JEE Physics Nuclei MCQs Set 04

Physics Objective Questions and Answers: Nuclei

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Multiple Choice Questions

Question 1: The Q value of a nuclear reaction A + b → C + d is Q = [mA + mb − mC − md]c². For the reaction \({}^{1}_{1}\mathrm{H}\) + \({}^{3}_{1}\mathrm{H}\) → \({}^{2}_{1}\mathrm{H}\) + \({}^{2}_{1}\mathrm{H}\) (atomic masses: \({}^{1}_{1}\mathrm{H}\) = 1.007825 u, \({}^{2}_{1}\mathrm{H}\) = 2.014102 u, \({}^{3}_{1}\mathrm{H}\) = 3.016049 u; 1 u = 931 MeV/c²), find Q and say whether the reaction is exothermic or endothermic.
(a) +4.03 MeV, exothermic
(b) −8.06 MeV, endothermic
(c) +8.06 MeV, exothermic
(d) −4.03 MeV, endothermic
Show Answer & Explanation

Answer: (d) −4.03 MeV, endothermic

Explanation:
1. Mass of reactants = 1.007825 + 3.016049 = 4.023874 u
2. Mass of products = 2 × 2.014102 = 4.028204 u
3. Δm = 4.023874 − 4.028204 = −0.004330 u, so Q = −0.004330 × 931 = −4.03 MeV
4. Q is negative, so energy must be supplied: the reaction is endothermic.

Teacher's Note:
1. The sign of Q decides the type: Q > 0 is exothermic and Q < 0 is endothermic.
2. Mass of the products larger than that of the reactants means energy is absorbed.

Question 2: The Q value of a nuclear reaction is Q = [mA + mb − mC − md]c². For the reaction \({}^{12}_{6}\mathrm{C}\) + \({}^{12}_{6}\mathrm{C}\) → \({}^{20}_{10}\mathrm{Ne}\) + \({}^{4}_{2}\mathrm{He}\), with atomic masses \({}^{12}_{6}\mathrm{C}\) = 12.000000 u, \({}^{20}_{10}\mathrm{Ne}\) = 19.992439 u and He = 4.0015 u (1 u = 931 MeV/c²), the Q value and type of reaction are
(a) −5.64 MeV, endothermic
(b) −2.82 MeV, endothermic
(c) +2.82 MeV, exothermic
(d) +5.64 MeV, exothermic
Show Answer & Explanation

Answer: (d) +5.64 MeV, exothermic

Explanation:
1. Mass of reactants = 2 × 12.000000 = 24.000000 u
2. Mass of products = 19.992439 + 4.0015 = 23.993939 u
3. Δm = 0.006061 u, so Q = 0.006061 × 931 = +5.64 MeV
4. Q is positive, so the reaction releases energy (exothermic).

Teacher's Note:
1. Atomic masses can be used directly when the number of electrons is the same on both sides.
2. Always subtract products from reactants to get the sign of Q right.

Question 3: The mass of a \({}^{35}_{17}\mathrm{Cl}\) nucleus is 34.98 u, the mass of a neutron is 1.008665 u and the mass of a proton is 1.007277 u. The binding energy of the nucleus (1 u = 931 MeV) is nearly
(a) 558 MeV
(b) 279 MeV
(c) 7.97 MeV
(d) 27.9 MeV
Show Answer & Explanation

Answer: (b) 279 MeV

Explanation:
1. Protons Z = 17 and neutrons N = 35 − 17 = 18
2. Mass of the separate nucleons = 17 × 1.007277 + 18 × 1.008665 = 35.27968 u
3. Mass defect Δm = 35.27968 − 34.98 = 0.29968 u
4. Binding energy = 0.29968 × 931 ≈ 279 MeV (about 7.97 MeV per nucleon).

Teacher's Note:
1. Binding energy = (mass of free nucleons − mass of nucleus) × 931 MeV.
2. Dividing by the mass number gives the binding energy per nucleon, a measure of stability.

Question 4: The energy released per fission of a \({}^{235}_{92}\mathrm{U}\) nucleus is 200 MeV. The energy released by the complete fission of 2 g of \({}^{235}_{92}\mathrm{U}\), in kWh, is nearly (NA = 6.02 × 10²³, 1 eV = 1.6 × 10⁻¹⁹ J)
(a) 1.64 × 10¹¹ kWh
(b) 4.55 × 10⁷ kWh
(c) 4.55 × 10¹ kWh
(d) 4.55 × 10⁴ kWh
Show Answer & Explanation

Answer: (d) 4.55 × 10⁴ kWh

Explanation:
1. Number of nuclei in 2 g = (2/235) × 6.02 × 10²³ = 5.12 × 10²¹
2. Energy = 5.12 × 10²¹ × 200 MeV = 1.02 × 10²⁴ MeV = 1.02 × 10²⁴ × 1.6 × 10⁻¹³ J = 1.64 × 10¹¹ J
3. 1 kWh = 3.6 × 10⁶ J, so energy = 1.64 × 10¹¹ / 3.6 × 10⁶ ≈ 4.55 × 10⁴ kWh.

Teacher's Note:
1. Convert MeV to joule using 1 MeV = 1.6 × 10⁻¹³ J and joule to kWh by dividing by 3.6 × 10⁶.
2. Note that 1.64 × 10¹¹ is the answer in joules, not in kWh.

Question 5: If the energy released in the fission of one nucleus is 3.2 × 10⁻¹¹ J, the number of nuclei that must undergo fission per second in a power plant of 16 kW is
(a) 5 × 10¹¹
(b) 5 × 10¹⁷
(c) 5 × 10¹⁴
(d) 3.2 × 10¹⁰
Show Answer & Explanation

Answer: (c) 5 × 10¹⁴

Explanation:
1. Power = (number of fissions per second) × (energy per fission)
2. Number per second = 16 × 10³ / 3.2 × 10⁻¹¹ = 5 × 10¹⁴.

Teacher's Note:
1. Power is energy per unit time, so the fission rate is power divided by energy per fission.
2. Keep power in watt (J/s), not kW, when dividing.

Question 6: The mass of uranium \({}^{235}_{92}\mathrm{U}\) that must undergo fission per day to generate 10 MW of power is nearly (200 MeV per fission, NA = 6.02 × 10²³)
(a) 2.4 g
(b) 1.05 g
(c) 10.5 g
(d) 105 g
Show Answer & Explanation

Answer: (c) 10.5 g

Explanation:
1. Energy needed per day = 10 × 10⁶ × 86400 = 8.64 × 10¹¹ J
2. Energy per fission = 200 × 1.6 × 10⁻¹³ = 3.2 × 10⁻¹¹ J, so fissions = 2.7 × 10²²
3. Mass = (2.7 × 10²² / 6.02 × 10²³) × 235 g ≈ 10.5 g.

Teacher's Note:
1. A day has 86400 s; convert power to energy first.
2. Atoms to mass: multiply moles by 235 g/mol.

Question 7: The mass defect in a nuclear fusion reaction is 0.3 percent. The energy liberated in the fusion of one kilogram of the reacting mass is
(a) 9 × 10¹⁶ J
(b) 2.7 × 10¹¹ J
(c) 2.7 × 10¹⁴ J
(d) 2.7 × 10¹³ J
Show Answer & Explanation

Answer: (c) 2.7 × 10¹⁴ J

Explanation:
1. Mass converted = 0.3% of 1 kg = 3 × 10⁻³ kg
2. E = Δm c² = 3 × 10⁻³ × (3 × 10⁸)² = 2.7 × 10¹⁴ J.

Teacher's Note:
1. Use E = Δm c² with the mass in kilogram.
2. The full 1 kg would give 9 × 10¹⁶ J, which is only for complete conversion.

Question 8: Two nuclei have mass numbers in the ratio 1 : 3. The ratio of their nuclear densities is
(a) 1 : 3
(b) 3 : 1
(c) 1 : 9
(d) 1 : 1
Show Answer & Explanation

Answer: (d) 1 : 1

Explanation:
1. Nuclear radius R = R₀A^(1/3), so volume is proportional to A
2. Mass is also proportional to A, so density = mass / volume is independent of A
3. Hence the ratio of densities is 1 : 1.

Teacher's Note:
1. Nuclear density (about 10¹⁷ kg/m³) is the same for all nuclei.
2. This shows that nucleons are packed with a constant spacing.

Question 9: Let Fpp, Fpn and Fnn denote the nuclear force between a proton–proton, proton–neutron and neutron–neutron pair respectively. When the separation is 1 fm,
(a) Fpp < Fpn = Fnn
(b) Fpp > Fpn = Fnn
(c) Fpp = Fpn = Fnn
(d) Fpp < Fpn < Fnn
Show Answer & Explanation

Answer: (c) Fpp = Fpn = Fnn

Explanation:
1. The nuclear force is independent of electric charge
2. So at the same separation (and in the same spin state) it is equal for p–p, p–n and n–n pairs.

Teacher's Note:
1. Charge independence is one of the key properties of the nuclear force.
2. The extra Coulomb repulsion between protons is a separate electric effect.

Question 10: A nuclear fission is given by A²⁴⁰ → B¹⁰⁰ + C¹⁴⁰ + Q (energy). The binding energy per nucleon of nuclei A, B and C is 7.6 MeV, 8.1 MeV and 8.1 MeV respectively. The value of Q is approximately
(a) 20 MeV
(b) 220 MeV
(c) 120 MeV
(d) 240 MeV
Show Answer & Explanation

Answer: (c) 120 MeV

Explanation:
1. Binding energy of A = 240 × 7.6 = 1824 MeV
2. Binding energy of B + C = 240 × 8.1 = 1944 MeV
3. Q = 1944 − 1824 = 120 MeV.

Teacher's Note:
1. Energy released = total binding energy of products − total binding energy of the parent.
2. Binding energy per nucleon times mass number gives total binding energy.

Question 11: The energy released when 10 kg of \({}^{235}_{92}\mathrm{U}\) is completely converted into energy is
(a) 5 × 10²⁷ MeV
(b) 5 × 10²⁴ MeV
(c) 9 × 10¹⁷ J
(d) all of these
Show Answer & Explanation

Answer: (c) 9 × 10¹⁷ J

Explanation:
1. Complete conversion of mass: E = mc² = 10 × (3 × 10⁸)² = 9 × 10¹⁷ J
2. In MeV this is 9 × 10¹⁷ / 1.6 × 10⁻¹³ = 5.6 × 10³⁰ MeV, so the first two values are wrong.

Teacher's Note:
1. When all the mass is converted, use E = mc² directly.
2. Fission converts only about 0.1% of the mass, so real fission would release much less.

Question 12: The energy released when 2 mole of \({}^{235}_{92}\mathrm{U}\) undergo fission (200 MeV per fission) is nearly
(a) 10²⁴ MeV
(b) 24 × 10²⁵ MeV
(c) 10²⁴ J
(d) 10²⁴ kWh
Show Answer & Explanation

Answer: (b) 24 × 10²⁵ MeV

Explanation:
1. Number of nuclei = 2 × 6.02 × 10²³ = 1.2 × 10²⁴
2. Energy = 1.2 × 10²⁴ × 200 MeV = 2.4 × 10²⁶ MeV = 24 × 10²⁵ MeV.

Teacher's Note:
1. Moles to nuclei: multiply by Avogadro's number.
2. Multiply by 200 MeV per fission for the total.

Question 13: As the mass number increases, the binding energy per nucleon
(a) increases
(b) decreases
(c) remains the same
(d) may increase or decrease
Show Answer & Explanation

Answer: (d) may increase or decrease

Explanation:
1. On the binding energy curve the value rises steeply for light nuclei up to about A = 56, then decreases slowly for heavy nuclei
2. So it can increase or decrease depending on the region.

Teacher's Note:
1. The curve has a broad maximum near iron (about 8.8 MeV per nucleon).
2. Fusion of light nuclei and fission of heavy nuclei both move towards the maximum.

Question 14: Possible forces on a proton by another proton in a nucleus are/is
(a) Coulomb force
(b) Nuclear force
(c) Gravitational force
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. Two protons repel each other by the Coulomb force, attract by the strong nuclear force, and also attract by the (very weak) gravitational force
2. So all three act.

Teacher's Note:
1. The nuclear force is much stronger than the Coulomb force at about 1 fm.
2. Gravity is negligible but is not zero.

Question 15: The energy radiated by a red giant star is produced by
(a) fission process
(b) fusion process
(c) chemical burning of hydrogen
(d) gravitational contraction
Show Answer & Explanation

Answer: (b) fusion process

Explanation:
1. Stars shine because of nuclear fusion of lighter nuclei (hydrogen, then helium and heavier elements) at very high temperature and pressure.

Teacher's Note:
1. The Sun fuses four hydrogen nuclei into one helium nucleus, releasing about 26 MeV.
2. Chemical burning cannot sustain a star for billions of years.

Question 16: In an atomic explosion, most of the fission energy is released in the form of
(a) γ-rays
(b) kinetic energy of products
(c) infra-red rays
(d) visible light
Show Answer & Explanation

Answer: (b) kinetic energy of products

Explanation:
1. About 93% of the energy released in fission appears as kinetic energy of the fission fragments and neutrons, and only a small part as γ-rays.

Teacher's Note:
1. Fission fragments carry away most of the 200 MeV.
2. This kinetic energy is quickly converted into heat.

Question 17: Which of the following nuclei is fissionable by slow neutrons?
(a) \({}^{238}_{92}\mathrm{U}\)
(b) \({}^{239}_{93}\mathrm{Np}\)
(c) \({}^{235}_{92}\mathrm{U}\)
(d) \({}^{4}_{2}\mathrm{He}\)
Show Answer & Explanation

Answer: (c) \({}^{235}_{92}\mathrm{U}\)

Explanation:
1. U-235 undergoes fission when it absorbs slow (thermal) neutrons of energy about 0.03 eV
2. U-238 needs fast neutrons and does not sustain a chain reaction; He-4 is stable.

Teacher's Note:
1. Natural uranium contains only 0.7% U-235.
2. A moderator is used to slow the neutrons down to thermal energies.

Question 18: An example of nuclear fusion is
(a) formation of barium and krypton from uranium
(b) formation of helium from hydrogen
(c) formation of plutonium-235 from uranium-235
(d) formation of water from hydrogen and oxygen
Show Answer & Explanation

Answer: (b) formation of helium from hydrogen

Explanation:
1. Fusion means joining light nuclei to form a heavier nucleus: hydrogen nuclei fuse to give helium
2. Barium and krypton formation is fission; water formation is a chemical reaction.

Teacher's Note:
1. Fusion needs temperatures of the order of 10⁷ K to overcome Coulomb repulsion.
2. The hydrogen bomb is based on fusion.

Question 19: An electron–positron pair can be created by γ-rays. The minimum energy of the γ-rays should be
(a) 5.0 MeV
(b) 4.02 MeV
(c) 15.0 MeV
(d) 1.02 MeV
Show Answer & Explanation

Answer: (d) 1.02 MeV

Explanation:
1. Rest energy of an electron or a positron = 0.51 MeV
2. Minimum photon energy = 2 × 0.51 = 1.02 MeV.

Teacher's Note:
1. Pair production takes place only in the field of a nucleus, which absorbs the momentum.
2. Any photon energy above 1.02 MeV is shared as kinetic energy of the pair.

Question 20: In the nuclear reaction \({}^{235}_{92}\mathrm{U}\) + n → \({}^{144}_{56}\mathrm{Ba}\) + ……… + 3n, the missing nucleus is
(a) \({}^{89}_{26}\mathrm{Kr}\)
(b) \({}^{89}_{36}\mathrm{Kr}\)
(c) \({}^{90}_{26}\mathrm{Sr}\)
(d) \({}^{89}_{38}\mathrm{Sr}\)
Show Answer & Explanation

Answer: (b) \({}^{89}_{36}\mathrm{Kr}\)

Explanation:
1. Mass number: 235 + 1 = 144 + A + 3, so A = 89
2. Charge: 92 = 56 + Z, so Z = 36
3. The nucleus is krypton-89.

Teacher's Note:
1. Balance A and Z separately; neutrons carry A = 1 and Z = 0.
2. Z = 36 identifies krypton.

Question 21: For the reaction \({}^{11}_{6}\mathrm{C}\) → \({}^{11}_{5}\mathrm{B}\) + β⁺ + X, the particle X is
(a) neutron
(b) anti-neutrino
(c) neutrino
(d) proton
Show Answer & Explanation

Answer: (c) neutrino

Explanation:
1. β⁺ emission (a proton changing into a neutron) is always accompanied by a neutrino, ν.

Teacher's Note:
1. β⁻ goes with an anti-neutrino, β⁺ with a neutrino.
2. The neutrino carries off the remaining energy and spin.

Question 22: In a breeder reactor, the useful fuel obtained from \({}^{238}_{92}\mathrm{U}\) is
(a) \({}^{233}_{89}\mathrm{Ac}\)
(b) \({}^{238}_{90}\mathrm{Th}\)
(c) \({}^{235}_{92}\mathrm{U}\)
(d) \({}^{239}_{94}\mathrm{Pu}\)
Show Answer & Explanation

Answer: (d) \({}^{239}_{94}\mathrm{Pu}\)

Explanation:
1. U-238 absorbs a neutron to form U-239, which changes by two β⁻ decays into Pu-239, a good fissile fuel.

Teacher's Note:
1. A breeder reactor produces more fissile material than it uses.
2. Pu-239 has a half-life of about 24000 years.

Question 23: Boron is used in an atomic reactor for
(a) absorption of neutrons
(b) absorption of α-particles
(c) speeding up the reaction
(d) changing the reaction
Show Answer & Explanation

Answer: (a) absorption of neutrons

Explanation:
1. Boron and cadmium are excellent neutron absorbers, so they are used as control rods to regulate the chain reaction.

Teacher's Note:
1. Pushing the rods in lowers the reproduction factor K; pulling them out raises it.
2. For steady power K = 1.

Question 24: Nuclear fission was discovered by
(a) Otto Hahn and Strassmann
(b) Fermi
(c) Bethe
(d) Rutherford
Show Answer & Explanation

Answer: (a) Otto Hahn and Strassmann

Explanation:
1. Hahn and Strassmann discovered fission of uranium in 1938–39; Meitner and Frisch explained it.

Teacher's Note:
1. Rutherford discovered the nucleus and the first artificial transmutation.
2. Fermi built the first nuclear reactor.

Question 25: Which one is the best neutron moderator in all respects?
(a) barium oxide
(b) water
(c) graphite
(d) heavy water
Show Answer & Explanation

Answer: (d) heavy water

Explanation:
1. A good moderator is light and absorbs very few neutrons
2. Heavy water (D₂O) slows neutrons well with extremely low absorption; ordinary water absorbs more, so it is less suitable for natural uranium.

Teacher's Note:
1. Graphite and heavy water are the common moderators.
2. The moderator should reduce neutron energy to thermal values without capturing them.

Question 26: In the nuclear reaction X(n, α) \({}^{7}_{3}\mathrm{Li}\), the element X is
(a) \({}^{10}_{5}\mathrm{B}\)
(b) \({}^{9}_{5}\mathrm{B}\)
(c) \({}^{11}_{4}\mathrm{Be}\)
(d) \({}^{4}_{2}\mathrm{He}\)
Show Answer & Explanation

Answer: (a) \({}^{10}_{5}\mathrm{B}\)

Explanation:
1. X + n → Li-7 + α (He-4)
2. A: AX + 1 = 7 + 4, so AX = 10
3. Z: ZX + 0 = 3 + 2, so ZX = 5 (boron).

Teacher's Note:
1. In X(a, b)Y notation, a is the projectile and b the emitted particle.
2. This reaction is also used to detect neutrons using boron.

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