AQA A-Level Physics Paper 1, June 2023: Question 1

6 marks · Medium difficulty · Short Answer

Deduce quark structures, interactions, antiparticles, and rest energies related to the neutral lambda particle and its decay.

Practise this question

Question

An exam question about the neutral lambda particle, Λ^0. Part 01.1 asks to deduce the quark structure of Λ^0 for 1 mark. Part 01.2 asks to state and explain the interaction involved in its decay for 2 marks. Part 01.3 asks to identify particle X when an antiparticle of Λ^0 decays, for 1 mark. Part 01.4 gives the frequency of a photon equivalent to the rest energy of Λ^0 and asks to determine its rest energy in MeV for 1 mark. Part 01.5 asks to suggest one reason why discovery of such particles is made by large international research teams, for 1 mark.
Question text

01 The neutral lambda particle Λ0 is a baryon with a strangeness of −1

One possible decay for a Λ0 is

Λ → π00 + n

01.1 Deduce the quark structure of a Λ0.

[1 mark]

01.2 State and explain which interaction is involved in this decay.

[2 marks]

01.3 An antiparticle of the neutral lambda particle decays into a neutral pion and particle X.

Identify X.

[1 mark]

01.4 The rest energy of a Λ0 is equal to the energy of a photon with a frequency

of 2.69 × 1023 Hz.

Determine, in MeV, the rest energy of a Λ0.

[1 mark]

rest energy = MeV

01.5 The discovery of particles such as the Λ0 is made by large international research

teams.

Suggest one reason for this.

[1 mark]

Mark scheme

Show the mark scheme The mark scheme providing correct answers for the neutral lambda particle questions. 01.1 requires 'uds', 01.2 requires 'weak' and explanation about strangeness change, 01.3 requires 'anti-neutron', 01.4 requires '1.11 x 10^3 MeV', and 01.5 accepts reasons such as expensive research or large data volume.

Question Answers Additional comments/Guidelines Mark AO

Do not accept D for d.

01.1 uds 1 AO3

Penalise extra particles

01.2 weak (interaction / force) MP2: 2 AO1

Reject negative arguments (eg ‘strangeness

strangeness changes (in this decay) (from -1 to 0 and

is conserved in a strong interaction’)

strangeness can only change in a weak interaction)

Reject the idea that strangeness always

changes in a weak interaction.

General statement of strangeness

conservation in the weak interaction on its

own is insufficient.

Accept “strangeness is not conserved (in this

decay)”.

Condone “strangeness is lost”.

Accept 𝑛𝑛�

01.3 anti-neutron 1 AO1

Reject ambiguous answers unless supported

by other evidence.

Do not accept answer solely in terms of

quarks

01.4 1.1(1) × 103 (MeV) Reject incorrectly rounded answers. 1 AO1

Accept: 1100 MeV (2sf) / 1110 MeV (3sf) /

1115 MeV (4sf) etc

Calculator value: 1114.66875 MeV

01.5 Any one from 1 AO1

(teams must be large and international) because:

• research is expensive / requires funding from many

countries Treat idea of peer review as neutral (this

• both scientists and engineers are required (because the argues for independent teams).

machines used for research are complex/large pieces of

civil engineering) Do not accept idea that it ‘avoids bias’ or

• research is multi-faceted / multi-disciplinary (because ‘reproducibility’.

computation/theory/ etc. is required)

• research is round-the-clock (so teams are large to work

on shift basis)

• they are needed to process the large amounts of data 7

produced

Total 6

How to answer it

Particles and Radiation: Lambda Particle Decay Study Guide

What this question tests

This exam question tests your core knowledge of particle physics classifications, conservation laws (specifically strangeness), antiparticle properties, photon energy calculations (E = hf), and the practical context of modern collaborative scientific research.

Question 01.1

Deduce the quark structure of a Λ⁰ particle.

✅ Correct Answer

uds

💡 Key Knowledge

  • A lambda particle (Λ⁰) is a baryon, meaning it consists of 3 quarks.
  • It has a charge of 0 and a strangeness of −1.
  • Combining an up ( u , charge +2/3, strangeness 0), down ( d , charge −1/3, strangeness 0), and strange ( s , charge −1/3, strangeness −1) quark gives total charge 0 and total strangeness −1.

❌ Common Errors

Students sometimes write lower-case and upper-case letters interchangeably or write capital D instead of d . Ensure you use standard lowercase quark symbols.

Mark allocation: [1 mark] Awarded for uds . Penalise extra particles.
Question 01.2

State and explain which interaction is involved in this decay.

✅ Correct Answer

Interaction: Weak interaction / force.

Explanation: Strangeness changes during this decay (from −1 in the Λ⁰ particle to 0 in the neutral pion and neutron), and strangeness can only change in a weak interaction.

🧠 Exam Technique

Two distinct marks are available here: one for identifying the force, and one for linking it explicitly to the change in strangeness. Vague statements like "strangeness is not conserved" get the mark, but always state the direction of change ( −1 to 0 ) for absolute safety.

❌ Common Errors

Examiners heavily penalised negative arguments such as claiming "strangeness is conserved in a strong interaction". You must directly address what happens to strangeness in this specific decay.

Mark allocation: [2 marks] Mark 1 for identifying the weak interaction; Mark 2 for explaining that strangeness changes (from −1 to 0).
Question 01.3

Identify particle X produced alongside a neutral pion from the decay of an antiparticle of the neutral lambda particle.

✅ Correct Answer

Anti-neutron (Accept symbol n̄ )

💡 Key Knowledge

Particle-antiparticle symmetry dictates that every particle decay has a corresponding antiparticle mirror decay. Since a lambda particle ( Λ⁰ ) decays into a neutral pion ( π⁰ ) and a neutron ( n ), its antiparticle ( Λ̄⁰ ) must decay into a neutral pion ( π⁰ ) and an anti-neutron ( n̄ ).

Mark allocation: [1 mark] Awarded for stating anti-neutron or n̄ .
Question 01.4

Determine, in MeV, the rest energy of a Λ⁰ particle given a photon frequency of 2.69 × 10²³ Hz.

📐 Step-by-Step Calculation

  1. Calculate photon energy in Joules:
    E = hf
    E = (6.63 × 10⁻³⁴ J s) × (2.69 × 10²³ Hz)
    E = 1.78347 × 10⁻¹⁰ J
  2. Convert Joules to electronvolts (eV):<
    Divide by the elementary charge (1.60 × 10⁻¹⁹ C):
    E = (1.78347 × 10⁻¹⁰) / (1.60 × 10⁻¹⁹) = 1.11467 × 10⁹ eV
  3. Convert eV to MeV and round appropriately:
    E = 1114.67 MeV = 1.11 × 10³ MeV (or 1110 MeV / 1115 MeV depending on data sheet constants used).

🧠 Exam Technique & Units

Always show your conversion clearly. Multiplying Planck's constant by frequency gives energy in Joules. Remember to divide by 1.60 × 10⁻¹⁹ to move to eV, and adjust the power of ten by 10⁶ for MeV.

Mark allocation: [1 mark] Awarded for correct value rounded to appropriate significant figures (e.g., 1.11 × 10³ MeV ).
Question 01.5

Suggest one reason why discoveries like the Λ⁰ particle are made by large international research teams.

✅ Correct Answer (Any one of):

  • Research is extremely expensive and requires pooled funding from multiple countries.
  • Both scientists and engineers are required because machinery and particle accelerators are massive, complex civil engineering projects.
  • Research is multi-faceted / multi-disciplinary (requiring specialists in computing, theoretical physics, hardware, etc.).
  • Teams must be large to work on round-the-clock shift bases.
  • Large teams are needed to process the massive amounts of data produced.

❌ Common Errors to Avoid

Examiner note: Treat the idea of peer review as neutral because it applies to independent teams. Do not accept vague answers like "it avoids bias" or "it ensures reproducibility".

Mark allocation: [1 mark] For any valid logistical, financial, or technical justification matching the mark scheme criteria.

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.