Edexcel IAL Physics: Nuclear Decay — 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 available free from your board.

Related: Nuclear Decay revision notes

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Section A

1. State what happens to the proton number and nucleon number in alpha, beta-minus and beta-plus decay. [3]

2. Define activity, decay constant and half-life. [3]

Section B

3. Complete and balance these decay equations:

(a) ²³⁸₉₂U → ? + ⁴₂He [2] (b) ¹⁴₆C → ¹⁴₇N + ? + ? [2]

4. A source has a decay constant of 2.3 × 10⁻³ s⁻¹.

(a) Calculate its half-life. [2] (b) A sample initially contains 5.0 × 10¹⁸ undecayed nuclei. Calculate its initial activity. [2] (c) Calculate the number of undecayed nuclei remaining after 900 s. [3]

5. A wooden artefact has a carbon-14 activity of 0.25 of that of living wood. The half-life of carbon-14 is 5730 years.

(a) Estimate its age without a calculator, explaining your method. [3] (b) Explain two limitations of carbon dating. [4]

6. Explain why radioactive decay is described as random and spontaneous, and explain how a half-life can be a reliable quantity despite this. [5]

7. State three precautions when handling a radioactive source in a school laboratory, with a reason for each. [6]

8. A nuclear reaction has a mass deficit of 3.2 × 10⁻²⁸ kg.

(a) Explain what is meant by mass deficit. [2] (b) Calculate the energy released. [2] (c) Explain why both fusion and fission release energy, referring to binding energy per nucleon. [3]

9. A student measures a count rate of 84 counts per minute from a source, with a background count rate of 12 counts per minute measured separately with the source removed.

(a) Calculate the true count rate due to the source. [1] (b) Explain why this correction is necessary. [2] (c) State one type of nuclear radiation that would be stopped by a sheet of paper, and one that would require thick lead or concrete to substantially reduce it. [2]

10. Describe how the absorption of gamma radiation by lead could be investigated experimentally. [3]

11. Compare alpha, beta-minus and gamma radiation in terms of nature, penetration and ionising power. [9]

12. A sample has an activity of 640 Bq. After 24 hours the activity is 40 Bq.

(a) Calculate the half-life. [3] (b) Calculate the decay constant. [2] (c) Calculate the activity after a further 12 hours. [2]

13. The mass deficit of a helium-4 nucleus is 5.05 × 10⁻²⁹ kg.

(a) Calculate the binding energy in joules and in MeV. [3] (b) Calculate the binding energy per nucleon in MeV. [2]

14. In beta-minus decay a neutron becomes a proton.

(a) Write the decay equation at the nucleon level, including the antineutrino. [2] (b) Explain why the antineutrino had to be proposed. [3] (c) (Background, not a specification outcome.) State which fundamental force is responsible. [1]

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Answers

1. Alpha — proton number decreases by 2, nucleon number decreases by 4 [1]. Beta-minus — proton number increases by 1, nucleon number unchanged [1]. Beta-plus — proton number decreases by 1, nucleon number unchanged [1].

2. Activity — the number of nuclei decaying per second, measured in becquerels [1]. Decay constant — the probability per unit time that a given nucleus will decay [1]. Half-life — the average time for half the undecayed nuclei in a sample to decay, or for the activity to halve [1].

3. (a) ²³⁴₉₀Th [1] [1]. (b) ⁰₋₁e (a beta-minus particle) [1] and an antineutrino, ν̄ₑ [1].

4. (a) T½ = ln 2 ÷ λ = 0.693 ÷ 2.3 × 10⁻³ [1] = 301 s [1]. (b) A = λN = 2.3 × 10⁻³ × 5.0 × 10¹⁸ [1] = 1.15 × 10¹⁶ Bq [1]. (c) N = N₀e^(−λt) [1]; λt = 2.3 × 10⁻³ × 900 = 2.07 [1]; N = 5.0 × 10¹⁸ × e^(−2.07) = 6.3 × 10¹⁷ [1].

5. (a) 0.25 is two halvings — 1 → 0.5 → 0.25 [1] [1]; so the age is 2 × 5730 = 11 460 years [1]. (b) Any two, 2 marks each: the method assumes the proportion of carbon-14 in the atmosphere has been constant, which is not exactly true — nuclear testing and fossil fuel burning have altered it, so calibration is required [1] [1]. It is only useful for a limited range of ages — beyond about 50 000 years the remaining activity is too low to measure reliably against background [1] [1]. It can only be used on material that was once living [1] [1]. Sample contamination by older or younger carbon distorts the result [1] [1].

6. Random means it is impossible to predict which nucleus will decay next or when [1]; each nucleus has the same probability of decaying in a given interval, independent of its neighbours [1]. Spontaneous means the decay is not affected by external conditions such as temperature, pressure or chemical state [1]. A half-life is nevertheless reliable because a sample contains an enormous number of nuclei — of the order of 10²³ [1], so although individual events are unpredictable, the statistical average over so many nuclei is highly consistent and reproducible [1].

7. Any three, 2 marks each: handle the source with long tongs [1], to increase the distance and reduce the dose, since intensity falls with the square of the distance [1]. Keep the source pointed away from people and never look directly along its axis [1], to avoid direct exposure of the eyes and body [1]. Store the source in a lead-lined container when not in use [1], because lead absorbs the radiation and shields those nearby [1]. Minimise the time the source is out of its container [1], since the dose received is proportional to exposure time [1]. Wash hands afterwards and never eat in the laboratory [1], to prevent ingestion of contamination [1].

8. (a) The mass deficit is the difference between the total mass of the separated, individual nucleons and the mass of the assembled nucleus — the “missing” mass is released as energy when the nucleus forms [2]. (b) ΔE = c²Δm = (3.0 × 10⁸)² × 3.2 × 10⁻²⁸ [1] = 2.88 × 10⁻¹¹ J [1]. (c) A graph of binding energy per nucleon against nucleon number peaks around iron [1]. Fusion (combining light nuclei) and fission (splitting heavy nuclei) both move nuclei towards this more stable, higher binding-energy-per-nucleon region [1], so both release energy overall, even though they act on opposite ends of the curve [1].

9. (a) 84 − 12 = 72 counts per minute [1]. (b) Background radiation (from cosmic rays, rocks and other environmental sources) is present in every measurement regardless of the source being tested, so it must be subtracted to find the count rate due to the source alone [2]. (c) Alpha particles are stopped by a sheet of paper (or a few cm of air) [1]; gamma rays require thick lead or concrete to be substantially reduced, since they are the most penetrating and weakly ionising of the three types [1].

10. Different thicknesses of lead are placed between a gamma source and a detector (such as a Geiger-Müller tube) [1]; the count rate is measured for each thickness, correcting for background radiation each time [1]; the results show how the intensity of gamma radiation decreases as absorber thickness increases, allowing the absorption to be characterised [1].

11. Alpha — a helium nucleus [1], stopped by paper or a few cm of air [1], most strongly ionising [1]. Beta-minus — a fast electron [1], stopped by a few mm of aluminium [1], moderately ionising [1]. Gamma — a high-energy electromagnetic photon [1], reduced by several cm of lead [1], least ionising [1].

12. (a) 640 → 320 → 160 → 80 → 40 is four halvings [1] in 24 hours [1], so the half-life is 6.0 hours [1]. (b) λ = ln 2 ÷ t½ = 0.693 ÷ 6.0 [1] = 0.116 h⁻¹ (3.2 × 10⁻⁵ s⁻¹) [1]. (c) 12 hours is two further half-lives [1], so activity = 40 ÷ 4 = 10 Bq [1].

13. (a) ΔE = c²Δm = (3.00 × 10⁸)² × 5.05 × 10⁻²⁹ [1] = 4.55 × 10⁻¹² J [1]; ÷ 1.60 × 10⁻¹³ J MeV⁻¹ = 28.4 MeV [1]. (b) 28.4 ÷ 4 [1] = 7.1 MeV per nucleon [1].

14. (a) ¹₀n → ¹₁p + ⁰₋₁e + ν̄ₑ [1], with nucleon and charge numbers balanced [1]. (b) The emitted beta particles had a continuous range of energies, not a single value [1], which appeared to violate the conservation of energy and momentum [1]. A third, almost undetectable particle carrying away the remaining energy was proposed to preserve those conservation laws [1]. (c) The weak nuclear force [1].

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Where marks are usually lost

• Forgetting the antineutrino in beta-minus decay.
• Using λ in per-hour with t in seconds.
• Saying half-life is the time for the source to become half as dangerous.
• Giving safety precautions without a reason.
• Dividing the binding energy by the proton number instead of the nucleon number when finding binding energy per nucleon.
• Saying only fission releases energy.