Session 2016-17 Kendriya Vidyalaya Sangathan · PDF fileSession 2016-17 Kendriya Vidyalaya Sangathan Regional Office ... one ovule after fertilisation forms one seed = ½ ... for identifying

STUDENT SPECIAL STUDY MATERIAL

Class XII

BIOLOGY

Session 2016-17

Kendriya Vidyalaya Sangathan

Regional Office

Guwahati

OUR SOURCE OF INSPIRATION

CHIEF PATRON

SHRI SANTOSH KUMAR MALL, IAS

COMMISSIONER KENDRIYA VIDYALAYA SANGATHAN

NEW DELHI

PATRONS

SHRI CHANDRA P NEELAP

DEPUTY COMMISSIONER KENDRIYA VIDYALAYA SANGATHAN

GUWAHATI REGION

SMT. ANJANA HAZARIKA

&

SHRI D. PATLE

ASSISTANT COMMISSIONERS KENDRIYA VIDYALAYA SANGATHAN

GUWAHATI REGION

CONVENOR

SHRI DHIRENDRA KUMAR JHA

PRINCIPAL

KV AFS BORJHAR

GUWAHATI

PREPARED BY: SMT. NG. SARJUBALA DEVI

PGT(BIOLOGY), KV AFS BORJHAR

This study module is aimed at ensuring at least pass mark in the board exams and is prepared using the

available study materials of KVS but in a concise manner. This also includes previous years CBSE questions

and marking scheme so that students will have idea on what type of questions can come from a particular

chapter and what points need to be in their answers to get marks. Hope this module will boost your

confidence both during the preparatory stage as well as during the examinations.

Students can follow these steps:

8, &10 carrying weightage of 42 marks, for complete and comprehensive revision

Make a target to learn 2 or 3 concepts a day and follow it up with weekly slip tests.

checking answers with the scoring key.

Practise these diagrams and flow charts in NCERT Textbook:

(i) 2.3 a-TS of young anther (ii) 2.5- enlarged view of pollen grain

(iii) 2.7(d)-Anatropous ovule (iv) 2.8 c-mature embryo sac

(v) 2.13 b-stages of embryo development (vi) 2.14-typical dicot, monocot embryo

(vii) 3.3b-Human female reproductive system (viii) 3.5: Sectional view of seminiferous tubule

(ix)3.6-human sperm (x) 3.7-human ovary section

(xi) 3.8-spermatogenesis & oogenesis (xii) 3.9-menstrual cycle

(xiii) 3.11-transport of embryo through fallopian tube (xiv) 6.4a-nucleosome

(xv) 6.8-DNA replication (xvi) 6.9-transcription unit

(xvii) 6.14-lac operon (xviii) 7.1-Millers experiment

(xix) 8.1- life cycle of Plasmodium (xx) 8.4-Antibody molecule

(xxi) 8.6-Life cycle of HIV (xxii) 10.8-Biogas plant

(xxiii) 11.2-rDNA technology (xxiv) 11.6-PCR

(xxv) 11.7-Bioreactor (xxvi) 13.3-organismic response representation

(xxvii) 13.4-human age pyramids (xxviii) 13.5-growth curves

(xxix) 14.4-ecological pyramids (xxx) 14.5-Primary succession

(xxxi) 14.6-carbon cycle (xxxii) 16.7-greenhouse gases

UNIT VI – REPORODUCTION

CHAPTER -1 REPRODUCTION IN ORGANISMS

Life Span: Period from birth till natural death.

Special Flowering: Bamboo- once in life, generally after 50-100 years.

Strobilanthus kunthiana (Neelakuranji) – flowers once in 12 years.

Dioecious: Only one type of reproductive structure in a plant. Eg. Papaya

Monoecious : Reproductive organs at different positions in same plant eg. Cucurbits, Maize.

Hermophrodite : Reproductive organs at different positions in same animal eg. Earthworm.

Cell division during gamete formation:

Haploid-parent (n) produces haploid gametes (n) by mitotic division, eg. Monera, fungi, algae and

bryophytes.

Diploid parent (2n) produces haploid gametes (n) by meiosis division (possess only one set of chromosomes)

and such specialized parent cell is called meiocyte or gamete mother cell (2n).

Parthenogenesis: Female gamete develops into new organism without fertilization. eg- Honey bee, turkey,

lizard, rotifers (Protozoans).

Seedless fruits formed by parthenocarpy.

Clone: A group of individuals of the same species that are morphologically and genetically similar to each

other & their parents.

PREVIOUS YEARS QUESTIONS & MARKING SCHEME

1. A male honey bee has 16 chromosomes whereas its female has 32 chromosomes. Give reason.1

Ans: Because male honey bee is produced by parthenogenesis, ie., it is developed from female gamete without

fertilization. 2. No organism is immortal, then why do we say there is no natural death in single – celled organisms?

Ans: Parent cell divides to give rise to new individuals

3. How many chromosomes do drones of honeybee possess? Name the type of cell division involved in

the production of sperms by them.

Ans. 16, Mitosis = ½ + ½

4. Meiosis is an essential event in the sexual life cycle of any organism. Give two reasons.

Ans. (i) Meiosis helps in formation of gametes by reductional division & maintains number of chromosomes

constant / maintains ploidy = ½

(ii) Recombination of genes in offsprings / brings variation = ½

5. Name any two organisms and the phenomenon involved where the female gamete

undergoes development to form new organisms without fertilization .

Ans. Rotifers / honeybees / some lizards / turkey (Any two) = ½ + ½; Parthenogenesis = 1

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CHAPTER : 2 SEXUAL REPRODUCTION IN FLOWERING PLANTS

FLOWERS : modified shoot, Site of sexual Reproduction.

POLLINATION: Transfer of pollen from anther to stigma

MICROSPOROGENESIS: The process of formation of micro spores from pollen mother cell (2n) through meiosis.

MEGASPOROGENESIS


1. Explain the process of pollination in Vallisneria. 2

Ans. The female flower reaches the surface of water by long stalk , male flower releases the pollen grains on

surface of water,pollen grains are carried by water currents,some of them reach the stigma and achieve

pollination. ½ x 4=2

2. Name the type of fruit apple is categorised under and why? Mention two other examples which belong to the

same category as apple. 2

Ans. False fruit , thalamus contributes to fruit formation : Strawberry, Cashew ½ x 4=2

3. Double fertilization is reported in plants of both, castor and groundnut. However, the mature seeds

of groundnut are non-albuminous and castor are albuminous. Explain the post fertilization events

that are responsible for it. 3

Ans. Development of endosperm (preceding the embryo) takes place in both , developing embryo derives

nutrition from endosperm = ½ + ½; Endosperm is retained / persists / not fully consumed in castor ,

endosperm is consumed in groundnut = 1 + 1

4. A flower of tomato plant following the process of sexual reproduction produces 240 viable

seeds. Answer the following questions giving reasons : [1 ×5 = 5 Marks]

(a) What is the minimum number of pollen grains that must have been involved in the pollination of its pistil ?

(b) What would have been the minimum number of ovules present in the ovary ?

(c) How many megaspore mother cells were involved ?

(d) What is the minimum number of microspore mother cells involved in the above case ?

(e) How many male gametes were involved in this case ?

Ans. (a) 240 , one pollen grain participates in fertilisation of one ovule = ½ + ½

(b) 240 , one ovule after fertilisation forms one seed = ½ + ½

(c) 240 , each MMC forms four megaspores out of which only one remain functional= ½ + ½

(d) 60, each microspore mother cell meiotically divides to form four pollen grains (240/4 = 60) ½, ½

(e) 480 , each pollen grain carries two male gametes (which participate in double

fertilisation) (240 × 2 = 480) = ½ + ½

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CHAPTER 3: HUMAN REPRODUCTION


1. Testes normally remain suspended in scrotum in mammals. Why? 1

Ans: Scrotum helps in maintaining low temperature , necessary for spermatogenesis

2. How many spermatozoa will be produced from 100 primary spermatocytes and how many ova will be

produced from 100 primary oocytes? 1

Ans: 400 spermatozoa , 100 eggs

3.Draw a sectional view of the ovary showing the different follicular stages of a humanfemale in her

preovulatory phase of menstrual cycle. 2

Ans: Primary, secondary, tertiary and graafian follicles(4 part to be labeled) ½X4=2

4. Draw the microscopic structure of human sperm and relate its different parts with their functions. 5

Ans: Plasma membrane-envelops the whole body of sperm; Acrosome – filled with enzymes that help in fertilization

if the ovum; Middle piece – Contains numerous mitochondria which produce energy for the movement of tail; Tail-

facilitates sperm motility essential for fertilization; Mitochondria – produce energy; Nucleus- Carries haploid set (n) of

genes (1 mark for diagram, 4 marks for function of any four parts)

5. During the reproductive cycle of a human female, when, where and how does a placenta develop? What is

the function of placenta during pregnancy and embryo development? 5

Ans. After implantation , uterus , chorionic villi and uterine tissue become interdigitated (physically fused) = 1 + 1 + 1

Placenta facilitates supply of oxygen / nutrients to the embryo = ½ Removal of carbon dioxide / waste material /

excretory material produced by the embryo = ½ Production of hCG / hPL / estrogens / progestogens (Any two) = ½ ×

2

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Chapter - 4: REPRODUCTIVE HEALTH


1. Explain how IUDs act as contraceptives. 3

A: IUDs . increase phagocytosis , suppress sperm motility fertilizing capacity of sperm uterus unsuitable for

implantation cervix hostile to sperms

2. (i) Expand MTP. (ii) Give two situations when MTP is advised. (iii) Write when amniocentesis and MTP can

be misused. 3

Ans. (i) Medical Termination of Pregnancy 1

(ii) (a) to get rid of unwanted pregnancies, (b)when continuation of pregnancy could be harmful or

fatal to mother/foetus/ both. ½ x2=1

(iii) for identifying the sex of the foetus 1

3. (a)HIV and Hepatitis –B are STDs. Mention the two other ways by which they can be transmitted to a

healthy person. (b)Why is early detection of STD essential? What can it lead to otherwise? Explain . 3

Ans. HIV and Hepatitis B can also be spread by- Infected needle, From infected mother to the child, Transfusion of

infected Blood (any 2) 1

b. Leads to complication in life later; pelvic inflammatory diseases /abortion/infertility/cancer (any 2)

4. Mrs. X was blamed for being childless though the problem was due to low sperm counts in the ejaculate of

her husband. Suggest a technique which could help the couple to have a child.4

Ans: IVF and its detail process and importance

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UNIT VII: GENETICS Chapter 5: PRINCIPLES OF INHERITANCE AND VARIATION

MENDEL’S LAW OF INHERITANCE

Mendel's first law (Law of dominance ) :

(i)Characters are controlled by discrete units called factors (genes).

(ii)Factors occur in pairs. (iii)In a dissimilar pair of factors one member of the pair dominates (dominant) the

other (recessive).

Mendel's Law of segregation (Purity of Gametes): The two alleles received, one from each parent,

segregate independently in gamete formation, so that each gamete receives one or the other with equal

probability. (Can be explained by monohybrid cross).

Mendel's law of Independent Assortment : Two characters determined by two unlinked genes are

recombined at random in gamete formation, so that they segregate independently of each other, each

according to the first law (note that recombination here is not used to mean crossing-over in meiosis). (Can be

explained by dihybrid cross).

Test Cross : Individual with dominant phenotype is crossed with homozygous recessive individuals to find the

homozygosity/heterozygosity .

Incomplete Dominance : Dominant gene is not fully expressed on recessive gene. So, the phenotype of hybrid do not

resemble with any of the parents. Eg- Antirrhinum majus (snapdragon), Mirabilis jalapa

Genotypic & phenotypic ratio- 1:2:1

Co-dominance : Both parental genes expressed in F1 progeny so the offspring shows resemblance with both the

parents. Eg- ABO blood group types in human.

Blood group shows 3 different alleles (IA, I

B, i) and 6 different possible genotypes.

Cross between IA i x I

B i shows the law of dominance, co-dominance & multiple alleles.

Mutation : Sudden changes in DNA. Mutagens : Chemicals/agents that caused mutation


1. Mention any two contrasting traits with respect to seeds in pea plant that were studied by Mendel.1 Ans.

Round/Wrinkled = ½, Yellow/ Green = ½

2. Give an example of a sex - linked recessive disorder in humans. 1

Ans. Colour blindness /any other correct example

3. How does the gene ‘I’ control ABO blood groups in humans ? Write the effect the gene has on the structure

of red blood cells. 2

Ans. – Gene ‘I’ has three different alleles IA , IB , i = ½

– IA produces A type of sugar / Antigen A group; IB produces B type of sugar / Antigen B group

– i - No sugar - O group = ½

– Structure - sugar polymers protrude from the surface of plasma membrane of RBCs = ½

4. Write the types of sex-determination mechanisms the following crosses show. Give an example of each type.

(i) Female XX with Male XO (ii) Female ZW with Male ZZ 2

Ans. (i) Male heterogamety , Grasshopper = ½ + ½ (ii) Female heterogamety , Birds = ½ + ½

5. In Snapdragon, A cross between true breeding red flower (RR) plants and true breeding white flower (rr)

plants showed a Progeny of plants with all pink flowers. (a) The appearance of pink flowers is not known as

blending. Why? (b) What is the phenomenon known as? 2

Ans. (a) R (dominant allele red colour) is not completely dominant over r (recessive allele white

colour) / r maintains its originality and reappear in F2 generation. = 1 (b) Incomplete dominance = 1

6. A colourblind child is born to a normal couple. Work out a cross to show how it is

possible. Mention the sex of this child.

Ans:

:, Male child-1.

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CHAPTER: 6 MOLECULAR BASIS OF INHERITANCE

Transcription in Eukaryotes:

Protein Synthesis:

PREVIOUS YEARS QUESTIONS & MARKING SCHEME:

1. A DNA segment has a total of 1000 nucleotides, out of which 240 of them are adenine containing nucleotides.

How many pyrimidine bases this DNA segment possesses? 1

2. Why does hnRNA need to undergo splicing ? Where does splicing occur in the cell? 2 Ans. hn RNA has both exons and introns, Introns are non-coding regions , which are

removed by the process called splicing ,splicing occurs in the nucleus. ½ x 4=2

3. (a) How many codons code for amino acids and how many are unable to do so? (b) Why are codes said to be

(i) degenerate and (ii) unambiguous? 2

Ans.(a) 61, 3 (½ + ½) (b) i. Degenerate-One amino acid may be coded by several codons 1

ii. unambiguous or specific- each codon codes for a specific amino acid. 1

4. Following the collision of two trains a large number of passengers are killed. A majority of them are beyond

recognition. Authorities want to hand over the dead to their relatives. Name a modern scientific method and

write the procedure that would help in the identification of kinship. 3

Ans. DNA fingerprinting (analysis) = ½

- Isolation and digestion of DNA by restriction endonuclease

- Separation of DNA fragments by electrophoresis and transferring them to synthetic membranes

/ nitrocellulose / nylon

- Hybridisation using labelled VNTR probe

- Detection of hybridised DNA fragments by autoradiography

- Matching banding pattern of DNA / DNA fingerprints / autoradiograms of the passengers

killed and that of relatives = ½ × 5

5. (a) Explain the role of regulatory gene, operator, promoter and structural genes in lac operon when E.Coli is

growing in a culture medium with the source of energy as lactose.

(b) Mention what would happen if lactose is withdrawn from the culture medium. 5

Ans. (a) Regulatory gene, codes for repressor of lac operon 1

Operator ; provides site for binding of repressor protein to prevent transcription 1

Promoter ; provides site for binding of RNA polymerase 1

Structural Genes; codes for enzymes / gene products required for metabolism of lactose 1

(b) If Lactose is withdrawn from the culture medium the operon is not induced or expressed 1

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CHAPTER - 7: EVOLUTION

Evolution: Process that results in heritable changes in a population spread over many generations (change in allele

frequencies over time) leading to diversity of organisms on earth.

Theory of Chemical Evolution:

hat the first form of life could have come from pre-existing

non-living organic molecules (e.g. RNA, protein, etc.) and that formation of life was preceded by chemical evolution.

onditions on earth – high temperature, volcanic storms, reducing atmosphere containing CH4, NH3, etc.

0 C.

ds. In similar experiments others observed, formation of sugars, nitrogen bases,

pigment and fats.

Evidences for Organic Evolution: Palaeontological evidences : fossils founds in rock and support the evolution.

Embryological Evidences

Comparative anatomy & morphology :

Anthropogenic Action: Industrial Melanism

Moths are able to camouflage themselves, i.e. hide in the background, survived. This type of evolution is due to

anthropogenic action. Lichen act as a industrial pollution indicator.

scale of months or years and not centuries.

Adaptive radiation The evolutionary process which produces new species from single point origin and spread to other geographical areas

(habitat) is called adaptive radiation.

Eg. Darwin finches found in Galapagos Island and Australian Marsupials.

MECHANISM OF EVOLUTION:

Hardy-Weinberg principle:

stable and is constant from generation to generation- genetic equilibrium-

Sum total of all the allelic frequencies is 1. p2 + 2pq + q

2 = 1 or, (p + q)

2 = 1

Five factors are known to affect Hardy-Weinberg equilibrium:


1. State a reason for the increased population of dark coloured moths coinciding with the loss of lichens (on tree

barks) during industrialization period in England. 1

Ans. Natural selection / survival of fittest / escaped predators due to camouflage

2. How can evolution by natural selection be explained by melanised moths before and after industrialisation in

England? 3

Ans. Before industrialization – mostly light moths (peppered moth) as they merged with the background of trees,

having lichen on their bark ,fewer black moths due to predation by birds as they were clearly visible against the grey

background ½ x 3= 1 ½

After Industrilisation – deposition of soot due to emission from factories made the tree barks black and the black

moths got camouflaged in the background , as the grey moths were now visible , they were predated upon by the birds

and their number got reduced . thus black moths selected over the grey ones. ½ x 3= 1 ½

3. How did S.L. Miller provide an experimental evidence in favour of Oparin and Haldane’s hypothesis?

Explain. 3

Ans. Miller created electric discharge in a closed flask, containing methane hydrogen ammonia and water vapour at

8000c, he observed formation of amino acids. 1x3=3

4. Explain adaptive radiation with the help of a suitable example. 3

Ans. Evolution of different species in a given geographical area starting from a point and literally radiating to other

geographical areas / habitat is called adaptive radiation = 1

A number of marsupials each different from other / Tasmanian wolf / Tiger Cat / Banded anteater /

Marsupial rat / Kangaroo/ Wombat / Bandicoot / Koala / Marsupial mole / Sugar glider (any two

or more) , evolved from an ancestral stock , but all within Australian continent = 1 + ½ + ½ = 2

// Darwin’s finches , from original seed eating features many other forms with altered beaks arose , enabling them to

become insectivorous / vegetarian finches on the same (Galapagos) islands = 1 + ½ + ½ = 2

5. (i) List any four evidences of evolution.

(ii) Explain any one of the evidences that helps to understand the concept of evolution. 5

i. Four evidence= Fossils/ comparative anatomy/ homologous organs / Analogous organs/ Bio- Chemical evidences /

embryological evidences (any four) ½ x 4=2

ii. Any one evidence explained, Definition/ concept 1, Example 1, how it explains evolution 1

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UNIT VIII: BIOLOGY IN HUMAN WELFARE

CHAPTER 8: HUMAN HEALTH AND DISEASES

Health- physical, mental and social well being

Common Human Diseases:

TYPES OF DRUGS OPIOIDS CANNABINOIDS COCA ALKALOIDS

SOURCE PLANT Poppy plant Papaver

somniferum

Cannabis sativa Erythroxylum coca

AFFECTED BODY

PART

CNS & GI Tract Cardiovascular system CNS

EXAMPLES Heroin, morphine Ganja, hashish,

marijuana

Cocaine, coke


1. Name any two secondary lymphoid organs in a human body and state the function of any of them. 2

Ans. Spleen, lymph nodes, tonsils, Peyer’s patches of small intestine, (vermiform) appendix (any two)½+ ½=1

They act as sites for interaction of lymphocytes with the antigen and cause immune response//

Function: Spleen: Trap blood-borne microorganisms thus filters blood/

Lymph nodes: Trap the microorganisms / antigens (which happen to get into the lymph and tissue fluid)

2. How are oncogenic viruses different from proto-oncogenes? 2

Oncogenic viruses are cancer causing viruses / external cancer causing factor

Proto-oncogenes- Identified in normal cells which when activated (under certain condition) could lead

to oncogenic transformation of cells / internal cancer causing factor 1+1=2

3. Why are adolescents especially advised not to smoke? How does smoking affect the functioning of the body?

2

Ans. Because smoking paves the way to hard drugs, causes increased chances of cancer, cause oxygen deficiency in

the body (any two); Nicotine (in cigarette) stimulates adrenal gland , which raises blood pressure / increased heart rate

. ½X4=2

4.(i) Which organ of the human body is initially affected when bitten by an infected female Anopheles? Name

the stage of the parasite that infects this organ. (ii)Explain the events that are responsible for chill and high

fever in the patient. 3

Ans.i. Liver cells, sporozoites ½+1/2=1

ii. Parasites reproduce asexually in RBC/multiply, Rapture of RBCs ,is associated with release of

toxic substance, haemozoin ½X4=2

5 (a) Why is mother’s milk considered very essential for the healthy growth of infants?

(b)What is the milk called that is produced in the initial days of lactation?

Ans.(a) The milk secreted has nutrients and contain antibodies Ig A , provide passives immunity 1

(b) Colostrum 1

6. Modern life style in big cities and towns is surely making the life more easy and comfortable for people. On

the contrary many more health issues and problems are on the rise and one of them is allergic reactions.

(a) Write any four steps you would suggest to minimise the cause of the above allergic responses.

(b) List any two allergens. How does the human body respond to them? Explain.

Ans. (a) Reduce air pollution , improve exposure /sensitivity of the children to the environment to

reduce vulnerability , improve resistance, improve food habits resulting in good health , introduce

physical exercise (any other appropriate measure ) (any four) ½X 4=2

(b) Mites in dust ,pollen, animal dander (any two) ½+ ½ =1

Immune system respond by producing antibodies of IgE type, inducing Mast cells release chemicals

like histamine/ serotonin in response to allergens 2

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CHAPTER-9: STRATEGIES FOR ENHANCEMENT IN FOOD PRODUCTION


1. How has mutation breeding helped in improving the production of mung bean crop ? 2

Ans. Produce disease resistant varieties, against yellow mosaic virus / powdery mildew = 1 + 1

2. (i) Why is inbreeding necessary? Give two reasons. (ii) What does continued inbreeding lead to? 3

Ans. Inbreedingis necessary if we want to evolve a pure line in any animal. 1

Inbreeding exposes harmful recessive genes that are eliminated by selection leading to accumulation of superior

genes, continued inbreeding reduces fertility and productivity( Inbreeding depression) 1+1=2

3. Give two reasons for keeping beehives in crop fields during flowering period. 3

Ans. Keeping beehives in crop fields during flowering period increases pollination efficiency, it improves yield of

crops and honey. 1+1=2

4. Explain how and why controlled breeding experiment is carried out in cattle. 3

Ans. Controlled breeding experiment are carried out using artificial insemination. Semen is collected from the male -

chosen as parent, injected into the reproduction tract of the selected female (cow), stored Semen may be used at a later

date and desirable matings are carried out. ½x4 =2

Helps overcome several problems of normal mating; improve the quality and quantity of desired yield 1

5. Enlist the steps involved in inbreeding of cattle. Suggest two disadvantages of this practice.

Ans. Inbreeding involves mating of closely related individuals within the same breed for 4-6 generations = ½

Superior males and superior females are identified and mated in pairs , the progeny are evaluated ,

superior males and females among them are selected for further mating = ½ × 3

Disadvantages : Inbreeding depression , reduction in fertility , reduction in productivity (any 2) = ½ × 2

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CHAPTER: 10 MICROBES IN HUMAN WELFARE Microbes in household products-

Microbes in industrial products-

Enzymes and other bioactive molecules-

Enzymes and their actions-

Microbes -- production of Bio-Gas- methanogens (e.g.) Methanobacterium.

Act on cellular material to produce methane, seen in anaerobic sludge digesters, rumen of cattle, flooded rice field &

in cattle dung


1. Name a free living and a symbiotic bacterium that serve as Biocontrol agents. Why are they so called?2

Ans. Free living-Azospirillum/Azotobacter ½ , Symbiotic- Rhizobium ½

Because they enrich the nutrient quality of the soil.1

2. Mention a product of human welfare obtained with the help of each one of the following microbes: 2

(a) LAB (b) Saccharomyces cerevisiae (c) Propionibacterium sharmanii (d) Aspergillus niger.

Ans. a) Milk to curd = ½ b) Bread / ethanol / alcoholic drinks / whiskey / brandy / beer / rum = ½

c) Swiss cheese = ½ d) Citric acid = ½

3. Name a genus of baculovirus. Why are they considered good biocontrol agents. 2

Ans. Nucleopolyhedrovirus 1, Species specific & narrow spectrum insecticidal applications 1

4. Choose any three microbes, from the following which are suited for organic farming which is in great

demand these days for various reasons. Mention one application of each one chosen. Mycorrhiza ; Monascus ;

Anabaena ; Rhizobium ; Methanobacterium ; Trichoderma. 3

Ans. Mycorrhiza : (Fungal symbiont of the association) Abosrb phosphorus from soil

Anabaena : Fix atmospheric nitrogen / Adds organic matter to the soil

Rhizobium : Fix atmospheric nitrogen (in leguminous plants)

Methanobacterium : They digest cellulosic material and the product / spent slurry can be used as fertiliser

Trichoderma : Biocontrol agent for several plant pathogens (Any 3 microbes = ½ × 3 = 1½)

(Any 3 corresponding roles = ½ × 3 = 1½)

5. Make a list of three household products along with the names of the micro-organism producing them. Ans.

Lactic acid bacteria – curd; Sacharomyces cerevisiae- bread; Propionibacterium sharmanii- swiss cheese

6x1/2=3

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UNIT IX: BIOTECHNOLOGY

CHAPTER -11: BIO TECHNOLOGY PRINCIPLES & PROCESSES


1. Mention the use of gel electrophoresis in biotechnology experiments. 1

Ans. Cut fragments of DNA can be segregated/separated

2. Name the technique that is used to alter the chemistry of genetic material (DNA, RNA) to obtain desired

result. 1 Ans. Genetic Engineering / Biochemical Engineering / Biotechnology

3. Write the functions of the following n biotechnology. 3

(a) Polymerase chain reaction technique (b) Restriction endonucleases (c) Thermus aquaticus

Ans a) Multiple copies of gene of interest can be obtained.

(b) They can cut DNA molecule at a particular point by recognizing a specific sequence of base pairs. Thus they are

useful in forming recombinant DNA.

(c) Thermus aquaticus – is the source of Taq- polymerase which remains active during high temperature induced

denaturation of DNA in PCR technique and therefore allows chain reaction to proceed .

4. (i) Why was a bacterium used in the first instance of the construction of an artificial recombinant DNA

molecule? (ii) Name the scientists who accomplished this and how?

Ans. (i) Bacterium has a plasmid in which the desired gene is introduced / the gene to betransferred is from a

bacterium –the anti-biotic resistance gene/ the host cell(bacterial cell) is required for gene cloning / bacteria produce

restriction endonucleases ½ + ½ (ii) Herbert Boyer and Stanley Cohen ½ + ½

Antibiotic resistant gene was isolated using restriction enzyme and introduced into the plasmid of bacterium –

Salmonella typhimurium ½ Later the recombinant plasmid was introduced into the bacterium – E.coli so that it could

make copies of gene. ½

5. Explain the mode of action of Eco RI.

Ans. EcoRI first inspects the length of a DNA sequence, then it binds with specific recognition sequence of DNA,

EcoRI cut each of the two strands of double helix at specific points in their sugar – phosphate backbones // diagram

with same value points to be accepted (1x3=3 Marks)

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CHAPTER- 12: BIO TECHNOLOGY & ITS APPLICATION

Application in agriculture : Genetically modified organisms (GMO)

Transgenic crops (GMO) -Crops contain or express one or more useful foreign genes.

Advantages -i) More tolerant to stresses (heat, cold, draught).

ii) Pest resistants GM crops, reduce the use of Chemical pesticides. Eg- BT-Cotton

iii) Reduced post harvest losses. iv) Enhance nutritional value of food.

PEST RESISTANT PLANTS: Bt- cotton

Protection of plants against nematodes –, Meloidogyne incognita infects tobacco plants & reduces yield. Specific

genes (DNA) from nematodes introduced into the plants using Agrobacterium tumifecians (soil bacteria). Genes

produce sense and antisense complementary RNA. Act as dsRNA and initiates RNAi ( RNA interference) and silences

the specific mRNA. Complementary RNA neutralizes the specific RNA of nematodes by a process called RNA

Interference and parasite cannot live in transgenic host.

In medicine:

Molecular diagnosis --

accurate detection of diseases can be done through : PCR (Polymerase chain reaction): Short stretches of

pathogenic genome is amplified for detection of suspected AIDS, Cancer or genetic disorder.

ELISA (Enzyme Linked Immunosorbent Assay) used to detect AIDS based on detection of antibodies produced

against antigen of pathogen.

Transgenic Animals 1. To know how genes contribute to development of disease.

2. To get biologically useful products . Eg. The first transgenic cow Rosie produced human protein enriched milk 3.

To verify vaccine and chemical safety.

Biropiracy -- Some organizations and multinational companies exploit or patent bioresources of other nations without

proper authorization. Indian patent bill is there to prevent such unauthorized exploitation.

GEAC- For validity of GM research and the safety of introducing GM organism.


1. People are quite apprehensive to use GM crops. Give three arguments in support of GM

crops so as to convince the people in favour of such crops. 3 Ans. i. Made crops more tolerant to abiotic stresses (cold, drought, salt, heat)

ii. Reduced reliance on chemical pesticides (pest- resistance crops)

iii. Helped to reduce post harvest losses.

iv. Increased efficiency of mineral usage by plants/this prevent early exhaustions of fertility of soil.

v. Enhanced nutritional value of food /vitamin ‘A’ enriched rice.

vi Create tailor made plants to supply alternative resources to industry (any three) 1X3=3

2. Explain how the company Eli Lilly was able to produce human insulin using recombinant DNA technique.

3

Ans.Eli lilly prepared two DNA sequences A and B, and introduced them in plasmids of E. Coli to produce insulin

chains , later extracted and combined by creating disulphide bonds. 1X3=3

4. How has the study of biotechnology helped in developing pest resistant cotton crop? Explain.3

Ans. Some strains of Bacillus thuringiensis produce proteins that kill insects (pests), these crystals contain a toxic

insecticidal protein, once the insect ingests this (inactive) toxin it is converted into an active form, due to alkaline pH

of the gut , activated toxin binds to surface of midgut epithelial cells and creates pores , causing swelling and lysis

leading to death of pest ½ x 6=3

5. Why is molecular diagnosis preferred over conventional methods? Name any two techniques giving one use

of each. 3

Ans. To allow early detection 1 Example –rDNA technology/ PCR / ELISA/ Probe (any two) ½ + ½

PCR- to detect low concentration of bacteria/ virus (HIV), ELISA- to detect antigens / to detect antibodies produce by

those antigens / to detect HIV, Probe- to detect a mutated gene (from a normal one) (any two corresponding functions)

½ + ½= 1

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UNIT - X ECOLOGY CHAPTER: 13 ORGANISMS AND POPULATIONS

Ecology: deals with the interaction among organisms between organisms & physical environment.

Biome: the largest ecological regions distinguishable by characteristic plants and animals.

There are six: tundra, conifer, deciduous forest, grassland, tropical, and desert.

RESPONSE TO ABIOTIC FACTORS

Regulation: Organisms maintain homeostasis achieved by physiological and behavioral means Thermo

regulation and osmoregulation.

Conformation Cannot maintain constant internal Environment # Body temperature and osmotic concentration

of body changes with ambient temperature and concentration of medium.-Thermo confirmer and osmo

confirmer

Migration : Organism moves away temporarily to another habitat in stressful condition

Suspension: Organisms suspend their metabolic activities during stressful condition

Resume their function at the return of favorable conditions.E.g. Hibernation (winter sleep) of Frog, Reptiles,

Polar Bear etc , Aestivation (summer sleep) in Snail and Fish.

Adaptation

Morphological, physiological and behavioral changes that enable organisms to adjust to the ever changing

environment . E.g. Kangaroo rat survives in desert conditions through internal oxidation of fat, removing

concentrated urine of limited quantity.

# Allen‘s rule-cold climate mammals have shorter ears and limbs to minimize heat loss.

# Polar mammals like seals have blubber to prevent heat loss.

# Burrowing habit to escape from heat

# Higher count of RBC, Hb(haemoglobin) at high altitudes.

Population attributes *Birth Rate/ NATALITY – Number of individuals born per thousand per year.

*Death Rate/MORTALITY – Number of individuals die per thousand per year.

*Sex Ratio – Ratio of male-female in the population.

Population density. - the number of individual organisms per unit area (appropriate measure – total number-

sometimes difficult to determine or meaningless because 4 factors N+I-M+E are concerned w.r.t habitat concerned

Age pyramids # Three ecological ages: Pre-reproductive, Reproductive and Post-Reproductive , High proportion pre-reproductive

individuals occur in Expanding population , Pre-reproductive and reproductive individuals are uniform in Stable

population and Pre-reproductive individuals are less in Declining population.

Population growth : Growth Models :

(i) Exponential growth: When resources in the habitat are unlimited.

(ii) Logistic growth: When resources are limited.

Density of population at any time at a given place depends on Natality, Mortality, Emigration Immigration

POPULATION INTERACTION


1. What is mutualism ? Mention any two examples where the organisms involved are commercially exploited in

agriculture. 2

Ans. Interaction between two species in which both are benefitted =1

i. Rhizobium in the roots (nodules) of legumes = ½ ii. Mycorrhiza / Glomus with the roots of higher plants = ½

2. How do snails, seeds, bears, zooplanktons, fungi and bacteria adapt to conditions unfavourable for their

survival ? 2

Ans. Snail-aestivation = ½; Seeds-dormancy/suspended metabolic activities = ½; Bear-Hibernation = ½; Zooplankton-

diapause/suspended development = ½; Fungi-Spore/Zygospore = ½ Bacteria-Cyst/spore = ½

3. Many fresh water animals can not survive in marine environment. Explain. 2

Ans. High salt concentration outside / hypertonic surroundings = 1

Loss of water from body / exosmosis from animal body / animal suffers osmotic problems = 1

4. (a) List the different attributes that a population has and not an individual organism.

(b) What is population density ? Explain any three different ways the population density can be measured, with

the help of an example each. 5

(a) Attributes of population: Birth rate , Death Rate , sex ratio,age pyramids/age distribution (any two) = ½ × 2

(b) Population density - Number of individuals per unit area at a given time / period = 1

1. Biomass / % Cover , e.g Hundred Parthenium plants and 1 huge banayan tree = ½×2

2. Relative Density , e.g Number of fish caught per trap from a lake =½×2

3. Numbers , e.g Human population = ½ × 2

4. Indirect estimation , e.g without actually counting/seeing them e.g tiger census based on

pugmarks and fecal pellets = ½ × 2

----------------------------------------------------------------------------------------

Chapter - 14: ECOSYSTEM The components of the ecosystem are seen to function as a unit: Productivity, Decomposition, Energy flow and

Nutrient cycle.

PRODUCTIVITY: Primary productivity:

o The amount of biomass or organic matter produced per unit area over a time period by plants during photosynthesis.

Gross primary productivity: (GPP) is the rate of production of organic matter during photosynthesis.

Net primary productivity: GPP – R = NPP.

Secondary productivity: is defined as the rate of formation of new organic matter by the consumer.

DECOMPOSITION:The process of decomposition completed in following steps:

o Fragmentation : Break down of detritus into smaller particles by detritivore (earthworm).

o Leaching: Water soluble inorganic nutrients go down into the soil horizon and get precipitated as unavailable salts.

o Catabolism : Bacterial and fungal enzymes degrade detritus into simple inorganic substances.

o Humification: Accumulation of dark coloured amorphous substances called humus.

o Mineralization: The humus is further degraded by some microbes and release of inorganic nutrients occur.

ENERGY FLOW IN ECOSYSTEM: Only 10% of energy transferred from one trophic level to other.

Food chain: Grazing food chain (GFC): it extends from producers through herbivore to carnivore.

Grass---- Grass hopper---- Frog----- Snake------- Hawk

Detritus food chain (DFC): Begins with dead organic matter (detritus) and pass through detritus feeding organism

in soil to organisms feeding on detritus-feeders.

Detritus--- Earthworm--- Bacteria/ Fungi ---- Plants--- Animals

Standing crop: each trophic level has a certain mass of living material at a particular time called as the standing

crop.

ECOLOGICAL PYRAMID: Three types : number, energy or biomass. In most ecosystems, all the pyramids, of number, of energy and biomass are

upright.

number in a tree ecosystem is inverted.

biomass in sea also inverted because the biomass of fishes is far exceeds that of phytoplankton.

energy is always upright, can never be inverted, because when energy flows from a particular trophic

level to the next, some energy is always lost as heat at each step.

ECOLOGICAL SUCCESSION: ecological succession.

the changes lead finally to a community that is in near equilibrium with the environment and that is called

climax community.

Primary succession: succession that starts where no living organisms are there- these could be areas where no

living organism ever existed may be a bare rock or new water body.

Secondary succession: succession that starts in areas that somehow, lost all the living organisms that existed there.

TYPES : Based on the nature of habitat – whether it is water or it is on very dry areas- succession of plants is called

hydrarch or xerarch.

Xerarch succession: Succession in bare rock: Lichen—Mosses—herbaceous plants--- shrubs--- trees

Hydrarch (succession in aquatic environment) Phytoplankton--- Zooplanktons --- rooted hydrophytes---- Sub merged and free-floating plant stage----- Reed-

swamp stage---- Marsh-meadow stage--- Shrub stage--- Trees--- the forest

NUTRIENT CYCLING: Of two types: Gaseous cycle & Sedimentary cycle.

Carbon Cycle rbon is fixed in the biosphere by photosynthesis, annually.

of Carbon is lost to sediments and removed from circulation.

releasing CO2 to atmosphere.

Phosphorus cycle: horus is the rock, which contain phosphorus in the form of phosphates.

Difference between Carbon and Phosphorus cycle: 1. No respiratory release of phosphorus

2. Reservoir for Carbon is atmosphere but for Phosphorus it is rocks.


1. ‘‘Man can be a primary as well as a secondary consumer.’’ Justify this statement. 1

Ans. Vegetarian diet - Primary consumer = ½ Non vegetarian diet - Secondary consumer = ½

2. How are productivity, gross productivity, net primary productivity and secondary productivity interrelated

?

Ans. Productivity is rate of biomass production = ½ GPP - R = NPP = 1

NPP - is biomass available to consumers for secondary productivity = ½

3. ‘‘It is often said that the pyramid of energy is always upright. On the other hand, the pyramid of biomass

can be both upright and inverted.’’ Explain with the help of examples and sketches. 5

Ans. Upright Pyramid of Energy diagram: e.g of any Grassland food chain depicting energy transfer at each trophic

level = 1+1

Upright Pyramid of Biomass: e.g grassland food chain- (Any other relevant example) =

1 for Diagram + ½ for example

Inverted Pyramid of Biomass: e.g aquatic ecosystem where small standing crop of phytoplanktons supports large

standing crop of zooplanktons =1 for Diagram + ½ for example

-------------------------------------------------------------------------------------

Chapter - 15: BIODIVERSITY AND CONSERVATION

Levels of biodiversity: Genetic diversity, Species diversity, Ecological diversity

Pattern of Biodiversity: Latitudinal gradients: Species diversity decreases as we move away from the equator towards the pole.

Reasons why tropical rain forest has greater biodiversity:

environments. Unlike temperate ones, are less seasonal, relatively more constant and predictable, promotes niche

specialization and lead to greater species diversity.

Species area relationship: Within a region species richness increased with increasing explored area but only up to a

limit.

Causes of biodiversity loss: Four major causes “The Evil Quartet”:

1) Habitat loss and fragmentation 2) Over-exploitation:

3) Alien species invasion: Nile perch introduced into Lake Victoria in east Africa led to extinction of 200 species of

cichlid fish in the lake. Parthenium, (carrot grass), Lantana, and water hyacinth (Eichornia) posed a thread to

Clarias gariepinus for aquaculture purposed is posing a threat to indigenous

catfishes in our rivers.

4) Co-extinction

BIODIVERSITY CONSERVATION: Reasons: Narrowly utilitarian, Broadly utilitarian and Ethical

Methods: 1. In situ conservation: When we conserve and protect the whole ecosystem, its biodiversity at all level is protected

– we save the entire forest to save the tiger

Biodiversity hot spot: regions with very high levels of species richness and high degree of endemism.(species

confined to that region and not found anywhere else)

Sacred groves: tract of forest were set aside, and all the trees and wildlife within were venerated and given total

protection.

2. Ex situ conservation: threatened animals and plants are taken out from their natural habitat and placed in special

setting where they can be protected and given special care.

cryopreservation, Genetic strains

are preserved in seed bank.

Convention on Biodiversity: “The Earth Summit” held in Rio de Janeiro, World Summit on Sustainable

development held in Johannesburg, South Africa


1. List any four techniques where the principle of ex-situ conservation of biodiversity has been employed. 2

Ans. Cryopreservation, in vitro fertilisation, micro propagation / tissue culture , sperm bank/ seed bank / gene bank ½

×4

2. Why are sacred groves highly protected? 2

Ans. Because they are the last refuges for a large number of rare and threatened plants.

3. What is meant by ‘alien species’ invasion? Name one plant and one animal alien species that are threat to

our Indian native species. 2

Ans. When alien species are introduced unintentionally or deliberately for whatever purpose, some of them turn

invasive, and cause decline or extinction of indigenous species. 1

Plant- Carrot grass/ lantana ½ ; Animal- African catfish Clarias gariepinus ½

4. Since the origin of life on the earth, there were five episodes of mass extinction of species. (i) How is the 'Sixth

Extinction', presently in progress, different from the previous episodes? (ii) Who is mainly responsible for the

'Sixth Extinction''? (iii) List any four points that can help to overcome this disaster.3

Ans. (i) The rates are faster / accelerated / current species extinction rate are estimated to be 100-1000 times faster

than in the pre-human times. = ½ (ii) Human activities. = ½ (iii) a. Preventing habitat loss and fragmentation b.

Checking overexploitation c. Preventing alien species invasion d. Preventing co-extinction e. Conservation /

Preservation of species.(any four) = ½×4 = 2 [1+2= 3 marks]

5. (a) Why is there a need to conserve biodiversity?

(b) Name and explain any two ways that are responsible for the loss of biodiversity. 5

Ans. (a) 1. to continue to get the products of human consumption

2. plays a major role in many eco system services that nature provides and that is invaluable

3. moral duty to pass on biological legacy in good order to future generations (Any two) = (1×2 = 2)

(b) 1. Habitat loss and fragmentation- large habitats when broken lead to loss of habitat for

animals needing large territories (are badly affected) – population decline

2. Overexploitation- leading to extinction of many, especially commercially important species

3. Alien species invasion - alien species when introduced may turn invasive causing decline

and extinction of indigenous species // explain with an example.

4. Coextinction- when one species become extinct , any other organism intimately associated

also becomes extinct. (any two) (1½ × 2) [2 + 3 = 5 marks]

----------------------------------------------------------------------------------------

CHAPTER-16: ENVIRONMENTAL ISSUES

GREEN HOUSE EFFECT AND GLOBAL WARMING: The greenhouse effect is a naturally occurring

phenomenon that is responsible for heating of Earth’s surface and atmosphere.

global warming or

enhanced green house effect. OZONE DEPLETION IN THE STRATOSPHERE: Thickness of ozone layer measured in Dobson units (DU)

Effects of UV rays:

– B radiation and high dose of UV – B causes inflammation of cornea called

snow-blindness, cataract etc.

Prevention: Montreal Protocol was signed at Montreal (Canada) in 1987 to control emission of ozone depleting substances.

PREVIOUS YEARS QUESTIONS & MARKING SCHEME 1. List two advantages of the use of unleaded petrol in automobiles as fuel. 1

Ans. (i) Allows the catalytic convertor to remain active = ½ (ii) Reduces air pollution = ½

2. State the cause of Accelerated Eutrophication 1

Ans. Pollutants from human activities /effluents from industries / effluents from home / sewage /

agricultural (chemical) wastes radically accelerate the ageing process

3. What is joint forest management? How can it help in conservation of forests ? 2

Ans. JFM - A programme (initiated by Govt. of India in 1980) where govt. works closely with local

communities for protecting & managing forests = 1

Forests are conserved by locals in a sustainable manner as locals are also benefitted with forest

products / (fruits / gum / rubber / medicines etc) = 1

4.“Determination of Biological Oxygen Demand (BOD) can help in suggesting the quality of a water body.”

Explain. 3

Ans. High BOD of a water body indicates more number of micro-organisms in water , resulting in bad

quality of water / death of aquatic creatures , more polluting potential = 1 × 3

// Lower BOD of water body indicates less number of micro-organisms in water , good quality of

water / aquatic life flourishes , less polluting potential = 1 × 3

5. With the help of a flow chart, show the phenomenon of biomagnification of DDT in an aquatic food chain.

BEST OF LUCK!


Class 12

Chemistry (Theory)

Session 2016-17


Regional Office

Guwahati

Kendriya Vidyalaya Sangathan Chemistry Study Material Guwahati Region

http://kvsroguwahati.org Page 2

Our Source of Inspiration

CHIEF PATRON

Shri. Santosh Kumar Mall

IAS Commissioner


New Delhi

PATRONS

Shri. Chandra P. Neelap

Deputy Commissioner


Guwahati Region

Smt. Anjana Hazarika & Shri. D. Patle

Assisstant Commissioners


Guwahati Region

CONVENOR

Shri. Dhirendra Kumar Jha

Principal

Kendriya Vidyalaya, Air Force Station, Borjhar

Guwahati

PREPARED BY:

Ashwajeet Dive

PGT Chemistry




Preface There is no substitute as such for hard work. However, planned study and a bit of

smart work can do the trick. With planning I mean prioritizing. When your days are

numbered you just can't go through everything. It is therefore advisable not to

panic and study steadily giving priority to the topics most likely to appear in the

examination.

When it comes to AISSCE, nothing is guaranteed. No one can predict anything

precisely. But, there exist concepts that can enable students to score more with

minimal of efforts.

One should NOT restrict his studies to this study material only. The content of this

material is something a student must not leave. It is designed especially for those

who are finding Chemistry difficult (and for those who are stressed by thoughts of

getting failed) at this time of the session. Different questions are frequently framed

based on these concepts. So, as a student if you are initiating your studies now, you

may take the content into consideration if you find it helpful. All the very best!

Feedback, Suggestions & Quarries: [email protected]

N.B. Please, bring corrections (if any) into notice.



Index

Sr.

No. Section Page Predicted

Marks

1. Structures (p-Block Elements) 5 2

2. Differentiating Tests 8

6 to 8 3. Name Reactions 12

4. Miscellaneous Reactions 19

5. Other Important Reactions 23 4 to 6

6. Exemplar Organic Conversions (involving Benzene) 29

7. Reaction Mechanisms 31 2

8. IUPAC Nomenclature 36 1 or 2

9. Biomolecules, Polymers, Chemistry in Everyday Life 36 10

10. Essentials from Other Chapters 38 8 to 10

TOTAL (lower limit count) >30



Structures

(The p-Block Elements)







Differentiating Tests



ALCOHOLS

1. Lucas Test

This test is based upon relative reactivity of various alcohols towards HCl in the

presence of ZnCl2. In this test, alcohol is treated with Lucas reagent (HCl+ZnCl2).

On reaction, alkyl chlorides are formed which being insoluble result in

cloudiness/turbidity in the solution.

If cloudiness appears immediately, tertiary (3˚) alcohol is indicated.

If cloudiness appears within 5-10 minutes, secondary (2˚) alcohol is indicated.

If cloudiness appears only upon heating, primary (1˚) alcohol is indicated.

PHENOLS

2. Ferric Chloride Test

Phenol gives a violet colored water soluble complex with ferric chloride

(FeCl3). The complex formation takes place in all compounds containing enolic group

(=C—OH). However, the colors of complexes are different such as green, blue, violet,

etc. and depend upon the structure of phenols.

Alcohols being weakly acidic DO NOT form such a complex and no change in

color is observed.

6C6H5—OH + FeCl3 [Fe(OC6H5)6]3– + 3H+ + 3HCl Phenol ferric chloride violet

CARBONYL (>C=O) COMPOUNDS

3. 2, 4-DNP Test

Carbonyl compounds (i.e. aldehydes and ketones) when treated with 2, 4-

Dinitrophenylhydrazine (2, 4-DNP) form yellow, orange or red precipitate.

No such precipitation occurs with other organic compounds.



ALDEHYDES

4. Tollen's Test (Silver Mirror Test)

Tollen’s reagent is ammonical solution of silver nitrate. On warming with this

reagent, aldehydes form a silver mirror on walls of the container.

R—CHO + 2[Ag(NH3)2]+ + 3OH– R—COO– + 2Ag↓ + 2H2O + 4NH3 Aldehyde Tollen’s reagent silver mirror

Ketones do not respond to this test with the exception of α-hydroxy ketones

(acyloins) which give this test positive.

Fructose (Monosaccharide) being α-hydroxy ketone gives this test positive.

Formic acid also gives silver mirror test positive.

4. Fehling's Test

Fehling’s solution is an alkaline solution of copper sulphate containing sodium

potassium tartarate (Rochelle salt) as a complexing agent. Aliphatic aldehydes on

warming with this solution, gives a reddish brown precipitate of cuprous oxide.

R—CHO + 2Cu2+ + 5OH– R—COO– + Cu2O↓ + 3H2O Aldehyde reddish brown

(Aliphatic)

Aromatic aldehydes DO NOT give this test and therefore this can also be used

to differentiate between aliphatic and aromatic aldehydes.

Monosaccharides respond to this test positively.

Formic acid also gives this test positive.

METHYL KETONES

5. Iodoform (or Haloform) Test

Iodoform test is given by acetaldehyde and methyl ketones. The reaction

involves their treatment with sodium hypoiodite (I2 + aq. NaOH). A yellow precipitate

of iodoform is obtained as a result.

NaOH + I2 NaOI + HI

Sodium hypoiodite

(or acetaldehyde) (Yellow)



AMINES

6. Carbylamine Test (Isocyanide Test) This test is employed to identify 1˚ amines. The compound is warmed with

chloroform in the presence of alcoholic solution of potassium hydroxide.

2˚and 3˚ amines do not respond to this test.

R—NH2 + CHCl3 + 3KOH R—NC + 3KCl + 3H2O 1˚ amine chloroform alkyl isocyanide

7. Hinsberg's Test

This test helps to differentiate between 1˚, 2˚ and 3˚ amines. The amine to be

tested is treated with benzenesulphonyl chloride, C6H5SO2Cl (Hinsberg's reagent) in

the presence of excess of aqueous potassium hydroxide.

A clear solution in aqueous KOH which on acidification gives an insoluble

substance indicates 1˚ amine.

A precipitate which is insoluble in KOH solution indicates 2˚ amine.

3˚ amines do not react with benzenesulphonyl chloride.

CARBOXYLIC ACIDS

8. Bicarbonate Test

Carboxylic acids react with hydrogen carbonates (bicarbonates) to produce brisk

effervescence due to the liberation of CO2 gas.

R—COOH + NaHCO3 R—COONa + CO2↑ + H2O Carboxylic acid sodium bicarbonate effervescence



Name Reactions



HALOALKANES AND HALOARENES

1. Finkelstein's Reaction

When alkyl chlorides or bromides are treated with sodium iodide (NaI) in the

presence of dry acetone yields alkyl iodides. This reaction is called Finkelstein's

reaction.

2. Swartz Reaction

The reaction in which alkyl fluorides are prepared by heating alkyl bromides or

chlorides in presence of metallic fluorides like AgF, CoF2, SbF3 or Hg2F2 are called

Swarts reaction.

3. Wurtz Reaction When alkyl halides react with sodium metal in dry ether medium to give

higher alkanes the reaction is called Wurtz reaction.

4. Fittig Reaction

Aryl halides when treated with sodium metal in dry ether, two aryl halides are

joined together. This is called Fittig reaction. The reaction is quiet useful for preparing

diphenyl.

5. Wurtz–Fittig Reaction

When the mixture alkyl and aryl halide is treated with Na metal in dry ether

medium alkyl benzene is obtained. This is called Wurtz-Fittig reaction.



6. Sandmeyer Reaction

The Sandmeyer reaction is a chemical reaction used to synthesise aryl halides from

aryl diazonium salts.

Aniline (aryl amines) is first converted to its diazonium salt (Ar—N2Cl) using

nitrous acid (HCl + NaNO2).

ALCOHOLS, PHENOLS AND ETHERS

7. Kolbe's Reaction

When sodium phenoxide is heated with CO2 at 400 K and at a pressure of 4-7

atm sodium salicylate is formed as the major product. This on acidification yields

salicylic acid. This is called Kolbe's reaction.

8. Reimer–Tiemann Reaction

Treatment of phenol with chloroform in the presence of aqueous alkali at 340 K

results in the formation of o-hydroxybenzaldehyde (salicylaldehyde) and p-

hydroxybenzaldehyde, the ortho isomer being the major product. This reaction is called

Reimer-Tiemann reaction.



9. Williamson's Synthesis

When sodium alkoxide is heated with alkyl halide, ethers are formed. This

reaction is called Williamson's synthesis.

Sodium alkoxide is prepared by the action of sodium on alcohol.

R—OH + Na RONa + ½H2 Alcohol sodium sodium alkoxide

R—X + NaOR' R—OR' + NaX Alkyl halide sodium alkoxide Ether

It is important to note that, the alkyl halide to be used in the Williamson's

synthesis should be 1 . This is because 3 alkyl halides have a strong tendency to

undergo elimination which results in the formation of alkene and not ether (refer page

338 for details).

ALDEHYDES, KETONES AND CARBOXYLIC ACIDS

10. Rosenmund's Reduction

Rosenmund’s reaction involves hydrogenation of acyl chloride (acid chloride)

over catalyst palladium on barium sulphate (Pd/BaSO4) to yield aldehydes.

11. Stephen's Reaction

Nitriles are reduced to corresponding imine hydrochloride by stannous

chloride (SnCl2) in presence of dil. HCl which on further acid hydrolysis gives

corresponding aldehyde. This reaction is called Stephen's reaction.

12. Etard's Reaction

Benzaldehyde can be prepared from toluene from this reaction. Etard's reaction

involves the oxidation of toluene with chromyl chloride (CrO2Cl2) in CCl4 or CS2.



13. Gattermann–Koch Reaction

This reaction involves the treatment of diazonium salts with Cu/HCl or Cu/HBr

to yield aryl chlorides or bromides respectively.

14. Cannizzaro's Reaction

Aldehydes which do not have α-hydrogen atom, such as formaldehyde and

benzaldehyde, when heated with concentrated (50%) alkali solution give a mixture of

alcohol and salt of carboxylic acid.

In this reaction, the aldehyde undergoes disproportionation. One molecule of

aldehyde is oxidized to (salt of) carboxylic acid while other one is reduced to

alcohol.

Ketones DO NOT give this reaction.

15. Clemmensen's Reduction

The carbonyl group (>C=O) can be reduced to methylene (>CH2) group resulting

in formation of alkanes by zinc amalgam and concentrated HCl (Zn-Hg/HCl). This

reaction is called is Clemmensen's reduction.



16. Wolff–Kishner's Reduction

The carbonyl group (>C=O) can be reduced to methylene (>CH2) group resulting

in formation of alkanes by hydrazine followed by heating with sodium or potassium

hydroxide in ethylene glycol. This reaction is called is Wolff-Kishner reduction.

17. Hell–Volhard–Zelinsky (HVZ) Reaction

When carboxylic acids are treated with Cl2 or Br2 in the presence of red

phosphorus, the α-hydrogen atoms of carboxylic acids are replaced by chlorine and

bromine.

AMINES

18. Gabriel–Phthalimide Synthesis

In this method phthalimide is first converted into potassium phthalamide by

reaction with KOH which on further treatment with alkyl halide gives N-alkyl

phthalimide. This on alkaline hydrolysis gives primary (1˚) amine.

By using this method, we can prepare only 1˚ aliphatic amines. Aromatic, 2˚ or

3˚ amines CANNOT be prepared by this method.



19. Hoffmann Bromamide Degradation

Primary (1˚) amides on reaction with Br2 in the presence of alkalis give 1˚

amines. It may be noted that the amine formed by this method has one carbon less

than the parent compound.

R—CONH2 + Br2 + 4NaOH R—NH2 + 2NaBr + Na2CO3 + H2O Amide 1˚ amine

20. Gattermann Reaction

This reaction involves the treatment of diazonium salts with Cu/HCl or Cu/HBr

to yield aryl chlorides or bromides respectively.



Miscellaneous Reactions



1. Aldol Condensation

Two molecules of an aldehyde or a ketone having at least one α-hydrogen atom

condense in the presence of dilute alkali to give β-hydroxy aldehyde (aldol) or β-

hydroxy ketone (ketol). This reaction is called aldol condensation.

2. Crossed Aldol Condensation

When aldol condensation takes place between two different aldehydes or ketones

then it is called crossed aldol condensation or mixed aldol condensation.

Crossed aldol condensation can also occur when one of the carbonyl molecule do

not contain α-hydrogen, with other molecule possessing α-H atom.



3. Coupling Reactions

Benzene diazonium chloride when reacts with compounds like phenol and

aniline form azo compounds. This reaction is called coupling reaction or azo coupling.

The azo compounds are coloured and many of them are used as dyes and

indicators.

4. Diazotization

Aryl amines (such as aniline) react with nitrous acid, HNO2 (HCl + NaNO2) at

low temperature to give diazonium salts. This reaction is known as diazotisation.

Nitrous acid being unstable is prepared in situ by the reaction of sodium nitrite and

dilute hydrochloric (mineral) acid.

5. Hydroboration–Oxidation Reaction

In this reaction alkene is treated with diborane (B2H6) followed by the

treatment with water in the presence of H2O2. Alcohol is obtained as a product.

3 CH3-CH=CH2 + (BH3)2 ————→ 3 CH3CH2CH2OH + B(OH)3 or H3BO3 Propene diborane propanol boric acid

1. Ozonolysis of Alkenes



Alkenes react with ozone to form ozonides which on subsequent reductive

cleavage with Zn dust and water or H2/Pd give carbonyl compounds (i.e. aldehydes or

ketones). In general, the reaction can be expressed as:

Zn dust removes H2O2 formed, which otherwise can further oxidise aldehydes

formed to acids. Thus, by starting with suitable alkene, the desired aldehyde or ketone

can be formed.

2. Decarboxylation

Sodium salts of carboxylic acids lose CO2 when heated with soda lime (NaOH +

CaO) and form alkane with one carbon less.

3. Esterification

The reaction involves treating an alcohol with carboxylic acid, acid chloride or

anhydride to form ester. In the reaction, O—H bond of ROH breaks, with —H getting

replaced with —COR. Therefore, the reaction is also referred to as acylation of

alcohol.



Other Important Reactions



1. Reduction of —CHO/>C=O group to 1 /2 alcohol.

2. Industrial/Commercial preparation of phenol.

3. Synthesis of aspirin.

4. Selective oxidation of 1 alcohol to aldehyde.

Where, CrO3 = chromium trioxide (in anhydrous medium)

PCC = Pyridinium chlorochromate (a complex of CrO3 with pyridine and HCl)

5. Passage of vapors of alcohol over heated Cu tube.

6. Reaction of phenol with Br2 in non-polar (CS2) and polar (H2O) media.



7. Manufacture of methanol (wood spirit).

8. Dehydration of alcohol at different temperatures.

9. Conversion of —CN and —COOR groups to —CHO group.

Where, (DIBAL-H) = Diisobutylaluminium hydride

10. Formation of acetals and ketals.



11. Reaction of aldehyde/ketone (>C=O) with derivatives of ammonia (Z—NH2).

12. Oxidation of alcohol to carboxylic acid by Jones reagent.



13. Preparation of benzoic acid from alkylbenzenes.

14. Preparation of phthalimide.

15. Reduction of —CN and —CONH2 to —CH2NH2.

16. Reaction of amines with nitrous acid, HNO2 (NaNO2 + HCl).





Exemplar Organic

Conversions

(Involving Benzene and its derivatives)





Reaction Mechanisms



1. SN1Mechanism

Reaction: (CH3)3C —Br + KOH (CH3)3C —OH + KBr 2-Bromo 2-methylpropane 2-Methyl propan-2-ol

Mechanism:

Preferred Alkyl Halide : Tertiary (3 )

Steps : Two

Molecularity of RDS : One (first order) i.e. (CH3)3C —Br

Attack : Front side as well as backside attack of nucleophile

Reaction Intermediate : Carbocation

Stereochemistry : Inversion as well as retention of configuration

SN1 : First Order (Unimolecular) Nucleophilic Substitution



2. SN2 Mechanism

Reaction: CH3—Br + KOH CH3—OH + KBr Methylbromide Methyl alcohol

Mechanism:

Preferred Alkyl Halide : Primary (1 )

Steps : One

Molecularity of RDS : Two (second order) i.e. CH3—Br & :OH−

Attack : Backside attack of nucleophile

Reaction Intermediate : Pentavalent C (simultaneous bond making/breaking)

Stereochemistry : Inversion of configuration (Walden inversion)

SN2 : Second Order (Bimolecular) Nucleophilic Substitution

3. Intramolecular Dehydration of Alcohol



4. Intermolecular Dehydration of Alcohol

2CH3—CH2—OH H+ CH3—CH2—O—CH2—CH3

Alcohol Ether

Mechanism:

5. Formation of Alcohol from Alkene



6. Esterification

R—OH + R—COOH H+ RCOOR + H2O Alcohol Carboxylic acid Ester



Other MUST DO from

Book 2



1. IUPAC Nomenclature (1 or 2 marks)

Assigning name to the give structure or vice versa is one of the most

commonly asked questions in AISSCE. Moreover, simple names or

structures are asked. Students are therefore advised to practice the

nomenclature.

2. Biomolecules, Polymers & Chemistry in Everyday Life (10

Marks)

These three chapters have a combined weightage of 10 marks.

Knowledge based questions are asked from these and students can very

well score full 10 marks provided they thoroughly prepare the contents.

Frequently asked questions include:

Classification (of carbohydrates, amino acids, vitamins, polymers,

etc.)

Reducing and non-reducing sugars

Vitamin deficiencies

Structures of glucose, fructose, sucrose, maltose, etc.

Monomers of given polymers (along with their structures)

Examples (of analgesics, antipyretics, tranquillizers, antiseptics,

artificial sweeteners, etc.)

Various terms (like peptide bond, denaturation of proteins,

copolymers, elastomers, thermoplastics and thermosetting plastics,

etc.)



Essentials from Other

Chapters



The Solid State

Fluids. Substances which are able to flow (i.e. liquids and gases).

Solid State. The state of a substance in which it has definite volume and definite

shape.

Solid Substances: Substances whose melting point is above the room temperature.

Crystalline Solids. The substance in which constituent particles have orderly

arrangement.

Amorphous Solids. The substance in which constituent particles do not have orderly

arrangement.

Crystalline Solids Amorphous Solids

1. Internal arrangement of particles is regular.

2. They have long range ordered arrangement of

particles.

3. They have sharp melting points.

4. They have characteristic heats of fusion.

5. They give a regular cut when cut with a sharp-edged

knife.

6. They are regarded as true solids.

7. They are anisotropic.

1. Internal arrangement of particles is irregular.

2. They have only short range ordered arrangement of

particles.

3. They do not have sharp melting points.

4. They do not have characteristic heats of fusion.

5. They give irregular cut.

6. They are regarded as pseudo solids or super cooled

liquids. 7. They are isotropic.

Isotropy. Phenomenon of showing same physical properties (such as refractive index,

conductivity, etc.) in all directions. It is caused by random arrangement of particles.

Anisotropy. Phenomenon of showing different physical properties in different

directions. It is caused by orderly arrangement of particles.

Polymorphs. Different crystalline forms of a substance. Diamond and graphite are

polymorphs of carbon. They are also known as polymorphic forms.

Classification of Crystalline Solids

Type Constituent Particles Binding Forces Examples General Properties

Molecular

Solids

Atoms or non-polar

molecules

London

(dispersion)

forces

Noble gases, H2,

Cl2, I2, dry ice

(solid CO2) Fairly soft, non-conductors of heat and

electricity, low to moderately high melting

points, generally exist as liquids or gases at

room temperature.

Polar molecules Dipole-dipole

interactions

Solid SO2 and

NH3

Polar hydrogen bonded

molecules Hydrogen bonds Ice

Ionic

Solids Cations and anions

Ionic bonds

or

electrostatic force

Salts

Hard and brittle, high melting points, high

heats of fusion, poor thermal and electrical

conductivity. However, conduct electricity in

molten or dissolved state.

Covalent

Solids or

Network

Atoms that are

connected in the

covalent bond network

Network of

covalent bonds

Diamond,

graphite, quartz,

silica

Very hard, very high melting points, poor

thermal and electrical conductivity. Graphite,

however, is an exception.



Solids

Metallic

Solids Cations in electron cloud Metallic bonds Metals

Soft to very hard, low to very high melting

points, excellent thermal and electrical

conductivity, malleable and ductile.

Crystal Lattice. Regular three-dimensional arrangement of identical points in space.

It is also called space lattice.

Unit Cell. Three-dimensional group of lattice points (particles) that generate the whole

lattice by translation or stacking.

It is simple (also called primitive or basic) when particles are present only at

the corners, face centred when particles are present at the centre of each face along

with the corners and body centred when particles are present at the centre of the body

along with the corners.

Draw table 1.3 Seven Primitive Unit Cells and their Possible Variations as Centred Unit Cells

Bravais Lattices. The 14 different types of lattices (as mentioned in the table above).

There are three types of cubic unit cells. Simple cubic cell has 1 particle in it, body

centred cubic (bcc) has 2 while face centred cubic (fcc) has a total of 4 particles in it.

For simple cubic cell, edge length (a) is related to radius (r) as, a 2r or r

For bcc, the two are related as, a

√ or r

√

a

For fcc, the relation is, a 2 √ r or r

√

Square Close Packing. The two-dimensional arrangement of particles in which each

sphere has the co-ordination number of four.

Hexagonal Close Packing. The two-dimensional arrangement of particles in which

each sphere has the co-ordination number of six.

Hexagonal Close Packing (hcp). The three-dimensional arrangement of particles

with hexagonal symmetry. In hcp the alternating layers are same (AB AB... type).

Cubic Close Packing (ccp). The three-dimensional arrangement of particles with

cubic symmetry. In ccp the first, second and third layers are all different (ABC ABC...

type). The cubic closed packed structure so obtained is face centred (fcc).

Co-ordination Number. The number of nearest neighbouring spheres or particles in

close packing.



Tetrahedral Voids. The vacant space between the four touching spheres, centres of

which are at the corners of a regular tetrahedron. The number of tetrahedral voids is

twice (2N) the number of spheres (N).

0.225

Octahedral Voids. The interstitial void formed by the combination of two triangular

voids of the first and second layer. The number of octahedral voids is same (N) as the

number of spheres (N).

0.414

Thus, octahedral voids are larger as compared to tetrahedral voids.

NaCl Structure. Cl– ions have ccp arrangement and Na+ ions occupy all the

octahedral voids. Co-ordination number if Na+ and Cl– is 6 : 6.

Zinc Blend Structure. S2– ions have ccp arrangement and Zn2+ ions occupy half the

alternate tetrahedral voids. Co-ordination number of Zn2+ and S2– is 4 : 4.

CsCl Structure. Cl– ions are in cubic arrangement and Cs+ ions occupy cubic voids.

Co-ordination number is 8 : 8.

Fluorite Structure. Ca2+ ions (cations) in ccp and F– ions (anions) occupy all

tetrahedral voids. Co-ordination number is 8 : 4.

Antifluorite Structure. Anions have ccp arrangement and cations occupy all the

tetrahedral voids. Co-ordination number is 4 : 8. For example, Na2O.

Packing Efficiency. For a particular unit cell, it is the per cent of total space occupied

by the particles (spheres). For simple cubic cell 52.4% of space is occupied, whereas

for bcc and fcc it is 68% and 74% respectively.

Density, d of the crystal is related to edge length, a and atomic mass (formula mass),

M as:

d (g cm–3)

or d (kg m–3)

or d (g cm–3)

Where, z is the number of particles in the unit cell and NA is Avogadro's number

. Further, mass of an atom, m

N.B. Numerical questions based on above formula are frequently asked for 3 marks in

the examination.



Imperfections in Solids. Any deviation from the perfect ordered arrangement

constitutes a defect or imperfection.

When there are irregularities or deviations from ideal arrangement around a point or

an atom it is considered as point defect, whereas if the irregularities are observed in

entire row of lattice points, then it is considered as line defect.

Stoichiometric Point Defects. The point defects that do not disturb the

stoichiometry (i.e. the ratio of cations and anions). They are also known as intrinsic or

thermodynamic defects. These are of following types:

Vacancy Defects. It is when some of the lattice sites of the crystal are vacant.

Interstitial Defects. It is when some constituent particles occupy the normally

vacant interstitial sites in the crystal. The particles occupying the interstitial

sites are called interstitials.

Schottky Defects. It is created when equal number of cations and anions are

missing from their respective positions leaving behind holes. These are more

common in ionic compounds with high co-ordination number and where the sizes

of cation and anion are almost equal. Examples, NaCl, KCl, CsCl, KBr and AgBr.

Frenkel Defects. It is created when an ion leaves its correct lattice site and

occupies an interstitial site. These are common in ionic compounds with low co-

ordination number and in which there is large difference in size of cations and

anions. Examples, ZnS, AgCl, AgBr and AgI. These are also known as

dislocation defects.

It must be noted that:

(i) Vacancy defects and Schottky defects decrease the density of the substance

while interstitial defects increase it and Frenkel defects have no effect on

density.

(ii) Vacancy defects and interstitial defects are generally observed in case of

non-ionic solids whereas Frenkel defects and Schottky defects are usually seen

in ionic solids.

Non-stoichiometric Defects. The point defects that disturb the stoichiometry of the

compound. These are of following types.

Metal Excess Defect due to Anionic Vacancies. It is when a compound has

excess cation due to the absence of an anion from its lattice site creating a 'hole'

(called F-centre or colour centre) which becomes occupied by electron to

maintain the electrical neutrality. F-centres are responsible for colour of the

compound (pink, yellow and violet colour of LiCl, NaCl and KCl respectively).

These types of defects are found in crystals which are likely to possess Schottky

defects.



Metal Excess Defects due to Interstitial Cations. It is due to the excess

cation accommodated in interstitial sites, with electrons trapped in the

neighbourhood. The yellow colour of non-stoichiometric ZnO (when it is heated)

and electrical conductivity is due to these trapped electrons. These types of

defects are found in crystals which are likely to possess Frenkel defects.

Metal Deficiency Defect. It is when the compound has metal deficiency due to

the absence of metal ion from its lattice site. The charge is balanced by an

adjacent ion hiving higher positive charge. Example, FeO.

Impurity Defects. It is when some foreign atoms (or ions) occupy interstitial or

substitutional sites in a crystal.

The conductivity of semiconductors and insulators increases with increase in

temperature while that of conductors decreases with an increase in temperature.

Conductors have partially filled or overlapping bands which is responsible for their

high electrical conductivity.

In case of insulators, the energy gap (called forbidden zone) is very large and therefore

electrons from valance band cannot be promoted to conduction band. Hence they have

low electrical conductivity.

In semiconductors the energy gap between valance and conduction band is small and

therefore some electrons from valance band can move into conduction band. This

results in some electrical conductivity.

The conduction by pure semiconductors such as Si and Ge is called intrinsic

conduction and these pure semiconductors exhibiting electrical conductivity is called

intrinsic semiconductors (also called undoped semiconductors or i-type

semiconductors).

Doping of Semiconductors. The process of increasing the conductivity of intrinsic

semiconductors (which is usually very low) by adding an appropriate amount of some

suitable impurity.

Group 14 elements (such as Si) doped with group 15 elements (such as As) behave as

n-type semiconductors while those doped with group 13 elements (such as B) behave

as p-type semiconductors.

Diamagnetic Substances. The substances which are weakly repelled by the external

magnetic field. They have all their electrons paired.

Paramagnetic Substances. The substances which are weakly attracted by the

external magnetic field. They have one or more unpaired electrons in them.



Ferromagnetic Substances. The substances which are strongly attracted by the

external magnetic field. These are permanently magnetised. In solid state, their ions

are grouped together into domains which act as tiny magnet.

Antiferromagnetic Substances. The substances (like MnO) whose domains are

oppositely oriented such that they cancel each other's magnetic moment.

Ferrimagnetic Substances. The substances (like Fe3O4, ferrites such as MgFe2O4 and

ZnFe2O4) in which the magnetic moment of domains are aligned in parallel and anti-

parallel directions in unequal numbers. These are weakly attracted by the magnetic

field as compared to ferromagnetic substances. They become paramagnetic on heating.

Piezoelectric Effect. Generation of electric current by applying pressure on a crystal.

Transition Temperature. Temperature at which substance starts behaving as super-

conductor.

Solutions

Solutions. A homogenous solid, liquid or gaseous mixture of two or more substances

whose concentration can be varied within certain limits.

Saturated Solution. A solution which cannot dissolve any more of the solute at a

particular temperature.

Solubility. The amount of solute present in 100 g of the solvent in a saturated solution

at particular temperature.

Super Saturated Solution. A solution in which the amount of solute present in 100 g

of the solvent at a particular temperature is more than its normal solubility at that

temperature.

Solubility of solids in liquids depend on:

Nature of Solute (like dissolves like).

Temperature. If the dissolution process is exothermic, the solubility decreases

with increase in temperature. And if the dissolution process is endothermic, the

solubility increases with increase in temperature (Le-chatelier's principle).

Solubility of gases in liquids depend on:

The nature of gas and the nature of solvent.

Temperature. Generally the solubility of gas decreases with increase in

temperature.



Pressure (Henry's Law). The solubility of gas at a given temperature is directly

proportional to the pressure at which it is dissolved.

P = KH . x1

Mass Per Cent (w/w). Mass of solute per 100 g of solution.

Volume Per Cent (V/V). Number of parts by volume of solute per hundred parts by

volume of solution.

Molarity (M). Number of moles of solute per litre of solution. Units, mol L-1.

Molality (m). Number of moles of solute per kilogram of solvent. It is independent of

temperature. It is temperature dependent. Units, mol kg-1.

Mole Fraction (x). Ration of number of moles of a component to total number of

moles. It has no units and is independent of temperature.

Parts Per Million (ppm). The number of parts by mass of solute per million parts by

mass of solution.

Vapour Pressure. The pressure developed above the liquid at particular temperature

at the equilibrium point.

Raoult's Law. The vapour pressure of a solution is equal to the product of mole

fraction of the solvent and its vapour pressure in pure state.

p1 = p1˚ x1 or p2 = p2˚ x2

Lowering of Vapour Pressure. Difference in the vapour pressure of the pure solvent

and that of solution.

Ideal Solution. The solution which obey Raoult's law at all concentrations and follow

the conditions, ∆Hmix = 0; ∆Vmix = 0.

Non-ideal Solutions. The solution which show positive or negative deviations from

Raoult's law. It does not obey the law at all concentrations and follow the conditions,

∆Hmix 0; ∆Vmix 0.

Azeotropes (Azeotropic Mixtures). The mixture of liquids which boils at constant

temperature like pure liquid and has same composition of component in liquid as well

as vapour phase.

Minimum Boiling Azeotrope. This type of azeotrope is formed by solutions showing

large positive deviations.



Maximum Boiling Azeotrope. This type of azeotrope is formed by solutions showing

large negative deviations.

Colligative Properties. The properties of the solution which are independent of

nature of solute but depend upon the concentration of solute particles.

Relative Lowering of Vapour Pressure. The ratio of lowering of vapour pressure to

the vapour pressure of pure solvent.

Boiling Point. The temperature at which the vapour pressure of the liquid becomes

equal to the atmospheric pressure.

Molal Elevation Constant (kb). The elevation in boiling point of the solution when its

molality is unity. Units, K kg mol-1. It is also called molal ebullioscopic constant.

Freezing Point. For a substance it is the temperature at which its solid and liquid

phases coexist. Scientifically, it is defined as the temperature at which substance's solid

and liquid phases have the same vapour pressure.

Molal Depression Constant (kf). The depression in freezing point when the molality

of the solution is unity. Units, K kg mol-1. It is also called molal cryoscopic constant.

Osmosis. The passage of solvent from pure solvent or solution of low concentration to

the solution of high concentration through semi-permeable membrane.

Osmotic Pressure (π). The excess pressure that must be applied to the solution side

to prevent the passage of solvent into it through semi-permeable membrane.

Isotonic Solutions. The solutions of same molar concentration and same osmotic

pressure at particular temperature.

A solution having higher osmotic pressure than some other solution is said to be

hypertonic with respect to the other solution.

A solution having lower osmotic pressure relative to some other solution is called

hypotonic with respect to the other solution.

Isopiestic Solutions. The solutions whose vapour pressures are equal at particular

temperature.

The abnormal value of molecular mass as calculated from any of the colligative

property is due to:

Association of solute molecules or

Dissociation of solute particles.



Van't Hoff Factor (i). It is the ratio of normal molecular mass to observed molecular

mass or the ratio of observed colligative property to normal colligative property.

IMPORTANT FORMULAE & RELATIONSHIPS

In the formulae given below; subscript 1 is used for solvent and 2 is used for solute. Also,

W1 = Mass of solvent in g; W2 = Mass of solute in g.

M1 = Molar/molecular mass of solvent; M2 = Molar/molecular mass of solute.

V1 = Volume of solvent; V2 = Volume of solute.

V = Volume of solution.

n1 = Number of moles of solvent; n2 = Number of moles of solute.

Mass % =

Volume % =

Molarity (M) =

Relationship between Molarity (M) and Mass Per Cent (%).

M =

; here d is the density of solution.

Molality (m) =

Relationship between Molarity (M) and Molality (m).

m =

– or

+

; here d is the density of solution.

Mole Fraction of Solvent, x1 =

Mole Fraction of Solute, x2 =

Also, (x1 + x2) = (

) = 1



COLLIGATIVE PROPERTIES

a) Molecular solutes which do not associate or dissociate

Relative Lowering of Vapour Pressure

∆p/p˚ = –

= x2

Elevation in Boiling Point

∆Tb =

; where W2, M2, W1 are expressed in g.

Depression in Freezing Point

∆Tf =


Osmotic Pressure

π =

; where W2, M2 are expressed in g.

b) Electrolytes or solutes undergoing association or dissociation in solution

Relative Lowering of Vapour Pressure

∆p/p˚ = –

= i x2

Elevation in Boiling Point

∆Tb =


Depression in Freezing Point

∆Tf =


Osmotic Pressure

π =

; where W2, M2 are expressed in g.

Relationship between Molal Elevation Constant (Kb) and Enthalpy of

Vaporisation (∆Hvap) of Solvent.

kb =



Relationship between Molal Depression Constant (Kf) and Enthalpy of Fusion

(∆Hfus) of Solvent.

kf =

Van't Hoff Factor

i

Electrochemistry

Electrochemistry. Branch of chemistry which deals with the study of relationship

between electrical energy and chemical energy and their interconversion.

Conductors. The substances which allow the passage of electricity through them.

Insulators (Non-conductors). The substances which do not allow the passage of

electricity through them.

Electronic Conductors. Substances which show conduction due to movement of

electrons. Example, metals, graphite, etc.

Electrolytes. Substance which allow the passage of electricity through their molten

state or through their aqueous solutions.

Strong Electrolytes. Electrolytes which are completely ionized in their aqueous

solution and has high conductivity.

Weak Electrolytes. Electrolytes which are ionized in their aqueous solution to a

smaller extent and has low conductivity. However, their conductivity increases with

dilution as it increases their degree of ionization (Ostwald's Dilution Law).

Non-electrolytes. Substance which do not allow the passage of electricity through

their molten state or through their aqueous solutions.

Resistance (Ohm's Law), R. R =

or I =

Resistivity (Specific Resistance), ρ R = ρ

Conductance, G. G =

Conductivity (Specific Conductance), κ. κ =

=

(

)



Molar Conductivity, Λm. Λm =

or Λm =

Cell Constant, G*. Ratio of

κ =

= G x G*

Property Unit S.I. Unit

Resistance (R) ohm (Ω) –

Resistivity (ρ) ohm-cm ohm-m

Conductance (G) ohm-1

S

Conductivity (κ) ohm-1

cm-1

S m-1

Molar Conductivity (Λm) ohm-1

cm2 mol

-1 S m

2 mol

-1

Cell Constant (G*) cm-1

m-1

Limiting Molar Conductivity, . Definite value attained by molar conductivity

when concentration approaches zero. It s the highest molar conductivity value for any

electrolyte.

Λm = – A c1/2 (Debye Huckel Onsager

Equation)

Kohlrausch's Law (of independent migration of ions). At infinite dilution, when

dissociation of electrolyte is complete, each ion makes a definite contribution of its own

towards the molar conductivity of electrolyte, irrespective of the nature of the other ion

with which it is associated.

= ν+λ˚+ + ν–λ˚–

Applications of Kohlrausch's law include the determination of:

Limiting molar conductivities of weak electrolytes.

Degree of dissociation of weak electrolytes, α =

Dissociation constant of weak electrolytes.

Solubility of sparingly soluble salts.

Ionic product of water.

Galvanic Cells. Device in which chemical energy is converted into electrical energy.

Anode. Electrode at which oxidation takes place. For galvanic cells it is the negative

electrode.

Cathode. Electrode at which reduction takes place. For galvanic cells it is the positive

electrode.

Ecell (EMF) = Eright – Eleft or Ecell (EMF) = Ecathode – Eanode



(Standard EMF) =

– or

(Standard EMF) = –

Reference Electrode. Electrode whose potential is arbitrarily fixed. Example,

Standard Hydrogen Electrode.

Electrochemical Series. The arrangement of various elements in the order of

decreasing values of standard reduction potentials.

Nernst Equation. EMn+

/ M = +

log

or EMn+

/ M = +

log

For a reaction, aA + bB ——→ cC + dD

Ecell = +

log

or Ecell =

–

log

N.B. Numerical questions based on above formula are frequently asked for 3 marks in

the examination.

Nernst Equation and Equilibrium Constant (Kc).

=

log Kc

Electrochemical Cell and Gibbs Energy.

∆G = –nFEcell

∆G˚ = –nF

Recharging of the Cell. Process in which a galvanic cell is connected with external

source that has higher potential than the cell. It involves the reversal of the net cell

reaction.

Primary Cells. A type of galvanic cells that become dead over a period of time and

cannot be recharged or reused again.

Arrangement of two or more galvanic cells connected in series is called a battery.

Dry Cell (Leclanche Cell).

Anode : Zn ——→ Zn2+

+ 2e–

Cathode : MnO2 + NH4+ + e

– ——→ MnO(OH) + NH3

Mercury Cell.

Anode : Zn(Hg) + 2OH– ——→ ZnO(s) + H2O(l) + 2e

–

Cathode : HgO(s) + H2O(l) + 2e– ——→ Hg(l) + 2OH

–

Net Reaction : Zn(Hg) + HgO(l) ——→ ZnO(s) + Hg(l)

Secondary Cells. Galvanic cells which can be recharged.



Lead Storage Battery.

Anode : Pb(s) + SO42–

(aq) ——→ PbSO4(s) + 2e–

Cathode : PbO2(s) + SO42–

(aq) + 4H+ + 2e

– ——→ PbSO4(s) + 2H2O(l)

Net Reaction : Pb(s) + PbO2(s) + 2H2SO4(aq) ——→ 2PbSO4(s) + 2H2O(l)

Nickel Cadmium Storage Cell (NiCad cells).

Cd(s) + 2Ni(OH)3(s) ——→ CdO(s) + 2Ni(OH)2(s) + H2O(l)

Fuel Cells. Cells which convert chemical energy of a fuel directly into electrical energy.

Advantages over traditional cells:

Pollution-free working.

High thermodynamic efficiency.

Continuous source of energy.

H2—O2 Fuel Cell (Bacon Cell).

Anode : 2H2(g) + 4OH–

(aq) ——→ 4H2O(l) + 4e–

Cathode : O2(g) + 2H2O(l) + 4e– ——→ 4OH

–(aq)

Net Reaction : 2H2(g) + O2(g) ——→ 2H2O(l)

Electrolysis. The process of chemical decomposition of the electrolyte by the passage

of electricity through its molten or dissolved state.

Electrolytic Cells. The device in which process of electrolysis is carried out and a non-

spontaneous chemical reaction is driven by the passage of electricity. It involves the

conversion of electrical energy into chemical energy.

Anode. Electrode at which oxidation takes place. For electrolytic cells it is the positive

electrode.

Cathode. Electrode at which reduction takes place. For electrolytic cells it is the

negative electrode.

Criteria of product formation in electrolysis.

At Cathode : Reduction reaction with higher reduction potential takes place.

At Anode : Oxidation reaction with higher oxidation potential (or lower

reduction potential) takes place.

Quantity of Charge in coulombs (Q) = Current (I) in amperes Time (t) in seconds

Q = I t

Faraday's First Law of Electrolysis. The mass of a substance liberated at the

electrode is directly proportional to the quantity of electricity passed.



Faraday's Second Law of Electrolysis. When same quantity of electricity is passed

through different electrolytes connected in series then the masses of the substances

liberated at the electrodes are proportional to their chemical equivalent weights.

Galvanic Cells Electrolytic Cells

1. Chemical energy is converted into

converted into electrical energy.

2. Reaction taking place is spontaneous.

3. The two half cells are kept in different

containers and are connected through salt

bridge or porous partitions.

4. Anode is negative and cathode is positive.

5. Electrons move from anode to cathode in

external circuit.

6. Used as a source of electricity.

1. Electrical energy is converted into chemical

energy.

2. Reaction taking place is non-spontaneous.

3. Both the electrodes are placed in solution or

molten electrolyte in the same container.

4. Anode is positive and cathode is negative.

5. Electrons are supplied by external source.

They enter through cathode and come out

through anode.

6. Used in electroplating, electro refining, etc.

Corrosion. The process of slow conversion of metals into their undesirable compounds

(usually oxides) by reaction with moisture and other gases present in the atmosphere.

Corrosion in Iron (Rusting).

Anode : 2Fe ——→ 2Fe2+

+ 4e–

Cathode : O2 + 4H+ + 4e

– ——→ 2H2O

Net Reaction : 2Fe + O2 + 4H+ ——→ 2Fe

2+ + 2H2O

2Fe2+

+ ½O2 + 2H2O ——→ Fe2O3 + 4 H+

Fe2O3 + xH2O ——→ Fe2O3.xH2O

(Rust)

Prevention of Rusting. Painting, alloy formation, galvanization, use of anti-rust

(some phosphate and chromate salts), solutions and cathodic protection.



Chemical Kinetics

Chemical Kinetics. The branch of chemistry which deals with the study of reaction

rates and their mechanism.

Rate of Reaction. The rate of change of concentration of any of the reactant or product

with time at any particular moment of time.

Instantaneous Rate. Decrease in concentration of any one of the reactants or increase

in concentration of any one of the products at particular instance of time for a given

temperature.

Factors affecting rate of reaction are:

Concentration of reactants,

Temperature of reactants,

Nature (reactivity) of reacting substance,

Presence of catalyst, and

Exposure to radiations.

Rate Constant (k). It is the rate of the reaction when concentration of each of reacting

species is unity. It is also called velocity constant or specific reaction rate of the

reaction.

Rate of Reaction Rate Constant

1. It is the speed at which the reactants are

converted into products at any moment of

time.

2. It depends on concentration of reactant

species at that moment of time.

3. It generally decreases with the progress of the

reaction.

4. It has the unit mol L–1

t–1

(atm t–1

for gaseous

reactions)

1. It is the constant of proportionality in the rate

law expression.

2. It refers to the rate of reaction at specific

point when concentration of every reacting

species is unity.

3. It is constant and does not depend on the

progress of the reaction.

4. Unit of rate constant depends on order of

reaction.

Rate Law. The mathematical expression based on experimental fact, which describes

the reaction rate in terms of concentration of reacting species. It cannot be written from

the balanced chemical equation.

Molecularity. The number of reacting particles which collides simultaneously to bring

about the chemical change. It is a theoretical concept.



Order of Reaction (x + y). The sum of the exponents of the concentration terms in the

experimental rate law of reaction. It can be zero, 1, 2, 3 or any fractional value.

Units,

Zero Order : mol L–1

s–1

First Order : s–1

Second Order : L mol–1

s–1

Third Order : L2 mol

–2 s

–1

In general, for nth

Order : (mol L–1

)1–n

s–1

For gaseous reactions : (atm)1–n

s–1

Molecularity Order

1. It is the number of reacting species

undergoing simultaneous collusion in the

reaction.

2. It is a theoretical concept.

3. It cannot be zero and can have integral values

only.

4. It does not change with change in

temperature and pressure.

1. It is the sum of powers of the concentration

terms in the rate law expression.

2. It is determined experimentally.

3. It can be zero and can have fractional values

also.

4. It changes with change in temperature and

pressure.

Elementary Reactions. Reactions involving single step.

Complex Reactions. Reactions involving more than one step.

Rate Determining Step. Slowest step of complex reaction. Also called rate

controlling step.

Pseudo First Order Reactions. Reactions of higher order that follow the kinetics of

first order under special conditions (when one of the reactants is taken in large

excess). They are also sometimes referred to as pseudo unimolecular reactions.

Half Life Period (t½). Time taken for the concentration of reactants to be reduced to

half of their initial concentration.

Activation Energy (Ea). The additional energy required by reacting species over and

above their average potential energy to enable them to cross the energy barrier

between reactants and products.



Activated Complex. The highly energetic arrangement of atoms formed during the

course of reaction which corresponds to the peak of curve in energy profile diagram for

the progress of reaction. Energy required to form this complex is equal to activation

energy.

Arrhenius Equation. For a reaction, it gives relationship between temperature and

rate constant.

Mechanism of Reaction. The sequence of elementary steps leading to overall

stoichiometry of reaction.

Threshold Energy. Minimum energy that a reacting species must possess in order to

undergo effective collisions.

Collision Frequency (Z). Number of collisions per second per unit volume of the

reaction mixture.

Effective Collisions. Collisions which facilitate breaking of bonds between reacting

species and formation of new bonds to form products.

Temperature Coefficient. Ratio of rate constant at 308 K and 298 K.

IMPORTANT FORMULAE

For the reaction, aA + bB ——→ cC + dD

Average Rate –

–

Instantaneous Rate –

–

Rate Law Rate k [A]x [B]

y (x & y are determined experimentally)

Order w.r.t. A x

Order w.r.t. B y

Overall Order x + y

Relationship between k and t

For zero order reactions, k –

For first order reactions, k

k

–



Half Life Period.

For zero order reactions, t½

thus, t½ [R]o

For first order reactions, t½

thus, t½ is independent of [R]o

Arrhenius Equation.

k A –

log

[

–

] where, T2 > T1

IMPORTANT GRAPHS

Graphs given on page 104, 106, 112, 113 and 115 in notebook

General Principles and Processes of Isolation of Elements

Minerals. Naturally occurring chemical substances in the earth's crust that contain

obtainable by mining. Generally it contains one or more metals.

Ore. Minerals which contain a high percentage of metal and from which metal can be

extracted profitably. All ores are minerals but all minerals are not ores.

Gangue. Contamination of earthy or undesirable materials such as silica, clay, etc.

Metallurgy. Scientific and technological process used for isolation of the metal from its

ore. Metal maybe isolated by heating (pyrometallurgy), by using electric discharge

(electrometallurgy) or by using suitable solvent, generally water (hydrometallurgy).

Principal Ores of Some Important Metals.(Draw table 6.1)

Steps for obtaining a pure metal from its ore:

1. Concentration 2. Conversion into oxide

3. Reduction of oxide to the metal 4. Refining

Concentration. Process of removal of unwanted materials (gangue) from the ore. It is

also called dressing or benefaction. It can be done by any one of these methods:

Hydraulic washing (gravity separation).

Magnetic separation.



Froth floatation method (used exclusively for sulphide ores). It works on the

principle that mineral particles become wet by oils while the gangue particles by

water. Collectors enhance non-wettability of mineral particles and froth

stabilizers stabilize the froth. Depressants may also be used to separate two

sulphide ores.

Leaching. Treating an ore with some suitable solvent in which the ore is soluble

(generally due to formation of a coordination complex) but gangue particles are

not. Important examples,

Al2O3(s) + 2NaOH(aq) + 3H2O(l) ——→ 2Na[Al(OH)4](aq)

Bauxite

2Na[Al(OH)4](aq) + CO2(g) ——→ Al2O3.xH2O + 2NaHCO3(aq)

Al2O3.xH2O(s) ——→ Al2O3(s) + xH2O(g)

Pure alumina

4M(s) + 8CN–

(aq) + 2H2O(aq) + O2(g) ——→ 4[M(CN)2](aq) + 4OH–

(aq)

2[M(CN)2](aq) + Zn(s) ——→ [Zn(CN)4]2–

(aq) + 2M(s) (M = Au or Ag)

Calcination. Heating the ore in the limited quantity of air so as to convert it into

metal oxide and eliminate the volatile matter. It is generally done when ore contains

appreciable amount of oxygen (maybe in the form of hydrated oxide, carbonate or

hydrogen carbonate). For example,

ZnCO3(s) ——→ ZnO(s) + CO2(g)

Roasting. Heating the ore below the melting point of metal in the excess or regular

supply of air so as to convert it into metal oxide. It is generally done when ore lacks

oxygen in it. For example,

ZnS(s) + 3O2(g) ——→ ZnO(s) + SO2(g)

Flux. Additional substance added during heating which combines with gangue and

convert it into easily separable material slag. For example,

FeO + SiO2 ——→ FeSiO3

Gangue flux slag

Ellingham Diagrams. The plot of change in standard Gibbs energy (∆G˚) versus

temperature (T) which enables the choice of proper reducing agent and also the

required temperature during the reduction of oxides into metals.

Pig Iron. Iron obtained from Blast furnace containing 4% of carbon and traces of

impurities.

Cast Iron. Extremely hard and brittle form of iron with slightly lower carbon content

than pig iron (~3%). It is prepared by melting pig iron with scrap iron and coke using

hot air blast.



Wrought Iron. Purest form of commercial iron. It is also called malleable iron and is

prepared from cast iron by oxidizing impurities in reverberatory furnace lined with

haematite.

Copper Matte. Copper obtained from reverberatory furnace when iron oxide is

removed as slag in the form of iron silicate. It contains Cu2S and FeS.

Blister Copper. Solidified copper obtained when copper matte is charged into silica

lined convertor. It has blister appearance due to evolution of SO2 gas.

Refining. Process of removal of fine impurities and obtaining metals of high purity.

Some important refining processes include:

Electrolytic refining. In this method, impure metal is made anode (–) and

same metal in pure form is made cathode (+). These are put in a suitable

electrolytic bath containing soluble salt of same metal.

Zone refining. It works on the principle that impurities are more soluble in

melt than in solid state of the metal.

Vapour phase refining. It involves the conversion of metal into its volatile

compound and then decomposing it to give pure metal.

Mond Process is used for refining nickel:

Ni + 4CO ——→ Ni(CO)4

Volatile complex

Ni(CO)4 ——→ Ni + 4CO

Pure nickel

Van Arkel Method is used for refining zirconium or titanium:

Zr + 2I2 ——→ ZrI4

Volatile complex

ZrI4 ——→ Zr + 2I2

Pure zirconium

Chromatographic methods. It works on the principle that different

components of a mixture are differently adsorbed on an adsorbent.

Chromatography, in general, involves the movement of mobile phase on a

stationary phase where different components get adsorb at different rates.



d- & f-Block Elements

d-Block Elements. Elements in which last electron enters in any of the d-orbital.

Transition Elements. The elements whose atoms or simple ions contain unpaired

electrons in the d-orbitals. Zn, Cd and Hg are not considered as transition elements.

The general electronic configuration of transition elements is (n-1) d1-10, ns1-2.

Electronic configuration of Cr (3d5, 4s1) and Cu (3d10, 4s1) is exceptional owing to the

fact that half filled and fully filled orbitals are extra stable.

When atoms of d-block elements change into cations, the electrons are removed from

ns-orbital first and then from (n-1) d-orbitals. For example,

26Fe: [Ar] 3d6, 4s2 26Fe2+: [Ar] 3d6

27Co: [Ar] 3d7, 4s2 27Co2+: [Ar] 3d7

Due to the presence of strong metallic bonds, the transition metals are hard,

possesses high densities and high enthalpies of atomization. Cr is the hardest

metal of 3d series and has highest melting point too. For 4d series it is Mo. And for 5d

series it is W. Os is the densest metal.

The melting points of transition elements are generally very high. This is due to strong

metallic bond and the presence of unpaired electrons in d-orbital in them. Due to these

unpaired electrons, some covalent bonds also exist between atoms of transition

elements resulting in stronger inter-atomic bonding which further results in high

melting and boiling points.

The ionization enthalpies of transition metals are higher than those of alkali metals

and alkaline earth metals. However, the relative difference of IE1 values of any two d-

block elements is much smaller. This is because, as these elements involve gradual

filling of (n-1) d-orbitals, the effect of increase in nuclear charge is partly cancelled by

the increase in shielding effect. Consequently, the increase in IE is very small.

Among the elements of particular transition series, as the atomic number increases,

atomic radii first decrease till the middle, become almost constant and then increases

towards the end of the period. This is because at first nuclear force of attraction is

dominant which attracts the electron towards the nucleus thereby decreasing the size.

However, in the middle of the series, it is cancelled by shielding or screening effect. Size

increases at the end as shielding effect exceeds nuclear force of attraction.

The elements of 4d and 5d series belonging to a particular group have almost equal

atomic radii because of lanthanoid contraction.



Transition elements show variable oxidation states. It is due to the participation of

ns and (n-1) d-electrons in bonding. Oxidation states of transition elements differ from

each other by unity, whereas for p-block elements it differs by two.

In each series, highest oxidation states increase with increase in atomic number,

reaches a maximum in the middle and then starts decreasing. This is because in the

beginning of the series elements have less number of electrons which they can lose or

contribute for sharing. Elements at the end of the series have too many d-orbitals and

hence have fewer vacant d-orbitals which can be involved in bonding.

For the elements of first transition series (except Sc) +2 oxidation state is most

common.

Cr and Cu can show the oxidation state of +1 also. Sc and Zn do not show variable

oxidation states. Most stable oxidation state is +3 for Cr, Fe and Co. It is +2 for Mn.

Elements in lower oxidation states form ionic compounds, whereas in higher oxidation

states they form covalent compounds.

Some transition metals also show oxidation state of zero in metal carbonyls, such as

Fe(CO)5 and Ni(CO)4.

Transition elements have high complex formation tendencies because of:

Their small size and high charge density of the ions of transition metals.

Presence of vacant orbitals of appropriate energy which can accept lone pair of

electrons from others (ligands).

The compounds of transition elements are usually brightly colored. Their colors are

explained on the basis of d-d transition of electrons and charge transfer spectra. d0 and

d10 configurations are colorless.

The transition metal ions generally contain one or more unpaired electrons in them and

hence, their complexes are generally paramagnetic. The magnetic moment is related

to the number of unpaired electrons according to the following (spin only) formula:

μ = √ BM (where, n is the number of unpaired

electrons)

Fe, Co and Ni in their elemental form are ferromagnetic.

Many transition metals and their compounds are known to act as catalysts. The

catalytic activity of transition metals is attributed to the following reasons:

Because of their variable oxidation states they can form unstable intermediate

compounds and provide a new path way with lower activation energy for the

reaction.



They can provide a suitable surface for the reactants to get adsorb and react

quickly.

Since the transition elements have comparable sizes, they are known to form a good

number of alloys.

They also form interstitial compounds as they are capable of entrapping smaller

atoms of other elements such as H, C and N. These compounds are hard and have high

tensile strength and melting points than pure elements. However, their chemical

reactivity is relatively low.

The oxides of transition metals are generally basic when the metal is in lower

oxidation state; acidic when it is in higher oxidation state and amphoteric in

intermediate oxidation state. ZnO and CuO are exceptionally amphoteric.

Sometimes a particular oxidation state becomes less stable relative to other oxidation

states, one lower and the other higher. In such a situation a part of the species

undergoes oxidation while a part undergoes reduction. Such is species is said to

undergo disproportionation.

For example,

VI VII IV

3MnO42− + 4H

+ ——→ 2 MnO4

− + MnO2 + 2H2O

(oxidized) (reduced)

Potassium dichromate (K2Cr2O7) is prepared from chromite ore (FeCr2O4). Various

steps involved are:

FeCr2O4 + 8Na2CO3 + 7O2 ——→ 8Na2CrO4 + 2Fe2O3 + 8CO2

2Na2CrO4 + 2H+ ——→ Na2Cr2O7 + 2Na

+ + H2O

Na2Cr2O7 + 2KCl ——→ K2Cr2O7 + 2NaCl

The dichromate ion (Cr2O72−) and chromate ion (CrO4

2−) exist in equilibrium with each

other at a pH of about 4. They are inter-convertible by changing the pH. CrO42− on

addition of acid changes into Cr2O72−, while Cr2O7

2− on addition of alkali change into

CrO42−. Dichromate is orange colored, while chromate is yellow colored. [Ref. textbook

for structures]

K2Cr2O7 acts as strong oxidizing agent in acidic medium.

Cr2O72−

+ 14H+ + 6e− ——→ 2Cr

3+ + 7H2O

It oxidizes:

1. Iodides to iodine: Cr2O72−

+ 14H+ + 6I− ——→ 2Cr

3+ + 3I2 + 7H2O

2. Ferrous to ferric: Cr2O72−

+ 14H+ + 6Fe

2+ ——→ 2Cr

3+ + 6Fe

3+ + 7H2O

3. Hydrogen sulphide to sulphur: Cr2O72−

+ 8H+ + 3H2S ——→ 2Cr

3+ + 3S + 7H2O

4. Stannous to stannic: Cr2O72−

+ 14H+ + 3Sn

2+ ——→ 2Cr

3+ + 3Sn

4+ + 7H2O

Potassium dichromate is used for volumetric estimation, in chromyl chloride test and

for cleansing glassware.



Potassium permanganate (KMnO4) is prepared from pyrolusite (MnO2). It is violet

crystalline solid. It acts as an oxidizing agent in acidic, neutral and alkaline media.

MnO4− + 8H

+ + 5e− ——→ Mn

2+ + 4H2O (in acidic medium)

MnO4− + 2H2O + 3e− ——→ MnO2 + 4OH− (in neutral and alkaline medium)

In acidic medium, it oxidizes:

1. Iodides to iodine: 2MnO4− + 16H

+ + 10I− ——→ 2Mn

2+ + 5I2

+ 8H2O

2. Ferrous to ferric: MnO4− + 8H

+ + 5Fe

2+ ——→ Mn

2+ + 5Fe

3+ + 4H2O

3. Oxalate ion or oxalic acid to CO2:

2MnO4− + 16H

+ + 5C2O4

2− ——→ 2Mn2+

+ 10CO2 + 8H2O

4. Sulphides to sulphur: 2MnO4− + 16H

+ + 5S

2− ——→ 2Mn2+

+ 5S + 8H2O

5. Sulphites to sulphates: 2MnO4− + 6H

+ + 5SO3

2− ——→ 2Mn2+

+ 5SO42−

+ 3H2O

6. Nitrites to nitrate: 2MnO4− + 6H

+ + 5NO2

− ——→ 2Mn2+

+ 5NO3−

+ 3H2O

In alkaline or fairly neutral medium, it oxidizes:

1. Iodides to iodate: 2MnO4− + H2O + I− ——→ 2Mn

2+ + 2OH−

+ IO3−

2. Thiosulphate to sulphate: 8MnO4− + 3S2O3

2− + H2O——→ 2MnO2 + 6SO42−

+ 2OH−

3. Managneous salts to MnO2: 2MnO4− + 3Mn2+ + 2H2O——→ 5MnO2 + 4H+

Potassium permanganate is for volumetric estimation and qualitative detection.

Alkaline solution of KMnO4 is called Baeyer's reagent and is used to detect

unsaturation.

The f-block elements consist of two series of inner transition elements i.e.

lanthanoids and actinoids. They are also called rare earth elements.

The general electronic configuration of f-block elements is (n-2) f1-14, (n-1) d0-1, ns1-2.

Lanthanoid contraction. The steady decrease in the atomic and ionic size of

lanthanoids with increase in atomic number. It is caused due to poor shielding effect

offered by 4f-electrons. Similarity in 4d and 5d transition series, difficulty in separation

of lanthanoids and decrease in basic strength from La(OH)3 to Lu(OH)3 are some of the

noteworthy consequences of lanthanoid contraction.

The lanthanoids exhibit a common oxidation state of +3. Ce and Tb also show +4

oxidation state. Ce4+ is good oxidizing agent.

Mischmetal, an alloy contains 95% lanthanoids (~40% Ce and ~44% La and Nd), 5%

iron and traces of S, C, Si, Ca and Al. It is pyrophoric and is used in cigarette and gas

lighters, flame throwing tanks, tracer bullets and shells.

*****


Class 12

PHYSICS (Theory)

Session 2016-17


Regional Office

Guwahati

Our Source of Inspiration

CHIEF PATRON

Shri. Santosh Kumar Mall

IAS Commissioner


New Delhi

PATRONS

Shri. Chandra P. Neelap

Deputy Commissioner


Guwahati Region

Smt. Anjana Hazarika &Shri. D. Patle

Assisstant Commissioners


Guwahati Region

CONVENOR

Shri. Dhirendra Kumar Jha

Principal


Guwahati

PREPARED BY:

Kiran Kumari Soren PGT, Physics

Devendra Kumar PGT, Physics


Preface There is no substitute as such for hard work. However, planned study and a bit of

smart work can do the trick. With planning I mean prioritizing. When your days are

numbered you just can't go through everything. It is therefore advisable not to panic

and study steadily giving priority to the topics most likely to appear in the

examination.

When it comes to AISSCE, nothing is guaranteed. No one can predict anything

precisely. But, there exist concepts that can enable students to score more with

minimal of efforts.

This study module is aimed at ensuring at least pass mark in the board exams and is prepared

using the available study materials of KVS but in a concise manner. This also includes previous

years CBSE questions and marking scheme so that students will have idea on what type of

questions can come from a particular chapter and what points need to be in their answers to get

marks. Hope this module will boost your confidence both during the preparatory stage as well as

during the examinations.

Students can go through in this way:

(i) Go through this material on this basis of weightage of the units like optics,

electrostatics, electromagnetic induction, AC and EM waves, Dual nature and

radiation, Which carrying weightage of around 35 marks.

(ii) Focus on the topics which you like most and Make the target of completing

those chapters by going through this material.

(iii) Make practice of diagrams and graphical representation during preparation

of topics.

(iv) Some common topics to score easy marks are (a) principles of the devices

involved (ii)Graphical variations specially(a) temperature vs resistance

graph, (b)V-I graph, (c)photocurrent vs potential graph ,(d)frequency vs

stopping potential

(v) Practice well on circuit diagrams in chapter 14 and block diagrams of

chapter 15 communication systems.

NOTE: It is advisible to practice those methods of answering which you are

already practiced,do not change the method at this moment of time.

While attempting numerical questions pay attention to (i) Write the given

part of question. Do not forget to write the formula .

GENERAL QUESTIONS

(I) Write the principle of

(a) Meter bridge

(b) Potentiometer

(c) Cyclotron

(d) Moving coil galvanometer

(e) AC Generator

(f) Transformer

(g) Compound microscope

(h) Telescope

(i) Rectifier

(II) Show the graphical variations of Electric field vs distance for (i) point

charge (ii) dipole (iii) linear charge (iv) surface charge (v) hollow

charged sphere

(III) Show the graphical variations of Electric Potential vs distance for (i)

point charge (ii) dipole (iii) hollow charged sphere

(IV) Show the variation of frequency vs reactance for a capacitor and

inductor.

(V) Draw a graph showing variation between current amplitude and

frequency also mark resonance frequency on the graph.

(VI) Draw I-V characteristics for (i) photodiode(ii) solar cell(iii)zener diode

and output characteristics of a transistor.

(VII) Draw input and output wave form of half weave and full wave rectifier.

(VIII) Draw input and output waveform of CE Amplifier.

UNIT – 1 ELECTROSTATICS

Q1. Write statement of coulumb’s law.

Ans. Coulumb’s Law:- It states that the electro-static force of attraction or repulsion between two charged bodies id

directly proportional to the product of their charges and varies inversely as the square of the distance between the two

bodies.

F = kq1 q2/r2

Here, k = 1/4πε 0 = 9 X 109 Nm2C2 ( in free space)

Q2 Write definition of electric field intensity.also write its unit.

Ans. Electric field Intensity or Electric field: Electric field strength at a point in an electric field is the electrostatic force

per unit positive charge acting on a small positive test charge placed at that point.

E = F / q0 it is a Vector quantity. Its unit is N/C.

Electric field intensity due to a point charge(Q)-

E=kQ/r2 Where k = 9 x 109 Nm2 / C2

Q3 Define electric field lines. Mention their properties.

Ans. Electric field lines – An electric line of force is defined as the path, straight or curved, along which a unit positive

charge is urged to move when free to do so in an electric field. The direction of motion of unit positive charge gives the

direction of line of force.

Properties of field lines :-

(a) The lines of force are directed away from a positive charge abd are directed towards a negative charge.

(b) A line of force starts from a positive charge and ends on a negative charge. This signifies line of force starts from

higher potential and ends on lower potential. These are continuous curves.

(c) Field lines do not cross each other because E cannot have two directions at a given point.

Q4 write formula for electric dipole moment. Calculate electric field due to an electric dipole at its axial position.

Ans. Electric dipole moment P = (2l)q; direction of P from negative to positive charge of dipole.

Electric field intensity due to a point charge:- E = (1/4πε0) (q/r2)

Electric field intensity due to an electric dipole at an axial point

Net electric field due to at point P is the vector sum of EA and Eb

E = (1/4πε0) q/(r-a)2 – (1/4πε0) q/(r+a)2 = (1/4πε0) 2p/r2 Where p = 2aq(assuming a<<r)

Q5 Calculate electric field due to an electric dipole at its equatorial position.

Ans. Electric field intensity due to at a point on the equatorial line (perpendicular bisector):

E+ = kq/(x+d)2 and E+ = kq/(x+d)2

Net electric field E = E+Cosθ

= 2E+Cosθ

=2 kq/(x+d)2 . a/(x+d)1/2

= (1/4πε0) p/x3 (assuming d<<x)

Q6 An electric dipole is placed in a uniform electric field. Calculate torque experienced by it . Also discuss its

potential energy for stable and unstable equillibrum.

Ans. Torque on an electric dipole in an electric field:-

Dipole in a uniform field

Magnitude of net torque = dFSinθ + dFSinθ

=2dFSinθ

=2dqESinθ

=pESinθ,

As the two forces are equal and opposite, net force on the dipole=O

Electric potential energy of an electric dipole in an electric field:-

Potential energy of an electric dipole, in an electrostatic field, is defined as the work done in rotating the dipole from

zero energy position (900) to the desired position(θ) in the electric field.

U=-pECosθ

(a) If θ = 900,then U = 0

(b) If θ = 00, then U = -Pe, (stable equillibrum)

(c) If θ = 1800,then U = Pe, (unstable equillibrum)

Q7 Write statement of Gauss’s law. Calculate electric field due to a linearly distributed charge using this law.

Ans. Gauss Law: Net electric flux through a closed surface = 𝒒

∈𝟎Electric firld due to a linear charge distribution

Net flux through the Gaussian surface =𝑞

∈0

Q8. Calculater electric field due to a uniformly charged spherical shell using Gauss’s Law.

Ans. Electric field due to charged spherical shell

Net flux through the Gaussian surface = 𝑞

∈0

E.4πr2=𝑞

∈0

Hence E=(1/4πε0) (q/r2) for r > R, and r= R

For a point inside the shell E=0, as charge enclosed is zero.

Q9. Calculate electric field due to a uniformly charged Plane sheet using law.

Ans. Electric field due to a plane sheet of charge.

Net flux = EA+EA

By Gauss’ Law the net flux = qenc/ε0

Short answer questions

1. An electric dipole is kept inside a cube. What is the net electric flux through the cube?

Ans- As the net change is zero, net electric flux is zero.

2. A test charge experiences a force F near a large plane sheet of chage. What will be the force experienced by the

charge if the distance is double ?

Ans- As the electric field near the sheet is constant; the force will remain the same .

3. A point charge is kept inside a spherical surface. If the radius of the surface is doubled, what will be the effect on

the flux passing through the surface?

Ans- the electric flux through the surface depends on the charge inside it. Hence the flux will be the same.

4. A short electric dipole produces an electric field E at a distance r on the equatorial line. What will be the electric

field if the distance is doubled?

Ans- It will be E/8.

5. Why two field lines cannot cross each other?

Ans- Because E cannot have two directions at a point.

6. Find the work done to rotate the dipole from the most stabe orientation to the most unstable orientation in an

external electric field?

Ans- W= final potential energy-initial potential energy.

=Pe-(-Pe)=2Pe

7. What is the effect on electric field due to presence of a dielectric medium?

Ans- electric field becomes E/K, where K is the dielectric constant of the medium.

8. Find the expression for E at the centre pf am e;ectric dipole of dipole length 2a.

E=kq/a2 + kq/a2 =2kq/a2

ELECTRIC POTENTIAL AND CAPACITANCE

Electric potential due to a point charge=work done in bringing a unit test charge from infinity to that point .

V=KQ/r It is a scalar . Its unit is J/C=Volt.

Electric potential due to an electric dipole at an axial point V= 1/4πε0(𝑞

𝑟1 -

𝑞

𝑟2)

Putting r1 =x-a and r2=r+a and simplifying above V=(1/4πε0)(q/r)

Electric potential due to an electric dipole at an equatorial point V = V++V-=0

ELECTRIC POTENTIAL ENERGY of a pair charges = work done in bringing the charges from infinity to their

respective locations.

U=kq1q2/r Here, k=1/4πε0 =9x109 Nm2C2(in free space)

CAPACITANCE

Capacitance=charge/Potential

Unit, 1 farad= 1 coulomb/1 volt

Capactance of a parallel plate capacitor

Inside the plates E=ϭ/2εo + ϭ/2εo=Q/Aεo

Potential difference V=Ed=Qd/AεO

Capacitance Q/V=Aεo/d

With dielectric filled inside the capacitor, the electric field decreases by k times hence, C=KAεo/d

Where k is the dielectric constant of the medium

Capacitance in parallel combination,1/CS=1/C1+1/C2

Capacitance in parallel combination, CP=C1+C2

Energy stored in a capacitor =work done to charge the capacitor=1/2CV2=1/2QV=1/2Q2/C

Short answer questions

1. What is an equipotential surface?

ANS-A surface of constant electric potential is called equipotential surface.

2. How much work is done to move a charge on an equipotential surface

ANS-Zero

3. What happens to electric potential due to a positive charge if the distance from the charge is increased?

ANS-As the value of v increases, becomes zero at infinity

4. What happens to electric potential due to a negative charge if the distance from the charge is increased?

ANS-The value of V increases, becomes zero at infinity.

5. A capacitor is charged and disconnected from the battery. It is now filled with a dielectric . What will be the

effect on the energy stored?

ANS-Capacitance becomes k times, charge is same. Hence energy will be reduced by k times.

CURRENT ELECTRICITY

Q.1. Write statement of Ohm’s law.

ANS- Ohm’s law: current through a conductor is directly propotional to the potential difference across the ends

of the conductor provided the physical conditions remains constant.

Mathematically V=IR , R is the resistance of the conductor.

Q.2. Write formula for resistivity or specific resistance of a conductor.

ANS-Resistance R=ρl/A where ρ is the resistivity of the material of the conductor .

Q.3. Draw V-I graph of ohmic and non ohmic conductors.

Q.4. A copper wire of resistance R and resistivity ρ is stretched to double its length. What will be the new

resistance and resistivity?

ANS. No change in resistivity, because it does not depend on length or radius. By stretching the wire its length

becomes double and area of cross section becomes half. Now according to formula. R = ρl/A , new resistance will

increase 4 times.

Q.5. What is drift velocity of electrons in a conductor.

ANS. Drift velocity is the average velocity of all electrons in the conductor under the influence of applied electric

field. Drift velocity vd = (e E /m)Ԏ ; where e is the charge of electron, E is electric field, m is mass of electron and

Ԏ is relaxation time.

Q.6. How does the drift velocity of electrons in a metallic conductor vary with increase in temperature?

ANS. Decrease with increase in temperature

Q.7. How is drift velocity changed if (i) length of the conductor is doubled ,(ii) Radius/area of cross section is

doubled?

ANS. According to formula drift velocity Vd = (e E /m)Ԏ

(i) If length is doubled then E will becomes half (because E= potential difference/ length) , therefore Vd

will also becomes half.

(ii) No change on drift velocity

.

Q.8. What is emf of a cell? Write points of comparision between emf and terminal potential difference of a cell.

The electro motive force is the maximum potential difference between the electrodes of the cell when no current is

drawn from the cell.

Comparision of EMF and P.D

EMF POTENTIAL DIFFERENCE

1 EMF is the maximum potential difference between the two electrodes of the cell when no current is drawn from the cell i.e when the circuit is open.

P.D is the difference of potentials between any two points in a closed circuit.

2 It is independent of the resistance of the circuit.

It is proportional to the ressistance between the given points.

3 The term ‘emf’ is used only for the source of emf.

It is measured between any two points of the circuit.

4 It is greater than the potential difference between any two points in a circuit

However p.d is greater than emf when the cell is being charged.

ELECTOMEGNETIC INDUCTION & ALTERNATING CURRENT

1. A bar magnet is moved in a direction indicated by the arrow between two coils PQ and CD. Predict the direction of the

induced current in each coil. (AI 2012 1 mark)

Ans : The direction of the current is from Q to P and from C to D .

This is due to the fact that the coil PQ will have its south pole at Q end and on the other hand the coil CD will have its

south pole near C end.

2. The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the

cause of this damping? ( A I 2013 , 1 mark)

Ans: This is due to the eddy current produced in the copper plate.

3. A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give reason.

( A I 2013 , 1 mark)

Ans : As the switch is on , eddy current is produced in the metal disc.

4. How does the mutual inductance of a pair of coil changes when

(i) Distance between the coil is increased and

(ii) Number of turns of the coil is increased ( A I 2013 , 1 mark)

Ans: (i) Decreased, as Ф = M I (flux decreased as distance is increased)

(ii) Increased, as M α n1 n2

5. What are eddy current? Write their two applications. ( A I 2012 , 2 mark)

Ans: Eddy current are the current induced in a conductor when placed in a changing magnetic field. The two applications

(i) Electromagnetic braking and

(ii) Induction furnace

6. A rectangular coil LMNO is placed in a uniform magnetic field of 0.5 T. The field is directed perpendicular to the plane of

the conductor. When the arm MN of length 20 cm is moved towards left with a velocity of 10 ms-1, calculate the emf

induced in the arm. Given the resistance of the arm to be 5 ohm ( assuming that the other arms are of negligible

resistance) find the value of the current in the arm. ( A I 2012)

Ans:

Given , B = 0.5 T ; L = 20 cm = 0.2 m ; v = 10 ms-1 ; R = 5 ohm ; I = ?

Using the expression,

Ε = - B l v = - 0.5 x 0.2 x 10 = - 1 V

I = E/R = 1/5 = 0.2 A

7. A wheel with 8 metallic spokes each 50 cm long is rotated with the speed of 120 rev/min in a plane normal to the

horizontal component of the earth’s magnetic field. The Earth’s magnetic field at the place is 0.4 G and the angle of dip is

60 0 . calculate the emf induced between the axle and the reem of the wheel. How will the value of emf be affected if the

number of spokes were increased? (2013)

Ans:

Given L = 50 cm = 0.5 m ; f = 120 rev/min = 120/60 rps = 2 rps ; B = 0.4 G = 0.4 x 10 -4 T

Dip = 60 0 ; B H = B cos 60 0 = 0.2 x 10 -4 T ; E = ?

Using the relation , E = ½ B w L2 , we have,

E = ½ 0.2 x 10 -4 x 2 x 3.14 x 2 x (0.5 )2

= 3.14 x 10 -5 V

8. (a) When a bar magnet is pushed towards ( or away) from the coil connected to a galvanometer, the pointer of the

galvanometer deflects. Identify the phenomenon causing this deflection and write the factors on which the amount and

direction of the deflection depends. State the law describing this phenomenon.

(b) Sketch the change in flux, emf and force when a conducting rod PQ of resistance R and length l moves freely to and fro

between A and C with speed v on a rectangular conductor placed in a uniform magnetic field as shown in the figure.

(AI 2016 , 5 marks)

Ans: (a) Phenomenon : Electromagnetic Induction

Factors : Strength of the magnetic field of the magnet

Speed of motion of the bar magnet

Direction depends on the north /south polarity of the magnet and the direction of motion

of the magnet .

(b)

9. (a) Define self inductance of a coil. Obtain the expression for the energy stored in a solenoid of self inductance ‘L’ when

the current through it grows from zero to ‘I’.

(b) A square loop MNOP of side 20 cm is placed horizontally in a uniform magnetic field acting vertically downwards as

shown in the figure. The loop is pulled with a constant velocity of 20 cm per second till it goes out of the field.

(i) Depict the direction of the induced current in the loop as it goes out of the field. For how long the current in the loop

persist?

(ii) Plot the graph showing the variation of the magnetic flux and induced emf as a function of time

Ans: (a) Self inductance is numerically equal to the magnetic flux linked with the coil when unit current flows through it.

We know that,

E = - L( dI/dt)

Or dW = E I dt = L I dI

Now ,

W = ∫dW = ∫ L I dI = ½ L I 2

This is the energy

(b) (i) Direction of the current is clockwise (MNOP) and the duration of the induced current is 1 s.

(ii) The graph is as shown

ELECTROMAGNETIC WAVE

1. What are the directions of electric and magnetic field vectors relative to each other and relative to the direction of

propagation of EMW? 1 (2012)

Ans- These waves are perpendicular to each other and perpendicular to the direction of propagation

2. Name the physical quantity which remains same for microwaves of wavelength 1mm and UV radiation of 1600𝐴° in

vacuum. 1 (2012)

Ans- velocity(3 × 108𝑚/𝑠) as both are electromagnetic wave

3. The speed of electromagnetic wave in a material medium is given by 𝑣 = 1

√𝜇𝜖.How does its frequency change?

Ans- does not change

4. Name the electromagnetic waves, which (i) maintain the earth’s warmth and (ii) are used in aircraft navigation.

Ans- (i) infrared (ii) Microwave

5. Welders wear special goggles or face masks with glass windows to protect their eyes from electromagnetic

radiations. Name the radiations and write the range of their frequency.

Ans- UV Radiation, frequency range (1015 − 1017) Hz

6. Why are infra-red radiations referred to as heat waves? Name the radiations in the electromagnetic spectrum

having (i) shorter wave length (ii) longer wave length

Ans- IR are produced by hot bodies and molecules.

Visible ,UV, X-ray ,𝛾 − 𝑟𝑎𝑦 (ii) Microwave, radio wave

7. To which part of the electromagnetic spectrum does a wave of frequency 5 × 1019𝐻𝑧 belong? Ans- X-ray or gamma ray

8. A capacitor, made of two parallel plates each of plate area A and separation d, is being charged by an external ac

source. Show that the displacement current inside the capacitor is the same as the current charging the capacitor.

Ans- 𝐼𝑑 =∈0𝑑∅𝐸

𝑑𝑡=∈0

𝑑

𝑑𝑡(

𝑞

∈0) =

𝑑𝑞

𝑑𝑡

9. Answer the following questions :

(a) Name the electromagnetic waves, which are produced during radioactive decay of a nucleus. Write their

frequency range.

(b) (b) Welders wear special goggles while working. Why? Explain.

(c) Why are infrared waves often called as heat waves? Give their one application

answer

(a) gamma rays , frequency range (1019 − 1023) Hz (b) to protect eyes from UV-radiations (c) IR are produced by hot bodies and molecules

And used in green houses to warm the plants

10. Answer the following

(a) Name the em waves which are used for the treatment of certain forms of cancer. write their frequency range.

(b) Thin ozone layer on top of stratosphere is crucial for human survival. Why?

(c) Why is the amount of the momentum transferred by the em waves incident on the surface so small?

Answer

(a)X-ray of gamma rays , range :1018 -1022

(b)Ozone layers absorb the ultraviolet radiation from the sun and prevent it from reaching the earth’s surface.

(c) Momentum transferred , p=u/c Where u=energy transferred , and c= speed of light due to the large value

of speed of light (c) the amount of momentum transferred by the em waves incident on the surface is small.

11. State two properties of electromagnetic waves. How can we show that electromagnetic waves carry momentum?

Ans-Transverse nature does not deflected by electric and magnetic fields, same speed in vacuum for all waves, no

material medium required for propagation and they get diffracted, refracted and polarized.

Momentum transferred, p=u/c Where u=energy transferred, and c= speed of light

12. Name the parts of electromagnetic spectrum which is

(i) suitable for radar systems used in aircraft navigation.

(ii) (ii) used to treat muscular strain.

(iii) (iii) used as a diagnostic stool in medicine.

Write in brief, how these waves can be produced.

Ans- (i) microwave (ii) infrared (iii) X-ray

Microwaves are produced by klystroms, magnetrons and gunn diodes

Infrared are produced by the vibrating molecules and atoms in hot bodies.

X-rays are produced by bombardment of high energy electrons on a metal target of high atomic weight like tungsten.

Communication system

VERY SHORT ANSWER TYPE QUESTIONS [1 MARK]

1. What is the function of a ‘Repeater’ in a communication system?

2. What is the function of a transmitter in a communication system?

3. What is the function of a Receiver in a communication system?

4. Which part of the electromagnetic spectrum is used in satellite communication?

5. What is sky wave propagation?

6. What is ground wave propagation?

7. What is space wave propagation?

8. State the reason why microwaves are best suited for long distance transmission of signals.

9. Give the reason why transmission of T.V. signals via sky waves is not possible.

10. What is the purpose of modulating a signal in transmission?

11. What should be the length of dipole antenna for a carrier wave of frequency 6x108 HZ ?

12. A T.V. tower has a height of 71 m. what is the maximum distance upto which T.V. transmission can be

received? Given that the radius of the earth= 6.4x 106 m

13. Suggest a possible communication channel for the transmission of a message signal which has a bandwidth

of 5 MHz.

14. Name the type of communication in which the signal is a discrete and binary coded version of the message

of information.

15. What is the length of a dipole antenna to transmit signals of frequency 200 MHz?.

16. Name the type of communication systems according to the mode of transmission.

17. Name an appropriate communication channel needed to send a signal of band-width 100 KHz over a

distance of 8 km.

18. What is transponder?

19. How does the effective power radiated by an antenna vary with wavelength?

SHORT ANSWER TYPE QUESTIONS [2 MRKS]

1. Distinguish between ‘Analog and Digital signals’

2. Mention the function of any two of the following used inm communication system:

(i) Transducer

(ii) Repeater

(iii) Transmitter

(iv) Bandpass Filter

3. (i) Define modulation index.

(ii) Why is the amplitude of modulating signal kept less than the amplitude of

carrier wave?

4. What is sky wave communication? Why is this mode of propagation restricted

to the frequencies only up to few MHz?

5. What is ground wave communication? On what factors does the maximum

range of propagation in this mode depend?

6. What is space wave communication? Write the range of frequencies suitable

for space wave communication.

7. For an amplitude modulated wave, the maximum amplitude is found to be 10

V while the minimum amplitude is 2 V. Calculate the modulation index. Why

is modulation index generally kept less than one ?

8. Draw a block diagram showing the important components in a communication

system. What is the function of a transducer?

9. ATV lower has a height of 80 m at a given place. Calculate the coverage range,

assuming the radius of the Earth to be 6400 km.

10. The transmission tower at a particular station has a height of 125 m. Calculate the

population covered by the transmission if the average population density around

the tower is 1000 km -2.

11. Explain the function of a repeater in a communication system.

12. What is range of frequencies used for TV transmission? What is common between

these waves and light waves?

13. Write two factors justifying the need of modulating a signal.

A carrier wave of peak voltage 12 V is used to transmit a message signal. What

should be the peak voltage of the modulating signal in order to have a modulation

index of 75% ?

14. In standard AM broadcast, what mode of propagation is used for transmitting a

signal? Why is this mode of propagation limited to frequencies upto a few MHz.

15. By what percentage will the transmission range of a TV tower be affected when

the height of the tower is increased by 21% ?

16. Why are high frequency carrier waves used for transmission?

17. What is meant by term ‘modulation’? Draw a block diagram of a simple modulator

for obtaining an AM signal.

18. Write the function of (i) Transducer and (ii) Repeater in the context of

communication system.

19. Write two factors justifying the need of modulation for transmission of a signal.

20. (i) What is line of sight communication?

(ii) Why is it not possible to use sky wavers propagation for transmission of TV

signals?

Optics

OPTICAL INSTRUMENTS

1.

2.

3.

4..

5.

6.

7. Which of the following waves can be polarized (i) heat waves (ii) sound waves? Give

reason to support your answer. (2013)

Ans: heat waves can be polarized as they are transverse in nature (1M)

8. What will be the effect on interference fringes if red light is replaced by blue light? What

will be the effect on interference fringes if red light is replaced by blue light?

(2013)

Ans: β=Dλ/d, the wavelength of blue light is less than that of red light, hence if red light

is replaced by blue light, the fringes width decreases i.e. fringes come closer.

9. How does the angular separation between fringes in single-slit diffraction experiment

change when the distance of separation between the slit and screen is doubled?

(2012)(1M)

ANs: Angular separation is θ= β/D = λ/d

Since θ is independent of D, angular separation would remain same.

10. In a single-slit diffraction experiment, the width of the slit is made double the original

width. how does this affect the size and intensity of the central diffraction band?

Ans: in single slit diffraction experiment fringe width is β=2Dλ/d. if d is doubled, the width

of central maxima is halved. Thus size of central maxima is reduced to half. Intensity of

diffraction pattern varies square of slit width. so when the slit gets doubled, it makes the

intensity four times.

11. Define the term ‘wavefront’.

Ans: the wavefront is defined as the locus of all the particles of a medium, which are

vibrating in the same phase. (1M)(2014)

12. Draw the shape of the wavefront coming out of a convex lens when a plane wave is

incident on it. (2014)(1M)

Ans:

13. Draw the shape of the wavefront coming out of a concave mirror when a plane wave is

incident on it. (2014)(1M)

Two mark questions:

14. Laser light of wavelength 640 nm incident on a pair of slits produces an interference

pattern in which the bright fringes are separated by 7.2 mm. calculate the wavelength of

another source of light which produces interference fringes separated by 8.1 mm using

same arrangement. Also find the minimum value of the order ‘n’ of the bright fringe of

shorter wavelength which coincides with that of the longer wavelength. (2012)

Ans: distance between two bright fringes= fringe width

Β=𝜆𝐷

𝑑

For same values of D and d, we have

β1/β2=λ1/λ2 or 7.2

8.1=

640

𝜆2

or 0.8λ2=576 or λ2=720 nm

Calculation of minimum value of order: for n to be minimum (n+1)th maxima of shorter

wavelength should coincide with the nth maxima of longer wavelength

(n+1)x 640= n x 720

n= 8

minimum order of shorter wavelength=(n+1)=(8+1)=9

15. Yellow light (λ=6000Å) illuminates a single slit of width 1x10-4m. Calculate (i) the distance

between the two dark lines on either side of the central maximum, when the diffraction

pattern is viewed on a screen kept 1.5m away from the slit, (i) the angular spread of the

first diffraction minimum. (2012)

Ans: (i) Distance between two dark lines, on either side of central maxima= 2𝜆𝐷

𝑑

=(2x60000x10-10x1.5)/91x10-4)= 18mm

(ii) Angular spread of the first diffraction minimum (on either side)

=θ =𝜆

𝑎=(6x10-7)/(1x10-4) = 6x 10-3 radians (2M)

16. (a) Write two characteristic features distinguishing the diffraction pattern from the

interference fringes obtained in Young’s double slit experiment. (2013)

Ans: Diffraction: (i) width of principal maxima is twice the width of the other fringes.

(ii)Intensity goes on decreasing as the order of the diffraction bands increasing

Interference: (i) width of all the fringes is the same. (ii) All the fringes are of same

intensity.

17. A parallel beam of light of 500nm falls on a narrow slit and the resulting diffraction

pattern is observed on a screen 1m away. It is observed that the first minimum is at a

distance of 2.5mm from the screen. Calculate the width of the slit. (2013) (2M)

Ans: xnd/D=nλ D=1 n=1

d= 2 x 10-4 m

18. A parallel beam of light of 600nm falls on narrow slit and the resulting diffraction

pattern is observed on a screen 1.2 m away. It is observed that the first minimum is

at a distance of 3mm from the centre of the screen. Calculate the width of the

slit.(2013)(2M)

Ans: λ=600nm D=1.2m

θ1=x1 /D θ1=2.5 x 10-3 rad

asin θ1=nλ

a=0.24 mm

19. Find an expression for the intensity of transmitted light when a Polaroid sheet is rotate

between two crossed polaroids. In which position of the Polaroid sheet will be

transmitted intensity be maximum.

Ans: Let the rotating Polaroid sheet make an angle θ with the first Polaroid.

The angle with the other Polaroid will be (90-θ)

Applying Malu’s law between

P1 and P3

I’=I0cos2θ

Between P3 and P2

I’’=(I0cos2θ) cos2 (90- θ)

I’’=I0/4xsin2θ

Θ=π/4

Three mark questions:

20. (a) Why photoelectric effect cannot be explained on the basis of wave nature of light?

Give reason.

(b) Write the basic features of photon picture of electromagnetic radiation on which

Einstein’s photoelectric equation is based. (2013)

Ans: (a) (i) The maximum kinetic energy of the emitted electron should be directly

proportional to the intensity of incident radiations but it is not observed experimentally.

Also maximum kinetic energy of the emitted electrons should not depend upon the

incident frequency according to wave theory, but it is not so.

(ii) According to wave theory, threshold frequency should not exist. Light of all

frequencies should emit electrons provided intensity of light is sufficient for electrons to

eject.

(iv) According to wave theory, photoelectric effect should not be instantaneous. Energy

of wave cannot be transferred to a particular electron but will be distributed to all

the electrons present in the illuminate portion. Hence, there has to be a time lag

between incident of radiation and the emission of electrons.

(b)

Basic features of photon picture of electromagnetic radiation:

(i) Radiation behaves as if it is made of particles like photons. Each photons has energy E

=hµ and the momentum p = h/λ.

(ii) Intensity of radiation can be understood in terms of number of photons falling per

second on the surface. Photon energy depends only on frequency and is independent of

intensity.

(iii)Photoelectric effect can be understood as the result of the one to one collision

between an electron and a photon.

(iv)When a photon of frequency (µ) is incident on a metal surface, a part of its energy is

used in overcoming the work function and other part is used in imparting kinetic energy,

so KE=h(µ-µo)

WAVE OPTICS QUESTIONS

1 What is the geometrical shape of the wave front when a plane wave passes through a

convex lens? 2008

2 How would the angular separation of interference fringes in Young’s double slit experiment

change when the distance between the slits and screen is doubled? 2009

3 How does the fringe width, in Young’s double-slit experiment, change when the distance of

separation between the slits and screen is doubled? 2012

4 Define the term ‘linearly polarised light’. When does the intensity of transmitted light

become maximum, when a Polaroid sheet is rotated between two crossed Polaroid? 2009

8 State clearly how an unpolarised light gets linearly polarised when passed through a

Polaroid.

a) Unpolarised light of intensity Io is incident on Polaroid P1 which is kept near another

Polaroid P2 whose pass axis is parallel to that of P1. How will the intensities I1 and I2,

transmitted by the Polaroids P1 and P2 respectively change on rotating P1 without

disturbing P2?

b) Write the relation between the intensities I1 and I2. 2015

9 Use Huygens` principle to show how a plane wave front propagates from a denser to rarer

medium. Hence verify snell`s law of refraction. 2015

10 Answer the following : (a) When a tiny circular obstacle is placed in the path of light from a

distance source, a bright spot is seen at the centre of the shadow of the obstacle. Explain,

why? (b) How does the resolving power of a microscope depend on (i) the wave length of

the light used and (ii) the medium between the object and the objective lens?

11 (a) State Huygens` principle. Using this principle explain how a diffraction pattern is

obtained on a screen due to a narrow slit on which a narrow beam coming from a

monochromatic source of light is incident normally.

(b) Show that the angular width of the first diffraction fringe is half of that of the central

fringe.

(c) If a monochromatic source of light is replaced by white light, what change would you

observe in the diffraction pattern?

14 (a) Using the phenomenon of polarisation, show how transverse nature of light can be

demonstrated.

(b)Two polaroids P1 and P2are placed with their pass axes perpendicular to each other.

Unpolarised light of intensity I0is incident on P1 . A third polaroid P3is kept in between

P1and P2 such that its pass axis makes an angle of 300 with that of P1. Determine the

intensity of light transmitted through P1,P2 and P3. 2014

15 (a) The light from a clear blue portion of the sky shows a rise and fall of intensity when

viewed through a Polaroid which is rotated. Describe, with the help of a suitable diagram,

the basic phenomenon/process which occurs to explain this observation.

(b) Show how light reflected from a transparent medium gets polarized. Hence deduce

Brewster’s law . 2015

16 (a) Define a wave front.

(b) Using Huygens` principle, draw diagrams to show the nature of the wave fronts when an

incident plane wave front gets

(i) reflected from a concave mirror,

(ii) refracted from a convex lens. 2015

17 18 State the condition under which the phenomenon of diffraction of light takes place.

Derive the expression for the width of the central maximum due to diffraction of light at a

single slit. A slit if width ‘a’ is illuminated by a monochromatic light of wavelength 700 nm at

normal incidence. Calculate the value of ‘a’ for position of (i) First minimum at an angle of

diffraction of 30o. (ii) First maximum at an angle of diffraction of 30o. 19 (a) In a single slit

diffraction experiment, a slit of which ‘d’ is illuminated by red light of wavelength 650 nm.

For what value of ‘d’ will: (i) The first minimum fall at an angle diffraction of 30o, and (ii)

The first maximum fall at an angle of diffraction 30o? (b) Why does the intensity of the

secondary maximum become less as compared to the central maximum? 2009

20 In Young’s double slit experiment, the two slits 0.15 mm apart are illuminated by

monochromatic light of wavelength 450 nm. The screen is 0.1 m away from the slits. (a) Find

the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum. (b)

How will the fringe pattern change if the screen is moved away from the slits? 2010

21 State the importance of coherent sources in the phenomenon of interference. In Young’s

double slit experiment to produce interference pattern, obtain the conditions for

constructive and destructive interference. Hence deduced the expression for the fringe

width. How does the fringe width get affected, if the entire experimental apparatus of

Young is immersed in water? 2011

22 1. How does an unpolarised light incident on a polaroid get polarised? Describe briefly, with

the help of the necessary diagram, the polarisation of light by refection from a transparent

medium. 2. Two polaroids ‘A’ and ‘B’ are kept in crossed position. How should a third

Polaroid ‘C’ be placed between them so that the intensity of polarised light transmitted by

Polaroid ‘B’ reduces to 1/8th of the intensity of unpolarised light incident on A?

23 (a) In Young’s double slit experiment, describe briefly how bright and dark fringes are

obtained on the screen kept in front of a double slit. Hence obtain the expression for the

fringe width.

(b) The ratio of the intensities at minima to the maxima in the Young’s double slit

experiment is 9:25. Find the ratio of the widths of the two slits. 2014

24 (a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow

slit illuminated by a mono-chromatic source of light. Hence obtain the conditions for the

angular width of secondary minima.

(b) Two wave lengths of sodium light of 590 nm and 596 nm are used in turn to study the

diffraction taking place at a single slit of aperture 2x 10-6 m. The distance between the slit

and the screen is 1.5 m. Calculate the separation between positions of first maxima of the

diffraction pattern obtained in the two cases. 2014

25 What is the effect on the interference fringes in Young’s double slit experiment when

(i) the width of the slit is increased ;

(ii) the monochromatic source of light is replaced by a source of white light?

26 (a) Using Huygens` construction of secondary wavelets explain how a diffraction pattern is


monochromatic source of light is incident normally.

(b) Show that the angular width of the first diffraction fringe is half of that of the central

fringe. (c) Explain why the maxima becomes weaker and weaker with increasing n. 2015

QUESTIONS HAVE BEEN ASKED TWO TIMES

1 In what way is diffraction from each slit related to the interference pattern in a double slit

experiment? [2013, 2015]

2 In Young’s double slit experiment, derive the condition for (i) Constructive interference and

(ii) Destructive interference at a point on the screen. [2011, 2012]

QUESTIONS HAVE BEEN ASKED THREE TIMES OR MORE

1 State Huygens’ principle. With the help of a suitable diagram, prove Snell’s law of refraction

using Huygens’ principle. 3 [2006 , 2013, 2015]

2 In Young’s double slit experiment, deduce the conditions for (i) constructive, and (ii)

destructive interference at a point on the screen. Draw a graph showing variation of the

resultant intensity in the interference pattern against position ‘x’ on the screen. 3 [2006 ,

2011, 2012]

WAVE OPTICS EXPECTED QUESTIONS FOR AISSCE 17

1 State Huygens’ principle. With the help of a suitable diagram, prove Snell’s law of refraction

using Huygens’ principle. 3

2 State Huygens’ principle. With the help of a suitable diagram, prove the laws of reflection

using Huygens’ principle. 3

3 In Young’s double slit experiment, deduce the conditions for (ii) constructive, and (ii)

destructive interference at a point on the screen. Draw a graph showing variation of the

resultant intensity in the interference pattern against position ‘x’ on the screen.

4 State the importance of coherent sources in the phenomenon of interference. In Young’s

double slit experiment to produce interference pattern, obtain the conditions for

constructive and destructive interference. Hence deduced the expression for the fringe

width. How does the fringe width get affected, if (i) the entire experimental apparatus of

Young is immersed in water? (ii) The wavelength of light is increased? (iii) Separation

between the two slits decreased? (iv) Monochromatic light is replaced by white light? (v)

Distance of the screen is increased? 5

5 (a) Using Huygens` construction of secondary wavelets explain how a diffraction pattern is


monochromatic source of light is incident normally. (b) Show that the angular width of the

first diffraction fringe is half of that of the central fringe.

(c) Explain why the maxima at becomes weaker and weaker with increasing n.

6. How does an unpolarised light incident on a polaroid get polarised?

Describe briefly, with the help of the necessary diagram, the polarisation of light by

refection from a transparent medium.

7. Two polaroids ‘A’ and ‘B’ are kept in crossed position. How should a third Polaroid ‘C’ be

placed between them so that the intensity of polarised light transmitted by Polaroid ‘B’

reduces to 1/8th of the intensity of unpolarised light incident on A?

8. (a) The light from a clear blue portion of the sky shows a rise and fall of intensity when

viewed through a Polaroid which is rotated. Describe, with the help of a suitable diagram,

the basic phenomenon/process which occurs to explain this observation.

(b) Show how light reflected from a transparent medium gets polarized. Hence deduce Brewster’s

law

8 (a) Define a wave front.

(b) Using Huygens` principle, draw diagrams to show the nature of the wave fronts when an

incident plane wave front gets

(i) reflected from a concave mirror,

(ii) refracted from a convex lens.

2016

Distinguish between polarized and unpolarized light. Does the intensity of polarized light

emitted by a polaroid depend on its orientation ? Explain briefly. The vibrations in a beam

of polarized light make an angle of 60° with the axis of the polaroid sheet. What percentage

of light is transmitted through the sheet ?

(i) State the essential conditions for diffraction of light.

(ii) Explain diffraction of light due to a narrow single slit and the formation of pattern of

fringes on the screen.

(iii) Find the relation for width of central maximum in terms of wavelength ‘λ’, width of slit

‘a’, and separation between slit and screen ‘D’.

(iv) If the width of the slit is made double the original width, how does it affect the size and

intensity of the central band ?

OR

(i) Draw a labelled schematic ray diagram of astronomical telescope in normal adjustment.

(ii) Which two aberrations do objectives of refracting telescope suffer from ? How are these

overcome in reflecting telescope ?

(iii) How does the resolving power of a telescope change on increasing the aperture of the

objective lens ? Justify your answer.

2015

(a) The ratio of the widths of two slits in Young’s double slit experiment is 4 : 1. Evaluate the

ratio of intensities at maxima and minima in the interference pattern. (b) Does the

appearance of bright and dark fringes in the interference pattern violate, in any way,

conservation of energy ? Explain.

Write the factors by which the resolving power of a telescope can be increased. (b)

Estimate the angular separation between first order maximum and third order minimum of

the diffraction pattern due to a single slit of width 1 mm, when light of wavelength 600 nm

is incident normal on it.

Good quality sun-glasses made of polaroids are preferred over ordinary coloured glasses.

Justifying your answer. (b) Two polaroids P1 and P2 are placed in crossed positions. A third

polaroid P3 is kept between P1 and P2 such that pass axis of P3 is parallel to that of P1. How

would the intensity of light (I2) transmitted through P2 vary as P3 is rotated ? Draw a plot of

intensity ‘I2’ Vs the angle ‘θ’, between pass axes of P1 and P3.

Define a wavefront. How is it different from a ray ? (b) Depict the shape of a wavefront in

each of the following cases. (i) Light diverging from point source. (ii) Light emerging out of

a convex lens when a point source is placed at its focus. (iii) Using Huygen’s construction of

secondary wavelets, draw a diagram showing the passage of a plane wavefront from a

denser into a rarer medium.

OR

(a) Draw a ray diagram showing the image formation by a compound microscope. Obtain

expression for total magnification when the image is formed at infinity. (b) How does the

resolving power of a compound microscope get affected, when (i) focal length of the

objective is decreased. (ii) the wavelength of light is increased ? Give reasons to justify

your answer.

2014

In Young’s double slit experiment, describe briefly how bright and dark fringes are obtained

on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.

(b) The ratio of the intensities at minima to the maxima in the Young’s double slit

experiment is 9 : 25. Find the ratio of the widths of the two slits.

OR

(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow

slit illuminated by a monochromatic source of light. Hence obtain the conditions for the

angular width of secondary maxima and secondary minima.

(b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the

– 6 m. The distance between the slit

and the screen is 1.5 m. Calculate the separation between the positions of first maxima of

the diffraction pattern obtained in the two cases.

2013

Which of the following waves can be polarized (i) Heat waves (ii) Sound waves?

(a) In what way is diffraction from each slit related to the interference pattern in a double

slit experiment?

(b) Two wavelength of sodium light 590 nm and 596 nm are used, in turn, to study the

diffraction taking place at a single slit of aperture 4 2 10− × m. The distance between the slit

and the screen is 1.5.m. Calculate the separation between the positions of the first maxima

of the diffraction pattern obtained in the two cases.

(a) State Huygen’s principle. Using this principle draw a diagram to show how a plane wave

front incident at the interface of the two media gets refracted when it propagates from a

rarer to a denser medium. Hence verify Snell’s law of refraction.

(b) When monochromatic light travels from a rarer to a denser medium, explain the

following, giving reasons:

(i) Is the frequency of reflected and reflected light same as the frequency of Incident light?

(ii) Does the decrease in speed imply a reduction in the energy carried by light wave?

2012

(a) In Young’s double slit experiment, derive the condition for (i) constructive interference

and (ii) destructive interference at a point on the screen. (b) A beam of light consisting of

two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young’s

double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28

mm, calculate the least distance from the central bright maximum where the bright fringes

of the two wavelengths coincide.

OR

(a) How does an unpolarised light incident on a polaroid get polarised? Describe briefly, with

the help of a necessary diagram, the polarisation of light by reflection from a transparent

medium.

(b) Two polaroids ‘A’ and ‘B’ are kept in crossed position. How should a third polaroid ‘C’ be

placed between them so that the intensity of polarised light transmitted by polaroid B

reduces to 1/8th of the intensity of unpolarised light incident on A?

How does the fringe width, in Young’s double-slit experiment, change when the distance of

separation between the slits and screen is doubled?

Ans- (distance between slits and screen) is doubled, then fringe width will be doubled

ATOMS AND NUCLIE Q. No.

Question and Answers.

Years

1. Define ionization energy. How would the ionization energy change when electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron but having the same charge?

2016

Ans: The ionisation energy is defined as the amount of energy required to remove the most loosely bound electron, the valence electron of an isolated gaseous atom to form a cation. Since, total energy is directly proportional to the mass of electron, so the ionisation energy becomes 200 times on replacing an electron by a particle of mass 200 times that of the electron but having the same charge.

2. Calculate the shortest wavelength of the spectral lines emitted in Balmer series. In which region (infrared, visible, ultraviolet) of hydrogen spectrum does this wavelength lie? (Given, Rydberg constant, R = 107𝑚−1 )

2016

Ans: The wavelength associated with Balmer series is: 1

= 𝑅 (

1

22 − 1

𝑛2). For shortest

wavelength, n = . So the shortest wavelength of the spectral lines emitted is: 1

=

𝑅

4

Or, = 4

𝑅 =

4

107 = 4 × 10−7m. = 4000 𝐴0.

It falls in visible light region of em wave spectrum.

3. Write the basic nuclear process involved in the emission of 𝛽+in a symbolic form by a radioactive nucleus.

2016

Ans: The basic nuclear process involved in the emission of 𝛽+in a symbolic form by a radioactive nucleus

𝑋𝑍𝐴 → 𝑌𝑍−1

𝐴 + 𝛽+ + 𝜈 During the emission of 𝛽+one proton within the nucleus of X changes to neutron, so that the total number of nucleon remains same but the number of proton decreases by one and neutron increases by one.

4. In the reactions given below, find the value of x, y and z and a, b and c. (a) 𝐶6

11 → 𝐵𝑦𝑧 + 𝑥 + 𝜈

(b) 𝐶612 + 𝐶6

12 → 𝑁𝑒𝑎20 + 𝐻𝑒𝑏

𝑐

2016

Ans: (a) According to equation, 𝐶6

11 → 𝐵𝑦𝑧 + 𝑥 + 𝜈

Comparing with 𝛽+decay equation: 𝑋𝑍𝐴 → 𝑌𝑍−1

𝐴 + 𝛽+ + 𝜈. We find – x is 𝛽+, y is 5 and z is 11.

(b) According to equation, 𝐶6

12 + 𝐶612 → 𝑁𝑒𝑎

20 + 𝐻𝑒𝑏𝑐 .

In the above nuclear reaction a = 10, b = 2 and c = 4.

5. A nucleus with the mass number A = 240 and BE/A = 7.6 MeV breaks into fragments each of A = 120 with BE/A = 8.5 MeV. Calculate the energy released.

2016

Ans: According to question: P → Q + Q BE/A of P = 7.6 MeV/A and mass number(A) of P = 240 So, BE of P = 7.6 × 240 MeV. = 1824 MeV. BE/A of Q = 8.5 MeV/A and mass number (A) of Q = 120 So, BE of Q = 8.5 × 120 MeV = 1020 MeV. Now, energy released = 2 (BE of Q) – BE of P. = 2 × 1020 – 1824 = 2040 – 1824 = 216 MeV.

6. Calculate the energy in the fusion reaction: 𝐻1

2 + 𝐻12 → 𝐻𝑒2

3 + 𝑛, where BE of 𝐻12 = 2.23𝑀𝑒𝑉 𝑎𝑛𝑑 𝑜𝑓 𝐻𝑒2

3 = 7.73 𝑀𝑒𝑉. 2016

Ans: According to question: 𝐻12 + 𝐻1

2 → 𝐻𝑒23 + 𝑛.

Energy of fusion = BE of 𝐻𝑒23 - 2 × 2 𝐵𝐸 𝑜𝑓 𝐻1

2 . = 7.73 – 2 × 2.23 = 3.27 MeV.

7. State Bohr’s quantisation condition for defining stationary orbits. How does de-Broglie’s hypothesis explain the stationary orbits?

2016

Ans: Bohr’s quantization principle states that electrons revolve in a stationary orbit of which energy and momentum are fixed. The momentum of electron in the fixed orbit

is given by 𝑛ℎ

2𝜋, where n is principal quantum number.

According to de-Broglie hypothesis, the electron is associated with wave character. Hence, a circular orbit can be taken to be a stationary energy state only if it contains an integral number of de-Broglie wavelengths, i.e.

2𝜋𝑟 = n.

Or, 2𝜋𝑟 = n × ℎ

𝑝,

Or, 2𝜋𝑟 = n× ℎ

𝑚𝑣.

Or, mv𝑟 = n× ℎ

2.

Or, L = n× ℎ

2.

This proves that angular momentum of the electron in a orbit is integral

multiple of ℎ

2.

8. Show that the radius of the orbit in hydrogen atom varies as 𝑛2, where ‘n’ is the principal quantum number of the atom.

2015

Ans: Electron revolves around the nucleus of an atom in circular orbit. The columbic force on the electron by the nucleus provides the required centripetal force. So we have- 𝑚𝑣2

𝑟 = k

𝑒2

𝑟2

Or, 1

𝑟 = k

𝑚𝑒2

𝑚2𝑟2𝑣2

Or, 1

𝑟 = k

𝑚𝑒2

𝐿2. ( L = mrv = n

ℎ

2𝜋 )

Or, r = 1

𝑘𝑚𝑒2 × 𝐿2 =

1

𝑘𝑚𝑒2 × (

𝑛ℎ

2𝜋)

2 =

1

𝑘𝑚𝑒2 ×

ℎ2

4𝜋2 × 𝑛2

Or, r = constant × 𝑛2 This shows that the radius of the orbit in hydrogen atom varies as 𝑛2, where ‘n’ is the principal quantum number of the atom.

9. In the study of Geiger-Marsden experiment on scattering of α – particles by a thin foil of gold, draw the trajectory of α – particles in the coulomb field of target nucleus. Explain briefly how one gets the information on the size of the nucleus from this study.

From the relation R = 𝑅0𝐴1

3⁄ , where 𝑅0 is constant and A is the mass number of the nucleus, show that nuclear matter density is independent of A.

2015

Ans: In the study of Geiger-Marsden experiment on scattering of α – particles by a thin foil of gold, the trajectory of α – particle in the coulomb field of target nucleus is as follows.

The fact that only a small fraction of the number of incident particles rebound back

indicates that the number of 𝛼-particles undergoing head on collision is small. This, in

turn, implies that the mass of the atom is concentrated in a small volume. Rutherford

scattering therefore, is a powerful way to determine an upper limit to the size of the

nucleus.

Nuclear matter density = mass of nucleus/ volume of nucleus. = mass of a nucleon × No. of nucleons/ volume of nucleus

= m × A/ 4

3 𝜋 𝑅3 = m × A/

4

3 𝜋 𝑅𝑜

3 A = 3𝑚

4𝑅𝑜3 = constant.

This show that nuclear matter density is independent of A.

10. Write the three characteristic properties of nuclear force. 2015

Ans: The three characteristic properties of nuclear force are: i) It is the strongest force in nature. ii) It is a short range force. iii) It is a saturated force.

11. Draw a plot of potential energy of a pair of nucleons as a function their separation. Write two important conclusions that can be drawn from the graph.

2015

Ans:

Potential energy of a pair of nucleons as a function of their separation.

Two important conclusions that can be drawn from the graph are: i) The nuclear force is the strongest force in nature which can produce

potential energy of range of MeV. ii) The nuclear force is attractive for small separation but for large separation

the force is repulsive.

12. Distinguish between nuclear fission and fusion. Show how in both these processes energy is released.

2015

Ans: Nuclear Fission Nuclear Fusion

It is the phenomenon of breaking of heavy nucleus to form two or more lighter nuclei.

It is the phenomenon of fusing two or more lighter nuclei to form a single heavy nucleus.

Radioactive waste are released. No radioactive waste are released.

Energy released is lesser than that released in nuclear fusion.

Energy released is greater than that released in nuclear fission.

Ex: 𝑛01 + 𝑈 92

235 → 𝑋𝑒54140 + 𝑆𝑟38

94 + 2 𝑛01

+ 200.4MeVn

Ex: 𝐻11 + 𝐻1

1 → 𝐻12 + 𝑒+ + 𝜈, +

0.42MeV

In both the processes, a certain mass (∆m) disappears, which appears in the form of

energy as per Einstein equation: E = (∆m)𝑐2.

13. Calculate the energy release in MeV in the deuterium-tritium fusion reaction: 𝐻1

2 + 𝐻13 → 𝐻2

4 + 𝑛 Using the data: m( 𝐻1

2 ) = 2.014102 u, m( 𝐻13 ) = 3.016049 u, m( 𝐻𝑒2

4 ) = 4.002603 u &𝑚𝑛 = 1.008665 u 1u = 931.5 MeV/𝑐2.

2015

Ans: According to question, the reaction is: 𝐻1

2 + 𝐻13 → 𝐻2

4 + 𝑛 Mass defect (∆m) = ((2.014102 + 3.016049) – (4.002603 + 1.008665))u = 0.018883u. Energy released = ∆m × 931.5 MeV = 0.018883 × 931.5 MeV = 17.589 MeV.

14. A 12.9 eV beam of electrons is used to bombard gaseous hydrogen at room temperature. Upto which energy level the hydrogen atoms would be excited? Calculate the wavelength of the first member of Lyman series and first member of Balmer series.

2014

Ans: The energy of gaseous hydrogen atom at room temperature are: 𝐸1 = −13.6 𝑒𝑉, 𝐸2 = −3.4 𝑒𝑉, 𝐸3 = −1.51 𝑒𝑉 and 𝐸4 = −0.85 𝑒𝑉. So the differences are: 𝐸2 − 𝐸1 = −3.4 + 13.6 = 10.2𝑒𝑉. 𝐸3 − 𝐸1 = −1.51 + 13.6 = 12.09 and 𝐸4 − 𝐸1 = −0.85 + 13.6 = 12.75 𝑒𝑉. As the energy given to the electron is more than 𝐸4 − 𝐸1, electron will jump to the 4th orbit.

For Lyman series, we have - 1

= 𝑅 (

1

12 − 1

𝑛2). For first member of the series n = 2.

Thus we get, 1

= 𝑅 (

1

12 − 1

22) = R × 3

4.

Or, = 4

3𝑅 =

4

3𝑋

1

1.097 × 107 = 1.215 × 10−7m.

For Balmer series, we have - 1

= 𝑅 (

1

22 − 1

𝑛2). For first member of the series n = 3.

Thus we get, 1

= 𝑅 (

1

2−

1

92) = R ×

5

36.

Or, = 36

5𝑅 =

36

5𝑋

1

1.097 × 107 = 6.56 × 10−7m.

15. Draw a plot of BE/A – mass number A for 2 ≤ 𝐴 ≤ 170. Use the graph to explain the release of energy in the process of nuclear fusion of two light nuclei and fission of a heavy nucleus into two lighter nuclei..

2014

Ans: A plot of BE/A – mass number A for 2 ≤ 𝐴 ≤ 170 is as follows:

Explanation of release of energy in the process of nuclear fusion of two light nuclei is as follows. Let us Consider two very light nuclei (A<10) joining to form a heavier nucleus. The

binding energy per nucleon of the fused heavier nuclei is more than the binding energy

per nucleon of the lighter nuclei. This means that the final system is more tightly

bound than the initial system. Again energy would be released in such a process of

fusion. A very heavy nucleus, say A = 240, has lower binding energy per nucleon compared to

that of a nucleus with A = 120. Thus if a nucleus A = 240 breaks into two A = 120

nuclei, nucleons get more tightly bound. This implies energy would be released in the

process. It has very important implications for energy production through

Fission.

16. Write the relation for Binding Energy (BE) (in MeV) of a nucleus of mass number (A) in terms of the masses of its constituents namely neutrons and protons.

2014

Ans: The relation is:

BE = [𝑍𝑚𝑝 + (𝐴 − 𝑍)𝑚𝑛 − 𝑀𝑍𝐴 ] × 𝑐2.

Where, M is mass of nucleus, 𝑚𝑝 mass of proton and 𝑚𝑝 mass of neutron.

17. Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?

2014

Ans: The electrostatic force of attraction, Fe between the revolving electrons and the

nucleus provides the requisite centripetal force (Fc) to keep them in their orbits. Thus,

for a dynamically stable orbit in a hydrogen atom.

Thus the relation between the orbit radius and the electron velocity is:

The kinetic energy (K) and electrostatic potential energy (U) of the electron in

hydrogen atom are

Thus the total energy E of the electron in a hydrogen atom is:

The total energy of the electron is negative. This implies the fact that the electron is

bound to the nucleus. If E were positive, an electron will not follow a closed orbit

around the nucleus.

18. Using Bohr’s postulate of the atomic model, derive the expression for the total energy of the electron in the stationary orbit of hydrogen atom.

2014

The electrostatic force of attraction, Fe between the revolving electrons and the



The relation between 𝑣𝑛and 𝑟𝑛 is:

Combining it with the eq:

,

Thus we get the following expressions for 𝑣𝑛and 𝑟𝑛.

and

The total energy of the electron in the stationary states of the hydrogen atom can be

obtained by substituting the value of orbital radius:

19. Using Bohr’s postulate of the atomic model, derive the expression for the radius of nth electron orbit. Hence, obtain the expression for Bohr’s radius.

2014






,


and

20. The value of ground state energy of hydrogen atom is -1.6 eV. (a) Find the energy required to move an electron from the ground state to the

first excited state of the atom. (b) Determine the kinetic energy and orbit radius in the first excited state of the

atom. ( Bohr’s radius is 0.53 𝐴0)

2014

Ans: a) Energy of first excited state is: 𝐸2 = −3.4 𝑒𝑉. So the energy required to move an electron from the ground state to the first excited state of the atom is: ∆𝐸 = 𝐸2 − 𝐸1 = −3.4 + 13.6 = 10.2𝑒𝑉.

b) Kinetic energy in the first excited state of the atom is 3.4eV and the orbit radius in the first excited state of the atom is 4 × 5.29 × 10−11𝑚.

21. The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10−11m. calculate its radius in n=3 orbit.

2014

Ans: The radius of the n=3 electron orbit of a hydrogen atom is 32 ×5.3 × 10−11m. That is 𝑟3= 47.5 × 10−11m.

22. The total energy of an electron in the first excited state of the hydrogen atom is -3.4 eV. Find out its kinetic and potential energy in this state.

2014

Ans: Kinetic energy = 3.4eV and potential energy = - 6.8 eV.

23. When is the 𝐻𝛼- line of the Balmer series in the emission spectrum of hydrogen atom obtained?

2013

Ans: The 𝐻𝛼- line of the Balmer series in the emission spectrum of hydrogen atom is obtained when the electron jumps from third orbit to second orbit of hydrogen atom.

24. In the ground state of hydrogen atom, its Bohr radius is 5.3 × 10−11m. The atom is excited such that the radius becomes 21.2 × 10−11m. Find (a) the value of the principle quantum number and (b) the total energy of the atom in this excited state.

2013

Ans: (a) The value of the principle quantum number 𝑛2 =21.2

5.3 = 4. So, n = 2

(b) the total energy of the atom in this excited state is E = −13.6

4 eV = - 3.4eV.

25. Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence, draw the energy level diagram showing how the spectra corresponding to Balmer series occur due to transition between energy levels.

2013






,


and

The total energy of the electron in the stationary states of the hydrogen atom can be

obtained by substituting the value of orbital radius:

The energy level diagram of hydrogen atom Different spectral series obtained are

shown below:

26. The number of nuclei of a given radioactive sample at time t=0 and t=T are 𝑁0 and 𝑁0 𝑛⁄ , respectively. Obtain an expression for the half – life (𝑁1

2⁄ ) of the nucleus in

terms of n and T.

203

Ans: According to law of radioactivity, we have-

N= 𝑁0𝑒−𝑡 and at t = T, N = 𝑁0

𝑛 .

So, 𝑁0

𝑛 = 𝑁0𝑒−𝑡. Or, n = 𝑒−𝑡.

Or, = 𝑙𝑜𝑔(𝑛)

𝑇.

Again, 𝑇12⁄ =

0.693

=

0.693 𝑇

𝑙𝑜𝑔(𝑛).

27. Draw a plot of potential energy between a pair of nucleons as a function of their separation. Mark the region where potential energy is (i) positive and (ii) negative.

2013

Ans: Graph showing the variation of potential energy between a pair of nucleons as a function of their separation

Here potential energy is positive in between 0 and the vertical dotted line and negative beyond that.

28. Write the basic nuclear process of nucleus undergoing 𝛽 − decay. Why is the detection of neutrinos found very difficult?

2013

Ans: During 𝛽 , either a proton or a neutron changes to a neutron or a proton respectively. The basic processes are as follows: In beta-minus decay, a neutron transforms into a proton within the nucleus according

to

whereas in beta-plus decay, a proton transforms into neutron (inside the nucleus) via

Neutrinos are neutral particles with very little mass, so they interact with matters very feebly. So detection of neutrinos found very difficult.

29. Why is the classic (Rutherford) model for an atom of electron orbiting around the nucleus not able to explain the atomic structure?

2012

Ans: According to classical electromagnetic theory, an accelerating charged particle emits

radiation in the form of electromagnetic waves. The energy of an accelerating electron

should therefore, continuously decrease. The electron would spiral inward and

eventually fall into the nucleus. Thus, such an atom cannot be stable. Further,

according to the classical electromagnetic theory, the frequency of the electromagnetic

waves emitted by the revolving electrons is equal to the frequency of revolution. As

the electrons spiral inwards, their angular velocities and hence their frequencies would

change continuously, and so will the frequency of the light emitted. Thus, they would

emit a continuous spectrum, in contradiction to the line spectrum actually observed.

Clearly Rutherford model tells only a part of the story implying that the classical ideas

are not sufficient to explain the atomic structure.

30. In hydrogen atom, an electron undergoes transition from second excited state to the first excited state and then to the ground state. Identify the spectral series to which these transitions belong. Find out the ratio of the wavelength of the emitted radiation in the two cases.

2012

Ans: Transition from second to first orbit results to Balmer series and again transition from first to ground state results to Lyman series.

The ratio of the wavelength of the two transition is: 1

2=

𝐸3− 𝐸2

𝐸2− 𝐸1 =

1.9

10.2 =

19

102

31. Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the nth orbital state in hydrogen atom is n-times the de-Broglie wavelength associated with it.

2012

Ans: From Bohr’s second postulate of quantization of orbital angular momentum we have-

L = nrv = n ℎ

2𝜋.

Or, 2𝜋𝑟 = n ℎ

𝑚𝑣 = n

ℎ

𝑝 = n .

This shows that the circumference of the electron in the nth orbital state in hydrogen atom is n-times the de-Broglie wavelength associated with it

32. The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?

2012

Ans: No. of spectral lines emitted is given by the formula: N = 𝑛(𝑛−1)

2.

Here n = 3, so, N = 3(3−1)

2 =

3 ×2

2 = 3.

Three spectral lines which can be emitted when it finally moves to the ground state.

33. The ground state energy of hydrogen atom is -13.6 eV. If the makes a transition from an energy level -0.86 eV to -1.51 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?

2012

Ans: We know, ℎ𝑐

= ∆𝐸 = (−0.86 + 1.51) × 1.6 × 10−19) = 0.69 × 1.6 × 10−19. .

Or, = 6.63 × 10−34 × 3 × 108

0.69 ×1.6 × 10−19 = 18.02 × 10−7m.

34. In the Geiger-Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when an ∝ -particle of 8 MeV energy impinges on it before it comes to momentarily rest and reverses its direction. How will the distance of closest approach be affected when the kinetic energy of the

2012

∝ -particle is doubled?

Ans: Thus the distance of closest approach d is given by-

The kinetic energy of -particles is 8 MeV or 12.8 × 10

–12 J. Since 1/4𝜀0= 9.0 × 10

9

N m2/C

2. Therefore with e = 1.6 × 10

–19 C, we have,

d = 2 ×80 ×1.6 ×1.6 × 10−38×9 × 109

12.8 × 10−12 = 3.19 × 10−14 𝑚 = 31.9 fm.

When the kinetic energy of the ∝ -particle is doubled the distance of closest approach reduces to half of its previous value.

35. In a given sample, two radio isotopes A and B are initially present in the ratio 1:4. The half-lives of A and B are 100 years and 50 years, respectively. Find the time after which the amount of A and B becomes equal.

2012

Ans: We have the condition: 𝑁𝐴 = 𝑁𝐵 → 𝑁0𝐴𝑒−𝐴𝑡 = 𝑁0𝐵𝑒−𝐴𝑡 → 1𝑒−𝐴𝑡 = 4 𝑒−𝐴𝑡

→ 4 = 𝑒−(𝐴− 𝐵)𝑡 → ln(4) = −(

𝐴− 𝐵)𝑡. ………….(i)

Again, 𝐴 = ln 2

𝑇 =

𝑙𝑛 2

100 𝑎𝑛𝑑 𝐵 =

ln 2

𝑇 =

ln 2

50

→ 𝐵 = 2 𝐴 ………………………………..(ii)

From equations (i) and (ii), we get- ln(4) = 𝐴𝑡 → 𝑡 = ln 4

ln 2× 100 = 200 𝑦𝑒𝑎𝑟𝑠.

36. Why is the binding energy per nucleon found to be constant for nuclei in the range of mass number (A) lying between 30 and 170?

2012

Ans: The constancy of the binding energy in the range 30 < A < 170 is a consequence of the

fact that the nuclear force is short-ranged. Let us consider a particular nucleon inside a

sufficiently large nucleus. It will be under the influence of only some of its neighbours,

which come within the range of the nuclear force. If any other nucleon is at a distance

more than the range of the nuclear force from the particular nucleon it will have no

influence on the binding energy of the nucleon under consideration. If a nucleon can

have a maximum of p neighbours within the range of nuclear force, its binding energy

would be proportional to p. Let the binding energy of the nucleus be pk, where k is a

constant having the dimensions of energy. If we increase A by adding nucleons they

will not change the binding energy of a nucleon inside. Since most of the nucleons in a

large nucleus reside inside it and not on the surface, the change in binding energy per

nucleon would be small. The binding energy per nucleon is a constant and is

approximately equal to pk. The property that a given nucleon influences only nucleons

close to it is also referred to as saturation property of the nuclear force.

37. When a heavy nucleus with mass number A=240 breaks into two nuclei, A=120, energy is released in the process.

2012

Ans: Let us take a nucleus with A = 240 breaking into two fragments each of A = 120.

Then 𝐸𝑏𝑛 for A = 240 nucleus is about 7.6 MeV,

𝐸𝑏𝑛 for the two A = 120 fragment nuclei is about 8.5 MeV.

Therefore, Gain in binding energy for nucleon is about 0.9 MeV.

Hence the total gain in binding energy is 240×0.9 or 216 MeV

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