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Mass-energy relation, mass defect and binding energy: NEET MCQs with answers

52 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 mass defect of a nucleus is 0.05 u. Taking 1 u = 931.5 MeV/c², its binding energy is:

  1. 46.6 MeV
  2. 23.3 MeV
  3. 0.466 MeV
  4. 18630 MeV
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Answer: (A) 46.6 MeV

BE = Δm × 931.5 = 0.05 × 931.5 = 46.6 MeV.

medium

2. Consider the following statements regarding mass defect and binding energy: Statement I: The mass defect of any stable bound nucleus is always strictly positive. Statement II: When individual nucleons combine to form a bound nucleus, energy is released, resulting in a system rest mass smaller than the sum of the constituent masses. In light of the above statements, choose the most appropriate answer:

  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.

For any stable bound nucleus, the energy of the bound state is less than that of free nucleons separated at infinity. Thus, energy is released upon formation, leading to a decrease in rest mass (E = mc²). Consequently, mass defect ΔM > 0 for all stable nuclei. Both statements are correct.

hard

3. If the velocity of light in vacuum c were reduced by a factor of 2, while the mass defect Δm for a given nucleus remained unaltered, the nuclear binding energy of that nucleus would:

  1. Increase by a factor of 2
  2. Decrease by a factor of 2
  3. Remain unchanged
  4. Decrease by a factor of 4
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Answer: (D) Decrease by a factor of 4

According to the mass-energy relationship, Binding Energy E_b = Δm·c². If the speed of light c is halved (c' = c/2), the binding energy becomes E_b' = Δm·(c/2)² = (Δm·c²)/4 = E_b / 4. Thus, it decreases by a factor of 4.

medium

4. A nuclear reactor generates useful electrical power of 450 MW at an overall efficiency of 25%. The rate of decrease of the rest mass of the fuel in the core is:

  1. 2.0 × 10⁻⁸ kg/s
  2. 5.0 × 10⁻⁹ kg/s
  3. 1.8 × 10⁻⁷ kg/s
  4. 2.0 × 10⁻⁵ kg/s
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Answer: (A) 2.0 × 10⁻⁸ kg/s

Thermal power generated = Electrical power / Efficiency = 450 MW / 0.25 = 1800 MW = 1.8 × 10⁹ W. From Einstein's mass-energy equivalence, P_thermal = (Δm/Δt)·c². Therefore, Δm/Δt = (1.8 × 10⁹) / (3 × 10⁸)² = (1.8 × 10⁹) / (9 × 10¹⁶) = 2.0 × 10⁻⁸ kg/s.

easy

5. A nucleus X has a mass number of 60 and its binding energy per nucleon is 8.5 MeV. What is the total energy required to completely break this nucleus into its constituent individual protons and neutrons?

  1. 141.2 MeV
  2. 1020.0 MeV
  3. 255.0 MeV
  4. 510.0 MeV
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Answer: (D) 510.0 MeV

The binding energy represents the total energy needed to separate a nucleus into its individual nucleons. Total binding energy = (Binding energy per nucleon) × A = 8.5 MeV × 60 = 510 MeV.

hard

6. A stationary nucleus with mass number A breaks into two fragments having mass numbers in the ratio 1 : 2. If the total energy released during the disintegration is Q and is entirely converted into kinetic energy of the fragments, the kinetic energy of the lighter fragment is:

  1. Q / 3
  2. 2Q / 3
  3. Q / 2
  4. 3Q / 4
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Answer: (B) 2Q / 3

By conservation of linear momentum, both fragments have equal and opposite momenta: p₁ = p₂ = p. Kinetic energy K = p² / (2m), so K₁ / K₂ = m₂ / m₁ = 2 / 1. Since K₁ + K₂ = Q, we have K₁ + K₁/2 = Q, which gives (3/2)K₁ = Q, hence K₁ = 2Q / 3.

easy

7. The energy equivalent of a rest mass of exactly 2 atomic mass units (2 u) is closest to:

  1. 931.5 MeV
  2. 2794.5 MeV
  3. 465.8 MeV
  4. 1863.0 MeV
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Answer: (D) 1863.0 MeV

One unified atomic mass unit (1 u) is equivalent to approximately 931.5 MeV of energy. Therefore, the energy equivalent of 2 u of rest mass is 2 × 931.5 MeV = 1863.0 MeV.

medium

8. An electron and a positron, each having a kinetic energy of 0.25 MeV, collide and annihilate each other head-on to produce two identical gamma-ray photons. Taking the rest mass energy of an electron as 0.51 MeV, the energy of each emitted photon is:

  1. 0.76 MeV
  2. 0.51 MeV
  3. 1.52 MeV
  4. 0.26 MeV
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Answer: (A) 0.76 MeV

Total initial energy of the electron-positron system = 2 × (m_e c² + K) = 2 × (0.51 + 0.25) = 2 × 0.76 MeV = 1.52 MeV. Since two identical photons are emitted in opposite directions to conserve momentum, the energy of each photon is 1.52 / 2 = 0.76 MeV.

medium

9. A hypothetical nucleus ₁₀X²⁴ has a total binding energy of 180 MeV. A second nucleus ₂₀Y⁴⁸ has a binding energy per nucleon that is 1.5 times that of nucleus X. What is the total binding energy of nucleus Y?

  1. 270 MeV
  2. 360 MeV
  3. 540 MeV
  4. 480 MeV
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Answer: (C) 540 MeV

The binding energy per nucleon of nucleus X is (E_bn)X = 180 MeV / 24 = 7.5 MeV/nucleon. For nucleus Y, (E_bn)Y = 1.5 × 7.5 MeV/nucleon = 11.25 MeV/nucleon. Therefore, the total binding energy of nucleus Y is 11.25 MeV/nucleon × 48 nucleons = 540 MeV.

easy

10. If the binding energy per nucleon of a nucleus with mass number A = 16 is 7.97 MeV, the total energy required to completely dissociate the nucleus into its individual constituent nucleons is:

  1. 63.76 MeV
  2. 255.04 MeV
  3. 127.52 MeV
  4. 31.88 MeV
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Answer: (C) 127.52 MeV

The total binding energy equals the binding energy per nucleon multiplied by the mass number A. Here, total binding energy = 16 × 7.97 MeV = 127.52 MeV. This is the exact energy needed to disassemble the nucleus completely into separate nucleons.

medium

11. In a fusion reactor, the reaction ₁H² + ₁H³ → ₂He⁴ + ₀n¹ takes place. The binding energies of ₁H², ₁H³, and ₂He⁴ are 2.22 MeV, 8.48 MeV, and 28.30 MeV, respectively. The kinetic energy carried away by the products per fusion event is:

  1. 22.04 MeV
  2. 17.60 MeV
  3. 19.82 MeV
  4. 14.40 MeV
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Answer: (B) 17.60 MeV

Energy released Q = [BE(₂He⁴)] - [BE(₁H²) + BE(₁H³)] = 28.30 MeV - (2.22 MeV + 8.48 MeV) = 28.30 - 10.70 = 17.60 MeV. A neutron is an unbound single nucleon, so its binding energy is zero.

medium

12. A nucleus ₆C¹² absorbs a gamma-ray photon and disintegrates into three identical alpha particles (₂He⁴). The minimum energy of the gamma-ray photon required for this process to occur is (take rest masses: M(₆C¹²) = 12.0000 u, M(₂He⁴) = 4.0015 u, and 1 u = 931.5 MeV/c²):

  1. 7.27 MeV
  2. 4.19 MeV
  3. 2.79 MeV
  4. 1.39 MeV
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Answer: (B) 4.19 MeV

Mass of products = 3 × 4.0015 u = 12.0045 u. Mass of reactant = 12.0000 u. The increase in mass Δm = 12.0045 - 12.0000 = 0.0045 u. The threshold photon energy is E = Δm × 931.5 MeV = 0.0045 × 931.5 ≈ 4.19 MeV.

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