AQA A-Level Physics Paper 1, June 2022: Question 2

7 marks · Medium difficulty · Short Answer

Deduce the quark change in beta-minus decay, explain how the kinetic energy distribution of beta particles supports the existence of the antineutrino, identify particle X in an antineutrino-proton interaction, and determine which gamma photon could be produced by positron annihilation.

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Question text

02 Carbon-14 decays into nitrogen-14 with the release of a beta (β−) particle and an

antineutrino ( ve ).

02.1 State the change of quark character in β− decay.

[1 mark]

02.2 Figure 2 shows the distribution of kinetic energies of β− particles from the decay of

carbon-14.

Figure 2

Explain how Figure 2 supports the existence of the antineutrino.

[2 marks]

The existence of the antineutrino was confirmed by experiments in which

antineutrinos interact with protons. The equation for this interaction is:

v + p → e+ + X

02.3 Identify particle X.

[1 mark]

02.4 The positron released in this interaction is annihilated when it encounters an electron.

A pair of gamma photons is then produced.

Particle X can be absorbed by a nucleus. This produces another gamma ray.

Table 1 contains data for three gamma photons detected during an

antineutrino–proton interaction experiment.

Table 1

Gamma photon Photon energy / J

G1 5.0 × 10−14

G2 6.6 × 10−14

G3 1.0 × 10−13

Deduce which of the three gamma photons could have been produced by positron

annihilation.

[3 marks]

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidelines Mark AO

02.1 Allow “d→u” 1

Condone udd→uud

down quark changes to up quark ✓ Condone U for u but not D for d. AO1

Do not accept answers with extra particles.

02.2 2 2 x AO1

Idea that (graph shows that) beta particles (from C-14)

have a range of (kinetic) energies ✓

There is a fixed/maximum/total amount of energy (released

A mention of conservation of energy on its

by C-14) so there must be another particle that carries the

own is insufficient for MP2.

energy differences/missing energy away ✓

– A-LEVEL PHYSICS – –

02.3 Condone “n” but not “N”. 1 AO1

neutron ✓

Do not allow “udd”.

Question Answers Additional comments/Guidelines Mark AO 9

02.4 Calculation of minimum energy produced in annihilation of 3 3 x AO3

positron and electron (from rest mass energy ×2) Calculation of the photon energy based on

E.g. 2 × 0.51 MeV = 1.6 × 10−13 J ✓ one particle can get MP2.

(2 photons produced so) energy per photon = 8.2 × 10−14 J

✓

Conclusion consistent with their calculated minimum The ‘correct’ answer would be a conclusion

energy. ✓ leading to G3 only.

ALTERNATIVE If no other mark awarded, award one mark for

determining rest energy of positron or

One calculation of mass equivalence of photon energy✓ electron in J.

Calculation of remaining mass equivalents Allow mass equivalent calculations in (M)eV

deduction about the other two photon energies ✓

Only G3 has sufficient energy to have been made in Allow explanation in terms of positron and

annihilation. ✓ electron for annihilation in alternative MP3

Total 7

How to answer it

Carbon-14 Decay, Antineutrinos and Annihilation Calculations

What this question tests

This sequence assesses your understanding of particle interactions, quark changes during beta-minus decay, historical evidence for the existence of the antineutrino via continuous beta energy spectra, conservation laws in particle reactions, and relativistic energy calculations involving pair annihilation and photons.

Question 02.1 (1 Mark)

Quark Character Change in Beta-Minus Decay

✅ Correct Answer

Down quark changes to an up quark ( d → u ).

💡 Key Knowledge

In β⁻ decay, a neutron transforms into a proton. Fundamentally, a neutron has quark composition udd and a proton has uud . Therefore, one down quark turns into an up quark, emitting a W⁻ boson which subsequently decays into an electron and an antineutrino.

❌ Common Errors

Students lose marks by writing full baryon symbols instead of specifying the quark change, or by including extra particles (like the W boson or leptons) in the quark transition description.

Question 02.2 (2 Marks)

Interpreting the Beta Kinetic Energy Distribution Graph

✅ Correct Answer

The graph shows that emitted beta particles have a range or distribution of kinetic energies (up to a maximum).
Since the total energy released in the decay is fixed, the missing energy for particles with less than maximum kinetic energy must be carried away by another particle (the antineutrino).

🧠 Exam Technique

This is a classic 2-mark explanation question. Mark 1 is awarded for stating what the graph displays (range of KE). Mark 2 requires connecting the conservation of energy to the variable energies observed, proving a second invisible particle must exist.

Examiner Note: Simply stating "conservation of energy" without explaining why it implies a second particle (due to the variable kinetic energies) will not score the second mark.

Question 02.3 (1 Mark)

Identifying Particle X

✅ Correct Answer

Neutron (accept n , do not allow N or quark symbols like udd ).

💡 Key Knowledge

Apply conservation laws to the equation v̅_e + p → e⁺ + X :

Lepton number: (-1) + 0 = (+1) + L(X) ⇒ L(X) = -2? Wait, let's check lepton numbers: antineutrino = -1, proton = 0, positron = -1. So (-1) + 0 = (-1) + 0. Lepton number of X is 0.
Charge: (-0) + (+1) = (+1) + Charge(X) ⇒ Charge(X) = 0.
Baryon number: 0 + 1 = 0 + Baryon(X) ⇒ Baryon(X) = 1.

A neutral baryon with zero lepton number is a neutron.

Question 02.4 (3 Marks)

Deducing the Correct Gamma Photon from Positron Annihilation

📐 Step-by-Step Calculation

Find minimum energy from annihilation: A positron and electron annihilate to produce a pair of gamma photons. Rest mass energy of an electron/positron = 0.51 MeV = 8.19 × 10⁻¹⁴ J.
Total minimum energy = 2 × 0.51 MeV = 1.02 MeV.
Convert to Joules:
1.02 MeV = 1.02 × 10⁶ × 1.6 × 10⁻¹⁹ J = 1.63 × 10⁻¹³ J.
Energy per photon (if two equal photons are produced) = 1.63 × 10⁻¹³ / 2 = 8.15 × 10⁻¹⁴ J (or minimum total energy required is 1.6 × 10⁻¹³ J ).
Compare with Table 1:
• G1 = 5.0 × 10⁻¹⁴ J (Too low)
• G2 = 6.6 × 10⁻¹⁴ J (Too low for total pair, or doesn't match single photon split)
• G3 = 1.0 × 10⁻¹³ J (Sufficient energy / matches required annihilation energy threshold).
Conclusion: Only G3 has sufficient energy to have been made in this annihilation.

❌ Common Calculation Traps

Students often calculate the rest mass energy for only one particle instead of multiplying by 2 for both the electron and positron participating in the annihilation event.

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.