Practice Questions
A Level Physics: Nuclear Physics — Practice Questions
Original exam-style practice questions with full worked answers on binding energy, mass defect, radioactive decay and fission for A Level Physics.
- Subject
- Physics
- Level
- A LEVEL
- Topic
- Nuclear physics
- Author
- Iftikhar Azeemi
- Updated
- Reviewed by
- Hassan (what this means)
Aligned to Cambridge A Level Physics (9702), 2025-2027. Official specification .
Syllabus page (what it covers and how it is assessed): Cambridge A Level Physics.
Syllabus points this page covers
9702 (A Level)
- 23.1 Mass defect and nuclear binding energy
- 23.2 Radioactive decay
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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 available free from your board.
Related: Nuclear Physics revision notes
Questions
1. Define mass defect and binding energy. [3]
2. Explain why both fission and fusion release energy, referring to the binding energy per nucleon curve. [4]
3. A helium-4 nucleus has mass 4.00150 u; a proton is 1.00728 u and a neutron 1.00867 u. (1 u = 931.5 MeV)
(a) Calculate the mass defect in u. [3] (b) Calculate the binding energy in MeV. [2] (c) Calculate the binding energy per nucleon. [1]
4. A radioactive source has a half-life of 8.0 days and an initial activity of 4.8 × 10⁵ Bq.
(a) Calculate the decay constant. [2] (b) Calculate the activity after 20 days. [3] (c) Calculate the number of undecayed nuclei initially present. [2]
5. Explain what is meant by saying radioactive decay is random and spontaneous. [2]
6. In a fission reactor, describe the function of:
(a) the moderator [2] (b) the control rods [2] (c) the coolant [1]
7. Define the decay constant, λ, and state the general exponential decay equation, noting the three quantities it can be applied to. [3]
8. A radioactive isotope has a decay constant of 0.020 s⁻¹. Calculate its half-life. [2]
9. A nucleus has a mass defect of 3.0 × 10⁻²⁸ kg. Calculate its binding energy in joules, using E = mc². [2]
10. The smooth exponential decay curve N = N₀e^(-λt) predicts that N never reaches exactly zero, however many half-lives have passed. Explain why this continuous model is only an approximation for a real, finite sample of nuclei. [2]
Answers
1. Mass defect — the difference between the mass of a nucleus and the total mass of its separate nucleons [1]. Binding energy — the energy required to separate a nucleus into its individual nucleons [1], equal to the mass defect multiplied by c² [1]. Equivalently, binding energy is the energy that would be released if the separate nucleons came together to form the nucleus.
2. Binding energy per nucleon peaks around iron-56 [1]. Fusion of light nuclei moves the product towards the peak, increasing binding energy per nucleon [1]. Fission of heavy nuclei also moves the products towards the peak [1]. In both cases the increase in binding energy per nucleon is released as energy [1].
3. (a) Mass of nucleons = 2(1.00728) + 2(1.00867) = 4.03190 u [1] [1] Δm = 4.03190 − 4.00150 = 0.03040 u [1]. (b) E = 0.03040 × 931.5 [1] = 28.3 MeV [1]. (c) 28.3 ÷ 4 = 7.08 MeV per nucleon [1].
4. (a) λ = ln2 ÷ t½ = 0.693 ÷ 8.0 [1] = 0.0866 day⁻¹ [1]. (b) A = A₀e^(−λt) = 4.8 × 10⁵ × e^(−0.0866 × 20) [1] = 4.8 × 10⁵ × e^(−1.732) = 4.8 × 10⁵ × 0.1769 [1] = 8.49 × 10⁴ Bq [1]. (c) A = λN, so N = A ÷ λ. Converting λ to s⁻¹: 0.0866 ÷ 86 400 = 1.002 × 10⁻⁶ s⁻¹ [1] N = 4.8 × 10⁵ ÷ 1.002 × 10⁻⁶ = 4.79 × 10¹¹ nuclei [1].
5. Random — it is impossible to predict which nucleus will decay or when [1]. Spontaneous — the decay is unaffected by external conditions such as temperature, pressure or chemical state [1].
6. (a) It slows the fast neutrons produced by fission [1] so they are more likely to be absorbed and cause further fission [1]. (b) They absorb neutrons [1], and are raised or lowered to control the rate of the chain reaction and keep it steady [1]. (c) It removes thermal energy from the core, transferring it to generate steam [1].
7. The decay constant is the probability that any individual nucleus decays in unit time [1]. The general equation x = x₀e^(−λt) [1] applies equally to the number of undecayed nuclei N, the activity A, or the measured count rate [1].
8. t½ = ln 2 ÷ λ = 0.693 ÷ 0.020 [1] ≈ 34.7 s [1].
9. E = mc² = 3.0 × 10⁻²⁸ × (3.0 × 10⁸)² [1] = 2.7 × 10⁻¹¹ J [1].
10. The exponential formula treats N as a continuous quantity that falls asymptotically towards zero without ever mathematically reaching it [1] — but a real sample contains a finite, whole number of nuclei, each decaying independently and probabilistically, so eventually every last nucleus can (and, given enough time, will) actually decay, taking the real activity to exactly zero [1]. The exponential curve is an excellent model for the expected behaviour of a large population, but it describes an idealised continuous average, not the discrete, probabilistic reality of a small or finite number of atoms.
Where marks are usually lost
- Saying the nucleus is heavier than its constituent nucleons.
- Explaining fission and fusion without reference to the binding energy per nucleon curve.
- Not converting the decay constant to consistent units before finding N.
- Confusing the roles of the moderator and the control rods.
- Explaining exponential decay by just citing “half-life” rather than its actual defining property — the rate of decay (or the rate of change of N) is proportional to N itself, so the same fraction decays in each equal time interval, however much is left.
- Forgetting that x = x₀e^(−λt) applies equally to N, activity, or measured count rate — not just to activity.
- Mixing up grams and kilograms when substituting mass defect into E = mc².
- Treating the smooth exponential curve as literal finite-sample behaviour — a real finite sample of nuclei can and eventually will fully decay, even though the continuous exponential model it approximately follows only approaches zero asymptotically.
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