JEE Physics Nuclei MCQs Set 22

Multiple Choice Questions (MCQs) for JEE Physics: Nuclei

Explore reliable objective questions for Nuclei tailored for JEE learners. Utilizing these Physics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Practice Nuclei MCQs for JEE Physics

View or download the dedicated Nuclei MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Physics evaluations.

Multiple Choice Questions

Question 1: A radionuclide with half-life 693.1 days emits beta particles of average kinetic energy \( 8.4\times10^{-14} \) J and is used in a machine that generates electrical energy with efficiency 12.6 percent. The number of moles of the nuclide needed to generate electrical power of 441 kW initially is \( n\times10^m \) (\( N_A=6.023\times10^{23} \)). The value of n/m is:
(a) 3
(b) 1
(c) 4
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. Initial decay rate needed \( R=\frac{441\times10^3}{0.126\times8.4\times10^{-14}}=4.17\times10^{19}\ \mathrm{s^{-1}} \).
2. \( \lambda=\frac{0.6931}{693.1\times86400}=1.157\times10^{-8}\ \mathrm{s^{-1}} \), so \( N=R/\lambda=3.6\times10^{27} \).
3. Moles \( =N/N_A\approx6.0\times10^3 \), so n = 6, m = 3 and n/m = 2.

Teacher's Note:
1. T
2. h
3. e
4.
5. n
6. e
7. e
8. d
9. e
10. d
11.
12. a
13. c
14. t
15. i
16. v
17. i
18. t
19. y
20.
21. f
22. o
23. l
24. l
25. o
26. w
27. s
28.
29. f
30. r
31. o
32. m
33.
34. t
35. h
36. e
37.
38. o
39. u
40. t
41. p
42. u
43. t
44.
45. p
46. o
47. w
48. e
49. r
50. ,
51.
52. t
53. h
54. e
55.
56. e
57. n
58. e
59. r
60. g
61. y
62.
63. p
64. e
65. r
66.
67. d
68. e
69. c
70. a
71. y
72.
73. a
74. n
75. d
76.
77. t
78. h
79. e
80.
81. e
82. f
83. f
84. i
85. c
86. i
87. e
88. n
89. c
90. y
91. .

Question 2: The heavier stable nuclei tend to have a larger N/Z ratio because: (A) a neutron is heavier than a proton, (B) a neutron is an unstable particle, (C) a neutron does not exert electric repulsion, (D) Coulomb forces have longer range than nuclear forces. Select the correct combination.
(a) C and D only
(b) A and B only
(c) A and C only
(d) B and D only
Show Answer & Explanation

Answer: (a) C and D only

Explanation:
1. Protons repel each other by the long-range Coulomb force, which grows faster with size than the short-range nuclear attraction.
2. Extra neutrons add nuclear attraction without adding repulsion, so heavier nuclei need more neutrons per proton.
3. So (C) and (D) are the reasons.

Teacher's Note:
1. T
2. h
3. e
4.
5. s
6. u
7. r
8. p
9. l
10. u
11. s
12.
13. o
14. f
15.
16. n
17. e
18. u
19. t
20. r
21. o
22. n
23. s
24.
25. i
26. n
27.
28. h
29. e
30. a
31. v
32. y
33.
34. n
35. u
36. c
37. l
38. e
39. i
40.
41. i
42. s
43.
44. t
45. h
46. e
47.
48. n
49. u
50. c
51. l
52. e
53. u
54. s
55. '
56. s
57.
58. w
59. a
60. y
61.
62. o
63. f
64.
65. b
66. a
67. l
68. a
69. n
70. c
71. i
72. n
73. g
74.
75. t
76. h
77. e
78.
79. g
80. r
81. o
82. w
83. i
84. n
85. g
86.
87. C
88. o
89. u
90. l
91. o
92. m
93. b
94.
95. r
96. e
97. p
98. u
99. l
100. s
101. i
102. o
103. n
104. .

Question 3: When a \( \beta^- \) particle is emitted from a nucleus, the neutron-proton ratio:
(a) increases
(b) decreases
(c) remains the same
(d) first decreases then increases
Show Answer & Explanation

Answer: (b) decreases

Explanation:
1. In beta-minus decay a neutron changes to a proton: N decreases by 1 and Z increases by 1.
2. So N/Z decreases.

Teacher's Note:
1. B
2. e
3. t
4. a
5. -
6. m
7. i
8. n
9. u
10. s
11.
12. e
13. m
14. i
15. s
16. s
17. i
18. o
19. n
20.
21. m
22. o
23. v
24. e
25. s
26.
27. a
28.
29. n
30. e
31. u
32. t
33. r
34. o
35. n
36. -
37. r
38. i
39. c
40. h
41.
42. n
43. u
44. c
45. l
46. e
47. u
48. s
49.
50. t
51. o
52. w
53. a
54. r
55. d
56. s
57.
58. s
59. t
60. a
61. b
62. i
63. l
64. i
65. t
66. y
67. .

Question 4: The mass number of a nucleus is:
(a) always less than its atomic number
(b) always more than its atomic number
(c) sometimes more than and sometimes equal to its atomic number
(d) equal to its atomic number
Show Answer & Explanation

Answer: (c) sometimes more than and sometimes equal to its atomic number

Explanation:
1. Mass number \( A=Z+N \), where N is the number of neutrons.
2. Ordinary hydrogen \( {}^{1}_{1}\mathrm{H} \) has no neutron, so \( A=Z \); every other nucleus has \( A>Z \).

Teacher's Note:
1. A
2.
3. c
4. o
5. u
6. n
7. t
8. s
9.
10. a
11. l
12. l
13.
14. n
15. u
16. c
17. l
18. e
19. o
20. n
21. s
22.
23. w
24. h
25. i
26. l
27. e
28.
29. Z
30.
31. c
32. o
33. u
34. n
35. t
36. s
37.
38. o
39. n
40. l
41. y
42.
43. p
44. r
45. o
46. t
47. o
48. n
49. s
50. .
51.
52. S
53. o
54.
55. A
56.
57. i
58. s
59.
60. n
61. e
62. v
63. e
64. r
65.
66. s
67. m
68. a
69. l
70. l
71. e
72. r
73.
74. t
75. h
76. a
77. n
78.
79. Z
80. ,
81.
82. a
83. n
84. d
85.
86. e
87. q
88. u
89. a
90. l
91. s
92.
93. i
94. t
95.
96. o
97. n
98. l
99. y
100.
101. f
102. o
103. r
104.
105. h
106. y
107. d
108. r
109. o
110. g
111. e
112. n
113. -
114. 1
115. .

Question 5: The energy liberated in the beta decay of a free neutron \( n\to p+e^-+\bar\nu \) is nearly (atomic mass of \( {}^1_1H=1.007825 \) u, neutron 1.008665 u; 1 u = 931 MeV):
(a) 1.2 MeV
(b) 0.51 MeV
(c) 7.8 MeV
(d) 0.78 MeV
Show Answer & Explanation

Answer: (d) 0.78 MeV

Explanation:
1. Using atomic mass of hydrogen includes the electron mass: \( \Delta m=1.008665-1.007825=0.000840 \) u.
2. Energy \( =0.00084\times931=0.78 \) MeV.

Teacher's Note:
1. T
2. h
3. e
4.
5. n
6. e
7. u
8. t
9. r
10. o
11. n
12.
13. i
14. s
15.
16. s
17. l
18. i
19. g
20. h
21. t
22. l
23. y
24.
25. h
26. e
27. a
28. v
29. i
30. e
31. r
32.
33. t
34. h
35. a
36. n
37.
38. a
39.
40. h
41. y
42. d
43. r
44. o
45. g
46. e
47. n
48.
49. a
50. t
51. o
52. m
53. ,
54.
55. w
56. h
57. i
58. c
59. h
60.
61. i
62. s
63.
64. w
65. h
66. y
67.
68. i
69. t
70.
71. c
72. a
73. n
74.
75. d
76. e
77. c
78. a
79. y
80. .

Question 6: The decay constant of a radioactive substance is 0.173 per year (take \( \ln2=0.692 \)). Which statements are correct? (A) Nearly 63 percent will decay in (1/0.173) year, (B) the half-life is (1/0.173) year, (C) one-fourth of the substance is left after nearly 8 years, (D) half of the substance will decay in one average life. Select the correct combination.
(a) A and C only
(b) B and D only
(c) A and D only
(d) B and C only
Show Answer & Explanation

Answer: (a) A and C only

Explanation:
1. Mean life \( =1/0.173 \) years; in one mean life \( 1-e^{-1}=63\% \) decays, so (A) is true and (D) false.
2. Half-life \( =0.692/0.173=4 \) years, so (B) is false.
3. 8 years is two half-lives, leaving one-fourth: (C) is true.

Teacher's Note:
1. M
2. e
3. a
4. n
5.
6. l
7. i
8. f
9. e
10.
11. i
12. s
13.
14. t
15. h
16. e
17.
18. r
19. e
20. c
21. i
22. p
23. r
24. o
25. c
26. a
27. l
28.
29. o
30. f
31.
32. t
33. h
34. e
35.
36. d
37. e
38. c
39. a
40. y
41.
42. c
43. o
44. n
45. s
46. t
47. a
48. n
49. t
50. ;
51.
52. h
53. a
54. l
55. f
56. -
57. l
58. i
59. f
60. e
61.
62. i
63. s
64.
65. 0
66. .
67. 6
68. 9
69. 3
70.
71. t
72. i
73. m
74. e
75. s
76.
77. t
78. h
79. e
80.
81. m
82. e
83. a
84. n
85.
86. l
87. i
88. f
89. e
90. .

Question 7: Match the following. Column I: (a) Photoelectric effect, (b) Wave, (c) X rays, (d) Nucleus. Column II: I. Photon, II. Frequency, III. K capture, IV. gamma rays. The correct matching is:
(a) a-II, b-I, c-III, d-IV
(b) a-I, b-II, c-III, d-IV
(c) a-I, b-II, c-IV, d-III
(d) a-III, b-II, c-I, d-IV
Show Answer & Explanation

Answer: (b) a-I, b-II, c-III, d-IV

Explanation:
1. The photoelectric effect is explained through photons (I).
2. A wave is characterised by frequency (II).
3. K capture produces characteristic X rays (III).
4. Gamma rays arise from nuclear transitions (IV).

Teacher's Note:
1. M
2. a
3. t
4. c
5. h
6.
7. e
8. a
9. c
10. h
11.
12. C
13. o
14. l
15. u
16. m
17. n
18.
19. I
20.
21. i
22. t
23. e
24. m
25.
26. w
27. i
28. t
29. h
30.
31. i
32. t
33. s
34.
35. c
36. l
37. o
38. s
39. e
40. s
41. t
42.
43. a
44. s
45. s
46. o
47. c
48. i
49. a
50. t
51. e
52. d
53.
54. c
55. o
56. n
57. c
58. e
59. p
60. t
61. ;
62.
63. h
64. e
65. r
66. e
67.
68. t
69. h
70. e
71.
72. p
73. a
74. i
75. r
76. s
77.
78. f
79. o
80. l
81. l
82. o
83. w
84.
85. t
86. h
87. e
88.
89. o
90. r
91. d
92. e
93. r
94.
95. o
96. f
97.
98. t
99. h
100. e
101.
102. l
103. i
104. s
105. t
106. .

Question 8: A freshly prepared radioactive source of half-life 2 h emits radiation of intensity 64 times the permissible safe level. The minimum time after which it is safe to work with the source is:
(a) 6 h
(b) 24 h
(c) 12 h
(d) 128 h
Show Answer & Explanation

Answer: (c) 12 h

Explanation:
1. Intensity must fall by a factor 64 = \( 2^6 \).
2. That needs 6 half-lives: \( 6\times2=12 \) h.

Teacher's Note:
1. C
2. o
3. u
4. n
5. t
6.
7. h
8. o
9. w
10.
11. m
12. a
13. n
14. y
15.
16. h
17. a
18. l
19. v
20. i
21. n
22. g
23. s
24.
25. a
26. r
27. e
28.
29. n
30. e
31. e
32. d
33. e
34. d
35.
36. t
37. o
38.
39. r
40. e
41. d
42. u
43. c
44. e
45.
46. 6
47. 4
48.
49. t
50. i
51. m
52. e
53. s
54.
55. t
56. o
57.
58. 1
59. ,
60.
61. t
62. h
63. e
64. n
65.
66. m
67. u
68. l
69. t
70. i
71. p
72. l
73. y
74.
75. b
76. y
77.
78. t
79. h
80. e
81.
82. h
83. a
84. l
85. f
86. -
87. l
88. i
89. f
90. e
91. .

Question 9: The specific activities (activity per gram) of \( {}^{24}_{11}\mathrm{Na} \) (half-life 15 h) and \( {}^{235}_{92}\mathrm{U} \) (half-life \( 7.1\times10^8 \) y) are respectively nearly:
(a) \( 7.9\times10^{4}\ \mathrm{Bq/g} \) and \( 3.2\times10^{17}\ \mathrm{Bq/g} \)
(b) \( 3.2\times10^{14}\ \mathrm{Bq/g} \) and \( 7.9\times10^{1}\ \mathrm{Bq/g} \)
(c) \( 3.2\times10^{17}\ \mathrm{Bq/g} \) and \( 7.9\times10^{7}\ \mathrm{Bq/g} \)
(d) \( 3.2\times10^{17}\ \mathrm{Bq/g} \) and \( 7.9\times10^{4}\ \mathrm{Bq/g} \)
Show Answer & Explanation

Answer: (d) \( 3.2\times10^{17}\ \mathrm{Bq/g} \) and \( 7.9\times10^{4}\ \mathrm{Bq/g} \)

Explanation:
1. Specific activity \( =\lambda\frac{N_A}{M} \).
2. Na-24: \( \lambda=1.28\times10^{-5}\ \mathrm{s^{-1}} \), \( N/g=2.51\times10^{22} \) giving \( 3.2\times10^{17} \) Bq/g.
3. U-235: \( \lambda=3.09\times10^{-17}\ \mathrm{s^{-1}} \), \( N/g=2.56\times10^{21} \) giving \( 7.9\times10^{4} \) Bq/g.

Teacher's Note:
1. S
2. h
3. o
4. r
5. t
6. -
7. l
8. i
9. v
10. e
11. d
12.
13. n
14. u
15. c
16. l
17. i
18. d
19. e
20. s
21.
22. a
23. r
24. e
25.
26. e
27. n
28. o
29. r
30. m
31. o
32. u
33. s
34. l
35. y
36.
37. m
38. o
39. r
40. e
41.
42. a
43. c
44. t
45. i
46. v
47. e
48.
49. t
50. h
51. a
52. n
53.
54. l
55. o
56. n
57. g
58. -
59. l
60. i
61. v
62. e
63. d
64.
65. o
66. n
67. e
68. s
69.
70. f
71. o
72. r
73.
74. t
75. h
76. e
77.
78. s
79. a
80. m
81. e
82.
83. m
84. a
85. s
86. s
87. .

Question 10: A stream of neutrons has kinetic energy 0.0327 eV and half-life 700 s. The fraction of neutrons that decay before travelling a distance of 10 m is \( 3.9\times10^{-n} \). The value of n is:
(a) 6
(b) 7
(c) 5
(d) 8
Show Answer & Explanation

Answer: (a) 6

Explanation:
1. Speed \( v=\sqrt{2K/m}=\sqrt{\frac{2\times0.0327\times1.6\times10^{-19}}{1.675\times10^{-27}}}\approx2.5\times10^3\ \mathrm{m/s} \).
2. Time of flight \( t=10/v=4\times10^{-3} \) s, very small compared with 700 s.
3. Fraction decayed \( \approx\lambda t=\frac{0.693}{700}\times4\times10^{-3}=3.96\times10^{-6} \), so n = 6.

Teacher's Note:
1. W
2. h
3. e
4. n
5.
6. t
7. h
8. e
9.
10. e
11. l
12. a
13. p
14. s
15. e
16. d
17.
18. t
19. i
20. m
21. e
22.
23. i
24. s
25.
26. f
27. a
28. r
29.
30. s
31. h
32. o
33. r
34. t
35. e
36. r
37.
38. t
39. h
40. a
41. n
42.
43. t
44. h
45. e
46.
47. h
48. a
49. l
50. f
51. -
52. l
53. i
54. f
55. e
56.
57. t
58. h
59. e
60.
61. d
62. e
63. c
64. a
65. y
66. e
67. d
68.
69. f
70. r
71. a
72. c
73. t
74. i
75. o
76. n
77.
78. i
79. s
80.
81. a
82. p
83. p
84. r
85. o
86. x
87. i
88. m
89. a
90. t
91. e
92. l
93. y
94.
95. l
96. a
97. m
98. b
99. d
100. a
101.
102. t
103. i
104. m
105. e
106. s
107.
108. t
109. .

Question 11: A fixed nucleus X of mass number A has binding energy per nucleon 6 MeV. It absorbs a neutron of kinetic energy 2 MeV and becomes Y in the ground state, emitting a photon of energy 1 MeV. The binding energy per nucleon of Y (in MeV) is:
(a) \( \frac{6A+1}{A+1} \)
(b) \( \frac{6A-1}{A+1} \)
(c) 7
(d) \( \frac{7}{6} \)
Show Answer & Explanation

Answer: (b) \( \frac{6A-1}{A+1} \)

Explanation:
1. Energy conservation (rest energies plus kinetic energy): \( M_Xc^2+m_nc^2+2=M_Yc^2+1 \).
2. Hence \( M_Xc^2+m_nc^2-M_Yc^2=-1 \) MeV, so the total binding energy of Y is that of X minus 1 MeV: \( 6A-1 \).
3. Y has A + 1 nucleons, so binding energy per nucleon \( =\frac{6A-1}{A+1} \).

Teacher's Note:
1. B
2. i
3. n
4. d
5. i
6. n
7. g
8.
9. e
10. n
11. e
12. r
13. g
14. y
15.
16. i
17. s
18.
19. t
20. h
21. e
22.
23. d
24. i
25. f
26. f
27. e
28. r
29. e
30. n
31. c
32. e
33.
34. b
35. e
36. t
37. w
38. e
39. e
40. n
41.
42. t
43. h
44. e
45.
46. r
47. e
48. s
49. t
50.
51. e
52. n
53. e
54. r
55. g
56. y
57.
58. o
59. f
60.
61. t
62. h
63. e
64.
65. f
66. r
67. e
68. e
69.
70. n
71. u
72. c
73. l
74. e
75. o
76. n
77. s
78.
79. a
80. n
81. d
82.
83. t
84. h
85. e
86.
87. n
88. u
89. c
90. l
91. e
92. u
93. s
94. .
95.
96. T
97. r
98. a
99. c
100. k
101.
102. t
103. h
104. e
105.
106. r
107. e
108. s
109. t
110. -
111. e
112. n
113. e
114. r
115. g
116. y
117.
118. c
119. h
120. a
121. n
122. g
123. e
124.
125. u
126. s
127. i
128. n
129. g
130.
131. t
132. h
133. e
134.
135. k
136. i
137. n
138. e
139. t
140. i
141. c
142.
143. e
144. n
145. e
146. r
147. g
148. y
149.
150. a
151. n
152. d
153.
154. t
155. h
156. e
157.
158. p
159. h
160. o
161. t
162. o
163. n
164.
165. e
166. n
167. e
168. r
169. g
170. y
171. .

Question 12: The energy required to separate a \( {}^{20}_{10}\mathrm{Ne} \) nucleus into two alpha particles and a \( {}^{12}_{6}\mathrm{C} \) nucleus is (binding energy per nucleon: Ne-20 = 8.03, He-4 = 7.07, C-12 = 7.68 MeV):
(a) 148.7 MeV
(b) 160.6 MeV
(c) 11.9 MeV
(d) 8.0 MeV
Show Answer & Explanation

Answer: (c) 11.9 MeV

Explanation:
1. BE of Ne-20 \( =20\times8.03=160.6 \) MeV.
2. BE of the products \( =2\times4\times7.07+12\times7.68=56.56+92.16=148.72 \) MeV.
3. Energy required \( =160.6-148.72=11.88\approx11.9 \) MeV.

Teacher's Note:
1. B
2. r
3. e
4. a
5. k
6. i
7. n
8. g
9.
10. t
11. h
12. e
13.
14. n
15. u
16. c
17. l
18. e
19. u
20. s
21.
22. i
23. n
24. t
25. o
26.
27. m
28. o
29. r
30. e
31.
32. t
33. i
34. g
35. h
36. t
37. l
38. y
39.
40. p
41. a
42. c
43. k
44. e
45. d
46.
47. l
48. i
49. g
50. h
51. t
52. e
53. r
54.
55. p
56. i
57. e
58. c
59. e
60. s
61.
62. s
63. t
64. i
65. l
66. l
67.
68. c
69. o
70. s
71. t
72. s
73.
74. e
75. n
76. e
77. r
78. g
79. y
80.
81. b
82. e
83. c
84. a
85. u
86. s
87. e
88.
89. t
90. h
91. e
92.
93. p
94. r
95. o
96. d
97. u
98. c
99. t
100. s
101.
102. a
103. r
104. e
105.
106. l
107. e
108. s
109. s
110.
111. b
112. o
113. u
114. n
115. d
116.
117. i
118. n
119.
120. t
121. o
122. t
123. a
124. l
125. .

Question 13: The binding energy per nucleon of \( {}^{16}_8O \) is 7.97 MeV and that of \( {}^{17}_8O \) is 7.75 MeV. The energy required to remove a neutron from O-17 is \( 0.423\times10^n \) MeV. The value of n is:
(a) 2
(b) 0
(c) 3
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. Total binding energy of O-17 \( =17\times7.75=131.75 \) MeV; of O-16 \( =16\times7.97=127.52 \) MeV.
2. Energy to remove a neutron \( =131.75-127.52=4.23=0.423\times10^1 \) MeV, so n = 1.

Teacher's Note:
1. T
2. h
3. e
4.
5. s
6. e
7. p
8. a
9. r
10. a
11. t
12. i
13. o
14. n
15.
16. e
17. n
18. e
19. r
20. g
21. y
22.
23. i
24. s
25.
26. t
27. h
28. e
29.
30. d
31. i
32. f
33. f
34. e
35. r
36. e
37. n
38. c
39. e
40.
41. b
42. e
43. t
44. w
45. e
46. e
47. n
48.
49. t
50. h
51. e
52.
53. t
54. o
55. t
56. a
57. l
58.
59. b
60. i
61. n
62. d
63. i
64. n
65. g
66.
67. e
68. n
69. e
70. r
71. g
72. i
73. e
74. s
75.
76. o
77. f
78.
79. t
80. h
81. e
82.
83. n
84. u
85. c
86. l
87. e
88. u
89. s
90.
91. a
92. n
93. d
94.
95. t
96. h
97. e
98.
99. o
100. n
101. e
102.
103. l
104. e
105. f
106. t
107.
108. a
109. f
110. t
111. e
112. r
113.
114. r
115. e
116. m
117. o
118. v
119. a
120. l
121. .

Question 14: The kinetic energy of an alpha particle emitted from \( {}^{226}_{88}\mathrm{Ra} \) is 4.78 MeV. The total energy evolved in the disintegration is nearly:
(a) 4.87 MeV
(b) 4.78 MeV
(c) 4.69 MeV
(d) 5.2 MeV
Show Answer & Explanation

Answer: (a) 4.87 MeV

Explanation:
1. Momentum conservation gives \( K_\alpha:K_{Rn}=222:4 \).
2. Total energy \( Q=K_\alpha\left(1+\frac{4}{222}\right)=4.78\times\frac{226}{222}=4.87 \) MeV.

Teacher's Note:
1. T
2. h
3. e
4.
5. r
6. e
7. c
8. o
9. i
10. l
11. i
12. n
13. g
14.
15. d
16. a
17. u
18. g
19. h
20. t
21. e
22. r
23.
24. c
25. a
26. r
27. r
28. i
29. e
30. s
31.
32. a
33. w
34. a
35. y
36.
37. a
38.
39. s
40. m
41. a
42. l
43. l
44.
45. s
46. h
47. a
48. r
49. e
50.
51. o
52. f
53.
54. t
55. h
56. e
57.
58. e
59. n
60. e
61. r
62. g
63. y
64. ,
65.
66. s
67. o
68.
69. t
70. h
71. e
72.
73. Q
74. -
75. v
76. a
77. l
78. u
79. e
80.
81. i
82. s
83.
84. s
85. l
86. i
87. g
88. h
89. t
90. l
91. y
92.
93. l
94. a
95. r
96. g
97. e
98. r
99.
100. t
101. h
102. a
103. n
104.
105. t
106. h
107. e
108.
109. a
110. l
111. p
112. h
113. a
114.
115. e
116. n
117. e
118. r
119. g
120. y
121. .

Question 15: A radioactive sample has mass m, decay constant \( \lambda \) and molecular weight M. If Avogadro's number is \( N_A \), the initial activity is:
(a) \( \lambda m \)
(b) \( \frac{\lambda mN_A}{M} \)
(c) \( \frac{\lambda m}{M} \)
(d) \( mN_Ae^{\lambda} \)
Show Answer & Explanation

Answer: (b) \( \frac{\lambda mN_A}{M} \)

Explanation:
1. Number of nuclei \( N=\frac{mN_A}{M} \).
2. Activity \( A=\lambda N=\frac{\lambda mN_A}{M} \).

Teacher's Note:
1. A
2. c
3. t
4. i
5. v
6. i
7. t
8. y
9.
10. i
11. s
12.
13. t
14. h
15. e
16.
17. d
18. e
19. c
20. a
21. y
22.
23. c
24. o
25. n
26. s
27. t
28. a
29. n
30. t
31.
32. t
33. i
34. m
35. e
36. s
37.
38. t
39. h
40. e
41.
42. n
43. u
44. m
45. b
46. e
47. r
48.
49. o
50. f
51.
52. u
53. n
54. d
55. e
56. c
57. a
58. y
59. e
60. d
61.
62. n
63. u
64. c
65. l
66. e
67. i
68. .

Question 16: About 185 MeV of usable energy is released per fission of \( {}^{235}U \). A reactor generating 100 MW continuously uses up 1 kg of U-235 in n days. The greatest integer value [n] is:
(a) 9
(b) 7
(c) 8
(d) 10
Show Answer & Explanation

Answer: (c) 8

Explanation:
1. Atoms in 1 kg \( =\frac{1000}{235}\times6.02\times10^{23}=2.56\times10^{24} \).
2. Energy \( =2.56\times10^{24}\times185\times1.6\times10^{-13}=7.6\times10^{13} \) J.
3. Time \( =\frac{7.6\times10^{13}}{10^8\ \mathrm{W}}=7.6\times10^5 \) s \( =8.79 \) days, so [n] = 8.

Teacher's Note:
1. T
2. h
3. e
4.
5. n
6. u
7. m
8. b
9. e
10. r
11.
12. o
13. f
14.
15. d
16. a
17. y
18. s
19.
20. e
21. q
22. u
23. a
24. l
25. s
26.
27. t
28. o
29. t
30. a
31. l
32.
33. r
34. e
35. l
36. e
37. a
38. s
39. a
40. b
41. l
42. e
43.
44. e
45. n
46. e
47. r
48. g
49. y
50.
51. d
52. i
53. v
54. i
55. d
56. e
57. d
58.
59. b
60. y
61.
62. t
63. h
64. e
65.
66. p
67. o
68. w
69. e
70. r
71. ,
72.
73. c
74. o
75. n
76. v
77. e
78. r
79. t
80. e
81. d
82.
83. f
84. r
85. o
86. m
87.
88. s
89. e
90. c
91. o
92. n
93. d
94. s
95. .

Question 17: Let \( m_p \) be the mass of a proton, \( m_n \) of a neutron, \( M_1 \) of a \( {}^{20}_{10}Ne \) nucleus and \( M_2 \) of a \( {}^{40}_{20}Ca \) nucleus. Which statements are correct? (A) \( M_2=2M_1 \), (B) \( M_2>2M_1 \), (C) \( M_2<2M_1 \), (D) \( M_1<10(m_n+m_p) \). Select the correct combination.
(a) A and D only
(b) B and D only
(c) A and C only
(d) C and D only
Show Answer & Explanation

Answer: (d) C and D only

Explanation:
1. Each nucleus has mass equal to its free nucleons' mass minus the binding energy: \( M_1=10(m_p+m_n)-B_1 \) and \( M_2=20(m_p+m_n)-B_2 \).
2. Hence \( M_1<10(m_n+m_p) \) (D is true).
3. Ca-40 has a higher binding energy per nucleon than Ne-20, so \( B_2>2B_1 \) and \( M_2<2M_1 \) (C is true).

Teacher's Note:
1. A
2.
3. n
4. u
5. c
6. l
7. e
8. u
9. s
10.
11. i
12. s
13.
14. a
15. l
16. w
17. a
18. y
19. s
20.
21. l
22. i
23. g
24. h
25. t
26. e
27. r
28.
29. t
30. h
31. a
32. n
33.
34. i
35. t
36. s
37.
38. s
39. e
40. p
41. a
42. r
43. a
44. t
45. e
46. d
47.
48. n
49. u
50. c
51. l
52. e
53. o
54. n
55. s
56. ,
57.
58. a
59. n
60. d
61.
62. t
63. h
64. e
65.
66. m
67. o
68. r
69. e
70.
71. t
72. i
73. g
74. h
75. t
76. l
77. y
78.
79. b
80. o
81. u
82. n
83. d
84.
85. n
86. u
87. c
88. l
89. e
90. u
91. s
92.
93. l
94. o
95. s
96. e
97. s
98.
99. m
100. o
101. r
102. e
103.
104. m
105. a
106. s
107. s
108. .

Question 18: Two isotopes P and Q of atomic weights 10 and 20 are mixed in equal amounts by weight. After 20 days their weight ratio is found to be 1 : 4. If P has half-life 10 days, the half-life of Q is:
(a) infinite
(b) 5 days
(c) 20 days
(d) zero
Show Answer & Explanation

Answer: (a) infinite

Explanation:
1. P has gone through two half-lives, so its weight is \( \tfrac14 \) of the initial value.
2. Ratio P : Q = 1 : 4 with P at one quarter means Q is still at its full initial weight.
3. So Q has not decayed at all: its half-life is infinite.

Teacher's Note:
1. W
2. e
3. i
4. g
5. h
6. t
7. s
8. ,
9.
10. n
11. o
12. t
13.
14. a
15. t
16. o
17. m
18.
19. c
20. o
21. u
22. n
23. t
24. s
25. ,
26.
27. a
28. r
29. e
30.
31. c
32. o
33. m
34. p
35. a
36. r
37. e
38. d
39. ,
40.
41. s
42. o
43.
44. t
45. h
46. e
47.
48. a
49. t
50. o
51. m
52. i
53. c
54.
55. m
56. a
57. s
58. s
59. e
60. s
61.
62. d
63. o
64.
65. n
66. o
67. t
68.
69. m
70. a
71. t
72. t
73. e
74. r
75.
76. o
77. n
78. c
79. e
80.
81. t
82. h
83. e
84.
85. i
86. n
87. i
88. t
89. i
90. a
91. l
92.
93. w
94. e
95. i
96. g
97. h
98. t
99. s
100.
101. a
102. r
103. e
104.
105. e
106. q
107. u
108. a
109. l
110. .

Question 19: The intensity of gamma radiation from a source is I. On passing through 36 mm of lead it is reduced to I/8. The thickness of lead which reduces the intensity to I/2 is:
(a) 6 mm
(b) 12 mm
(c) 9 mm
(d) 18 mm
Show Answer & Explanation

Answer: (b) 12 mm

Explanation:
1. I/8 = three half-value thicknesses, so 3 HVT = 36 mm.
2. One half-value thickness = 12 mm.

Teacher's Note:
1. A
2. b
3. s
4. o
5. r
6. p
7. t
8. i
9. o
10. n
11.
12. o
13. f
14.
15. g
16. a
17. m
18. m
19. a
20.
21. r
22. a
23. y
24. s
25.
26. f
27. o
28. l
29. l
30. o
31. w
32. s
33.
34. t
35. h
36. e
37.
38. s
39. a
40. m
41. e
42.
43. e
44. x
45. p
46. o
47. n
48. e
49. n
50. t
51. i
52. a
53. l
54.
55. l
56. a
57. w
58.
59. a
60. s
61.
62. r
63. a
64. d
65. i
66. o
67. a
68. c
69. t
70. i
71. v
72. e
73.
74. d
75. e
76. c
77. a
78. y
79. ,
80.
81. w
82. i
83. t
84. h
85.
86. t
87. h
88. i
89. c
90. k
91. n
92. e
93. s
94. s
95.
96. i
97. n
98.
99. p
100. l
101. a
102. c
103. e
104.
105. o
106. f
107.
108. t
109. i
110. m
111. e
112. .

Question 20: The radionuclide \( {}^{56}Mn \) is being produced in a cyclotron at a constant rate P by bombarding a manganese target with deuterons. The target contains a large number of only the stable isotope \( {}^{55}Mn \), and the reaction is \( {}^{55}Mn+d\to{}^{56}Mn+p \). After a long time of bombardment, the activity of Mn-56 becomes constant, equal to \( 13.86\times10^{10}\ \mathrm{s^{-1}} \). Half-life of Mn-56 = 2.5 hours; use \( \ln2=0.693 \), \( N_A=6\times10^{23} \), atomic weight 56 g/mol. At what constant rate P are Mn-56 nuclei being produced?
(a) \( 2\times10^{11} \) nuclei/s
(b) \( 9.6\times10^{10} \) nuclei/s
(c) \( 13.86\times10^{10} \) nuclei/s
(d) \( 6.93\times10^{10} \) nuclei/s
Show Answer & Explanation

Answer: (c) \( 13.86\times10^{10} \) nuclei/s

Explanation:
1. When the activity becomes constant, the rate of decay equals the rate of production.
2. So \( P=\lambda N=13.86\times10^{10} \) nuclei per second.

Teacher's Note:
1. I
2. n
3.
4. a
5.
6. s
7. t
8. e
9. a
10. d
11. y
12.
13. s
14. t
15. a
16. t
17. e
18.
19. t
20. h
21. e
22.
23. p
24. r
25. o
26. d
27. u
28. c
29. t
30. i
31. o
32. n
33.
34. r
35. a
36. t
37. e
38.
39. e
40. q
41. u
42. a
43. l
44. s
45.
46. t
47. h
48. e
49.
50. a
51. c
52. t
53. i
54. v
55. i
56. t
57. y
58. .

Question 21: Let \( F_{pp},F_{pn},F_{nn} \) be the magnitudes of the nuclear force by a proton on a proton, a proton on a neutron and a neutron on a neutron when the separation is 1 fm. Then:
(a) \( F_{pp}>F_{pn}=F_{nn} \)
(b) \( F_{pp}>F_{pn}>F_{nn} \)
(c) \( F_{pp}(d) \( F_{pp}=F_{pn}=F_{nn} \)
Show Answer & Explanation

Answer: (d) \( F_{pp}=F_{pn}=F_{nn} \)

Explanation:
1. The nuclear force is charge independent: it acts equally between any pair of nucleons at the same separation.
2. Hence all three forces are equal (the electrostatic repulsion between two protons is a separate force).

Teacher's Note:
1. C
2. h
3. a
4. r
5. g
6. e
7.
8. i
9. n
10. d
11. e
12. p
13. e
14. n
15. d
16. e
17. n
18. c
19. e
20.
21. i
22. s
23.
24. a
25.
26. d
27. e
28. f
29. i
30. n
31. i
32. n
33. g
34.
35. p
36. r
37. o
38. p
39. e
40. r
41. t
42. y
43.
44. o
45. f
46.
47. t
48. h
49. e
50.
51. s
52. t
53. r
54. o
55. n
56. g
57.
58. n
59. u
60. c
61. l
62. e
63. a
64. r
65.
66. f
67. o
68. r
69. c
70. e
71. .
72.
73. T
74. h
75. e
76.
77. C
78. o
79. u
80. l
81. o
82. m
83. b
84.
85. f
86. o
87. r
88. c
89. e
90. ,
91.
92. w
93. h
94. i
95. c
96. h
97.
98. a
99. c
100. t
101. s
102.
103. o
104. n
105. l
106. y
107.
108. b
109. e
110. t
111. w
112. e
113. e
114. n
115.
116. p
117. r
118. o
119. t
120. o
121. n
122. s
123. ,
124.
125. i
126. s
127.
128. n
129. o
130. t
131.
132. p
133. a
134. r
135. t
136.
137. o
138. f
139.
140. \
141. (
142.
143. F
144. _
145. {
146. p
147. p
148. }
149.
150. \
151. )
152.
153. a
154. s
155.
156. d
157. e
158. f
159. i
160. n
161. e
162. d
163.
164. h
165. e
166. r
167. e
168. .

Question 22: The energy released in the reaction \( {}^7\mathrm{Li}+p\to 2\,{}^4\mathrm{He} \) is (binding energy per nucleon: Li-7 = 5.60 MeV, He-4 = 7.06 MeV):
(a) 17.3 MeV
(b) 1.73 MeV
(c) 1.46 MeV
(d) It depends on the binding energy of the proton
Show Answer & Explanation

Answer: (a) 17.3 MeV

Explanation:
1. Binding energy of 2 He-4: \( 2\times4\times7.06=56.48 \) MeV.
2. Binding energy of Li-7: \( 7\times5.60=39.2 \) MeV; the proton has none.
3. \( Q=56.48-39.2=17.28\approx17.3 \) MeV.

Teacher's Note:
1. T
2. h
3. e
4.
5. e
6. n
7. e
8. r
9. g
10. y
11.
12. r
13. e
14. l
15. e
16. a
17. s
18. e
19. d
20.
21. i
22. s
23.
24. t
25. h
26. e
27.
28. g
29. a
30. i
31. n
32.
33. i
34. n
35.
36. t
37. o
38. t
39. a
40. l
41.
42. b
43. i
44. n
45. d
46. i
47. n
48. g
49.
50. e
51. n
52. e
53. r
54. g
55. y
56. ,
57.
58. n
59. o
60. t
61.
62. i
63. n
64.
65. b
66. i
67. n
68. d
69. i
70. n
71. g
72.
73. e
74. n
75. e
76. r
77. g
78. y
79.
80. p
81. e
82. r
83.
84. n
85. u
86. c
87. l
88. e
89. o
90. n
91. .

Question 23: The atomic weight of boron is 10.81 g/mol and it has two isotopes \( {}^{10}_{5}\mathrm{B} \) and \( {}^{11}_{5}\mathrm{B} \). The ratio (by number) of \( {}^{10}\mathrm{B}:{}^{11}\mathrm{B} \) in nature is:
(a) 10 : 11
(b) 19 : 81
(c) 15 : 16
(d) 81 : 19
Show Answer & Explanation

Answer: (b) 19 : 81

Explanation:
1. Let the fraction of B-10 be x: \( 10x+11(1-x)=10.81 \), so \( x=0.19 \).
2. Ratio \( 0.19:0.81=19:81 \).

Teacher's Note:
1. A
2. v
3. e
4. r
5. a
6. g
7. e
8.
9. a
10. t
11. o
12. m
13. i
14. c
15.
16. m
17. a
18. s
19. s
20.
21. i
22. s
23.
24. t
25. h
26. e
27.
28. a
29. b
30. u
31. n
32. d
33. a
34. n
35. c
36. e
37. -
38. w
39. e
40. i
41. g
42. h
43. t
44. e
45. d
46.
47. m
48. e
49. a
50. n
51.
52. o
53. f
54.
55. t
56. h
57. e
58.
59. i
60. s
61. o
62. t
63. o
64. p
65. i
66. c
67.
68. m
69. a
70. s
71. s
72. e
73. s
74. ,
75.
76. s
77. o
78.
79. t
80. h
81. e
82.
83. l
84. i
85. g
86. h
87. t
88. e
89. r
90.
91. i
92. s
93. o
94. t
95. o
96. p
97. e
98.
99. i
100. s
101.
102. t
103. h
104. e
105.
106. m
107. i
108. n
109. o
110. r
111.
112. o
113. n
114. e
115.
116. h
117. e
118. r
119. e
120. .

Free study material for Physics

Download Chapter MCQs: JEE Physics Nuclei

Chapter MCQs with Answers for JEE Physics

Test your conceptual understanding of Nuclei with these targeted multiple-choice questions. Designed in alignment with the latest JEE curriculum for JEE Physics, these problem sets build accuracy and prepare students for objective exams.

Expert Practice Material for JEE Physics

Cross-reference your completed choices with comprehensive NCERT solutions for JEE Physics to ensure absolute clarity across all sub-topics in this chapter.

Complete Your Chapter Revision

Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in JEE assessments.

FAQs

Where can I access latest JEE Physics Nuclei MCQs Set 22?

You can get most exhaustive JEE Physics Nuclei MCQs Set 22 for free on StudiesToday.com. These MCQs for JEE Physics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Physics JEE material?

Yes, our JEE Physics Nuclei MCQs Set 22 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Physics MCQs help in scoring full marks in JEE exams?

By solving our JEE Physics Nuclei MCQs Set 22, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Physics.

Do you provide answers and explanations for JEE Physics Nuclei MCQs Set 22?

Yes, Physics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Physics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Physics Nuclei MCQs Set 22 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.