JEE Chemistry Nuclear Chemistry MCQs Set 02

Practice MCQs for JEE Chemistry Nuclear Chemistry

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Multiple Choice Questions

Question 1: On bombarding \( \ce{^{14}_{7}N} \) with alpha particles, a proton is released in \( \ce{^{14}_{7}N + ^{4}_{2}He -> ^{A}_{Z}X + ^{1}_{1}H} \). The product nucleus \( \ce{^{A}_{Z}X} \) is
(a) \( \ce{^{17}_{8}O} \)
(b) \( \ce{^{18}_{9}F} \)
(c) \( \ce{^{17}_{9}F} \)
(d) \( \ce{^{18}_{8}O} \)
Show Answer & Explanation

Answer: (a) \( \ce{^{17}_{8}O} \)

Explanation:
1. Mass number: 14 + 4 = A + 1, so A = 17.
2. Atomic number: 7 + 2 = Z + 1, so Z = 8.
3. X is \( \ce{^{17}_{8}O} \).

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 2: A positron has nearly the same mass as
(a) an alpha particle
(b) a proton
(c) a neutron
(d) an electron
Show Answer & Explanation

Answer: (d) an electron

Explanation:
1. A positron is the antiparticle of the electron: equal mass, opposite (positive) charge.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 3: In the reaction \( \ce{^{6}_{3}Li + ? -> ^{4}_{2}He + ^{3}_{1}H} \) the missing particle is
(a) electron
(b) neutron
(c) proton
(d) deuteron
Show Answer & Explanation

Answer: (b) neutron

Explanation:
1. Mass number: 6 + A = 4 + 3, so A = 1.
2. Atomic number: 3 + Z = 2 + 1, so Z = 0.
3. The particle with A = 1, Z = 0 is a neutron.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 4: In the upper atmosphere \( \ce{^{14}_{6}C} \) is generated by the nuclear reaction
(a) \( \ce{^{14}_{7}N + ^{1}_{1}H -> ^{14}_{6}C + ^{0}_{1}e} \)
(b) \( \ce{^{14}_{7}N -> ^{14}_{6}C + ^{0}_{-1}e} \)
(c) \( \ce{^{14}_{7}N + ^{1}_{0}n -> ^{14}_{6}C + ^{1}_{1}H} \)
(d) \( \ce{^{14}_{7}N + ^{3}_{1}H -> ^{1}_{0}n + ^{14}_{6}C + ^{4}_{2}He} \)
Show Answer & Explanation

Answer: (c) \( \ce{^{14}_{7}N + ^{1}_{0}n -> ^{14}_{6}C + ^{1}_{1}H} \)

Explanation:
1. Cosmic-ray neutrons strike atmospheric nitrogen.
2. Check option (c): mass 14 + 1 = 14 + 1 and charge 7 + 0 = 6 + 1, so it is balanced.
3. Option (a) has mass 15 on the left and 14 on the right; option (b) has charge 7 on the left and 5 on the right; option (d) is not balanced either.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 5: Deuterons bombarded on a nuclide produce \( \ce{^{38}_{18}Ar} \) and neutrons. The target nuclide is
(a) \( \ce{^{35}_{17}Cl} \)
(b) \( \ce{^{27}_{19}K} \)
(c) \( \ce{^{37}_{17}Cl} \)
(d) \( \ce{^{39}_{19}K} \)
Show Answer & Explanation

Answer: (c) \( \ce{^{37}_{17}Cl} \)

Explanation:
1. Reaction: \( \ce{^{2}_{1}D + ^{A}_{Z}X -> ^{38}_{18}Ar + ^{1}_{0}n} \).
2. A = 38 + 1 - 2 = 37 and Z = 18 + 0 - 1 = 17.
3. The target is \( \ce{^{37}_{17}Cl} \).

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 6: The radionuclide \( \ce{^{234}_{90}Th} \) undergoes two successive beta decays followed by one alpha decay. The atomic number and mass number of the resulting nuclide are
(a) 92 and 234
(b) 94 and 230
(c) 90 and 230
(d) 92 and 230
Show Answer & Explanation

Answer: (c) 90 and 230

Explanation:
1. Two beta decays: Z = 90 + 2 = 92; A stays 234.
2. One alpha decay: Z = 92 - 2 = 90; A = 234 - 4 = 230.
3. The result is atomic number 90 and mass number 230 (an isotope of thorium).

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 7: In the nuclear reaction \( \ce{^{130}_{52}Te + ^{2}_{1}H -> ^{131}_{53}I + ?} \) the other product is
(a) a positron
(b) an alpha particle
(c) one neutron
(d) a proton
Show Answer & Explanation

Answer: (c) one neutron

Explanation:
1. Mass number: 130 + 2 - 131 = 1.
2. Charge: 52 + 1 - 53 = 0.
3. A particle with A = 1 and Z = 0 is a neutron.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 8: The reaction \( \ce{^{8}_{5}B -> ^{8}_{4}Be + ^{0}_{1}e} \) is due to
(a) loss of alpha particles
(b) loss of beta particles
(c) loss of positron
(d) electron loss
Show Answer & Explanation

Answer: (c) loss of positron

Explanation:
1. Atomic number decreases by 1 with no change in mass number.
2. The emitted particle has charge +1 and negligible mass: a positron.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 9: Positronium is the name given to an atom-like combination formed between
(a) a positron and a proton
(b) a positron and a neutron
(c) a positron and an alpha particle
(d) a positron and an electron
Show Answer & Explanation

Answer: (d) a positron and an electron

Explanation:
1. Positronium is a short-lived bound state of a positron and an electron.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 10: In the nuclear reaction \( \ce{^{24}_{12}Mg + ^{4}_{2}He -> ^{1}_{0}n + ?} \) the product nucleus is
(a) \( \ce{^{27}_{13}Al} \)
(b) \( \ce{^{27}_{14}Si} \)
(c) \( \ce{^{28}_{13}Al} \)
(d) \( \ce{^{25}_{12}Mg} \)
Show Answer & Explanation

Answer: (b) \( \ce{^{27}_{14}Si} \)

Explanation:
1. Mass number: 24 + 4 - 1 = 27.
2. Atomic number: 12 + 2 - 0 = 14.
3. The product is \( \ce{^{27}_{14}Si} \).

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 11: In the nuclear reaction \( \ce{^{238}_{92}U -> ^{206}_{82}Pb + x\,^{4}_{2}He + y\,^{0}_{-1}e} \) the values of x and y are respectively
(a) 8, 6
(b) 6, 4
(c) 6, 8
(d) 8, 10
Show Answer & Explanation

Answer: (a) 8, 6

Explanation:
1. Mass: 238 - 206 = 32 = 4x, so x = 8.
2. Charge: 92 = 82 + 2(8) - y, so y = 6.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 12: If an isotope of hydrogen has two neutrons, its atomic number and mass number respectively are
(a) 2 and 1
(b) 3 and 1
(c) 1 and 1
(d) 1 and 3
Show Answer & Explanation

Answer: (d) 1 and 3

Explanation:
1. All hydrogen isotopes have Z = 1.
2. Mass number = 1 proton + 2 neutrons = 3.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 13: Which of the following nuclear transformations is of the (n, p) type?
(a) \( \ce{^{7}_{3}Li + ^{1}_{1}H -> ^{7}_{4}Be + ^{1}_{0}n} \)
(b) \( \ce{^{75}_{33}As + ^{4}_{2}He -> ^{78}_{35}Br + ^{1}_{0}n} \)
(c) \( \ce{^{209}_{83}Bi + ^{2}_{1}H -> ^{210}_{84}Po + ^{1}_{0}n} \)
(d) \( \ce{^{45}_{21}Sc + ^{1}_{0}n -> ^{45}_{20}Ca + ^{1}_{1}H} \)
Show Answer & Explanation

Answer: (d) \( \ce{^{45}_{21}Sc + ^{1}_{0}n -> ^{45}_{20}Ca + ^{1}_{1}H} \)

Explanation:
1. An (n, p) reaction has a neutron as the projectile and a proton as the ejected particle.
2. Only option (d) absorbs a neutron and emits a proton.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 14: In the reaction \( \ce{^{239}_{93}Np -> ^{239}_{94}Pu + ?} \) the missing particle is
(a) proton
(b) positron
(c) electron
(d) neutron
Show Answer & Explanation

Answer: (c) electron

Explanation:
1. Mass number is unchanged and Z rises by 1.
2. This is beta-minus decay, so an electron is emitted.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 15: According to the nuclear reaction \( \ce{^{A}_{4}Be + ^{4}_{2}He -> ^{12}_{6}C + ^{1}_{0}n} \), the mass number of Be is
(a) 4
(b) 9
(c) 7
(d) 6
Show Answer & Explanation

Answer: (b) 9

Explanation:
1. A + 4 = 12 + 1, so A = 9.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 16: The composition of tritium \( \ce{^{3}_{1}H} \) is
(a) 1 electron, 1 proton, 1 neutron
(b) 1 electron, 2 protons, 1 neutron
(c) 1 electron, 1 proton, 2 neutrons
(d) 1 electron, 1 proton, 3 neutrons
Show Answer & Explanation

Answer: (c) 1 electron, 1 proton, 2 neutrons

Explanation:
1. Z = 1 gives 1 proton and 1 electron in the neutral atom.
2. Neutrons = 3 - 1 = 2.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 17: Identify X in \( \ce{^{32}_{16}S + X -> ^{30}_{15}P + ^{4}_{2}He} \)
(a) \( \ce{^{1}_{1}H} \)
(b) \( \ce{^{2}_{1}D} \)
(c) \( \ce{^{1}_{0}n} \)
(d) an electron
Show Answer & Explanation

Answer: (b) \( \ce{^{2}_{1}D} \)

Explanation:
1. Mass: 32 + A = 30 + 4, so A = 2.
2. Charge: 16 + Z = 15 + 2, so Z = 1.
3. X is deuterium \( \ce{^{2}_{1}D} \).

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 18: How many neutrons are present in the nucleus of \( \ce{^{226}_{88}Ra} \)?
(a) 88
(b) 226
(c) 140
(d) 138
Show Answer & Explanation

Answer: (d) 138

Explanation:
1. Neutrons = A - Z = 226 - 88 = 138.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 19: In the equation \( \ce{^{23}_{11}Na + ^{1}_{1}H -> ^{23}_{12}Mg + x} \), x represents
(a) a neutron
(b) a deuteron
(c) an alpha particle
(d) a positron
Show Answer & Explanation

Answer: (a) a neutron

Explanation:
1. Mass: 23 + 1 - 23 = 1; charge: 11 + 1 - 12 = 0.
2. So x is a neutron.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 20: In the reaction Po \( \xrightarrow{-\alpha} \) Pb \( \xrightarrow{-\beta} \) Bi, if Bi belongs to group 15, Po belongs to group
(a) 14
(b) 15
(c) 13
(d) 16
Show Answer & Explanation

Answer: (d) 16

Explanation:
1. Bi (group 15) has Z = 83; beta emission raises Z by 1, so Pb has Z = 82 (group 14).
2. Alpha emission lowers Z by 2, so Po has Z = 84.
3. Z = 84 is in group 16.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 21: In the nuclear reaction \( \ce{^{9}_{4}Be(p,\alpha)X} \), X is
(a) \( \ce{^{4}_{2}He} \)
(b) \( \ce{^{6}_{3}Li} \)
(c) \( \ce{^{7}_{3}Li} \)
(d) \( \ce{^{8}_{4}Be} \)
Show Answer & Explanation

Answer: (b) \( \ce{^{6}_{3}Li} \)

Explanation:
1. Be-9 + p gives A = 10, Z = 5; subtracting the alpha (A = 4, Z = 2) gives A = 6, Z = 3.
2. X is \( \ce{^{6}_{3}Li} \).

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 22: Which of the following does NOT contain a number of neutrons equal to that of \( \ce{^{40}_{18}Ar} \)?
(a) \( \ce{^{41}_{19}K} \)
(b) \( \ce{^{43}_{21}Sc} \)
(c) \( \ce{^{42}_{21}Sc} \)
(d) \( \ce{^{42}_{20}Ca} \)
Show Answer & Explanation

Answer: (c) \( \ce{^{42}_{21}Sc} \)

Explanation:
1. Ar-40 has 40 - 18 = 22 neutrons.
2. K-41: 22; Sc-43: 22; Sc-42: 21; Ca-42: 22.
3. Sc-42 differs.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 23: In the reaction \( \ce{^{64}_{29}Cu -> ^{64}_{28}Ni + x} \), x will be
(a) a proton
(b) an electron
(c) a neutron
(d) a positron
Show Answer & Explanation

Answer: (d) a positron

Explanation:
1. Mass number unchanged; atomic number falls by 1.
2. This is positron emission (or electron capture).

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 24: For the nuclear reaction \( \ce{^{24}_{12}Mg + ^{2}_{1}D -> \alpha + ?} \) the missing nuclide is
(a) \( \ce{^{22}_{11}Na} \)
(b) \( \ce{^{23}_{11}Na} \)
(c) \( \ce{^{23}_{12}Mg} \)
(d) \( \ce{^{26}_{12}Mg} \)
Show Answer & Explanation

Answer: (a) \( \ce{^{22}_{11}Na} \)

Explanation:
1. Mass: 24 + 2 - 4 = 22.
2. Charge: 12 + 1 - 2 = 11.
3. The nuclide is \( \ce{^{22}_{11}Na} \).

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

Question 25: An element \( \ce{^{227}_{96}X} \) emits 4 alpha and 5 beta particles to form a new element Y. The atomic number and mass number of Y are
(a) 93; 211
(b) 211; 93
(c) 212; 88
(d) 88; 211
Show Answer & Explanation

Answer: (a) 93; 211

Explanation:
1. Z = 96 - 4(2) + 5 = 93.
2. A = 227 - 4(4) = 211.
3. Y has atomic number 93 and mass number 211.

Teacher's Note:
1. Balance the mass number (top) and the atomic number (bottom) separately on both sides.
2. Alpha emission lowers A by 4 and Z by 2; beta emission raises Z by 1 with no change in A.

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Multiple Choice Questions (MCQs) for JEE Chemistry Nuclear Chemistry

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