 10^-9 kg and υ = 2.2 m/s
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{1 \times 10^{-9} \times 2.2}\) = 3.0 × 10^-25 m.

Question 16.
An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Solution:
Given, λ = 1 mm = 10^-9m; h = 6.63 × 10^-34 J-S; c = 3 × 10⁸ m/S

Question 17.
(a) For what kinetic energy of a neutron will the associated de Brogue wavelength be 1.40 × 10^-10 m?
(b) Also find the de Brogue wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.
Solution:
(a) Given, for neutron, λ = 1.40 × 10^-10 m and m = 1.675 × 10^-27 kg
KE = \(\frac{\mathrm{P}^2}{2 \mathrm{~m}}\) = \(\frac{h^2}{2 \mathrm{~m} \lambda^2}\) = \(\frac{\left(6.63 \times 10^{-34}\right)^2}{2 \times\left(1.40 \times 10^{-10}\right)^2 \times 1.675 \times 10^{-27}}\) ∴ KE = 6.686 × 10^-21J

b) Given, T = 300 k and K = 1.38 × 10^-23 J/K
KE = \(\frac{3}{2}\) KT = \(\frac{3}{2}\) × 1.38 × 10^-21 × 300 = 6.21 × 10^-21 J
λ = \(\frac{h}{\sqrt{2 m(K E)}}\) = \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}}}\) ∴ λ = 1.45 × 10^-10m = 1.45 A

Question 18.
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Solution:
The momentum of a photon of frequency v, wavelength λ is given by p = \(\frac{\mathrm{hv}}{\mathrm{c}}\) = \(\frac{\mathrm{h}}{\lambda}\)
λ = \(\frac{\mathrm{h}}{\mathrm{p}}\) ⇒ de-Broglie wavelength of photon, λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{h}{p}\) = \(\frac{\frac{\mathrm{h}}{\mathrm{hv}}}{\mathrm{c}}\) = \(\frac{\mathrm{c}}{\mathrm{v}}\)
Thus, the wavelength of electromagnetic radiation is equal to the de-Broglie wavelength.

Question 19.
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K ? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Solution:
Given, T = 300 k; K = 1.38 × 10^-23 J/k; m = 28.0152u = 28.0152 × 1.67 × 10^-27 kg;
h = 6.63 × 10^-34 Js; Mean KE of molecules \(\frac{1}{2}\) mυ² = \(\frac{3}{2}\) KT
υ = \(\sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}}\) = \(\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{28.0152 \times 1.66 \times 10^{-27}}}\)
∴ υ = 516.78 m/s
de-Broglie wavelength, λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{28.0152 \times 1.66 \times 10^{-27} \times 516.78}\) = 2.75 × 10^-11 m
∴ λ = 0.0275 × 10^-19 m = 0.028 nm.

Additional Exercises

Question 1.
(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 10¹¹ C kg^-1.
(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong ? In what way is the formula to be modified ?
Solution:
a) Given, V = 500 V, \(\frac{\mathrm{e}}{\mathrm{m}}\) = 1.76 × 10¹¹ C/kg; KE = \(\frac{1}{2} \mathrm{mv}^2\) = eV

b) V = 10 MV = 10⁷ V; υ = \(\sqrt{\frac{\mathrm{e}}{\mathrm{m}} \times 2 \mathrm{~V}}\) = \(\sqrt{1.76 \times 10^{11} \times 2 \times 10^7}\) ∴ υ = 1.8762 × 10⁹ m/s
This speed is greater than speed of light, which is not possible. As o approaches to c, then mass m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)

Question 2.
(a) A monoenergetic electron beam with electron speed of 5.20 × 10⁶ m s^-1 is subject to
a magnetic field of 1.30 × 10^-4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 10¹¹ C kg^-1.
(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam ? If not, in what way is it modified ?
[Note : Exercises 20(b) and 21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]
Solution:
a) Given, υ = 5.20 × 10⁶ m/s; B = 1.30 × 10^-4 T; \(\frac{\mathrm{e}}{\mathrm{m}}\) = 1.76 × 10¹¹ C/kg
Since centripetal force is balanced by Force due to magnetic field, \(\frac{\mathrm{m} v^2}{\mathrm{r}}\) = Bυ
[∵ (\(\vec{v} \times \vec{B}\)) = e υ B sin 90° = Beυ]

b) Given, E = 20 MeV = 20 × 1.6 × 10^-13J; m_e = 9.1 × 10^-31 kg
E = \(\frac{1}{2} \mathrm{mv}^2\)
⇒ v = \(\sqrt{\frac{2 E}{m}}\) = \(\sqrt{\frac{2 \times 20 \times 1.6 \times 10^{-13}}{9.1 \times 10^{-32}}}\) ∴ v = 2.67 × 10⁹ m/s

As υ > C, the formula used in (a) r = \(\frac{\mathrm{mv}}{\mathrm{eB}}\) is not valid for calculating the radius of path of 20 MeV electron beam because electron with such a high energy has velocity in relatistic domain i.e., comparable with the velocity of light and the mass varies with the increase in velocity but we have taken it a constant.
∴ m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\) ⇒ Thus, the modified formula will be r = \(\frac{\mathrm{mv}}{\mathrm{eB}}\) = \(\left[\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\right] \frac{v}{e B}\)

Question 3.
An electron gun with its collector at a potential of 100V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~ 10^-2 mm of Hg). A magnetic field of 2.83 × 10^-4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons and emitting light by electron capture; this method is known as the fine beam tube’ method.) Determine e/m from the data.
Solution:
Given, V = 100 V; B = 2.83 × 10^-4 T; m = 9.1 × 10^-31 kg; e = 1.6 × 10^-19 C;
r = 12 cm = 0.12m; KE = \(\frac{1}{2} \mathrm{mv}^2\) = eV ⇒ \(\frac{1}{2}\) × 9.1 × 10^-31 × υ² = 1.6 × 10^-19 × 100
υ² = \(\frac{2 \times 1.6 \times 10^{-17}}{9.1 \times 10^{-3.1}}\) = 3.516 × 10¹³ ∴ υ = \(\sqrt{3.516 \times 10^{13}}\) = 5.93 × 10⁶ m/s
Specific charge of electron, \(\frac{\mathrm{e}}{\mathrm{m}}\) = \(\frac{v}{r B}\) [∵ \(\frac{\mathrm{mv}^2}{\mathrm{r}}\) = Beυ] = \(\frac{5.93 \times 10^6}{2.83 \times 10^{-4} \times 0.12}\)
∴ \(\frac{\mathrm{e}}{\mathrm{m}}\) = 1.74 × 10¹¹ C/kg.

Question 4.
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 A. What is the maximum energy of a photon in the radiation ?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube ?
Solution:
a) Given, λ = 0.45 A = 0.45 × 10^-10 m; E = \(\frac{\mathrm{hc}}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{0.45 \times 10^{-10} \times 1.6 \times 10^{-19}}\) eV
∴ E = 27.6 × 10³ eV = 27.6 KeV

b) In X-ray tube, accelerating voltage provides the energy to the electrons which produce X-rays. For getting X-rays, photons of 27.51 KeV is required that the incident electrons must posess kinetic energy of 27.61 KeV.
Energy = eV = E; eV = 27.6 KeV; V = 27.6 KV .
So, the order of accelerating voltage is 30 KV.

Question 5.
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray ? (1 BeV = 10⁹ eV)
Solution:
Given, energy of 2 γ-rays, 2E = 10.2 BeV

⇒ 2\(\frac{\mathrm{hc}}{\lambda}\) = 10.2 BeV [∵ E = \(\frac{\mathrm{hc}}{\lambda}\)] ⇒ λ = \(\frac{2 \mathrm{hc}}{10.2 \mathrm{BeV}}\)
Here h = 6.63 × 10^-34 J-S; c = 3 × 10⁸m/s, 1 BeV = 10⁹ eV = 10⁹ × 1.6 × 10^-19J
⇒ λ = \(\frac{2 \times 6.63 \times 10^{-34} \times 3 \times 10^8}{10.2 \times 10^9 \times 1.6 \times 10^{-19}}\) ∴ λ = 2.436 × 10^-16 m

Question 6.
Estimating the following two numbers should bé interesting. The first number will tell you why radio engineers do not need to worry much about photons ! The second number tells you why our eye can never count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~ 10^-10 W m^-2). Take the area of the pupil to be about 0.4 cm², and the average frequency of white light to be about 6 × 10¹⁴ Hz.
Solution:
a) Given, P = 10kW = 10 × 10³ W; λ = 500m; h = 6.63 × 10^-34 J – s; C = 3 × 10⁸
The no. of photons emitted per second, N = \(\frac{\mathrm{P}}{\mathrm{E}}\) = \(\frac{\mathrm{P}}{\frac{\mathrm{hc}}{\lambda}}\) = \(\frac{\mathrm{p} \lambda}{\mathrm{hc}}\) = \(\frac{10 \times 10^3 \times 500}{6.63 \times 10^{-34} \times 3 \times 10^8}\)
∴ N = 2.51 × 10³¹ photons/s

b) Given, v = 6 × 10^-4 Hz; I = \(\frac{E}{A-t}\) = 10^-10 W/m²; Area of pupil, A = 0.4 cm² = 0.4 × 10^-4 m².
Total energy falling on pupil in unit time, E’ = IA = 10^-10 × 0.4 × 10^-4 ∴ E’ = 4 × 10^-155 J/s
Energy of each photon, E” = hv = 6.63 × 10^-34 × 6 × 10¹⁴ = 3.978 × 10^-19 J
No. of photons per second, N = \(\frac{E^{\prime}}{E^{\prime \prime}}\) = \(\frac{4 \times 10^{-15}}{3.978 \times 10^{-19}}\) = 1.206 × 10⁴ photons/s
As this number is not so large a: in part (a), so it is large enough for us never to sense the individual photons by our eye.

Question 7.
Ultraviolet light of wavelength 2271 A from a 1oo W mercury source irradiates a photo cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (-10⁵ W m^-2) red light of wavelength 6328 A produced by a He-Ne laser?
Solution:
Given, for UV light, λ = 2271A = 2271 × 10^-10 m
V₀ = 1.3 V; P = 100W; h = 6.63 × 10^-34 J-s; c = 3 × 10⁸ m/s
From Einstein’s equation E = KE + ϕ₀, hυ = eV₀ + ϕ₀
ϕ₀ = \(\frac{\mathrm{hc}}{\lambda}\) – eV₀ = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2271 \times 10^{-10}}\) – 1.6 × 10^-19 × 1.3 = 8.758 × 10^-19 – 2.08 × 10^-19
ϕ₀ = \(\frac{6.678 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV = 4.17 eV ∴ ϕ₀ = 4.2 eV
Given, for red light, λ = 6328Å = 6328 × 10^-10m
E = \(\frac{\mathrm{hc}}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6328 \times 10^{-10}}\) = \(\frac{3.143 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV ∴ E = 1.96 eV
Here, E < ϕ₀, So, the photocell will not respond to this red light. (It is independent of intensity).

Question 8.
Monochromatic radiation of wavelength 640.2 nm (1 nm = 10^-9 m) from a neon, lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Solution:
Given, for Neon X = 640.2 nm = 640.2 × 10^-9 m ; V₀ = 0.54 V
h = 6.63 × 10^-34 J-s; c = 3 × 10⁸ m/s; e = 1.6 × 10^-19 C
ϕ = \(\frac{\mathrm{hc}}{\lambda}\) – eV₀ = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{640.2 \times 10^{-9}}\) – 1.6 × 10^-19 × 0.54
= 3.1 × 10^-19 – 0.864 × 10^-19 = 2.236 × 10^-19J = \(\frac{2.236 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV ∴ ϕ = 1.4 eV
For iron; given ϕ₀ = 1.4eV; λ = 427.2 nm = 427.2 × 10^-9 m
Let V₀ be the new stopping potential, eV₀ = \(\frac{\mathrm{hc}}{\lambda}\) – ϕ₀
eV₀’ = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{427.2 \times 10^{-9} \times 1.6 \times 10^{-19}}\) – 1.4 = 1.51 eV. Required stopping potential V₀‘ = 1.51 V.

Question 9.
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:
λ₁ = 3650Å, λ₂ = 4047Å, λ₃ = 4358Å, λ₄ = 5461 Å, λ₅ = 6907Å,
The stopping voltages, respectively, were measured to be:
V₀₁ = 1.28 V, V₀₂ = 0.95 V, V₀₃ = 0.74 V, V₀₄ = 0.16 V, V₀₅ = 0V.
Determine the value of Plancks constant h, the threshold frequency and work function
for the material.
[Note : You will notice that, to get h from the data, you will need to know e(which you can take to be 1.6 × 10^-19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]
Solution:
Given λ₁ = 3650 A = 3650 × 10^-10 m
λ₂ = 4047 A = 4047 × 10^-10 m
λ₃ = 4358 A = 4358 × 10^-10 m
λ₄ = 5461 A = 5461 × 10^-10 m
λ₅ = 6907 A = 6907 × 10^-10 m
V₀₁ = 1.28V, V₀₂ = 0.95, V₀₃ = 0.74 V; V₀₅ = 0

a) v₁ = \(\frac{\mathrm{c}}{\lambda_1}\) = \(\frac{3 \times 10^8}{3650 \times 10^{-10}}\) = 8.219 × 10¹⁴ Hz
v₂ = \(\frac{\mathrm{c}}{\lambda_2}\) = \(\frac{3 \times 10^8}{4047 \times 10^{-10}}\) = 7.412 × 10¹⁴ Hz
v₃ = \(\frac{\mathrm{c}}{\lambda_3}\) = \(\frac{3 \times 10^8}{4358 \times 10^{-10}}\) = 6.884 × 10¹⁴ Hz
v₄ = \(\frac{\mathrm{c}}{\lambda_4}\) = \(\frac{3 \times 10^8}{5461 \times 10^{-10}}\) = 5.493 × 10¹⁴ Hz
v₅ = \(\frac{\mathrm{c}}{\lambda_5}\) = \(\frac{3 \times 10^8}{6907 \times 10^{-10}}\) = 4.343 × 10¹⁴ Hz
As the graph between V₀ and frequency v is a straight line.
The slope of this graph gives the values of \(\frac{\mathrm{h}}{\mathrm{e}}\)

∴ \(\frac{\mathrm{h}}{\mathrm{e}}\) = \(\frac{V_{01}-V_{04}}{v_1-v_4}\) = \(\frac{1.28-0.16}{(8.219-5.493) \times 10^{14}}\)
h = \(\frac{1.12 \times 1.6 \times 10^{-19}}{2.726 \times 10^{14}}\) = 6.674 × 10^-34 J . s

b) ϕ₀ = hv₀ = 6.574 × 10^-34 × 5 × 10¹⁴
= 32.870 × 10^-20 J = \(\frac{32.870 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}\)
= 2.05 eV

Question 10.
The work function for the following metals is given:
Na : 2.75 eV; K: 2.30 eV; Mo : 4.17 eV; Ni : 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 A from a He-Cd laser placed 1 m away from the photocell ? What happens if the laser is brought nearer and placed 50 cm away?
Solution:
Given λ = 3300 A = 3300 × 10^-10 m
Energy of incident photon, E = \(\frac{\mathrm{hc}}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3300 \times 10^{-10} \times 1.6 \times 10^{-19}}\) ∴ E = 3.75 eV
Here Na, K has lesser work function than 3.75 eV. So, they produce photoelectric effect. If the laser is brought nearer then only the intensity change or the number of photoelctrons change.

Question 11.
Light of intensity 10^-5 W m^-2 falls on a sodium photo-cell of surface area 2 cm². Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Solution:
Given, I = 10^-5 W/m²; A = 2 cm² = 2 × 10^-4 m²; ϕ₀ = 2eV
Let t be the time.
The effective atomic area of Na = 10^-20 m² and it contains one conduction electron per
atom.
No. of conduction electrons m five layers

We know that sodium has one free electron (or conduction electron) per atom.
Incident power on the surface area of photocell
= Incident intensity × Area on the surface area of photo cell
= 10^-5 × 2 × 10^-4 = 2 × 10^-9 W.
The electron present in all the 5 layers of sodium will share the incident energy equally.
Energy absorbed per second per electron, E =
= \(\frac{2 \times 10^{-9}}{10^{17}}\) = 2 × 10^-26 W.
Time required for emission by each electron,
which is about 0.5 yr.
The answer obtained implies that the time of emission of electron is very large and is not agreement with the observed time of emission. There is no time lag between the incidence of light and the emission of photoelectron.
Thus, it is implied that the wave theory cannot be applied in this experiment.

Question 12.
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative. comparison. take the wavelength of the probe equal to 1 A, which is of the order of interatomic spacing in the lattice) (m_e = 9.11 × 10^-31 kg).
Solution:
Given λ = 1 A = 10^-10 m ; m_e = 9.11 × 10^-31 kg; h = 6.63 × 10^-34 J – s; c = 3 × 10⁸ m/s

Thus, for the same wavelength a X-ray photon has much KE than an electron.

Question 13.
(a) Obtain the de Brogue wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable ? Explain. (m_n = 1.675 × 10^-27 kg)
(b) Obtain the de Brogue wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffráction experiments.
Solution:
a) Given, KE = 150 eV; m = 1.675 × 10^-27 kg

The interatomic spacing is 10^-10 m, which is greater than this wavelength. So, neutron beam of 150 eV is not suitable for diffraction experiment.

b) T = t + 273 = 27 + 273 = 300 K; K = 1.38 × 10^-23 J/mol/K

This wavelength is order of interatomic spacing. So, the neutron beam first thermalised and then used for diffraction.

Question 14.
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
Solution:
Given, V = 50 KV s 50000 V
λ =

= 0.055 A ⇒ λ = 5.5 × 10^-12 m; For yellow light (λ) = 5.9 × 10^-7m
As resolving power (RP) ∝ \(\frac{1}{\lambda}\)
⇒

Question 15.
The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10^-15 m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.5 11 MeV.)
Solution:
Given λ = 10^-15 m; E = 0.5 11 MeV; P = \(\frac{\mathrm{h}}{\lambda}\) = \(\frac{6.63 \times 10^{-34}}{10^{-15}}\) = 6.63 × 10^-19 kg^m/s
Rest mass energy; E₀ = m₀c² = 0.511 MeV = 0.511 × 1.6 × 10^-13 T.
From relativistic theory, E² = p²c² + \(m_0^2 c^4\)
= (3 × 10⁸ × 6.63 × 10^-19)² + (0.511 × 10^-13 × 1.6)² = 9 × (6.63)² × 10^-22.
As the rest mass energy is negligible ∴ Energy E = \(\sqrt{p^2 c^2}\) = pc = 6.63 × 10^-19 × 3 × 10⁸
= \(\frac{1.989 \times 10^{-10}}{1.6 \times 10^{-19}}\)eV = 1.24 × 10⁹ eV = 1.24 BeV
Thus, to energies the electron beam, the energy should be of the order of BeV.

Question 16.
Find the typical de Brogue wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure; and compare It with the mean separation between two atoms under these conditions.
Solution:
Given T = 27 + 273 = 300 K; K = 1.38 × 10^-23 J/mol/K; p = 1 atm = 1.01 × 10⁵ Pa

We can see that the wave length with mean separation r, it can be observed (r >> λ) that separation is larger than wave length.

Question 17.
Compute the typical de Broglie wavelength of an electron in a metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10^-10 m.
[Note : Exercise 35 and 36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.]
Solution:
Given, T = 27 + 273 = 300 K; r = 2 × 10^-10m
Momentum, P = \(\sqrt{3 \mathrm{mKT}}\) = \(\sqrt{3 \times 9.11 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}\) = 1.06 × 10^-25 kg-m/s
λ = \(\frac{\mathrm{h}}{\mathrm{p}}\) = \(\frac{6.63 \times 10^{-34}}{1.06 \times 10^{-25}}\) = 62.6 × 10^-10m; Mean separation, r = 2 × 10^-10 m
\(\frac{\lambda}{r}\) = \(\frac{62.6 \times 10^{-10}}{2 \times 10^{-10}}\) = 31.3
We can see that de-Broglie wavelength is much greater than the electron separation.

Question 18.
Answer the following questions :
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e; (-1/3)e]. Why do they not show up in Millikan’s oil-drop experiment ?
Solution:
The quarks have fractional charges. These quarks are bound by forces. These forces become stronger when the quarks are tried to be pulled apart. That is why, the quarks always remain’ together. It is due to this reason that tough fractional charges exists in nature but the observable charges are always integral multiple of charge of electron.

(b) What is so special about the combination e/m ? Why do we not simply talk of e and m separately ?
Solution:
The motion.of electron in electric and magnetic fields are governed by these two equations.
\(\frac{1}{2} \mathrm{mv}^2\) = eV or Beυ = \(\frac{m v^2}{\mathrm{r}}\)
In these equations, e and m both are together i.e. there is no equation in which e or m are alone. So, we always take e/m.

(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures ?
Solution:
At ordinary pressure, only very few positive ions and electrons are produced by the ionisation of gas molecules. They are not able to reach the respective electrodes and becomes insulators. At low pressure, density decreases and the mean free path becomes large. So, at high voltage, they acquire sufficient amount of energy and they collide with molecules for further ionisation. Due to this, the number of ions in a gas increases and it becomes a conductor.

(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic ? Why is there an energy distribution of photoelectrons ?
Solution:
Because all the electrons in the metal do not belong to same level but they occupy a continuous band of levels, therefore for the given incident radiation, electrons come out from different levels with different energies.

(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations :
E = h v, p = \(\frac{\mathbf{h}}{\lambda}\)
But while the value of λ is physically significant, the value of v (and therefore, the value of the phase speed v λ) has no physical significance. Why ?
Solution:
As λ = \(\frac{\mathrm{h}}{\mathrm{p}}\) = p = \(\frac{h}{\lambda}\) ⇒ E = hv = \(\frac{\mathrm{hc}}{\lambda}\)
Energy of moving particle E’ = \(\frac{p^2}{2 m}\) = \(\frac{1}{2} \frac{\left(\frac{h}{\lambda}\right)^2}{m}\) = \(\frac{1}{2} \frac{h^2}{\lambda^2 \mathrm{~m}}\). For the relation of E and p, we note that there is a physical significance of λ but not for frequency v.