Edexcel IAL Physics: Nuclear and Particle Physics — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Nuclear and Particle Physics revision notes

Scope note: this paper is filed under Unit 4 sub-topic 4.5. Questions 1–6 and 9–10 (radioactive decay types, half-life, decay constant, beta decay, mass defect/binding energy, fission/fusion, particle classification) sit substantially in Unit 5 sub-topic 5.4, and are included here as useful overlap/context alongside the genuine Unit 4.5 material. Questions 11–12 test the Unit 4.5 outcomes proper (thermionic emission, accelerators, MeV/MeV-c² conversions).

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Section A

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

2. State the quark composition of a proton and a neutron. [2]

Section B

3. 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]

4. 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) State which fundamental force is responsible. [1]

5. The mass defect 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] (c) Explain, using a binding energy per nucleon curve, why both fission and fusion release energy. [3]

6. Check whether this interaction is possible, testing charge, baryon number and lepton number: p + p → p + n + e⁺ + ν_e. [4]

7. State what the Geiger-Marsden alpha-particle scattering experiment provided evidence for, and what model of the atom it replaced. [2]

8. State the equation for the radius of a charged particle’s circular path in a magnetic field, and explain how this can be used to identify a particle’s momentum and the sign of its charge. [3]

9. An electron and a positron, each of rest mass 9.11 × 10⁻³¹ kg, annihilate at rest, producing two identical gamma-ray photons. Calculate the energy of each photon. [3]

10. Classify each of the following as a baryon, a meson or a lepton: proton, pion, electron. [3]

11. Describe how thermionic emission is used to produce a beam of electrons for injection into a linear accelerator (linac), and explain briefly how a cyclotron differs from a linac in the way it accelerates particles. [4]

12. A particle has a rest mass of 1.67 × 10⁻²⁷ kg. (a) Convert this mass to units of MeV/c². (b) A different particle has a rest energy of 105 MeV. State its mass in MeV/c², and explain why MeV/c² is a convenient unit for particle rest mass. [4]

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Answers

1. 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].

2. Proton = uud [1]; neutron = udd [1].

3. (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].

4. (a) ¹₀n → ¹₁p + ⁰₋₁e + ν̄_e [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].

5. (a) E = Δmc² = 5.05 × 10⁻²⁹ × (3.00 × 10⁸)² [1] = 4.55 × 10⁻¹² J [1]; ÷ 1.60 × 10⁻¹³ = 28.4 MeV [1]. (b) 28.4 ÷ 4 [1] = 7.1 MeV per nucleon [1]. (c) The curve peaks near iron-56 [1]. Fusion of light nuclei moves the product up the curve towards the peak, increasing binding energy per nucleon and releasing energy [1]. Fission of very heavy nuclei also moves the products towards the peak from the right, so energy is again released [1].

6. Charge: left 2, right 1 + 0 + 1 + 0 = 2 ✓ [1]. Baryon number: left 2, right 1 + 1 + 0 + 0 = 2 ✓ [1]. Lepton number: left 0, right 0 + 0 + (−1) + (+1) = 0 ✓ [1]. All three quantities are conserved, so the interaction is possible [1].

7. It provided evidence for a small, dense, positively charged nucleus [1] (from the observation of large-angle scattering), replacing the earlier “plum pudding” model of the atom [1].

8. r = p ÷ BQ [1]. The radius of the curved path reveals the particle’s momentum (from r, B and Q) [1], and the direction of curvature reveals the sign of its charge [1].

9. Total mass converted: Δm = 2 × 9.11 × 10⁻³¹ = 1.822 × 10⁻³⁰ kg [1]. Total energy released: ΔE = c²Δm = (3.00 × 10⁸)² × 1.822 × 10⁻³⁰ = 1.640 × 10⁻¹³ J [1]. Since two identical photons share this energy equally, each photon has energy = 1.640 × 10⁻¹³ ÷ 2 = 8.20 × 10⁻¹⁴ J (about 0.51 MeV) [1].

10. Proton — baryon (three quarks) [1]; pion — meson (a quark and an antiquark) [1]; electron — lepton (a fundamental particle) [1].

11. A heated filament (cathode) releases electrons from its surface by thermionic emission [1]; these electrons are then accelerated by a potential difference towards an anode to form the initial beam injected into the linac [1]. In a linac, the particle is accelerated in a straight line through a series of tubes with an alternating field switching sign as it crosses each gap [1]. In a cyclotron, the particle instead spirals outward inside a magnetic field, crossing the same alternating-field gap repeatedly as its radius grows [1].

12. (a) E = mc² = 1.67 × 10⁻²⁷ × (3.00 × 10⁸)² = 1.503 × 10⁻¹⁰ J [1]; ÷ 1.60 × 10⁻¹³ = 939 MeV, so mass = 939 MeV/c² [1]. (b) Mass = 105 MeV/c² [1]. MeV/c² is convenient because rest energy in MeV can be converted to mass by simply dividing by c², avoiding repeated use of very small SI masses and the large factor of c² in every calculation [1].

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Where marks are usually lost

• Giving the charge of a positron as −1.
• Forgetting that an antineutrino has lepton number −1.
• Using grams instead of kilograms in E = mc².
• Saying only fission releases energy.
• Forgetting that the top quark’s existence was predicted by the symmetry of the standard model before it was observed.
• Confusing a baryon (three quarks) with a meson (a quark and an antiquark).