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Nuclear fission and fusion: NEET MCQs with answers

49 questions on this topic of Atoms and Nuclei (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. The energy released per fission of a uranium-235 nucleus is about:

  1. 20 keV
  2. 2000 MeV
  3. 200 MeV
  4. 2 MeV
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Answer: (C) 200 MeV

When ²³⁵U absorbs a neutron and splits into intermediate-mass fragments (e.g. Ba and Kr), about 200 MeV of energy is released along with 2-3 neutrons.

NCERT Class 12

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2. Consider the following two statements regarding nuclear reactions: Statement I: Fast neutrons are more effective than thermal neutrons in causing fission of ²³⁵₉₂U. Statement II: Prompt neutrons are released practically instantaneously at the moment of nuclear fission. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (C) Statement I is incorrect but Statement II is correct

Statement I is incorrect because ²³⁵U has a much higher fission cross-section for slow (thermal) neutrons than for fast neutrons. Statement II is correct because prompt neutrons are emitted within about 10⁻¹⁴ seconds of fission.

hard

3. Why can ²³⁵₉₂U undergo fission with thermal (zero kinetic energy) neutrons, whereas ²³⁸₉₂U requires fast neutrons with kinetic energy greater than 1 MeV to undergo fission?

  1. ²³⁵U has a higher binding energy per nucleon than ²³⁸U in its ground state.
  2. ²³⁸U has a much smaller geometric cross-section than ²³⁵U.
  3. The neutron binding energy added to ²³⁵U forms an even-even compound nucleus ²³⁶U, providing extra pairing energy that exceeds the fission barrier.
  4. The nuclear force in ²³⁸U is purely repulsive for slow incoming neutrons.
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Answer: (C) The neutron binding energy added to ²³⁵U forms an even-even compound nucleus ²³⁶U, providing extra pairing energy that exceeds the fission barrier.

When ²³⁵₉₂U (an even-odd nucleus) absorbs a neutron, it becomes ²³⁶₉₂U (an even-even nucleus). The pairing energy released raises the excitation energy of ²³⁶U to ~6.5 MeV, which is greater than its fission barrier (~5.7 MeV). For ²³⁸₉₂U, absorbing a neutron forms ²³⁹₉₂U (even-odd), releasing only ~4.8 MeV of excitation, which is below its ~5.8 MeV barrier; hence an extra ~1 MeV of kinetic energy is necessary.

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4. An unmoderated fast neutron with an initial kinetic energy of 2.0 MeV undergoes an elastic, head-on collision with a stationary carbon nucleus (¹²₆C). Assuming non-relativistic mechanics, what fraction of its initial kinetic energy does the neutron retain after the collision?

  1. 0.284
  2. 0.716
  3. 0.852
  4. 0.545
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Answer: (B) 0.716

For an elastic head-on collision of mass m₁ with a stationary mass m₂, the final velocity of m₁ is v₁' = [(m₁ - m₂) / (m₁ + m₂)] v₁. The fraction of retained kinetic energy is (v₁'/v₁)² = [(m₁ - m₂) / (m₁ + m₂)]². Here m₁ ≈ 1 u and m₂ ≈ 12 u, so fraction retained = [(1 - 12) / (1 + 12)]² = (-11/13)² = 121 / 169 ≈ 0.716.

easy

5. In a hypothetical fission reaction, a nucleus of mass number A = 240 with binding energy per nucleon of 7.6 MeV splits into two fragments of mass number A = 120 each, having binding energy per nucleon of 8.5 MeV. The net energy released per fission is:

  1. 108 MeV
  2. 180 MeV
  3. 216 MeV
  4. 90 MeV
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Answer: (C) 216 MeV

Total initial binding energy = 240 × 7.6 MeV = 1824 MeV. Total final binding energy = 2 × (120 × 8.5 MeV) = 2040 MeV. The energy released Q = BE(final) - BE(initial) = 2040 - 1824 = 216 MeV.

hard

6. In magnetic confinement fusion (tokamak), the Lawson criterion for achieving ignition in a D-T plasma specifies that the product of plasma ion density n and energy confinement time τ_E must satisfy:

  1. n τ_E² ≥ 1.5 × 10²⁴ m⁻³ s² at T ≈ 1 MeV
  2. n τ_E ≤ 1.5 × 10¹⁴ m⁻³ s at T ≈ 100 eV
  3. n / τ_E ≥ 1.5 × 10²⁰ m⁻³ s⁻¹ at T ≈ 1 keV
  4. n τ_E ≥ 1.5 × 10²⁰ m⁻³ s at T ≈ 10–20 keV
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Answer: (D) n τ_E ≥ 1.5 × 10²⁰ m⁻³ s at T ≈ 10–20 keV

The Lawson criterion for D-T fusion requires n τ_E ≥ 1.5 × 10²⁰ s/m³ (or 10¹⁴ s/cm³) at temperatures of 10 to 20 keV (around 10⁸ K) so that the fusion power output exceeds the plasma energy losses.

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7. Two deuterons (²₁H) undergo nuclear fusion to form a helium nucleus (⁴₂He). The binding energy per nucleon of ²₁H is 1.1 MeV and that of ⁴₂He is 7.0 MeV. The total energy liberated in this fusion reaction is:

  1. 23.6 MeV
  2. 11.8 MeV
  3. 5.9 MeV
  4. 28.0 MeV
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Answer: (A) 23.6 MeV

Total binding energy of reactants (two ²₁H nuclei) = 2 × (2 × 1.1 MeV) = 4.4 MeV. Total binding energy of product (⁴₂He) = 4 × 7.0 MeV = 28.0 MeV. Energy released = BE(products) - BE(reactants) = 28.0 - 4.4 = 23.6 MeV.

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8. The energy released in the fusion of 1.0 kg of deuterium (²₁H) via the reaction 2 ²₁H → ⁴₂He + 24.0 MeV is denoted by E_fusion. The energy released in the complete fission of 1.0 kg of ²³⁵U via a typical fission process (200 MeV per event) is E_fission. The ratio E_fusion / E_fission is approximately:

  1. 14.2
  2. 3.5
  3. 7.0
  4. 0.28
Show answer and explanation

Answer: (C) 7.0

For 1 kg of deuterium, number of pairs = (1000 / 4) × N_A = 250 N_A. Thus E_fusion = 250 N_A × 24.0 MeV = 6000 N_A MeV. For 1 kg of ²³⁵U, number of fissions = (1000 / 235) N_A fissions. E_fission = (1000 / 235) N_A × 200 MeV ≈ 851 N_A MeV. Therefore, E_fusion / E_fission = 6000 / 851 ≈ 7.05 ≈ 7.0.

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9. Consider the following statements regarding nuclear fission and fusion: Statement I: The energy released per nucleon in nuclear fusion of light nuclei is generally greater than that in nuclear fission of heavy nuclei. Statement II: Both fission and fusion processes result in daughter nuclei with a higher binding energy per nucleon compared to the parent reactants.

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

Statement I is correct because fusion yields around 6 to 7 MeV per nucleon (e.g., in D-T fusion), whereas fission yields roughly 0.85 MeV per nucleon. Statement II is correct because both processes move nuclei toward the peak of the binding energy per nucleon curve (near Fe-56), releasing energy as mass defect.

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10. In a binary nuclear fission event, a parent nucleus ²³⁶₉₂U at rest splits into two fragments of mass numbers A₁ = 95 and A₂ = 139 along with two free neutrons. Neglecting the kinetic energy carried by the two neutrons, what is the approximate ratio of the kinetic energy of fragment 1 to that of fragment 2 (K₁ / K₂)?

  1. 1.46
  2. 0.68
  3. 2.14
  4. 1.00
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Answer: (A) 1.46

By conservation of linear momentum for the two fragments, p₁ = p₂. The kinetic energy is related to momentum by K = p² / (2m). Since mass is proportional to mass number A, K₁ / K₂ = m₂ / m₁ = A₂ / A₁ = 139 / 95 ≈ 1.46.

easy

11. In the carbon-nitrogen (C-N) cycle responsible for energy generation in stars hotter than the Sun, carbon nuclei fundamentally act as:

  1. Nuclear fuel consumed to form magnesium
  2. Coolants to dissipate stellar radiative pressure
  3. Moderators to absorb excess fast neutrons
  4. Nuclear catalysts that are regenerated in the cycle
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Answer: (D) Nuclear catalysts that are regenerated in the cycle

In the C-N cycle, carbon nuclei absorb protons in sequential steps and eventually regenerate a ¹²₆C nucleus while releasing a helium nucleus (⁴₂He), thereby acting as a nuclear catalyst.

easy

12. In the nuclear fission reaction ²³⁵₉₂U + ¹₀n → ¹⁴⁴₅₆Ba + ⁸⁹₃₆Kr + x ¹₀n, the number x of neutrons released is:

  1. 1
  2. 2
  3. 4
  4. 3
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Answer: (D) 3

Conserving mass number across the reaction gives 235 + 1 = 144 + 89 + x(1). Thus, 236 = 233 + x, which yields x = 3. Atomic numbers are already balanced (92 + 0 = 56 + 36 + 0).

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