OCR A Level Physics A: Mass-energy, binding energy, fission and fusion (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 6.4.4 (Nuclear fission and fusion), outcomes (a) to (k), of the OCR A Level Physics A (H556) specification, Version 3.0 (March 2026). Learn the content from the study guide and revision notes. See also the course hub, checklist and free diagnostics.

Data: c = 3.00 × 10⁸ m s⁻¹; 1 u = 1.661 × 10⁻²⁷ kg, equivalent to 931.5 MeV; 1 MeV = 1.60 × 10⁻¹³ J; e = 1.60 × 10⁻¹⁹ C; h = 6.63 × 10⁻³⁴ J s; k = 1.38 × 10⁻²³ J K⁻¹; 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻²; mass of an electron = 9.11 × 10⁻³¹ kg.

Questions

1. Define the mass defect and the binding energy of a nucleus. [2]

2. In a nuclear reaction the total mass of the products is 3.6 × 10⁻²⁹ kg less than the total mass of the reactants. Calculate the energy released, in joules and in MeV. [3]

3. A photon passes close to a nucleus and may produce an electron-positron pair.

(a) Calculate the energy of a photon of wavelength 1.5 × 10⁻¹² m. [2] (b) Show that this photon cannot produce an electron-positron pair. [2] (c) A photon of wavelength 0.80 × 10⁻¹² m does produce a pair. Assuming the nucleus takes negligible energy and the two particles share the rest equally, calculate the kinetic energy of each particle. [3]

4. The atomic mass of aluminium-27 is 26.981539 u. A hydrogen atom has mass 1.007825 u and a neutron 1.008665 u. Aluminium-27 has 13 protons. Calculate the binding energy per nucleon of aluminium-27 in MeV. [4]

5. Describe the main features of the curve of binding energy per nucleon plotted against nucleon number, and use it to explain why both fusion and fission release energy. [5]

6. A fission reaction is ²³⁵₉₂U + ¹₀n → ¹⁴⁰₅₅Cs + ⁹³₃₇Rb + x ¹₀n. The binding energy per nucleon is 7.59 MeV for U-235, 8.26 MeV for Cs-140 and 8.52 MeV for Rb-93.

(a) Find x. [1] (b) Calculate the energy released by one fission, in MeV. [4] (c) Express this energy in joules. [1]

7. Complete the equations.

(a) ²³⁹₉₂U → ?Np + ⁰₋₁e + antineutrino, giving the proton number of neptunium. [2] (b) ²₁H + ³₂He → ⁴₂He + ? [2]

8. This question is about a thermal fission reactor.

(a) Describe the function of the moderator and name a suitable material. [2] (b) Describe the function of the control rods and how they are used to reduce the reaction rate. [2] (c) Explain why, in a steady reactor, each fission must lead on average to exactly one further fission. [2]

9. A reactor has a thermal power of 2.4 × 10⁹ W. Each fission releases 190 MeV.

(a) Calculate the energy released per fission in joules. [1] (b) Calculate the number of fissions per second. [2] (c) Calculate the mass of uranium-235 fissioned in one day (1 day = 8.64 × 10⁴ s). Take the mass of one nucleus as 235 u. [3]

10. Discuss the environmental impact of nuclear waste from fission reactors. [5]

11. Two deuterium nuclei, each of charge +e, are brought head-on to a separation of 3.0 × 10⁻¹⁵ m.

(a) Calculate their electric potential energy. [2] (b) Assuming each nucleus supplies half this energy as kinetic energy, estimate the temperature at which the mean kinetic energy of a nucleus is (3/2)kT. [3] (c) The temperature at the centre of the Sun is about 1.5 × 10⁷ K. Comment on your answer. [1]

12. Deuterium and helium-3 can fuse: ²₁H + ³₂He → ⁴₂He + ¹₁H. Atomic masses: H-2 2.014102 u, He-3 3.016029 u, He-4 4.002603 u, H-1 1.007825 u. Total binding energies: H-2 2.22 MeV, He-3 7.72 MeV, He-4 28.3 MeV.

(a) Show that the equation balances. [1] (b) Calculate the energy released using the masses. [3] (c) Calculate the energy released using the binding energies. [2] (d) Explain why this reaction needs a very high temperature although energy is released. [2]

Answers

1. Mass defect: the total mass of the free protons and neutrons less the mass of the nucleus [1]. Binding energy: the energy required to pull the nucleus apart into free nucleons (equal to the mass defect × c²) [1]. [2] Examiner insight: Say “free nucleons” and “pull apart”; a vague phrase such as “energy holding the nucleus together” is easy to misread.

2. E = Δmc² = 3.6 × 10⁻²⁹ × (3.00 × 10⁸)² [1]; = 3.24 × 10⁻¹² J [1]; ÷ 1.60 × 10⁻¹³ = 20 MeV (20.25) [1]. [3] Examiner insight: Show the conversion from joules to MeV as a separate line; the unit is part of the answer.

3. (a) E = hc/λ [1] = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 1.5 × 10⁻¹² = 1.33 × 10⁻¹³ J [1]. [2] (b) Minimum energy needed = 2mc² = 2 × 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 1.64 × 10⁻¹³ J [1]. The photon has less than this (1.33 × 10⁻¹³ J), so it cannot create a pair [1]. [2] (c) E = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 0.80 × 10⁻¹² = 2.49 × 10⁻¹³ J [1]; energy left for motion = 2.49 × 10⁻¹³ − 1.64 × 10⁻¹³ = 8.5 × 10⁻¹⁴ J [1]; each particle 4.2 × 10⁻¹⁴ J (about 0.26 MeV) [1]. [3] Examiner insight: In a “show that” question, state the threshold value and make the comparison explicitly; a calculation without the conclusion leaves the point unmade.

4. Mass of constituents = 13 × 1.007825 + 14 × 1.008665 = 13.101725 + 14.12131 = 27.223035 u [1]. Mass defect = 27.223035 − 26.981539 = 0.241496 u [1]. Binding energy = 0.241496 × 931.5 = 225 MeV [1]. Per nucleon = 225 / 27 = 8.33 MeV [1]. [4] Examiner insight: Pair hydrogen atom masses with an atomic mass of the nuclide; mixing nuclear and atomic masses gives a wrong mass defect.

5. Binding energy per nucleon rises steeply for small nucleon numbers, with a local peak at helium-4 [1]. It reaches a maximum of about 8.8 MeV near A = 56 (iron) [1]. It then falls slowly for heavier nuclei [1]. In fusion, light nuclei join to give a nucleus with a higher binding energy per nucleon [1]. In fission, a heavy nucleus splits into fragments nearer the peak with a higher binding energy per nucleon; in both cases total binding energy increases, so energy is released [1]. [5] Examiner insight: Link the direction of movement on the curve to the release of energy; naming the peak alone does not explain the process.

6. (a) 235 + 1 = 140 + 93 + x, so x = 3 [1]. [1] (b) Cs-140: 140 × 8.26 = 1156.4 MeV [1]; Rb-93: 93 × 8.52 = 792.4 MeV [1]; U-235: 235 × 7.59 = 1783.7 MeV [1]; energy released = 1948.8 − 1783.7 = 165 MeV [1]. [4] (c) 165 × 1.60 × 10⁻¹³ = 2.64 × 10⁻¹¹ J [1]. [1] Examiner insight: Say which total is subtracted from which; a negative energy released should prompt a check.

7. (a) Proton number rises by 1 [1]; neptunium has proton number 93, ²³⁹₉₃Np [1]. [2] (b) Nucleon number 5 − 4 = 1 and proton number 3 − 2 = 1 [1]; the missing particle is a proton, ¹₁H [1]. [2] Examiner insight: Show both the nucleon number and proton number checks in your working.

8. (a) It slows fast neutrons [1] by collisions with its light nuclei, so they are more likely to cause fission; water or graphite [1]. [2] (b) They absorb neutrons [1]; lowering them further into the core absorbs more neutrons and reduces the rate [1]. [2] (c) If more than one, the number of fissions grows and the reaction runs away [1]; if fewer than one, the chain dies away, so exactly one keeps the power constant [1]. [2] Examiner insight: Keep the roles separate: “slows neutrons” belongs to the moderator and “absorbs neutrons” to the control rods.

9. (a) 190 × 1.60 × 10⁻¹³ = 3.04 × 10⁻¹¹ J [1]. [1] (b) Rate = power / energy per fission [1] = 2.4 × 10⁹ / 3.04 × 10⁻¹¹ = 7.9 × 10¹⁹ s⁻¹ [1]. [2] (c) Mass of nucleus = 235 × 1.661 × 10⁻²⁷ = 3.90 × 10⁻²⁵ kg [1]; mass per second = 7.89 × 10¹⁹ × 3.90 × 10⁻²⁵ = 3.08 × 10⁻⁵ kg [1]; per day = 3.08 × 10⁻⁵ × 8.64 × 10⁴ = 2.7 kg [1]. [3] Examiner insight: Carry extra digits and round only at the end.

10. Spent fuel and fission products are radioactive and emit ionising radiation [1]; some nuclides have half-lives of thousands of years or more, so waste must be kept secure for a very long time [1]; management includes cooling ponds, sealing in glass and deep underground storage in stable rock [1]; a benefit is that the waste volume is small and fission releases no carbon dioxide in operation [1]; a reasoned judgement that weighs the long-term risk against these benefits [1]. [5] Examiner insight: A discussion question rewards both sides and a conclusion; a list of hazards alone is one-sided.

11. (a) U = (1/4πε₀) × e² / r [1] = 8.99 × 10⁹ × (1.60 × 10⁻¹⁹)² / 3.0 × 10⁻¹⁵ = 7.7 × 10⁻¹⁴ J [1]. [2] (b) Each nucleus has 7.67 × 10⁻¹⁴ / 2 = 3.84 × 10⁻¹⁴ J [1]; (3/2)kT = 3.84 × 10⁻¹⁴ [1]; T = 2 × 3.84 × 10⁻¹⁴ / (3 × 1.38 × 10⁻²³) = 1.9 × 10⁹ K [1]. [3] (c) The estimate is far higher than the Sun’s core temperature, because only the faster nuclei need this energy, so fusion occurs even though the mean energy is lower [1]. [1] Examiner insight: Compare the two numbers and offer a reason.

12. (a) Nucleon number 2 + 3 = 5 = 4 + 1; proton number 1 + 2 = 3 = 2 + 1 [1]. [1] (b) Before = 2.014102 + 3.016029 = 5.030131 u [1]; after = 4.002603 + 1.007825 = 5.010428 u [1]; Δm = 0.019703 u, energy = 0.019703 × 931.5 = 18.4 MeV [1]. [3] (c) Binding energy before = 2.22 + 7.72 = 9.94 MeV; after = 28.3 MeV (a proton has none) [1]; released = 28.3 − 9.94 = 18.4 MeV, agreeing with (b) [1]. [2] (d) Both nuclei are positively charged, so they repel [1]; they need very high kinetic energy, and so a very high temperature, to come close enough for the strong nuclear force to act [1]. [2] Examiner insight: When two methods are asked for, state that they agree; it shows the answer has been checked.

Where marks are usually lost

• Forgetting to convert u or grams to kilograms before using ΔE = Δmc².
• Giving one photon instead of two for annihilation, or not showing the 2mc² threshold.
• Dividing binding energy by proton number instead of nucleon number.
• Mixing atomic and nuclear masses in a mass-defect sum.
• Subtracting in the wrong order, giving a negative energy released.
• Confusing the roles of moderator and control rods.
• Writing a waste answer with only hazards, or with no reason for the claim.

Next steps

• Revision notes
• Study guide
• OCR A Level Physics hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OCR Level 3 Advanced GCE in Physics A (H556), Version 3.0 (March 2026), first assessment 2017, published by OCR. Section 6.4.4, outcomes (a) to (k).