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A2 · Practice questions · Mass defect and binding energy

The missing mass is the glue.

Six original Cambridge-style questions on mass defect, binding energy and E = Δmc², binding energy per nucleon as the measure of stability, and the energy released in fusion and fission.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
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Mass defect times c² is binding energy.

01
Analysis
[2 marks]

Define the mass defect of a nucleus and its binding energy, and state the relation between them.

  • The mass defect is the difference between the total mass of the separate nucleons (protons and neutrons) and the mass of the nucleus, the nucleus being the lighter. ✓
  • The binding energy is the minimum energy needed to separate the nucleus completely into its individual nucleons (equivalently the energy released when it forms), and it equals the mass defect times c²: E = Δmc². ✓
02
Calculation
[3 marks]

A carbon-12 nucleus contains 6 protons and 6 neutrons and has a mass of 11.99671 u. The mass of a proton is 1.00728 u and the mass of a neutron is 1.00867 u. Calculate the mass defect of the nucleus in u, its binding energy in MeV, and its binding energy per nucleon. (1 u is equivalent to 931 MeV.)

  • Δm = 6(1.00728) + 6(1.00867) − 11.99671 = 12.09570 − 11.99671 = 0.09899 u. ✓
  • Binding energy = 0.09899 × 931 = 92.2 MeV. ✓
  • Per nucleon: 92.2 / 12 = 7.68 MeV. ✓
03
Calculation
[3 marks]

The mass defect of an oxygen-16 nucleus is 0.1370 u. Working in SI units, calculate the binding energy of the nucleus in joules, convert it to MeV, and find the binding energy per nucleon. (1 u = 1.66 × 10⁻²⁷ kg.)

  • Δm = 0.1370 × 1.66 × 10⁻²⁷ = 2.274 × 10⁻²⁸ kg, so E = Δmc² = 2.274 × 10⁻²⁸ × (3.00 × 10⁸)² = 2.05 × 10⁻¹¹ J. ✓
  • E = 2.05 × 10⁻¹¹ / 1.60 × 10⁻¹³ = 128 MeV. ✓
  • Per nucleon: 128 / 16 = 8.0 MeV. ✓
04
Analysis
[3 marks]

The graph of binding energy per nucleon against nucleon number rises steeply for light nuclei, reaches a maximum near A = 56 and falls slowly for heavier nuclei. Use the graph to explain why energy is released when a uranium-235 nucleus undergoes fission, and explain why fusing two iron-56 nuclei together would not release energy.

  • Uranium lies on the slowly falling part of the curve to the right of the peak, so its binding energy per nucleon is lower than that of the medium-mass fragments it splits into. ✓
  • The fragments are more tightly bound: the total binding energy of the products exceeds that of the uranium nucleus, and the difference is released (the products have less total mass). ✓
  • Iron-56 is at the peak, so a nucleus of A = 112 formed from two of them would have a lower binding energy per nucleon than iron; its total binding energy would be smaller, so energy would have to be supplied rather than released. ✓
05
Calculation
[3 marks]

Two deuterium nuclei fuse to form a helium-3 nucleus and a neutron: ²₁H + ²₁H → ³₂He + ¹₀n. The masses are: ²₁H 2.01410 u, ³₂He 3.01603 u, neutron 1.00867 u. Calculate the energy released in one fusion, in MeV and in joules. (1 u is equivalent to 931 MeV.)

  • Δm = 2(2.01410) − (3.01603 + 1.00867) = 4.02820 − 4.02470 = 0.00350 u. ✓
  • E = 0.00350 × 931 = 3.26 MeV. ✓
  • E = 3.26 × 10⁶ × 1.60 × 10⁻¹⁹ = 5.2 × 10⁻¹³ J. ✓
06
Calculation
[4 marks]

One possible fission of uranium-235 is ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n. The masses are: U-235 235.0439 u, Ba-141 140.9144 u, Kr-92 91.9262 u, neutron 1.00867 u. (1 u is equivalent to 931 MeV.)

(a) Calculate the decrease in mass in this fission, in u. (b) Calculate the energy released, in MeV and in joules. (c) A reactor produces 1.0 GW of thermal power from this reaction alone. Calculate the number of fissions per second.

  • (a) Reactants: 235.0439 + 1.00867 = 236.05257 u; products: 140.9144 + 91.9262 + 3(1.00867) = 235.86661 u; Δm = 0.1860 u. ✓
  • (b) E = 0.1860 × 931 = 173 MeV. ✓
  • E = 173 × 10⁶ × 1.60 × 10⁻¹⁹ = 2.77 × 10⁻¹¹ J. ✓
  • (c) Fissions per second = P / E = 1.0 × 10⁹ / 2.77 × 10⁻¹¹ = 3.6 × 10¹⁹ s⁻¹. ✓

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