OxfordAQA International AS and A-level Physics: Induced fission, reactor safety and nuclear fusion (9630) – 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.

For mass-energy background see the nuclear radius and binding energy guide. These questions cover sections 3.12.3 to 3.12.5 of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, for exams from May/June 2020 onwards. Section 3.12 is International A-level only. Use k = 1.38 × 10⁻²³ J K⁻¹, e = 1.60 × 10⁻¹⁹ C, ε₀ = 8.85 × 10⁻¹² F m⁻¹, 1 MeV = 1.60 × 10⁻¹³ J, 1 u = 1.661 × 10⁻²⁷ kg = 931.5 MeV, R₀ = 1.2 × 10⁻¹⁵ m and c = 3.00 × 10⁸ m s⁻¹ unless a question gives other values.

Links: study guide, revision notes, the OxfordAQA Physics hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. Neutrons in a reactor coolant at 310 K are in thermal equilibrium. The mass of a neutron is 1.675 × 10⁻²⁷ kg. (a) Calculate the mean kinetic energy of a neutron in eV. [2] (b) Calculate the corresponding speed. [1]

2. A reactor has k = 1.02, with 1000 fissions in one generation. (a) Calculate the number of fissions per generation after a further 50 generations. [2] (b) Explain how the operator brings k back to 1. [2]

3. A neutron of kinetic energy 1.5 MeV collides elastically and head-on with a helium-4 nucleus at rest, taken as 4 times the neutron mass. (a) Show that the neutron keeps 0.36 of its kinetic energy. [3] (b) Calculate the number of such collisions needed to reduce the energy to 0.030 eV. [2]

4. A gas-cooled reactor produces 600 MW of thermal power. The gas has specific heat capacity 1.1 × 10³ J kg⁻¹ K⁻¹ and warms by 200 K. (a) Calculate the mass of gas that must pass through the core each second. [2] (b) Give one property that makes carbon dioxide suitable as a coolant and one drawback shown by your answer. [2]

5. A power station has electrical output 900 MW at 33% efficiency. Each fission releases 190 MeV. (a) Calculate the thermal power. [1] (b) Calculate how many fissions occur each second. [2] (c) Find the mass of uranium-235 (235 u per nucleus) that undergoes fission in one day. [2]

6. A reactor operates at 2400 MW thermal. (a) Explain why spent fuel is moved by remote handling under water. [2] (b) The control rods are fully inserted in an emergency. Immediately afterwards, decay heat is 4.0% of the full power. Calculate this power and explain why it still needs cooling. [3]

7. A store holds waste with a strontium-90 activity of 3.6 × 10¹⁵ Bq. The half-life is 28.8 years. (a) Calculate the activity after 100 years. [2] (b) Calculate the time for the activity to fall to 1.0% of its initial value. [2] (c) State one way this class of waste is stored. [1]

8. A country plans to replace coal-fired stations with fission reactors. Evaluate this plan. [6]

9. Compare two fusion reactions: deuterium with tritium, and helium-4 with helium-4. Treat each pair as touching spheres of radius R₀A^(1/3). (a) Calculate the electrical potential energy for deuterium-tritium at closest approach. [2] (b) Calculate the same energy for two helium-4 nuclei. [2] (c) Explain why hydrogen isotopes are the chosen fuel. [2]

10. A star radiates 1.6 × 10²⁷ W by turning hydrogen into helium. Atomic masses: H-1 1.007825 u and He-4 4.002603 u. (a) Calculate the energy released, in MeV, when four hydrogen atoms become one helium atom. [2] (b) Calculate the mass of hydrogen consumed per second. Proton mass 1.673 × 10⁻²⁷ kg. [2] (c) Calculate the percentage of the hydrogen mass converted to energy. [2]

11. In a magnetic confinement reactor a deuteron of kinetic energy 20 keV (mass 3.344 × 10⁻²⁷ kg) moves at right angles to a field of 4.0 T. (a) Calculate the radius of its path. [3] (b) Calculate the temperature for which the mean kinetic energy is 20 keV. [1] (c) State two problems that must be solved to make a practical fusion reactor. [2]

Answers

1. (a) KE = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 310 = 6.42 × 10⁻²¹ J [1]. Divide by 1.60 × 10⁻¹⁹ to get 0.040 eV [1]. (b) v = √(2 × 6.42 × 10⁻²¹ ÷ 1.675 × 10⁻²⁷) = 2.8 × 10³ m s⁻¹ [1]. Examiner insight: A bare answer in joules when eV is asked is a common slip, so write the conversion line.

2. (a) Fissions = 1000 × 1.02⁵⁰ [1] = 2.7 × 10³ [1]. (b) Control rods are pushed further into the core [1]. They absorb neutrons, so fewer cause fission and k falls to 1 [1]. Examiner insight: Say that the rods absorb neutrons. Saying they slow them is a different component’s job.

3. (a) Momentum: m u = m v + 4m V [1]. Elastic: speed of approach equals speed of separation, u = V − v [1]. Then v = −0.6u, and KE kept = (0.6)² = 0.36 [1]. (b) n = ln(1.5 × 10⁶ ÷ 0.030)/ln(1/0.36) = 17.73 ÷ 1.0217 = 17.35 [1], so 18 collisions [1]. Examiner insight: A “show that” answer tends to need each line of the derivation, and a number of collisions is rounded up.

4. (a) P = (m/t) c ΔT [1]. m/t = 600 × 10⁶ ÷ (1.1 × 10³ × 200) = 2.7 × 10³ kg s⁻¹ [1]. (b) It absorbs few neutrons (or does not corrode the core) [1]. Its specific heat capacity is low, so a very large mass flow is needed [1]. Examiner insight: The drawback should be tied to the value you calculated, not stated generally.

5. (a) 900 ÷ 0.33 = 2.73 × 10³ MW [1]. (b) Energy per fission = 190 × 1.60 × 10⁻¹³ = 3.04 × 10⁻¹¹ J [1]. Rate = 2.73 × 10⁹ ÷ 3.04 × 10⁻¹¹ = 9.0 × 10¹⁹ s⁻¹ [1]. (c) Mass of one nucleus = 235 × 1.661 × 10⁻²⁷ = 3.90 × 10⁻²⁵ kg. Mass = 8.97 × 10¹⁹ × 86 400 × 3.90 × 10⁻²⁵ [1] = 3.0 kg [1]. Examiner insight: Efficiency applies to the electrical output. Dividing by it gives thermal power, and multiplying is a classic error.

6. (a) The fuel is intensely radioactive, so people must be kept away [1]. Water absorbs radiation and removes the heat the fuel still produces [1]. (b) 0.040 × 2400 MW = 96 MW [1]. Fission products continue to decay [1], releasing energy that would melt the fuel without cooling [1]. Examiner insight: A good answer links decay heat to fission products, not to the chain reaction still running.

7. (a) A = 3.6 × 10¹⁵ × (1/2)^(100 ÷ 28.8) [1] = 3.2 × 10¹⁴ Bq [1]. (b) (1/2)ⁿ = 0.010, n = ln 100/ln 2 = 6.64 [1]. Time = 6.64 × 28.8 = 1.9 × 10² years [1]. (c) High-level waste is cooled, sealed in glass and steel and stored deep underground [1]. Examiner insight: Show the number of half-lives. It is easy to check, and it protects against an arithmetic slip.

8. Benefits: far more energy per kilogram of fuel [1]; no carbon dioxide from the reactor in normal operation [1]. Drawbacks: accidents are rare but severe and widespread [1]; long-lived radioactive waste needs safe storage for a very long time [1]. Weighing: the risk is probability multiplied by consequence, and strict safety systems reduce the probability [1]. Judgement: a reasoned conclusion, for example that the plan is justified if waste storage and safety are guaranteed, or that it needs careful long-term planning [1]. Examiner insight: Extended answers tend to earn credit for balance and a clear conclusion, not for a long list.

9. (a) r = 1.2 × 10⁻¹⁵ × (2^⅓ + 3^⅓) = 3.24 × 10⁻¹⁵ m [1]. Ep = (1.60 × 10⁻¹⁹)² ÷ (4π × 8.85 × 10⁻¹² × 3.24 × 10⁻¹⁵) = 7.1 × 10⁻¹⁴ J [1]. (b) r = 1.2 × 10⁻¹⁵ × 2 × 4^⅓ = 3.81 × 10⁻¹⁵ m [1]. Ep = 4 × (1.60 × 10⁻¹⁹)² ÷ (4π × 8.85 × 10⁻¹² × 3.81 × 10⁻¹⁵) = 2.4 × 10⁻¹³ J [1]. (c) The helium barrier is about 3.4 times larger [1]. Hydrogen nuclei have one proton each, so less repulsion and a lower temperature is needed [1]. Examiner insight: For helium, remember each nucleus has charge 2e, so the charge product is 4e² and not 2e².

10. (a) Δm = 4 × 1.007825 − 4.002603 = 0.028697 u [1]. Energy = 0.028697 × 931.5 = 26.7 MeV [1]. (b) Energy per helium = 26.7 × 1.60 × 10⁻¹³ = 4.28 × 10⁻¹² J. Reactions per second = 1.6 × 10²⁷ ÷ 4.28 × 10⁻¹² = 3.74 × 10³⁸ [1]. Hydrogen = 4 × 3.74 × 10³⁸ × 1.673 × 10⁻²⁷ = 2.5 × 10¹² kg s⁻¹ [1]. (c) Fraction = 0.028697 ÷ (4 × 1.007825) [1] = 0.00712, which is 0.71% [1]. Examiner insight: Keep the full masses until the subtraction, because rounding early destroys a small difference.

11. (a) KE = 20 × 10³ × 1.60 × 10⁻¹⁹ = 3.2 × 10⁻¹⁵ J, so v = √(2 × 3.2 × 10⁻¹⁵ ÷ 3.344 × 10⁻²⁷) = 1.38 × 10⁶ m s⁻¹ [1]. Bqv = mv²/r so r = mv/(Bq) [1]. r = 3.344 × 10⁻²⁷ × 1.38 × 10⁶ ÷ (4.0 × 1.60 × 10⁻¹⁹) = 7.2 × 10⁻³ m [1]. (b) T = 2KE/(3k) = 2 × 3.2 × 10⁻¹⁵ ÷ (3 × 1.38 × 10⁻²³) = 1.5 × 10⁸ K [1]. (c) Any two: heating the plasma to about 10⁸ K without touching walls; confining it long enough at high density; neutron damage and activation of the structure; tritium supply [2]. Examiner insight: Name the problems precisely. “It is hard” earns nothing without the physics reason.

Where marks are usually lost

• Mixing up moderator, control rods and coolant.
• Not rounding up a number of collisions.
• Forgetting the 1/2 when two nuclei share the Coulomb energy.
• Using Celsius for temperature in (3/2)kT.
• Using charge 1 for helium nuclei.
• Not converting MeV to joules, or hours and years to seconds.
• Describing waste without naming a class.
• Listing risks and benefits without any judgement.

Next steps

• Revision notes
• Study guide
• OxfordAQA Physics hub and printable checklist
• All free 10-minute diagnostics
• Book a free trial class: /trial/

Official syllabus

OxfordAQA International AS and A-level Physics (9630), specification version 4.4, for exams from May/June 2020 onwards, published by Oxford International AQA Examinations: sections 3.12.3 Induced fission, 3.12.4 Safety aspects of nuclear reactors and 3.12.5 Nuclear fusion (International A-level only).