JEE Physics Nuclei MCQs Set 24

Answer: (a) P-2, Q-1, R-4, S-3

Explanation:
1. U-238 to Th-234 loses 4 mass units and 2 charge units: alpha decay, so P-2.
2. O-15 to N-15 keeps A and lowers Z by one: beta-plus decay, so Q-1.
3. Pu-239 to La-140 is a big split into medium nuclei: fission, so R-4.
4. Bi-185 to Pb-184 loses one proton: proton emission, so S-3.

Teacher's Note:
1. I
2. d
3. e
4. n
5. t
6. i
7. f
8. y
9.
10. e
11. a
12. c
13. h
14.
15. p
16. r
17. o
18. c
19. e
20. s
21. s
22.
23. b
24. y
25.
26. t
27. h
28. e
29.
30. c
31. h
32. a
33. n
34. g
35. e
36.
37. i
38. n
39.
40. A
41.
42. a
43. n
44. d
45.
46. Z
47. .

Answer: (b) 3

Explanation:
1. Power available falls as \( P_0(1/2)^{t/T} \).
2. It stays enough until \( (1/2)^{t/T}=0.125=(1/2)^3 \), i.e. \( t=3T \), so \( n=3 \).

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

Answer: (c) 3.42 fm

Explanation:
1. B.E. of N-15 \( =(7\times1.007825+8\times1.008665-15.000109)\times931.5=115.49\ \mathrm{MeV} \); B.E. of O-15 \( =111.96\ \mathrm{MeV} \); the difference is 3.536 MeV.
2. Coulomb energies: \( \frac35\frac{1.44}{R}[8\times7-7\times6]=\frac{3\times1.44\times14}{5R}=\frac{12.1}{R}\ \mathrm{MeV} \).
3. Equating, \( R=12.1/3.536=3.42 \) fm.

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

Answer: (a) Deuteron and alpha particle can undergo complete fusion

Explanation:
1. Li-6 to He-4 + H-2: products \( 4.002603+2.014102=6.016705 \) u exceed \( 6.015123 \) u, so alpha emission is not possible.
2. Po-210 to Bi-209 + H-1: products \( 208.980388+1.007825=209.988213 \) u exceed \( 209.982876 \) u, so proton emission is not possible.
3. H-2 + He-4 forming Li-6: \( 6.016705>6.015123 \), so fusion is allowed.

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

Answer: (a) \( 3.56\times10^{5} \)

Explanation:
1. \( N_Y=\frac{\lambda_{Sr}N_{Sr}}{\lambda_Y-\lambda_{Sr}}\left(1-e^{-(\lambda_Y-\lambda_{Sr})t}\right) \) approximately \( \frac{\lambda_{Sr}N_{Sr}}{\lambda_Y}(1-e^{-\lambda_Yt}) \) because \( \lambda_Y\gg\lambda_{Sr} \).
2. With \( \lambda_Y=0.693/64=0.01083\ \mathrm{h^{-1}} \) and \( \lambda_{Sr}=0.693/(28\times8760)=2.83\times10^{-6}\ \mathrm{h^{-1}} \), at t = 1 h: \( \frac{N_{Sr}}{N_Y}=\frac{\lambda_Y-\lambda_{Sr}}{\lambda_{Sr}(1-e^{-\lambda_Y t})}\approx3.56\times10^{5} \).

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

Answer: (b) 1 mg

Explanation:
1. Activity \( A=N/\tau \), so \( N=A\tau=10^{10}\times10^{9}=10^{19} \) atoms.
2. Mass \( =10^{19}\times10^{-25}=10^{-6}\ \mathrm{kg}=1\ \mathrm{mg} \).

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

Answer: (c) 2/3

Explanation:
1. Let the neutron have mass m and speed u, the deuteron 2m. After the collision the neutron moves at right angles with speed v, and the deuteron with speed w at angle \( \theta \).
2. Momentum: \( 2mw\cos\theta=mu,\ 2mw\sin\theta=mv \) and energy \( u^2=v^2+2w^2 \).
3. Solving gives \( v^2=u^2/3 \), so the neutron retains 1/3 of its energy and loses 2/3, \( =\frac{2m}{m+M} \).

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

Answer: (d) 17.28 MeV

Explanation:
1. Total B.E. of Li-7 \( =7\times5.60=39.2 \) MeV; of two He-4 \( =2\times4\times7.06=56.48 \) MeV.
2. Energy released \( =56.48-39.2=17.28 \) MeV.

Teacher's Note:
1. E
2. n
3. e
4. r
5. g
6. y
7.
8. r
9. e
10. l
11. e
12. a
13. s
14. e
15. d
16.
17. e
18. q
19. u
20. a
21. l
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. .

Answer: (a) 11.48 MeV

Explanation:
1. \( {}^{42}_{20}\mathrm{Ca} \) \( \to \) \( {}^{41}_{20}\mathrm{Ca} \) + n.
2. Energy \( =(40.962278+1.008665-41.958622)\times931.5=0.012321\times931.5=11.48\ \mathrm{MeV} \).

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

Answer: (b) the energy delivered by radiation to a target

Explanation:
1. The rate of decay is measured in becquerel or curie, ionisation in air in roentgen, and biological effect in rem.
2. The rad is the unit of absorbed dose: 1 rad is 0.01 J of energy absorbed per kilogram.

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

Answer: (c) 108

Explanation:
1. Required reduction factor 64 \( =2^6 \), so 6 half-lives are needed.
2. Time \( =6\times18=108 \) days.

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

Answer: (d) 10.27 MeV

Explanation:
1. \( {}^{42}_{20}\mathrm{Ca} \) \( \to \) \( {}^{41}_{19}\mathrm{K} \) + p (using atomic masses with hydrogen).
2. Energy \( =(40.961825+1.007826-41.958622)\times931.5=0.011029\times931.5=10.27\ \mathrm{MeV} \).

Teacher's Note:
1. R
2. e
3. m
4. o
5. v
6. i
7. n
8. g
9.
10. a
11.
12. p
13. r
14. o
15. t
16. o
17. n
18.
19. t
20. a
21. k
22. e
23. s
24.
25. l
26. e
27. s
28. s
29.
30. e
31. n
32. e
33. r
34. g
35. y
36.
37. t
38. h
39. a
40. n
41.
42. r
43. e
44. m
45. o
46. v
47. i
48. n
49. g
50.
51. a
52.
53. n
54. e
55. u
56. t
57. r
58. o
59. n
60.
61. h
62. e
63. r
64. e
65.
66. b
67. e
68. c
69. a
70. u
71. s
72. e
73.
74. t
75. h
76. e
77.
78. p
79. r
80. o
81. t
82. o
83. n
84. '
85. s
86.
87. e
88. l
89. e
90. c
91. t
92. r
93. o
94. s
95. t
96. a
97. t
98. i
99. c
100.
101. r
102. e
103. p
104. u
105. l
106. s
107. i
108. o
109. n
110.
111. a
112. l
113. r
114. e
115. a
116. d
117. y
118.
119. l
120. o
121. w
122. e
123. r
124. s
125.
126. i
127. t
128. s
129.
130. b
131. i
132. n
133. d
134. i
135. n
136. g
137. .

Answer: (a) \( t=T\,\frac{\log1.3}{\log2} \)

Explanation:
1. \( N_A=N_0e^{-\lambda t},\ N_B=N_0(1-e^{-\lambda t}) \).
2. \( \frac{N_B}{N_A}=e^{\lambda t}-1=0.3 \), so \( e^{\lambda t}=1.3 \).
3. \( t=\frac{\ln1.3}{\lambda}=T\frac{\ln1.3}{\ln2} \).

Teacher's Note:
1. E
2. v
3. e
4. r
5. y
6.
7. B
8.
9. n
10. u
11. c
12. l
13. e
14. u
15. s
16.
17. c
18. o
19. m
20. e
21. s
22.
23. f
24. r
25. o
26. m
27.
28. a
29.
30. d
31. e
32. c
33. a
34. y
35. e
36. d
37.
38. A
39.
40. n
41. u
42. c
43. l
44. e
45. u
46. s
47. ,
48.
49. s
50. o
51.
52. \
53. (
54.
55. N
56. _
57. B
58. =
59. N
60. _
61. 0
62. -
63. N
64. _
65. A
66.
67. \
68. )
69. .

Answer: (b) \( E_2>E_1 \)

Explanation:
1. The parent has more mass than the daughters, \( \Delta m>0 \), so energy is released in the decay.
2. Energy is released only when the daughters are more tightly bound per nucleon, so \( E_2>E_1 \).

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

Answer: (c) 5319

Explanation:
1. \( Q=(209.982876-205.974455-4.002603)\times931.5=0.005818\times931.5=5.42\ \mathrm{MeV} \).
2. Momentum conservation shares Q inversely with mass: \( K_\alpha=Q\times\frac{206}{210}=5.32\ \mathrm{MeV}\approx5319\ \mathrm{keV} \).

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

Answer: (d) x = n, y = n, \( K_{Sr}=129 \) MeV, \( K_{Xe}=86 \) MeV

Explanation:
1. Charge: 92 = 54 + 38, so x and y carry no net charge; mass number: 236 = 140 + 94 + 2, so both are neutrons.
2. Energy released \( Q=140(8.5)+94(8.5)-236(7.5)=219 \) MeV; neutrons take 4 MeV, leaving 215 MeV for Xe and Sr.
3. Equal momenta: \( K\propto1/m \), so \( K_{Sr}=215\times\frac{140}{234}=129 \) MeV and \( K_{Xe}=215\times\frac{94}{234}=86 \) MeV.

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

Answer: (a) \( 0\le K<0.8\times10^6\ \mathrm{eV} \)

Explanation:
1. The electron can still have almost no energy, so the lower limit stays at 0.
2. With a massive anti-neutrino, at least \( 3\ \mathrm{eV} \) of rest energy must be created, so the electron can no longer reach the full \( 0.8\times10^6\ \mathrm{eV} \): the upper limit is not attained.

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

Answer: (b) \( N_\alpha=5 \) and \( N_\beta=2 \) together

Explanation:
1. Mass number change \( 232-212=20 \), so \( N_\alpha=20/4=5 \).
2. Charge: \( 90-2\times5=80 \) but the final Z is 82, so \( N_\beta=2 \).

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

Answer: (c) 5 : 4

Explanation:
1. A: 80 min is 4 half-lives, so the fraction decayed is \( 1-\frac1{16}=\frac{15}{16} \).
2. B: 2 half-lives, fraction decayed \( 1-\frac14=\frac34 \).
3. Ratio \( \frac{15/16}{3/4}=\frac54 \).

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

Answer: (d) 7

Explanation:
1. All the initial kinetic energy turns into electrostatic potential energy at closest approach: \( K=\frac{(120e)(e)}{4\pi\varepsilon_0 r}=\frac{9\times10^9\times120\times(1.6\times10^{-19})^2}{10^{-14}}=2.76\times10^{-12}\ \mathrm{J} \).
2. \( p=\sqrt{2mK}=\sqrt{2\times\frac53\times10^{-27}\times2.76\times10^{-12}}=9.6\times10^{-20}\ \mathrm{kg\,m/s} \).
3. \( \lambda=h/p=\frac{6.72\times10^{-34}}{9.6\times10^{-20}}=7\times10^{-15}\ \mathrm{m}=7\ \mathrm{fm} \).

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

Answer: (a) 0.89

Explanation:
1. For a head-on elastic collision the fractional loss is \( \frac{4mM}{(m+M)^2} \).
2. With \( m=1,\ M=2 \): \( \frac{8}{9}=0.89 \).

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

Answer: (b) \( 3.8\times10^{3} \)

Explanation:
1. After 10 years Y is in transient equilibrium: \( e^{-\lambda_Yt}\approx0 \).
2. \( \frac{N_{Sr}}{N_Y}=\frac{\lambda_Y-\lambda_{Sr}}{\lambda_{Sr}}\approx\frac{T_{Sr}}{T_Y}=\frac{28\times8760}{64}=3.8\times10^3 \).

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

Answer: (c) (i) and (iv)

Explanation:
1. Energy is released when the products have a higher B.E. per nucleon than the reactants.
2. (i) light nuclei fuse into a nucleus near the peak: positive. (iv) a heavy nucleus splits into nuclei near the peak: positive.
3. (ii) and (iii) go away from the peak, so they need energy.

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

Answer: (d) gamma photons

Explanation:
1. \( {}^7_3\mathrm{Li}+{}^1_1\mathrm{H}\to{}^8_4\mathrm{Be}+X \): mass number 8 = 8 and charge 4 = 4, so X has zero mass number and zero charge.
2. That is a gamma photon.

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

Answer: (a) \( 3.6\times10^{-7} \)

Explanation:
1. At secular equilibrium the activities are equal: \( \lambda_{Ra}N_{Ra}=\lambda_UN_U \).
2. So \( \frac{N_{Ra}}{N_U}=\frac{T_{Ra}}{T_U}=\frac{1620}{4.5\times10^{9}}=3.6\times10^{-7} \).

Teacher's Note:
1. I
2. n
3.
4. e
5. q
6. u
7. i
8. l
9. i
10. b
11. r
12. i
13. u
14. m
15.
16. t
17. h
18. e
19.
20. n
21. u
22. m
23. b
24. e
25. r
26.
27. o
28. f
29.
30. n
31. u
32. c
33. l
34. e
35. i
36.
37. i
38. s
39.
40. p
41. r
42. o
43. p
44. o
45. r
46. t
47. i
48. o
49. n
50. a
51. l
52.
53. t
54. o
55.
56. t
57. h
58. e
59.
60. h
61. a
62. l
63. f
64. -
65. l
66. i
67. f
68. e
69. .