Veer

Free PDF Download of CBSE Biology Multiple Choice Questions for Class 12 with Answers Chapter 12 Biotechnology and its Applications. Biology MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Biology Biotechnology and its Applications MCQs Pdf with Answers to know their preparation level.

Biotechnology and its Applications Class 12 Biology MCQs Pdf

1. The genes crylAb and cryllAb produce toxins against _______ and _______, respectively.
(a) cotton bollworms, com borer
(b) nematode, cotton bollworm
(c) com borer, cotton bollworm
(d) com borer, nematodes

Answer

Answer: c

---

2. Which among the following is based on antigen-antibody interaction?
(a) PCR
(b) Electrophoresis
(c) ELISA
(d) All of these.

Answer

Answer: c

---

3. Which among the following is not allowed to take place in the case of RNA interference employed in making tobacco plants resistant to the nematode, Meloidegyne incognitia?
(a) Transcription of mRNA
(b) Translation of mRNA
(c) Replication of DNA
(d) Maturation of hn RNA.

Answer

Answer: b

---

4. Night blindness can be prevented by use of
(a) golden rice
(b) transgenic tomato
(c) transgenic maize
(d) Bt brinjal.

Answer

Answer: a

---

5. The T_i plasmid used for producing transgenic plants is found in
(a) Azotobacter
(b) Rhizobium
(c) Azospirillum
(d) Agrobacterium

Answer

Answer: d

---

6. a-1 antitrypsin is: [NCERT Exemplar]
(a) an antacid
(b) an enzyme
(c) used to treat arthritis
(d) used to treat emphysema.

Answer

Answer: d

---

7. The trigger for activation of toxin of Bacillus thuringiensis is [NCERT Exemplar]
(a) acidic pH of stomach.
(b) high temperature.
(c) alkaline pH of gut
(d) mechanical action in the insect gut.

Answer

Answer: c

---

8. In RNAi, genes are silenced using [NCERT Exemplar]
(a) ss DNA
(b) ds DNA
(c) ds RNA
(d) ss RNA

Answer

Answer: c

---

9. C-peptide of human insulin is [NCERT Exemplar]
(a) a part of mature insulin molecule
(b) responsible for formation of disulphide bridges
(c) removed during maturation of pro-insulin to insulin
(d) responsible for its biological activity.

Answer

Answer: c

---

10. Silencing of a gene could be achieved through the use of [NCERT Exemplar]
(a) short interfering RNA (RNAi)
(b) antisense RNA
(c) by both
(d) none of the above.

Answer

Answer: c

---

11. GM crops help to reduce the ______ losses.

Answer/Explanation

Answer:
Explaination: Post-harvest.

---

12. The alkaline pH in the stomach of insect larvae triggers the activation of ______ .

Answer/Explanation

Answer:
Explaination: Bt-toxin.

---

13. ______ is the compound obtained from transgenic animal that is used to treat emphysema.

Answer/Explanation

Answer:
Explaination: α-1-antitrypsin.

---

14. In RNAi, the silencing of mRNA is carried out by using ______ .

Answer/Explanation

Answer:
Explaination: ds RNA.

---

15. The Bt toxin does not kill the bacterium producing it, because it is produced in an _________ state.

Answer/Explanation

Answer:
Explaination: Inactive.

---

16. The site of production of ADA in the body is ______ .

Answer/Explanation

Answer:
Explaination: Lymphocytes.

---

17. ______ of human insulin is removed during maturation process.

Answer/Explanation

Answer:
Explaination: C-peptide.

---

18. Using ______ vectors, nematode-specific genes are introduced into host plants.

Answer/Explanation

Answer:
Explaination: Agrobacterium.

---

19. The protein coded by the gene, crylAb provides resistance to ______ .

Answer/Explanation

Answer:
Explaination: Com borer.

---

20. An American company obtained patent rights on ______ rice through US patent and Trademark Office.

Answer/Explanation

Answer:
Explaination: Basmati.

---

21. Match the items in Column I with those in Column II.

Column I	Column II
A. Rosie	1. Polio vaccine safety
B. T. plasmid	2. Human alpha-lactalbumin
C. RNAi	3. Agrobacterium tumefaciens
D. ELISA	4. Meloidegyne incognitia
E. Transgenic mice	5. Antigen-antibody interaction

Answer/Explanation

Answer:
Explaination: A – 2, B – 3, C – 4, D – 5, E – 1

---

22. Match the terms in Column I with those in Column II.

Column I	Column II
A. Gene therapy	1. Human insulin
B. Cotton bollworm	2. Biopiracy
C. EliLilly	3. Emphysema
D. Basmati Rice	4. ADA deficiency
E. α -1 antitrypsin	5. Lepidopteran

Answer/Explanation

Answer:
Explaination: A – 4, B – 5, C – 1, D – 2, E – 3

---

23. The disorder ADA deficiency can be cured by gene therapy only. [True/False]

Answer/Explanation

Answer:
Explaination: False.

---

24. Transgenic mice are used to test the safety of polio vaccine. [True/False]

Answer/Explanation

Answer:
Explaination: True.

---

25. The recombinant therapeutics induce unwanted immunological responses. [True/False]

Answer/Explanation

Answer:
Explaination: False.

---

26. Bt toxins provide resistance to all types of insect pests. [True/Faise]

Answer/Explanation

Answer:
Explaination: False.

---

27. ELISA is based on introducing a functional gene is place by a defective gene. [True/False]

Answer/Explanation

Answer:
Explaination: False.

---

Directions (Q28 to Q30): Mark the odd one in each of the following groups.
28. PCR, Widal, rDNA technology, ELISA.

Answer/Explanation

Answer:
Explaination: Widal.

---

29. Tetanus taxoid, a-1 antitrypsin, Hepatitis B vaccine, Humulin.

Answer/Explanation

Answer:
Explaination: Tetanus taxoid.

---

30. Agrobacterium, RNA/, crylAc, Meloidegyne.

Answer/Explanation

Answer:
Explaination: crylAc

---

31. What do the letters Bt stand for, in Bt cotton plants?

Answer/Explanation

Answer:
Explaination: Bt stand for Bacillus thuringiensis.

---

32. What are cry genes? In which organism are they present? [AI2017]

Answer/Explanation

Answer:
Explaination:
– The genes which code for the Bt toxin proteins, are called cry genes.
– They are present in the bacterium, Bacillus thuringiensis.

---

33. Mention the source organism of the gene crylAc and its target pest. [Foreign 2011]

Answer/Explanation

Answer:
Explaination:
– Bacillus thuringiensis is the source organism.
– Cotton bollworm is its target pest.

---

34. Mention the source organism of the gene, crylAb and its target pest. [Foreign 2011]

Answer/Explanation

Answer:
Explaination:
– Bacillus thuringiensis is the source 42. organism.
– Com borer is its target pest.

---

35. List the type of cry genes that provide resistance to com plants and cotton plants, respectively, against lepidopterans. [Foreign 2017]

Answer/Explanation

Answer:
Explaination:
– crylAb controls com borer and crylAc and cryllAb control cotton bollworm.

---

36. Why are certain cotton plants called Bt- cotton plants?

Answer/Explanation

Answer:
Explaination: They are the transgenic cotton plants with genes from Bacillus thuringiensis’, they are resistant to bollworms.

---

37. Name the specific type of gene that is incorporated in a cotton plant to protect the plant against cotton bollworm infestation. [AI 2017]

Answer/Explanation

Answer:
Explaination: crylAc and cryllAb.

---

38. Bt toxins are released as inactive crystals in the bacterial body. What happens to it in the cotton bollworm body that it kills the bollworm? [AI 2017]

Answer/Explanation

Answer:
Explaination:
– It becomes activated in the alkaline pH of the gut of the insect.
– The activated toxin binds to the surface of midgut epithelial cells, create pores that cause swelling and lysis of the cells and eventually cause death of the insect.

---

39. What is the significance of the process of RNA interference (RNAi) in eukaryotic organisms?

Answer/Explanation

Answer:
Explaination: RNA interference is a method of cellular defense in all eukaryotic organisms.

---

40. Write the possible source of RNA interference (RNAi) gene. [Delhi 2013C]

Answer/Explanation

Answer:
Explaination:
(i) Infection by a virus having RNA genome.
(ii) Mobile genetic elements, called transposons that replicate via an RNA intermediate.

---

41. How does dsRNAgain entry into eukaryotic cell to cause RNA interference? [Delhi 2011C]

Answer/Explanation

Answer:
Explaination:
(i) Infection by viruses with RNAgenome.
(ii) Transposons, the mobile genetic elements, that can replicate via an RNA intermediate.

---

42. How does silencing of specific mRNA in RNA interference prevent parasitic infestation?

Answer/Explanation

Answer:
Explaination: The nematode (parasite) cannot live in the transgenic host that expresses RNA interference.

---

43. State the role of transposons in silencing of mRNA in eukaryotic cells. [AI 2013]

Answer/Explanation

Answer:
Explaination: Transposons are the source of complementary RNA that binds to and prevents the translation of specific mRNA.

---

44. How are tobacco plants benefitted when nematode-specific genes are introduced into them using certain vectors? Name the vector used.

Answer/Explanation

Answer:
Explaination:
– The tobacco plants show resistance to nematode attack; the nematode cannot survive in such plants, as they show RNA interference.
– The vector used is Agrobacterium.

---

45. Mention the chemical change that proinsulin undergoes, to be able to act as mature insulin. [CBSE 2018]

Answer/Explanation

Answer:
Explaination:
– The C-peptide is removed.
– The peptides A and B are joined by disulphide bridges.

---

46. State the role of C-peptide in human insulin. [AI 2014]

Answer/Explanation

Answer:
Explaination:
The C-peptide joins the A-peptide with B-peptide in the proinsulin; it is removed during the processing of proinsulin into insulin.

---

47. How are the two short polypeptide chains of insulin linked together?

Answer/Explanation

Answer:
Explaination:
They are linked together by creating disulphide bonds.

---

48. How is proinsulin different from the functional insulin in humans? [AI 2012C]

Answer/Explanation

Answer:
Explaination:
– Proinsulin contains an additional polypeptide chain called C-peptide.
– In the functional insulin, C-peptide is absent, as it is removed during processing.

---

49. A boy has been diagnosed with ADA deficiency. Suggest any one possible treatment. [Delhi 2014 C]
Or
Suggest any two possible treatments that can be given to a patient exhibiting adenosine deaminase deficiency. [AI 2015]

Answer/Explanation

Answer:
Explaination:
(i) Gene therapy.
(ii) Enzyme replacement therapy.
(iii) Bone marrow transplantation.

---

50. State the cause of adenosine deaminase enzyme deficiency. [AI 2015]

Answer/Explanation

Answer:
Explaination: It is caused due to deletion of the gene coding for the enzyme, adenosine deaminase.

---

51. Why do children cured by enzyme replacement therapy of adenosine deaminase deficiency, need periodic treatment?[AI 2015]

Answer/Explanation

Answer:
Explaination: The lymphocytes are not immortal, but have a life span; hence with the formation of new lymphocytes, every time, the enzyme has to be injected.

---

52. Mention an example of stem cell therapy.

Answer/Explanation

Answer:
Explaination: Bone marrow transplantation.

---

53. Suggest a molecular diagnostic procedure that detects HTV in a suspected AIDS patient. [Foreign 2016]

Answer/Explanation

Answer:
Explaination: Enzyme-linked immunosorbent assay (ELISA)

---

54. Name any two techniques that serve the purpose of early diagnosis of some bacterial/ viral human diseases. [Foreign 2011]
Or
Name a molecular diagnostic technique to detect the presence of a pathogen in its early stage of infection. [Delhi 2010]

Answer/Explanation

Answer:
Explaination:
(i) Polymerase chain reaction (PCR).
(ii) Enzyme-linked immunosorbent assay (ELISA).

---

55. State the principle on which ELISA works.

Answer/Explanation

Answer:
Explaination: ELISA works on the principle of antigen- antibody interaction.

---

56. What are transgenic animals? Give an example.

Answer/Explanation

Answer:
Explaination:
– Transgenic animals are those animals which have their DNA manipulated to possess and express one/more foreign genes.
e.g. – the transgenic cow, Rosie, possesses the gene for human alpha- lactalbumin.

---

57. Name the protein produced in transgenic animals that is used to treat emphysema.

Answer/Explanation

Answer:
Explaination: α-1-antitrypsin.

---

58. What is biopiracy? [Delhi 2017, AI 2016, Delhi 2015]

Answer/Explanation

Answer:
Explaination: Biopiracy refers to the use of bioresources and traditional knowledge related to bioresources for commercial benefit by certain organisations or multi-national companies without proper consent from the countries and compensatory payment to the people concerned.

---

59. Name the Indian variety of rice patented by an American company.

Answer/Explanation

Answer:
Explaination: Basmati variety.

---

60. Mention two objectives of setting up GEAC by our government. [AI 2016]
Or
State the purpose for which the Indian Government has set up GEAC. [Foreign 2013]

Answer/Explanation

Answer:
Explaination:
The objectives of setting up GEAC are:
(a) to have some ethical standards to evaluate the morality of human activities that might help or harm living organisms.
(b) to have a regulation, as genetic modifications of organisms may have unpredictable results when such organisms are introduced into the ecosystem.

---

We hope the given Biology MCQs for Class 12 with Answers Chapter 12 Biotechnology and its Applications will help you. If you have any query regarding CBSE Class 12 Biology Biotechnology and its Applications MCQs Pdf, drop a comment below and we will get back to you at the earliest.

Free PDF Download of CBSE Biology Multiple Choice Questions for Class 12 with Answers Chapter 11 Biotechnology: Principles and Processes. Biology MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Biology Biotechnology: Principles and Processes MCQs Pdf with Answers to know their preparation level.

Biotechnology: Principles and Processes Class 12 Biology MCQs Pdf

1. Biolistics (gene gun) is suitable for
(a) introducing rDNA into plant cells
(b) introducing rDNA into animal cells
(c) disarming the pathogen vectors
(d) DNA fingerprinting.

Answer

Answer: a

---

2. In genetic engineering experiments, restriction enzymes are used for
(a) viral DNA
(b) bacterial DNA
(c) eukaryotic DNA
(d) any type of DNA.

Answer

Answer: b

---

3. The DNA fragments produced by the use of restriction endonucleases can be separated by
(a) polymerase chain reaction
(b) gel electrophoresis
(c) density gradient centrifugation
(d) any of the above.

Answer

Answer: b

---

4. Plasmids in bacterial cells are
(a) extra-chromosomal DNA, which cannot replicate
(b) extra-chromosomal DNA, which can . self-replicate
(c) extra DNA associated with the genome
(d) extra DNA, associated with the genome, but cannot replicate.

Answer

Answer: b

---

5. The DNA polymerase enzyme used in PCR is obtained from
(a) Thermus aquaticus
(b) Escherichia coli
(c) Agrobacterium tumefaciens
(d) Salmonella typhimurium.

Answer

Answer: a

---

6. While isolating DNA from bacteria, which of the following enzymes is not used? [NCERT Exemplar]
(a) Lysozyme
(b) Ribonuclease
(c) Deoxyribonuclease
(d) Protease

Answer

Answer: a

---

7. Significance of ‘heat shock’ method in bacterial transformation is to facilitate [NCERT Exemplar]
(a) binding of DNA to the cell wall
(b) uptake of DNA through membrane transport proteins
(c) uptake of DNA through transient pores in the bacterial cell wall
(d) expression of antibiotic resistance gene

Answer

Answer: c

---

8. Which of the following steps are catalysed by Taq polymerase in a PCR reaction? [NCERT Exemplar]
(a) Denaturation of template DNA
(b) Annealing of primers to template DNA
(c) Extension of primer end on the template DNA
(d) All of the above

Answer

Answer: c

---

9. Which of the given statement is correct in the context of observing DNA separated by agarose gel electrophoresis? [NCERT Exemplar]
(a) DNA can be seen in visible light
(b) DNA can be seen without staining in visible light
(c) Ethidium bromide stained DNA can be seen in visible light
(d) Ethidium bromide stained DNA can be seen under exposure to UV light

Answer

Answer: d

---

10. ‘Restriction’ in Restriction enzyme refers to [NCERT Exemplar]
(a) cleaving ofphosphodiester bond in DNA by the enzyme
(b) cutting of DNA at specific position only
(c) prevention of the multiplication of bacteriophage in bacteria
(d) all of the above

Answer

Answer: c

---

11. The role of DNA ligase in the construction of a recombinant DNA molecule is [NCERT Exemplar]
(a) formation of phosphodiester bond between two DNA fragments
(b) formation of hydrogen bonds between sticky ends of DNA fragments
(c) ligation of all purine and pyrimidine bases
(d) none of the above

Answer

Answer: b

---

12. In an experiment, recombinant DNA bearing ampicillin-resistance gene is transferred into E.coli cells. The host cells are then cultured on a medium containing ampicillin. The result will be
(a) both transformants and non-transformants cannot survive.
(b) both transformants and non-transformants can survive.
(c) transformants only and not the non-transformants can survive.
(d) transformants cannot survive, but non-transformants can not.

Answer

Answer: c

---

13. The _______ in a vector helps in identifying the transformants and eliminating the non-transformants.

Answer/Explanation

Answer:
Explaination: Selectable marker.

---

14. When the enzyme _______ is inactivated in E.coli, the transformants do not produce any colour in the presence of a chromogenic substrate.

Answer/Explanation

Answer:
Explaination: β-galactosidase.

---

15. _______ is the method of bombarding high velocity microparticles of gold or tungsten coated with DNA, into plant cells.

Answer/Explanation

Answer:
Explaination: Biolistics/gene gun.

---

16. To isolate DNA from fungal cells for biotechnology experiments, enzyme _______ is necessary.

Answer/Explanation

Answer:
Explaination: Chitinase.

---

17. _______ is used as cloning vector for transformation in plant cells.

Answer/Explanation

Answer:
Explaination: Agmbacterium.

---

18. Downstream processing involves separation and _______ .

Answer/Explanation

Answer:
Explaination: Purification.

---

19. _______ is the process of extraction of DNA from the separated bands of DNA in agarose gel.

Answer/Explanation

Answer:
Explaination: Elution.

---

20. _______ gene in the E.coli vector, pBR 322 codes for enzymes/proteins involved in the replication of the plasmid.

Answer/Explanation

Answer:
Explaination: Rop.

---

21. _______ are the E.coli enzymes that remove the nucleotides from the ends of DNA strands.

Answer/Explanation

Answer:
Explaination: Exonucleases.

---

22. _______ is an autonomously replicating, circular, extra-chromosomal DNA in bacterial cells.

Answer/Explanation

Answer:
Explaination: Plasmid.

---

23. Match the items in Column I with those in Column II.

Column I	Column II
A. Sea weeds	1. Gel electrophoresis
B. Staining of DNA	2. Source of Agarose
C. Separation of DNA fragments	3. Isolation of DNA from the gel
D. Elution	4. Ethidium bromide

Answer/Explanation

Answer:
Explaination: A – 2, B – 4, C – 1, D – 3

---

24. Match the Column l with the Column II.

Column I	Column II
A. Competent host	1. Separation and purification
B. Cloning vector	2. Culturing of cells in large volumes
C. Downstream processing	3. Taq polymerase
D. PCR	4. Divalent cation (Ca2+)
E. Bioreactor	5. pBR 322
	6. Gel electrophoresis

Answer/Explanation

Answer:
Explaination: A – 4, B – 5, C – 1, D – 3, E – 2

---

25. Since DNA fragments are positively charged they move towards the anode. [True/False]

Answer/Explanation

Answer:
Explaination: False.

---

26. Agarose, the most commonly used matrix in gel electrophoresis is obtained from fungi. [True/False]

Answer/Explanation

Answer:
Explaination: False.

---

27. The DNA fragments separate according to their size through agarose gel during electrophoresis. [True/False]

Answer/Explanation

Answer:
Explaination: True.

---

28. The cloning vector pBR 322 has three antibiotic-resistance genes. [True/False]

Answer/Explanation

Answer:
Explaination: False.

---

29. The vector DNA and foreign DNA are cut by the same restriction endonuclease. [True/False]

Answer/Explanation

Answer:
Explaination: True.

---

Directions (Q30 to Q32): Mark the odd one in each of the following groups.
30. Cellulose, Lysozyme, Chitinase, Endonuclease

Answer/Explanation

Answer:
Explaination: Endonuclease.

---

31. Hind III, EcoRI, Sal I, Rop

Answer/Explanation

Answer:
Explaination: Rop.

---

32. Denaturation, Elution, Annealing, Extension

Answer/Explanation

Answer:
Explaination: Elution.

---

33. How has European Federation of Biotechnology (EFB) defined biotechnology?

Answer/Explanation

Answer:
Explaination: The European Federation of Biotechnology defines biotechnology as the integration of natural science and organisms, cells or parts thereof and molecular analogues, for products and services; it encompasses both .traditional and modem molecular biology.

---

34. Name the technique that is used to alter the chemistry of genetic material (DNA/RNA) to obtain the desired result. [Delhi 2016C]

Answer/Explanation

Answer:
Explaination: Genetic engineering.

---

35. What is meant by gene cloning?

Answer/Explanation

Answer:
Explaination: Gene cloning refers to the process in which a numer of identical copies of a gene of interest are made by introducing the gene into an appropriate host using a suitable Vector.

---

36. Why is it not possible for an alien DNA to become part of a chromosome anywhere along its length and replicate normally? [AI 2014]

Answer/Explanation

Answer:
Explaination: The alien DNA has to be linked to a specific sequence of DNA, called origin of replication, on the chromosome; this origin of replication is responsible for initiation of replication.

---

37. Name the specific sequence of DNA in a plasmid that the ‘gene of interest’ ligates with, to enable it to replicate. [AI 2013C]

Answer/Explanation

Answer:
Explaination: Origin of replication (Ori)

---

38. Write the two components of the first artificial recombinant DNA molecule constructed by Cohen and Boyer. [Foreign 2014]

Answer/Explanation

Answer:
Explaination: An antibiotic-resistance gene and the plasmid of Salmonella typhimurium.

---

39. Name the two enzymes that are essential for constructing a recombinant DNA. [Delhi 2017C]

Answer/Explanation

Answer:
Explaination: Restriction endonuclease and DNA ligase.

---

40. In the year 1963, two enzymes responsible for restricting the growth of bacteriophage in E.coli were isolated. How did the enzymes act to restrict the growth of the bacteriophage? [AI 2011C]

Answer/Explanation

Answer:
Explaination:
– One of them (exonucleases) added methyl groups to DNA.
– The other (endonucleases) cut the DNA at specific points within the DNA.

---

41.Name the enzyme that helps to join DNA fragments. [AI 2017C]

Answer/Explanation

Answer:
Explaination: DNA ligase

---

42. Mention the role of Restriction enzyme in Recombinant DNA technology. [Delhi 2017C]

Answer/Explanation

Answer:
Explaination: They act as molecular scissors to cut DNA at specific locations.

---

43. What does ‘R’ stand for, in the restriction endonuclease, EcoRI?

Answer/Explanation

Answer:
Explaination: R stands for RY 13, the strain of E.coli bacterium.

---

44. A plasmid and a DNA sequence in a cell need to be cut for producing recombinant DNA. Name the enzyme that acts as molecular scissors to cut the DNAs.

Answer/Explanation

Answer:
Explaination: Restriction endonucleases.

---

45. Suggest a technique to a researcher who needs to separate fragments of DNA. [Delhi 2016]

Answer/Explanation

Answer:
Explaination: Gel electrophoresis.

---

46. Name the material used as matrix in gel electrophoresis and mention its role. [AI 2014C]

Answer/Explanation

Answer:
Explaination: Agarose; it provides the sieving effect for the DNA to resolve according to their size.

---

47. Why do DNA-fragments move towards the anode during gel electrophoresis? [CBSE 2018C, Delhi 2011C; HOTS]

Answer/Explanation

Answer:
Explaination: DNA-fragments are negatively charged and hence, move towards the anode.

---

48. What is the role of ethidium bromide during agarose-gel electrophoresis of DNA fragments?

Answer/Explanation

Answer:
Explaination: The gel is stained by ethidium bromide, to view the separated DNA bands when exposed to UV light.

---

49. Which main technique and instrument is used to isolate DNA from a plant cell? [CBSE Sample Paper 2013, 2014]

Answer/Explanation

Answer:
Explaination:
– Centrifugation is the technique.
– Centrifuge is the instrument.

---

50. Name two commonly used vectors for genetic engineering.

Answer/Explanation

Answer:
Explaination: Plasmids and bacteriophages.

---

51. Why are engineered vectors preferred these days?

Answer/Explanation

Answer:
Explaination:
(i) Engineered vectors facilitate easy linking of foreign DNA.
(ii) They help in selection of recombinants.

---

52. Mention the uses of cloning vectors in biotechnology. [Delhi 2011]

Answer/Explanation

Answer:
Explaination:
(i) Cloning vectors are used to make multiple copies of the desired DNA/ gene.
(ii) They are used to transfer the gene of interest to the host cell.

---

53. Why is ‘plasmid’ an important tool in biotechnology experiments? [AI2013C]

Answer/Explanation

Answer:
Explaination: Since plasmids can replicate within the bacterial cell independently of the genomic DNA, any alien DNA ligated to it will also multiply, i.e. it is used as a vector as well as in gene cloning.

---

54. Why is it essential to have a ‘selectable marker’ in a cloning vector? [AI 2010; HOTS]

Answer/Explanation

Answer:
Explaination: It helps in identifying the recombinants from the non-recombinants.

---

55. Why are antibiotic-resistance genes used as markers in E.coli? [HOTS]

Answer/Explanation

Answer:
Explaination: It is because the normal E. coli cells do not have resistance to any such antibiotics.

---

56. Name two antibiotic-resistance genes in the pBR 322 of E.coli plasmid.

Answer/Explanation

Answer:
Explaination: Ampicillin-resistance gene and tetracyclin- resistance gene.

---

57. A plasmid without a selectable marker was chosen as vector for cloning a gene. How does this affect the experiment? [HOTS]

Answer/Explanation

Answer:
Explaination: In the absence of a selectable marker, it will not be possible to identify the recombinants from the non-recombinants.

---

58. State what happens when an alien gene is ligated at Sal I site of pBR 322 plasmid. [Delhi 2013C]

Answer/Explanation

Answer:
Explaination: The transformant loses the tetracycline- resistance.

---

59. State what happens when an alien gene is ligated at Pvu I site of pBR 322 plasmid. [AI 2013C]

Answer/Explanation

Answer:
Explaination: The transformant loses ampicillin-resistance.

---

60. How is Agrobacterium tumefaciens able to transform a normal plant cell into a tumour? [Delhi 2013C]
Or
Biotechnologists refer to Agrobacterium tumefaciens as a natural genetic engineer of plants. Give reasons to support the statement. [AI 2011]

Answer/Explanation

Answer:
Explaination: When it infects a plant cell, it delivers a part of its DNA, called ‘T-DNA’ (or Tumour- inducing (Ti) plasmid) into the plant cell and transforms it into a tumour cell.

---

61. How can retroviruses be used efficiently in biotechnology experiments in spite of them being disease causing? [AI 2013C]

Answer/Explanation

Answer:
Explaination: Retroviruses are disarmed (diseases-causing gene is removed/inactivated) and used as vectors to deliver the recombinant/alien DNA into animal cells.

---

62. Name the commonly used vector for transformation in plant cells.

Answer/Explanation

Answer:
Explaination: Agrobacterium tumefaciens.

---

63. How does an alien DNA gain entry into a plant cell by ‘biolistic’ method? [Foreign 2013]

Answer/Explanation

Answer:
Explaination: The cells are bombarded with high velocity micro-particles of gold or tungsten, coated with the alien DNA.

---

64. Mention the type of host cells suitable for the gene guns to introduce an alien DNA. [Delhi 2014]

Answer/Explanation

Answer:
Explaination: Plant cells.

---

65. Name the host cells in which micro-injection technique is used to introduce an alien DNA. [Foreign 2014]

Answer/Explanation

Answer:
Explaination: Animal cells.

---

66. Why is the enzyme cellulase needed for isolating genetic material from plant cells and not from animal cells? [Delhi 2013, 2010; HOTS]

Answer/Explanation

Answer:
Explaination: Cellulase is used to digest the cell wall (cellulose) of plant cells; animal cells have no cell wall and hence, it is not needed.

---

67. How can bacterial DNA be released from the bacterial cell for biotechnology experiments? [Delhi 2011]

Answer/Explanation

Answer:
Explaination: The bacterial cell has to be treated with lysozyme, to digest the cell wall to release DNA.

---

68. To use the restriction enzyme to cut the DNA, the DNA must be in a pine form, i.e. free from proteins and RNAs associated with it. How is it achieved?

Answer/Explanation

Answer:
Explaination:
– Proteins are removed by using proteases.
– RNAs are removed by using ribonucleases (RNases).

---

69. Write the names of the enzymes that are used for isolation of DNA from bacterial and fungal cells, respectively for Recombinant DNA technology. [AI, Foreign 2014]

Answer/Explanation

Answer:
Explaination:
– Lysozymes for bacterial cells.
– Chitinase for fungal cells.

---

70. Write the importance of the bacterium Thermus aquaticus in polymerase chain reaction. [Delhi 2013C]

Answer/Explanation

Answer:
Explaination: It is the source of thermostable DNA- polymerase enzyme.

---

We hope the given Biology MCQs for Class 12 with Answers Chapter 11 Biotechnology: Principles and Processes will help you. If you have any query regarding CBSE Class 12 Biology Biotechnology: Principles and Processes MCQs Pdf, drop a comment below and we will get back to you at the earliest.

Free PDF Download of CBSE Physics Multiple Choice Questions for Class 12 with Answers Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits. Physics MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Semiconductor Electronics: Materials, Devices and Simple Circuits MCQs Pdf with Answers to know their preparation level.

Semiconductor Electronics: Materials, Devices and Simple Circuits Class 12 Physics MCQs Pdf

Class 12 Physics MCQ Question 1. In the given figure V0 is the potential barrier across a p-n junction, when no battery is cdnnected across the junction. [NCERT Exemplar]

(a) 1 and 3 both correspond to forward bias of junction.
(b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction.
(c) 1 corresponds to forward bias and 3 corresponds to reverse bias of junction.
(d) 3 and 1 both correspond to reverse bias of junction.

Answer/Explanation

Answer: b
Explanation:
(b) Height of potential barrier is decreases when p-n junction is forward bias.

---

Physics MCQs Pdf Question 2. In figure given, assuming the diodes to be ideal [NCERT Exemplar]

(a) D₁ is forward biased and D₂ is reverse biased and hence current flows from A to B.
(b) D₂ is forward biased and D₁ is reverse biased and hence no current flows from B to A and vice versa.
(c) D₁ and D₂ are both forward biased and hence current flows from A to B.
(d) D₁ and D₂ are both reverse biased and hence no current flows from A to B and vice versa.

Answer/Explanation

Answer: b
Explaination:
(b) Explanation:
10 V is the lower voltage in the circuit. Now p side of the p-n junction diode D₁ is connected to lower voltage and n- side of D₁ to higher voltage , thus D₁, is reverse biased. In D₂ p-side of p-n junction diode is at higher potential and n-side is at lower potential, therefore D₂ is forward biased. Hence current flows through junction from B to A.

---

3. The output of the given circuit in figure is given below. [NCERT Exemplar]

(a) would be zero at all times.
(b) would be like a half-wave rectifier with positive cycles in output.
(c) would be like a half-wave rectifier with negative cycles in output.
(d) would be like that of a full-wave rectifier.

Answer/Explanation

Answer: c
Explaination:
(c) Explanation:
When the diode is forward biased, the resistance of pn junction diode will be low then current in the circuit is maximum. In this situation, a maximum potential difference will appear across resistance connected in a series of circuit. This result into zero

---

Semiconductor Numericals Class 12 Question 4. Electrical conductivity of a semiconductor
(a) decreases with the rise in its temperature.
(b) increases with the rise in its temperature.
(c) does not change with the rise in its temperature.
(d) first increases and then decreases with the rise in its temperature

Answer/Explanation

Answer:
Explaination:
(b) With temperature rise, the conductivity of semiconductor increases.

---

5. The forbidden energy band gap in conductors, semiconductors and insulators are EG₁,EG₂ and EG₃ respectively. The relation among them is
(a) EG₁ = EG₂ = EG₃
(b) EG₁ < EG₂ < EG₃
(c) EG₁ > EG₂ > EG₃
(d) EG₁ < EG₂ > EG₃

Answer/Explanation

Answer:
Explaination:
(b) In insulators, the forbidden energy gap is very large, in case of semiconductor it is moderate and in conductors the energy . gap is zero.

---

MCQ Physics Class 12 Question 6. In an insulator, the forbidden energy gap between the valence band and the conduction band is of the order of __________ .

Answer/Explanation

Answer:
Explaination: 5 eV

---

Semiconductor Class 12 Question 7. A n-type semiconductor is
(a) negatively charged.
(b) positively charged.
(c) neutral.
(d) none of these

Answer/Explanation

Answer: c
Explaination:
(c) n-type semiconductors are neutral because neutral atoms are added during doping.

---

8. In the half-wave rectifier circuit shown. Which one of the following waveforms is true for V_CD the output across C and D?

Answer/Explanation

Answer: b
Explaination:
(b) Half wave rectifier rectifies only the half cycle of the input ac signal and it blocks the other half.

---

9. A full-wave rectifier circuit along with the input and output voltages is shown in the figure The contribution to output voltage from diode 2 is

(a) A, C
(b) B, D
(c) B, C
(d) A, D

Answer/Explanation

Answer: b
Explaination:
(b) In the positive half cycle of input ac signal diode D₁ is forward biased and D₂ is reverse biased so in the output voltage signal, A and C are due to D₁. In negative half cycle of input ac signal, D2 conducts, hence output signals B and D are due to D₂

---

10. A 220 V AC supply is connected between points A and B (figure). What will be the potential difference V across the capacitor? [NCERT Exemplar]

(a) 200 V
(b) 110 V
(c) 0 V
(d) 220√2 V

Answer/Explanation

Answer: d
Explaination:
(d) As p-n junction diode will conduct during positive half cycle only and during negative half cycle diode is reverse biased. During this diode will not give any output. So potential difference across the capacitor C = peak voltage of the given AC voltage.
V_o=V_rms√2 = 200√2 V.

---

MCQ Semiconductor Physics Question 11. In the circuit shown in figure below, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is [NCERT Exemplar]

(a) 1.3 V
(b) 2.3 V
(c) 0
(d) 0.5 V

Answer/Explanation

Answer: b
Explaination:
(b) Suppose the potential difference be¬ tween A and B is V_AB.
Then,V_AB – 0.3
= [(r₁ + r₂)10³] × (0.2 × 10^-3) [∵ V_AB=ir]
= [(5 + 5)10³] × (0.2 × 10^-3)
= 10 × 10³ × 0.2 × 10^-3 = 2
⇒ V_AB = 2 + 0.3 = 2.3V

---

12. When an electric field is applied across a semicoriductor [NCERT Exemplar]
(a) holes move from lower energy level to higher energy level in the conduction band.
(b) electrons move from higher energy level to lower energy level in the conduction band.
(c) holes in the valence band move from higher energy level to lower energy level.
(d) holes in the valence band move from lower energy level to higher energy level.

Answer/Explanation

Answer: c
Explaination:
(c) When electric field is applied across a semiconductor, the electrons in the con-duction band move from lower energy level to higher energy level. While the holes in valence band move from higher energy level to lower energy level, where they will be having more energy.

---

13. At absolute zero, Si acts as a
(a) metal
(b) semiconductor
(c) insulator
(d) none of these

Answer

Answer: c

---

14. In good conducrors of electricity the type of bonding that exist is
(a) Van der Walls
(b) covalent
(c) ionic
(d) metallic

Answer

Answer: d

---

15. The manifestation of band structure in solids is due to
(a) Heisenberg uncertainty priniciple
(b) Pauli’s exclusion principle
(c) Bohr’s correspondence principle
(d) Boltzmann law

Answer

Answer: b

---

16. The probability of electrons to be found in the conduction band of an intrinsic semiconductor of finite temperature
(a) increases exponentially with increasing band gaP
(b) decreases exponentially with increasing band gap
(c) decreases with increasing temperature.
(d) is independent of the temperature and band gap

Answer

Answer: b

---

17. In an n-type silicon, which of the following statements is true.
(a) Electrons are majority carriers and trivalent atoms are the dopants’
(b) Electrons are minority carriers and pentava- lent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.

Answer

Answer: c

---

18. If a small amount of antimony is added to germanium crystal
(a) its resistance is increased
(b) it becomes a p-type semiconductor
(c) there will be more free electrons than holes in the semiconductor,
(d) none of these.

Answer

Answer: c

---

Semiconductor Electronics Question 19. The dominant mechanism for motion of charge carriers in forward and reverse biased silicon p-n junction are
(a) drift in forward bias, diffusion in reverse bias
(b) diffusion in forward bias, drift in reverse bias
(c) diffusion in both forward and reverse bias
(d) drift in both forward and reverse bias

Answer

Answer: b

---

20. In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them
(b) they move across the junction by the potential difference
(c) hole concentration in p-region is more as compared to u-region.
(d) all of these

Answer

Answer: c

---

21. Region without free electrons and holes in a p-n junction is
(a) n-region
(b) p-region
(c) depletion region
(d) none of these

Answer

Answer: c

---

22. Which of the following statements is incorrect for the depletion region of a diode?
(a) There the mobile charges exist.
(b) Equal number of holes and electrons exist, making the region neutral.
(c) Recombination of holes and electrons has taken place.
(d) None of these

Answer

Answer: a

---

23. Potential barrier developed in a junction diode opposes the flow of
(a) minority carrier in both regions only
(b) majority carriers only
(c) electrons in p region
(d) holes in p region

Answer

Answer: b

---

24. The breakdown in a reverse biased p-n junction diode is more likely to occur due to
(a) large velocity of the minority charge carriers if the doping concentration is small
(b) large velocity of the minority charge carriers if the doping concentration is large
(c) strong electric field in a depletion region if the doping concentration is small
(d) none of these

Answer

Answer: b

---

25. What happens during regulation action of a Zener diode?
(a) The current through the series resistance (Rs) changes.
(b) The resistance offered by the Zener changes.
(c) The Zener resistance is constant.
(d) Both (a) and (b)

Answer

Answer: d

---

26. A zener diode is specified as having a breakdown voltage of 9.1 V, with a maximum power dissipation of 364 mW. What is the maximum current the diode can handle?
(a) 40 mA
(b) 60 mA
(c) 50 mA
(d) 45 mA

Answer

Answer: a

---

27. In a half wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple would be
(a) 25 Hz
(b) 50 Hz
(c) 70.7 Hz
(d) 100 Hz

Answer

Answer: b

---

28. In the circuit shown if current for the diode is 20 μA, the potential difference across the diode is

(a) 2 V
(b) 4.5 V
(c) 4 V
(d) 2.5 V

Answer

Answer: c

---

29. Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (E^)c, (Eg)Si and (Eg)Ge. Which of the following statements is true?
(a) (E_g)_Si < (E_g)_Ge < (E_g)_C
(b) (E_g)_C< (E_g)_Ge < (E_g)_Si
(c) (E_g)_C < (E_g)_Si < (E_g)_Ge
(d) (E_g)_C = (E_g)_Si < (E_g)_Ge

Answer

Answer: c

---

30. If the energy of a photon of sodium light (A = 589 nm) equals the band gap of semiconductor, the minimum energy required to create hole electron pair
(a) 1.1 eV
(b) 2.1 eV
(c) 3.2 eV
(d) 1.5 eV

Answer

Answer: b

---

31. If in a n-type semiconductor when all donor states are filled, then the net charge density in the donor states becomes
(a) 1
(b) > 1
(c) < 1, but not zero
(d) zero

Answer

Answer: b

---

Physics MCQs for Class 11 with Answers Pdf Question 32. The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?

(a) 2.0 A
(b) 1.33 A
(c) 1.73 A
(d) 2.31 A

Answer

Answer: c

---

33. The conductivity of a semiconductor increases with increase in temperature, because
(a) number density of free current carriers increases.
(b) relaxation time increases.
(c) both number density of carriers and relaxation time increase.
(d) number density of carriers increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density. [NCERT Exemplar]

Answer

Answer: d

---

34. What happens during regulation action of a Zener diode? [NCERT Exemplar]
(a) The current in and voltage across the Zener remains fixed.
(b) The current through the series resistance (R_s) does not change.
(c) The Zener resistance is constant.
(d) The resistance offered by the Zener changes.

Answer

Answer: d

---

35. Silicon is a semiconductor. If a small amount of As is added to it, then its electrical conductivity _________ .

Answer/Explanation

Answer:
Explaination: increases

---

36. When the electrical conductivity of a semiconductor is due to the breaking of its covalent bonds, then the semiconductor is said to be ___________ .

Answer/Explanation

Answer:
Explaination: intrinsic

---

37. Why are the elemental dopants mainly taken from 13th and 15th group, for doping Silicon or Germanium?

Answer/Explanation

Answer:
Explaination:
The dopant has to be such that it does not distort the original pure semiconductor lattice. So that the sizes of the dopant and the semiconductor atoms should be nearly the same.

---

38. What is a hole? What is its physical significance?

Answer/Explanation

Answer:
Explaination:
Hole is the vacancy of electron in valence band. The vacancy with the hole behaves as an apparent free particle with effective positive charge.

---

39. Name one impurity each, which when added, to pure Si, produces
(i) n-type, and
(ii)p-type semiconductor.

Answer/Explanation

Answer:
Explaination:
(i) for n-type, arsenic.
(ii) for p-type, Indium.

---

40. Why is the conductivity of n-type semi-conductor greater than that of the p-type semi-conductor even when both of these have same level of doping?

Answer/Explanation

Answer:
Explaination:
In n-type semiconductor charge carriers are electrons and mobility of electrons is more than that of holes.

---

41. Name two factors on which electrical conductivity of a pure semiconductor at a given temperature depends.

Answer/Explanation

Answer:
Explaination:
(i) Band gap
(ii) Biasing.

---

42. What is an ideal diode?

Answer/Explanation

Answer:
Explaination:
It is a p-n junction diode which offer zero resistance in forward biasing and infinite re-sistance in reverse biasing, i.e. current flows through it in one direction only.

---

43. What happens to the width of depletion layer of ap-n junction when it is
(i) forward biased,
(ii) reverse biased? [AI2011]

Answer/Explanation

Answer:
Explaination:
(i) The width of depletion layer decreases.
(ii) The width of depletion layer increases.

---

44. What do you understand by a dynamic resistance of p-n junction diode.

Answer/Explanation

Answer:
Explaination:
Dynamic resistance is the ratio of a small change in voltage ΔV to a small change in current ΔI, i.e.\(\frac{\Delta V}{\Delta I}\).

---

45. Name the junction diode whose I – V characteristics are drawn below. [Delhi 2017]

Answer/Explanation

Answer:
Explaination: Solar cell.

---

46. State the reason, why GaAs is most commonly used in making of a solar cell.

Answer/Explanation

Answer:
Explaination: It has higher absorption coefficient.

---

47. Name the type of biasing of a p-n junction diode so that the junction offers very high resistance.

Answer/Explanation

Answer:
Explaination: Reverse biasing.

---

48. What is internal field emission?

Answer/Explanation

Answer:
Explaination:
The emission of electrons from the host atoms present in the p-n junction due to the high electric field is known as internal field emission or field ionisation.

---

Semiconductor Class Question 49. Why is a typical solar cell drawn in fourth quadrant?

Answer/Explanation

Answer:
Explanation:
I – V characteristics of solar cell is drawn in the fourth quadrant because a solar cell does not draw current but supplies the same to the load.

---

50. Why are materials like CdS or CdSe (E_g ~ 2.4 eV) not used for the fabrication of a solar cell?

Answer/Explanation

Answer:
Explaination:
Materials like CdS or CdSe (E_g ~ 2.4 eV) are not used because they will use only the high energy component of the solar energy for photo-conversion and a significant part of energy will be of no use.

---

We hope the given Physics MCQs for Class 12 with Answers Chapter 13 Semiconductor Electronics: Materials, Devices and Simple Circuits will help you. If you have any query regarding CBSE Class 12 Physics Semiconductor Electronics: Materials, Devices and Simple Circuits MCQs Pdf, drop a comment below and we will get back to you at the earliest.

Free PDF Download of CBSE Physics Multiple Choice Questions for Class 12 with Answers Chapter 13 Nuclei. Physics MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Nuclei MCQs Pdf with Answers to know their preparation level.

Nuclei Class 12 Physics MCQs Pdf

Atoms and Nuclei Class 12 MCQ Question 1. When a β-particle is emitted from a nucleus then its neutron-proton ratio
(a) increases
(b) decreases
(c) remains unchanged.
(d) may increase or decrease depending upon the nucleus.

Answer/Explanation

Answer: b
Explaination:
(b) In P-decay neutron converts to proton with emission of electron and neutrino.

---

Physics MCQs with Answers for Class 12 Question 2. The relation between half-life T_1/2 of a radioactive sample and its mean life x is

Answer/Explanation

Answer: a
Explaination:

---

3. The quantity which is not conserved in a nuclear reaction is
(a) momentum.
(b) charge.
(c) mass.
(d) none of these.

Answer/Explanation

Answer: c
Explaination: (c) Energy equivalent to mass detect is released.

---

4. The half-life of a radioactive nucleus is 3 hours. In 9 hours, its activity will be reduced to a factor of

Answer/Explanation

CBSE Class 12 Physics MCQs with Answer: d
Explaination:

---

MCQs Of Physics 2nd Year with Answers Chapter 13 Question 5. A radioactive element has half-life period 1600 years. After 6400 years what amount will remain?

Answer/Explanation

Answer: b
Explaination:

---

Class 12 Physics MCQs Pdf Question 6. Ratio of the radii of the nuclei with mass numbers 8 and 27 would be

Answer/Explanation

Answer: c
Explaination:

---

7. A radioactive nucleus emits a beta particle. The parent and daughter nuclei are
(a) isotopes
(b) isotones
(c) isomers
(d) isobars

Answer/Explanation

Answer: d
Explaination:
(d) Isobars have the same atomic mass but 1 different atomic number.

---

Physics MCQs for Class 11 Chapter wise Pdf Question 8. In the disintegration series

the values of Z and A respectively will be
(a) 92, 236
(b) 88, 230
(c) 90, 234
(d) 91, 234

Answer/Explanation

Answer: d
Explaination:

---

CBSE Class 12 Physics MCQ Pdf Question 9. A nucleus \(_{Z}^{\mathbf{A}} \mathbf{X}\) emits an α-particle. The resultant nucleus emits a β-particle. The respective atomic and mass numbers of the daughter nucleus will be
(a) Z – 3, A – 4
(b) Z – 1, A – 4
(c) Z – 2, A – 4
(d) Z, A – 2

Answer/Explanation

Answer: b
Explaination:

---

10. In the nuclear reaction

What does X stand for?
(a) Electron
(b) Proton
(c) Neutron
(d) Neutrino

Answer/Explanation

Answer: d
Explaination:
(d) By conservation of mass A = 0, and by conservation of charge Z = 0, Hence X is neutrino.

---

11. The set which represent the isotope, isobar, and isotone respectively is

Answer

Answer: d

---

12. The mass number of iron nucleus is 56 the nuclear density is
(a) 2.29 × 10¹⁶ kg m^-3
(b) 2.29 × 10¹⁷ kg m^-3
(c) 2.29 × 10¹⁸ kg m^-3
(d) 2.29 × 10¹⁵ kg m^-3

Answer

Answer: b

---

13. Order of magnitude of density of uranium nucleus is
(a) 10²⁰ kg m^-3
(b) 10¹⁷ kg m^-3
(c) 10¹⁴ kg m^-3
(d) 10¹¹ kg m^-3

Answer

Answer: b

---

14. The radius of a spherical nucleus as measured by electron scattering is 3.6 fm. What is the mass number of the nucleus most likely to be?
(a) 27
(b) 40
(c) 56
(d) 120

Answer

Answer: a

---

15. The half life of a radioactive susbtance is 30 days. What is the time taken to disintegrate to 3/4^th of its original mass?
(a) 30 days
(b) 15 days
(c) 60 days
(d) 90 days

Answer

Answer: c

---

16. The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an
(a) isomer of parent
(b) isotone of parent
(c) isotope of parent
(d) isobar of parent

Answer

Answer: c

---

17. During negative β-decay, an antineutrino is also emitted along with the emitted electron. Then,
(a) only linear momentum will be conserved
(b) total linear momentum and total angular momentum but not total energy will be conserved
(c) total linear momentum, and total energy but not total angular momentum will conserved
(d) total linear momentum, total angular momentum and total energy will be conserved

Answer

Answer: d

---

18. Consider α and β particles and γ-rays each having an energy of 0.5 MeV. In the increasing order of penetrating power, the radiation are respectively
(a) α, β, γ
(b) α, γ, β
(c) β, γ, α
(d) γ β, α

Answer

Answer: a

---

19. An electron emitted in beta radiation originates from
(a) inner orbits of atom
(b) free electrons existing in the nuclei
(c) decay of a neutron in a nuclei
(d) photon escaping from the nucleus

Answer

Answer: c

---

20. Complete the series 6He → e^– + ⁶Li +
(a) neutrino
(b) antineutrino
(c) proton
(d) neutron

Answer

Answer: b

---

21. An element A decays into an element C by a two step process A → B+ ₂He⁴ and B → C + 2e^–. Then,
(a) A and C are isotopes
(b) A and C are isobars
(c) B and C are isotopes
(d) A and B are isobars

Answer

Answer: a

---

22. The equation 4₁ ¹H⁺ → 2⁴He²⁺ + 2e^– + 26 MeV
represents
(a) β-decay
(b) γ-decay
(c) fusion
(d) fission

Answer

Answer: c

---

23. Light energy emitted by star is due to
(a) breaking of nuclei
(b) joining of nuclei
(c) burning of nuclei
(d) reflection of solar light

Answer

Answer: b

---

24. In nuclear reaction, there is conservation of
(a) mass only
(b) energy only
(c) momentum only
(d) mass, energy and momentum

Answer

Answer: d

---

25. In nuclear reactors, the control rods are made of
(a) cadmium
(b) graphite
(c) krypton
(d) plutonium

Answer

Answer: a

---

26. Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half-life of 1 year. After 1 year, [NCERT Exemplar]
(a) all the containers will have 5000 atoms of the material.
(b) all the containers will contain the same number of atoms of the material but that number will only be approximately 5000.
(c) the containers will in general have different numbers of the atoms of the material but their average will be close to 5000.
(d) none of the containers can have more than 5000 atoms.

Answer

Answer:c

---

27. Jhe gravitational force between a H-atom and another particle of mass m will be given by Newton’s law: [NCERT Exemplar]

= magnitude of the potential energy of electron in the H-atom).

Answer

Answer: b

---

28. When a nucleus in an atom undergoes a radioactive decay, the electronic energy levels of the atom [NCERT Exemplar]
(a) do not change for any type of radioactivity.
(b) change for α and β radioactivity but not for γ-radioactivity.
(c) change for α-radioactivity but not for others.
(d) change for β-radioactivity but not for others.

Answer

Answer: b

---

29. M_x and M_y denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The Q-value for a β^– decay is Q₁ and that for a β⁺ decay is Q₂ If m_e denotes the mass of an electron, then which of the following statements is correct? [NCERT Exemplar]

Answer

Answer: a

---

30. Tritium is an isotope of hydrogen whose nucleus Triton contains 2 neutrons and 1 proton. Free neutrons decay into p + \(\bar{e}+\bar{v}\). If one of the neutrons in Triton decays, it would transform into He³ nucleus. This does not happen. This is because [NCERT Exemplar]
(a) Triton energy is less than that of a He³\ nucleus.
(b) the electron created in the beta decay process cannot remain in the nucleus.
(c) both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a He³ nucleus.
(d) because free neutrons decay due to external perturbations which is absent in a triton nucleus.

Answer

Answer: a

---

31. Heavy stable nuclei have more neutrons than protons. This is because of the fact that [NCERT Exemplar]
(a) neutrons are heavier than protons.
(b) electrostatic force between protons are repulsive.
(c) neutrons decay into protons through beta decay.
(d) nuclear forces between neutrons are weaker than that between protons.

Answer

Answer: b

---

32. Samples of two radioactive nuclides A and B are taken λ_A and λ_B are the disintegration constants of A and B respectively. In which of the following cases, the two samples can simultaneously have the same decay rate at any time?
(a) Initial rate of decay of A is twice the » initial rate of decay of B and λ_A = λ_B.
(b) Initial rate of decay of A is less than the initial rate of decay of B and λ_A < λ_B.
(c) Initial rate of decay of B is twice the initial rate of decay of A and λ_A > λ_B.
(d) Initial rate of decay of B is same as the rate of decay of A at t = 2h and λ_B = λ_A.

Answer

Answer: d

---

33. The variation of decay rate of two radioactive samples A and B with time is shown in figure. Which of the following statements are true?

(a) Decay constant of A is greater than that of B, hence A always decays faster than B.
(b) Decay constant of B is greater than that of A but its decay rate is always smaller than that of A.
(c) Decay constant of A is equal to that ofB.
(d) Decay constant of B is smaller than that of A but still its decay rate becomes equal to that of A at a later instant.

Answer

Answer: d

---

34. Radioactivity is the phenomenon associated with
(a) decay of nucleus.
(b) production of radio waves.
(c) transmission of radio waves.
(d) reception of radio waves.

Answer

Answer: a

---

35. Which of the following are not emitted by radioactive substances?
(a) Electrons
(b) Protons
(c) Gamma rays
(d) Helium nuclei

Answer

Answer: b

---

36. In an α-decay
(a) the parent and daughter nuclei have some number of protons.
(b) the daughter nucleus has one proton more than parent nucleus.
(c) the daughter nucleus has two protons less than parent nucleus.
(d) the daughter nucleus has two nucleus more than parent nucleus.

Answer

Answer: c

---

37. When a radioactive nucleus emits a (β-particle, the mass number of the atom:
(a) increases by one.
(b) remains the same.
(c) decreases by one.
(d) decreases by four.

Answer

Answer: b

---

38. In a β-decay
(a) the parent and daughter nuclei have the same number of protons.
(b) the daughter nucleus has one proton less than parent nucleus.
(c) the daughter nucleus has one proton more than the parent nucleus.
(d) the daughter nucleus has one neutron more than the parent nucleus.

Answer

Answer: c

---

39. P-rays emitted by a radioactive material are
(a) electromagnetic radiations.
(b) electrons orbiting around the nucleus.
(c) neutral particles.
(d) charged particles emitted by nucleus.

Answer

Answer: d

---

40. During a mean life of a radioactive element the fraction that disintegrates is

Answer

Answer: c

---

41. γ-rays are originated from
(a) nucleus.
(b) outermost shell of nucleus.
(c) innermost shell of nucleus.
(d) outermost shell of atom.

Answer

Answer: a

---

42. Binding energy per nucleon of a stable nucleus is
(a) 8 eV
(b) 8 KeV
(c) 8 MeV
(d) 8 BeV

Answer

Answer: c

---

43. Sun’s radiant energy is due to
(a) nuclear fission.
(b) nuclear fusion.
(c) photoelectric effect.
(d) spontaneous radioactive decay.

Answer

Answer: b

---

44. Average binding energy is maximum for
(a) C¹²
(b) Fe⁵⁶
(c) U²³⁵
(d) Po²¹⁰

Answer

Answer: b

---

45. A nucleus undergoes γ-decay due to
(a) excess of protons.
(b) excess of neutrons.
(c) large mass.
(d) its excited state.

Answer

Answer: d

---

46. The decay constant of a radioactive substance is X. Its half-life and mean life, respectively are

Answer

Answer: b

---

47. Neutrino is a particle, which is chargeless and has spin.

Answer/Explanation

Answer:
Explaination: ±\(\frac{1}{2}\)

---

48. Isotones have the same number of _________ .

Answer/Explanation

Answer:
Explaination: neutrons

---

49. Packing fraction of a nucleus is its _________ its per nucleon.

Answer/Explanation

Answer:
Explaination: mass defect

---

50. How is the radius of a nucleus related to its mass number? [Panchkula 2019] [AI2011C]

Answer/Explanation

Answer:
Explaination:
The radius if of a nucleus of mass number A is related as if = R₀A^1/3, where R₀ is a constant.

---

51. Two nuclei have mass numbers in the ratio 27 : 125. What is the ratio of their nuclear radii?

Answer/Explanation

Answer:
Explaination:

---

52. Two nuclei have mass numbers in the ratio 2 : 5. What is the ratio of their nuclear densities?

Answer/Explanation

Answer:
Explaination:
Nuclear density is independent of mass number. So, the ratio will be 1 : 1.

---

53. What is the relation between the binding energy per nucleon and stability of a nucleus?

Answer/Explanation

Answer:
Explaination:
The larger the binding energy per nucleon, the more stable is the nucleus.

---

54. Write any two characteristic properties of nuclear force. [Chennai-2019] [AI 2012,13]

Answer/Explanation

Answer:
Explaination:
The following are the two characteristic properties:
(i) Nuclear force is a short range force.
(ii) Nuclear forces show the saturation effect.

---

55. How is the mean life of a radioactive sample related to its half-life? [Foreign 2011]

Answer/Explanation

Answer:
Explaination:

---

56. Define the activity of a given radioactive substance. Write its SI unit.

Answer/Explanation

Answer:
Explaination:
The rate of disintegration or count rate of sample of radioactive material is called activity. The SI unit of activity is becquerel (Bq).

---

57. The radioactive isotope D decays according to the sequence

If the mass number and atomic number of D₂ are 176 and 71 respectively, what is
(i) the mass number
(ii) atomic number of D?

Answer/Explanation

Answer:
Explaination:

(i) Mass number of D = 180
(ii) Atomic number of D = 72

---

58. A nucleus \(_{n} X^{m}\) emits one a-particle and one β-particle. What is the mass number and atomic number of the product nucleus?

Answer/Explanation

Answer:
Explaination:

Hence, the mass number of product is m – 4, and the atomic number of product is n – 1.

---

59. In both β^– and β⁺ decay processes, the mass number of a nucleus remains same, whereas the atomic number Z increases by one in β^– decay and decreases by one in β⁺ decay. Explain, giving reason. [Foreign 2014]

Answer/Explanation

Answer:
Explaination:
In β^– decay, one neutron inside the nucleus decays into one proton and one electron (β^–). The proton remains inside the nucleus and the electron is ejected out.

In β⁺ decay, the conversion of proton into neutron and position (β⁺) takes place.

---

60. Which nucleus has greater mean life, A or B?

Answer/Explanation

Answer:
Explaination:

Since, slope of A is more than slope of B.
Therefore, X is high and mean life e = \(\frac{1}{\lambda}\) for A is small.

---

61. Why is a free neutron unstable but a free proton is a stable particle?

Answer/Explanation

Answer:
Explaination:
A free neutron is unstable outside the nucleus with an average life of 1000 s. It decays into a proton and emits a β^– particle, i.e.

---

62. A neutron strikes a nucleus of \(_{5}^{\mathbf{10}} \mathbf{B}\) and emits an alpha particle. Write down the nuclear reaction for it.

Answer/Explanation

Answer:
Explaination:

---

63. Write the necessary condition required for fusion reaction.

Answer/Explanation

Answer:
Explaination:
(i) Nuclear fusion will occur when the kinetic energy of colliding nuclei is enough to overcome the strong electrostatic forces of repulsion between the protons. For this, high temperature is required.
(ii) The density of nuclei should also be very high to increase the number of collisions.

---

64. Out of \(_{14}^{30} X ;_{3}^{6} Y \text { and }_{40} Z^{130}\), which is more likely to undergo the nuclear fusion?

Answer/Explanation

Answer:
Explaination:
A lighter unstable nucleus \(_{3}^{\mathbf{6}} \mathbf{Y}\) can undergo the nuclear fusion.

---

65. What is the effect of temperature on radioactivity?

Answer/Explanation

Answer:
Explaination:
No effect. Radioactivity is independent of temperature.

---

66. What is the difference between an electron and a β-particle?

Answer/Explanation

Answer:
Explaination:
An electron resides outside the nucleus, whereas β-particle is an electron like particle of nuclear origin.

---

67. What is the source of stellar energy?

Answer/Explanation

Answer:
Explaination: Nuclear fusion reactions.

---

68. Four nuclei of an element fuse together to form a heavier nucleus. If the process is accompanied by the release of energy, which of the two—the parent or the daughter nucleus would have a higher binding energy/ nucleons? [CBSE 2018]

Answer/Explanation

Answer:
Explaination: Daughter nucleus.

---

69. You are given two nuelei \(_{3}^{7} X \text { and }_{3}^{4} Y\), which one of the two is likely to be more stable? Give reason.

Answer/Explanation

Answer:
Explaination: \(_{3}^{\mathbf{7}} \mathbf{X}\) is more stable, as it contains more neutrons than protons.

---

70. The \(_{10}^{\mathbf{23}} \mathbf{Ne}\) decays by β^– emission into \(_{11}^{\mathbf{23}} \mathbf{Na}\). Write down the β^– decay equation.

Answer/Explanation

Answer:
Explaination:

---

We hope the given Physics MCQs for Class 12 with Answers Chapter 13 Nuclei will help you. If you have any query regarding CBSE Class 12 Physics Nuclei MCQs Pdf, drop a comment below and we will get back to you at the earliest.

Free PDF Download of CBSE Physics Multiple Choice Questions for Class 12 with Answers Chapter 12 Atoms. Physics MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Atoms MCQs Pdf with Answers to know their preparation level.

Atoms Class 12 Physics MCQs Pdf

1. If 13.6 eV energy is required to ionise the hydrogen atom, then energy required to remove an electron from n = 2 is
(a) 10.2 eV
(b) 0 eV
(c) 3.4 eV
(d) 6.8 eV.

Answer/Explanation

Answer: c
Explaination:

---

2. In Bohr’s model of an atom which of the following is an integral multiple of \(\frac{h}{2 \pi}\)?
(a) Kinetic energy
(b) Radius of an atom
(c) Potential energy
(d) Angular momentum

Answer/Explanation

Answer:
Explaination:
(d) Angular momentum L = mvr = \(\frac{n h}{2 \pi}\)

---

3. In Bohr’s model, the atomic radius of the first orbit is rQ. Then, the radius of the third orbit is
(a) r₀/9
(b) r₀
(c) 9r₀
(d) 3r₀

Answer/Explanation

Answer: c
Explaination:
r_n = r_nn²
∴ r₃ = 9r₀

---

4. The K.E. of the electron in an orbit of radius r in hydrogen atom is proportional to

Answer/Explanation

Answer: b
Explaination:
(b) \(\frac{e^{2}}{2 r}\), Scince K.E = \(\frac{k e^{2}}{2 r}\)

---

5. The ratio between Bohr radii is
(a) 1 : 2 : 3
(b) 2 : 4 : 6
(c) 1 : 4 : 9
(d) 1 : 3 : 5

Answer/Explanation

Answer: c
Explaination:
(c) 1 : 4 : 9, In Bohr’s atomic model, r_n n²

---

6. The longest wavelength in Balmer series of hydrogen spectrum will be
(a) 6557 Å
(b) 1216 Å
(c) 4800 Å
(d) 5600 Å

Answer/Explanation

Answer: a
Explaination:
(a) 6557 Å
For longest wavelength in Balmer series n₁ = 2 and n₂ = 3

---

7. In terms of Rydberg constant R, the wave number of the first Balmer line is
(a) R
(b) 3R
(c) \(\frac{5 R}{36}\)
(d) \(\frac{8 R}{9}\)

Answer/Explanation

Answer: c
Explaination:

---

8. The ionisation energy of hydrogen atom is 13.6 eV. Following Bohr’s theory the energy corresponding to a transition between 3rd and 4th orbits is
(a) 3.40 eV
(b) 1.51 eV
(c) 0.85 eV
(d) 0.66 eV

Answer/Explanation

Answer: d
Explaination:

---

9. The energy of hydrogen atom in the nth orbit is En, then the energy in the nth orbit of single ionised helium atom is

Answer/Explanation

Answer: c
Explaination:
(c) As energy E ∝ Z²
For hydrogen atom Z = 1,
for Helium Z = 2
E_He = 4E_n.

---

10. On moving up in the energy states of a H-like atom, the energy difference between two consecutive energy states
(a) decreases.
(b) increases.
(c) first decreases then increases.
(d) first increases then decreases.

Answer/Explanation

Answer: a
Explaination:
(a) As, E_n = \(\frac{-13.6}{n^{2}}\)

---

11. The transition of electron from n = 4, 5, 6, ………. to n = 3 corresponds to
(a) Lyman series
(b) Balmer series
(c) Paschen series
(d) Brackettseries

Answer/Explanation

Answer: c
Explaination:
(c) In transition from n₁ = 3 and n₂ = 4, 5, 6,….
Infrared radiation of Paschen spectral is emitted.

---

12. As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of double ionized Li atom (Z = 3) is
(a) 1.51 eV
(b) 13.6 eV
(c) 40.8 eV
(d) 122.4 eV

Answer/Explanation

Answer: d
Explaination:
(d) Since energy of electron in nth state of hydrogen like atom is

---

13. Which of the following spectral series in hydrogen atom gives spectral line of 4860 A?
(a) Lyman
(b) Balmer
(c) Paschen
(d) Brackett

Answer/Explanation

Answer: b
Explaination:
(b) Since spectral line of wavelength 4860 A
lies in the visible region of the spectrum which is Balmer series of the spectrum.

---

14. In terms of Rydberg constant R, the shortest wavelength in Balmer series of hydrogen atom spectrum will have wavelength

Answer/Explanation

Answer: b
Explaination:
(b) For shortest wavelength n₁ =∞, n₂ = 2

---

15. The first model of atom in 1898 was proposed by
(a) Ernst Rutherford
(b) Albert Einstein
(c) J.J. Thomson
(d) Niels Bohr

Answer

Answer: c

---

16. In Geiger-Marsden scattering experiment, the trajectory traced by an a-particle depends on
(a) number of collision
(b) number of scattered a-particles
(c) impact parameter
(d) none of these

Answer

Answer: c

---

17. In the Geiger-Marsden scattering experiment the number of scattered particles detected are maximum and minimum at the scattering angles respectively at
(a) 0° and 180°
(b) 180° and 0°
(c) 90° and 180°
(d) 45° and 90°

Answer

Answer: a

---

18. In the Geiger-Marsden scattering experiment, is case of head-on collision the impact parameter should be
(a) maximum
(b) minimum
(c) infinite
(d) zero

Answer

Answer: a

---

19. Rutherford’s experiments suggested that the size of the nucleus is about
(a) 10^-14 m to 10^-12 m
(b) 10^-15 m to 10^-13 m
(c) 10^-15 m to 10^-14 m
(d) 10^-15 m to 10^-12 m

Answer

Answer: c

---

20. Which of the following spectral series falls within the visible range of electromagnetic radiation?
(a) Lyman series
(b) Balmer series
(c) Paschen seriee
(d) Pfund series

Answer

Answer: b

---

21. The first spectral series was discovered by
(a) Balmer
(b) Lyman
(c) Paschen
(d) Pfund

Answer

Answer: a

---

22. Which of the following postulates of the Bohr model led to the quantization of energy of the hydrogen atom?
(a) The electron goes around the nucleus in circular orbits.
(b) The angular momentum of the electron can only be an integral multiple of h/2π.
(c) The magnitude of the linear momentum of the electron is quantized.
(d) Quantization of energy is itself a postulate of the Bohr model.

Answer

Answer: b

---

23. The Bohr model of atoms
(a) assumes that the angular momentum of elec-trons is quantized.
(b) uses Einstein’s photoelectric equation.
(c) predicts continuous emission spectra for at-oms.
(d) predicts the same emission spectra for all types of atoms.

Answer

Answer: a

---

24. If tt is the orbit number of the electron in a hydrogen atom, the correct statement among the following is
(a) electron energy increases as n increases.
(b) hydrogen emits infrared rays for the electron transition from n = to n = 1
(c) electron energy is zero for n = 1 (<0 electron energy varies as n2.

Answer

Answer: a

---

25. If the radius of inner most electronic orbit of a hydrogen atom is 5.3 * 10~n m, then the radii of n = 2 orbits is
(a) 1.12 Å
(b) 2.12 Å
(c) 3.22 Å
(d) 4.54 Å

Answer

Answer: b

---

26. The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

(a) I
(b) II
(c) III
(d) IV

Answer

Answer: c

---

27. In a hydrogen atom, the radius of n^th Bohr orbit is r_n. The graph between log(r_n/r₁) and log n will be

Answer

Answer: a

---

28. The transition from the state n = 5 to n = 1 in a hydrogen atom results in UV radiation. Infrared radiation will be obtained in the transition
(a) 2 → 1
(b) 3 → 2
(c) 4 → 3
(d) 6 → 2

Answer

Answer: c

---

29. The hydrogen atom can give spectral lines in the Lyman, Balmer and Paschen series. Which of the following statement is correct?
(a) Lyman series is in the infrared region.
(b) Balmer series is in the visible region.
(c) Paschen series is in the visible region.
(d) Balmer series is in the ultraviolet region.

Answer

Answer: b

---

30. Which of the relation is correct between time period and number of orbits while an electron is resolving in an orbit?

Answer

Answer: c

---

31. Energy of an electron in the second orbit of hydrogen atom is E and the energy of electron in 3rd orbit of He will be

Answer/Explanation

Answer: b
Explaination:

---

32. The spectral lines in the Brackett series arise due to transition of electron in hydrogen atom from higher orbits to the orbit with
(a) n = 1
(b) n = 2
(c) n = 3
(d) n = 4

Answer

Answer: d

---

33. Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its ground state, on the basis of Bohr’s model, will be about [NCERT Exemplar]
(a) 53 pm
(b) 27 pm
(c) 18 pm
(d) 13 pm

Answer

Answer: c

---

34. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because [NCERT Exemplar]
(a) of the electrons not being subject to a central force.
(b) of the electrons colliding with each other.
(c) of screening effects.
(d) the force between the nucleus and an electron will no longer be given by Coulomb’s law.

Answer

Answer: a

---

35. For the ground state, the electron in the H-atom has an angular momentum = h, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true, [NCERT Exemplar]
(a) because Bohr model gives incorrect values of angular momentum.
(b) because only one of these would have a minimum energy.
(c) angular momentum must be in the direction of spin of electron.
(d) because electrons go around only in horizontal orbits.

Answer

Answer: a

---

36. 02 molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms [NCERT Exemplar]
(a) is not important because nuclear forces are short-ranged.
(b) is as important as electrostatic force for binding the two atoms.
(c) cancels the repulsive electrostatic force between the nuclei.
(d) is not important because oxygen nucleus has equal number of neutrons and protons.

Answer

Answer: a

---

37. Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is [NCERT Exemplar]
(a) 10.20 eV
(b) 20.40 eV
(c) 13.6 eV
(d) 27.2 eV

Answer

Answer: a

---

38. A set of atoms in an excited state decays. [NCERT Exemplar]
(a) in general to any of the states with lower energy.
(b) into a lower state only when excited by an external electric field.
(c) all together simultaneously into a lower state.
(d) to emit photons only when they collide.

Answer

Answer: a

---

39. An ionised H-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state, [NCERT Exemplar]
(a) the electron would move in circular orbits.
(b) the energy would be (2)⁴ times that of a H-atom.
(c) the electrons, orbit would go around the protons.
(d) the molecule will soon decay in a proton and a H-atom.

Answer

Answer: c

---

40. The Balmer series for the H-atom can be observed
(a) if we measure the frequencies of light emitted when an excited atom falls to the ground state.
(b) if we measure the frequencies of light emitted due to transitions between excited states and the first excited state.
(c) in any transition in a H-atom.
(d) as a sequence of frequencies with the lower frequencies getting closely packed.

Answer

Answer: b

---

41. Let \(E_{n}=\frac{-1}{8 \varepsilon_{0}^{2}} \frac{m e^{4}}{n^{2} h^{2}}\) be the energy of the nth level of H-atom. If all the H-atoms are in the ground state and radiation of frequency (E₂ – E₁)/h falls on it,
(a) it will not be absorbed at all.
(b) some of atoms will move to the first excited state.
(c) all atoms will be excited to the n = 2 state.
(d) all atoms will make a transition to the n = 3 state.

Answer

Answer: d

---

42. The Bohr model of an atom
(a) assumes that the angular momentum of electrons is quantised.
(b) uses Einstein’s potoelectric equation.
(c) predicts continuous emission spectra for atoms,
(d) predicts the same emission spectra for all types of atoms.

Answer

Answer: a

---

43. For ionising an exicited hydrogan atom, the energy required (in eV) will be
(a) a little less than 13.6 eV
(b) 13.6 eV
(c) more than 13.6 eV
(d) 3.4 or less

Answer/Explanation

Answer: d
Explaination:
(d) As the energy of the electron is – -3.4 eV in first excited state and magnitude is less for higher excited state.

---

44. The electrons in the Bohr’s orbit have
(a) K.E. greater than P.E.
(b) P.E. greater than K.E.
(c) the same values
(d) none of these

Answer/Explanation

Answer: a
Explaination:

---

45. The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is
\(B=-\frac{m e^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}\). (m = electron mass)
If one decides to work in a frame of reference where the electron is at rest, the proton would be moving arround it. By similar arguments, the binding energy would be
\(B=-\frac{M e^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}\)(M= proton mass)
This last expression is not correct because [NCERT Exemplar]
(a) n would not be integral.
(b) Bohr-quantisation applies only to electron.
(c) the frame in which the electron is at rest is not inertial.
(d) the motion of the proton would not be in circular orbits, even approximately.

Answer

Answer: c

---

46. Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom,
(a) because of energy loss.
(b) without simultaneously releasing energy in the from of radiation.
(c) because of momentum conservation.
(d) because of angular momentum conservation.

Answer

Answer: b

---

47. The Bohr model for the spectra of a H-atom
(a) will be applicable to hydrogen in the molecular from.
(b) will not be applicable as it is for a He- atom.
(c) is valid only at room temperature.
(d) predicts continuous as well as discrete spectral lines.

Answer

Answer: b

---

48. Non-radiating electron orbits in an atom are called __________ orbits.

Answer/Explanation

Answer:
Explaination: stationary

---

49. At distance of closest approach, kinetic energy of a-particle is __________ .

Answer/Explanation

Answer:
Explaination: zero

---

50. The centripetal force required for revolution of electron in an orbit is provided by __________ between the electron and the nucleus.

Answer/Explanation

Answer:
Explaination: electrostatic attraction

---

51. __________ is the perpendicular distance of the velocity vector of the a-particle from the centre of the nucleus.

Answer/Explanation

Answer:
Explaination: Impact parameter

---

52. The number of waves, contained in unit length of the medium is called __________ .

Answer/Explanation

Answer:
Explaination: wave number

---

53. Angular momentum and energy of an electron in an atom is __________ .

Answer/Explanation

Answer:
Explaination: quantised.

---

54. Number of possible spectral lines emitted on dexcitation of electron from energy level n to ground state is equal to __________ .

Answer/Explanation

Answer:
Explaination: \(\frac{n(n-1)}{2}\)

---

55. In the Rutherford scattering experiment, die distance of closest approach for an a-particle is dQ. If an a-particle is replaced by a proton, how much kinetic energy in comparison to a-particle will it require to have the same distance of closest approach d₀?

Answer/Explanation

Answer:
Explaination:
At distance of closest approach, the K.E. with the charged particle is converted into electrostatic P.E.
As q is half with a proton in comparison to a-particle, for same d, energy E has to be made half.

---

56. What is the ratio of radii of the orbits corresponding to first excited state and ground state in a hydrogen atom?

Answer/Explanation

Answer:
Explaination:

---

57. Write the expression for Bohr’s radius in hydrogen atom.

Answer/Explanation

Answer:
Explaination:
Bohr’s radius in hydrogen atom,

---

58. Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its
(i) second permitted energy level to the first level, and
(ii) the highest permitted energy level to the first permitted level.

Answer/Explanation

Answer:
Explaination:

---

59. What is the distance of closest approach?

Answer/Explanation

Answer:
Explaination:
The minimum distance up to which an energetic a-particle travelling directly towards a nucleus can reach.

---

60. An electron in a hydrogen atom is revolving round a positively charged nucleus. Which two physical quantities explain the orbit of an electron?

Answer/Explanation

Answer:
Explaination:
Two physical quantities are:
(i) angular momentum, and
(ii) total energy of electron.

---

61. What will happen if an electron instead of revolving becomes stationary in H-atom?

Answer/Explanation

Answer:
Explaination:
Then the electrostatic field of the nucleus will attract the electron into the nucleus itself.

---

62. Calculate the speed of electron revolving around the nucleus of a hydrogen atom in order that it may not be pulled into the nucleus by electrostatic attraction.

Answer/Explanation

Answer:
Explaination:
It is only possible when the centripetal force is equal to electrostatic force of attraction.

---

63. What is the value of ionization energy for a hydrogen atom?

Answer/Explanation

Answer:
Explaination: 13.6

---

We hope the given Physics MCQs for Class 12 with Answers Chapter 12 Atoms will help you. If you have any query regarding CBSE Class 12 Physics Atoms MCQs Pdf, drop a comment below and we will get back to you at the earliest.