AS Physics: 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: Particle Physics revision notes

Syllabus note: Cambridge 9702 (2025–2027) Topic 11 outcomes cover quarks, hadrons, leptons and the quark-level description of β decay, but do not name exchange bosons (gluon, photon, W/Z, graviton) or require baryon/lepton/ strangeness conservation checks as explicit AS outcomes. Questions 3, 5(b) and 6(a) below go beyond the stated AS boundary as useful extension material — they build genuine physical understanding, but are not guaranteed AS exam content in their own right.

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Section A

1. State which of the following are fundamental: proton, neutron, electron, quark, neutrino. [2]

2. Distinguish between a baryon and a meson. [2]

3. Name the exchange particle for each of the four fundamental forces. [4]

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Section B

4. A neutron consists of the quarks u, d, d.

(a) Show that this gives the correct charge. [2]

(b) In beta-minus decay a neutron becomes a proton. State the quark change involved and name the force responsible. [2]

(c) Write the full equation for beta-minus decay, including the neutrino species. [2]

(d) Explain why the neutrino was postulated before it was detected. [3]

5. Consider the proposed interaction:

p + p  ->  p + n + pi+

(a) Check whether charge is conserved. [1]

(b) Check whether baryon number is conserved. [2]

(c) State whether this interaction is possible, justifying your answer. [1]

6. Consider a second proposed interaction:

p  ->  n + e+ + neutrino(e)

(a) Check charge, baryon number and lepton number. [3]

(b) State whether this can occur for a free proton, and explain. [2]

7. An electron and a positron, both initially at rest (negligible kinetic energy), annihilate.

(a) State how many photons are produced and why. [2]

(b) Calculate the energy of each photon. (m_e = 9.11 × 10⁻³¹ kg; c = 3.00 × 10⁸ m s⁻¹) [2]

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Section C

8. A nucleus of radium-226 (₈₈²²⁶Ra) undergoes alpha decay to form radon (Rn).

(a) Write the balanced nuclear equation for this decay. [2]

(b) State the two quantities that must be equal on each side of a nuclear equation, and explain why. [2]

9. A sample contains a mixture of a beta emitter and an alpha emitter that decays via a single, dominant alpha transition. A detector plots the number of particles received against their kinetic energy.

(a) Describe the difference in the shape of the two energy spectra. [2]

(b) Explain, in terms of the number of bodies sharing the released energy, why beta particles show a continuous range of energies while alpha particles do not. [3]

10. A nuclide undergoes β⁺ decay, emitting a positron.

(a) State the composition, relative mass and charge of the emitted particle. [2]

(b) State which other particle must also be emitted, identifying it as a particle or antiparticle. [2]

(c) State the quark-level change that occurs inside the nucleus during this decay. [2]

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Answers

1. Fundamental: electron, quark, neutrino [1]. Not fundamental: proton and neutron, which are composed of quarks [1].

2. A baryon consists of three quarks [1]; a meson consists of one quark and one antiquark [1].

3. Strong — gluon [1]. Electromagnetic — photon [1]. Weak — W⁺, W⁻ or Z⁰ boson [1]. Gravitational — graviton (hypothetical) [1].

4. (a) u = +2/3, d = −1/3 each [1]. Total = (+2/3) + (−1/3) + (−1/3) = 0 [1].

(b) A down quark changes to an up quark [1]; the weak force is responsible [1].

(c) n → p + e⁻ + ν̄ₑ [1], with the electron antineutrino correctly identified [1].

(d) The emitted beta particles had a range of energies rather than the single fixed value expected from a two-body decay [1]. This appeared to violate conservation of energy and momentum [1]. Rather than abandon those laws, an undetected particle carrying the balance was proposed [1].

5. (a) Left: +1 + 1 = +2. Right: +1 + 0 + 1 = +2. Conserved [1].

(b) Left: 1 + 1 = 2 [1]. Right: 1 + 1 + 0 = 2 (the pion is a meson, baryon number 0) [1]. Conserved.

(c) Possible [1] — all the relevant quantities are conserved.

6. (a) Charge: +1 → 0 + 1 + 0 = +1 ✓ [1]. Baryon number: 1 → 1 + 0 + 0 = 1 ✓ [1]. Lepton number: 0 → 0 + (−1) + (+1) = 0 ✓ [1].

(b) No [1]. Although all the conservation laws are satisfied, the neutron is more massive than the proton, so the decay would require an input of energy and cannot occur spontaneously for a free proton [1]. It does occur for protons bound within a nucleus, where the surrounding binding energy supplies the difference.

7. (a) Two photons [1], travelling in opposite directions to conserve momentum — since both particles start at rest, the initial total momentum is zero, so a single photon (which would carry away nonzero momentum) would violate conservation [1]. This two-photon result assumes annihilation from rest; a fast-moving pair need not produce exactly two photons travelling in exactly opposite directions.

(b) Total energy = 2m_ec² = 2 × 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 1.64 × 10⁻¹³ J [1]. Each photon = 1.64 × 10⁻¹³ ÷ 2 = 8.20 × 10⁻¹⁴ J (≈ 0.511 MeV) [1].

8. (a) ₈₈²²⁶Ra → ₈₆²²²Rn + ₂⁴α [1]. Nucleon number: 226 = 222 + 4 ✓; proton number: 88 = 86 + 2 ✓ [1].

(b) Nucleon number and charge (proton number) must each balance separately [1] — both mass-energy and charge are conserved quantities, so neither can be created or destroyed across the decay [1].

9. (a) For a single alpha transition, alpha particles are emitted with a single, discrete energy, appearing as a sharp line on the spectrum [1]; beta particles instead show a continuous spread of energies up to a fixed maximum [1]. (A real source can have more than one alpha transition, or a mixture of alpha-emitting isotopes, giving several discrete lines rather than just one — but each individual transition still gives a single sharp energy, in contrast with beta’s continuous spectrum.)

(b) In alpha decay, the released energy is shared between only two bodies — the alpha particle and the recoiling nucleus — so conservation of energy and momentum together fix a single value for each [1]. In beta decay, the energy is shared between three bodies — the beta particle, the recoiling nucleus and the accompanying (anti)neutrino — in a variable proportion each time, so the beta particle can carry anywhere from close to zero up to the maximum available energy [2].

10. (a) A positron has the same mass as an electron (~0 on the nucleon scale) and a charge of +1 [2].

(b) An (electron) neutrino is also emitted [1] — the particle form, not the antiparticle, since lepton number must balance against the positron rather than an electron [1].

(c) An up quark changes to a down quark [1], converting a proton within the nucleus into a neutron [1].

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Where marks are usually lost

• Calling protons or neutrons fundamental.
• Forgetting that mesons have baryon number 0.
• Not stating which conservation law is violated when rejecting an interaction.
• Forgetting to check energy as well as the quantum numbers.
• Saying annihilation produces one photon.
• Balancing only nucleon number, or only charge, in a decay equation rather than checking both separately.
• Mixing up which quark change (up→down or down→up) belongs to β⁻ decay and which belongs to β⁺ decay.