AQA AS Level Physics Paper 1, June 2023: Question 1

13 marks · Hard difficulty · Short Answer

Analyze a particle interaction involving a negative kaon and a proton, complete property tables, calculate rest energies and photon energies, deduce quark structures using conservation laws, and identify decay products.

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Question text

01 A strong interaction between a negative kaon (K−) and a proton (p) produces

an omega-minus (Ω−) particle, a neutral kaon (K0) and an unidentified particle Y.

The interaction is:

K− + p → Ω− + K0 + Y

Table 1 contains information on the particles in this interaction.

Table 1

K− p Ω− K0 Y

Rest energy / MeV 493.8 938.3 1672 497.8 493.8

Baryon number +1 +1 0

Charge −1e +1e −1e 0

Strangeness −1 0 −3 +1

01.1 Complete Table 1.

[2 marks]

01.2 Calculate, in J, the rest energy of the Ω−.

[2 marks]

3 rest energy = J

01.3 Suggest how energy is conserved in this interaction.

Refer to the rest energies of the particles in Table 1.

[2 marks]

The quark structure of the Ω− particle is sss.

The Ω− is unstable. It decays into a proton through a series of decays:

Ω− → Ξ0 + π−

followed by

Ξ0 → Λ0 + π0

followed by

Λ0 → p + π−

The Ξ0 and Λ0 are both hadrons.

01.4 Deduce the quark structure of the Λ0 particle.

[4 marks]

quark structure of4 Λ0 =

The products of the decay series include π0 and π− particles. These particles are

unstable and decay.

01.5 The π0 decays into gamma photons. Each gamma photon has a wavelength

*03* of 1.25 × 10−14 m.

Calculate the energy of one of these photons.

[2 marks]

energy of photon = J

01.6 The negative pion π− decays.

Which row shows the particles that could be created in this decay?

Tick ( ) one box.

[1 mark]

μ− + ν

e− + v

e− + ν

e− + e+ + e−

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidance Mark AO

01.1 2 rows correct 2 AO2.1f

3 rows correct AO2.1e

Y’s charge: Allow 1 or +1 or +1e

Y’s strangeness: Allow +1 or 1

6 –19 MP1 allow POT error in attempted conversion of eV

01.2 1672 × 10 × 1.6(0) × 10 OR to J where 1672 × 1.6 2 AO1.1a

is seen.

Correct conversion of: AO1.1b

1672 MeV to 1.672 × 109 (eV) OR

Condone correct conversion of 1672 MeV or 1 MeV

Correct conversion of: seen in an otherwise incorrect expression.

1 MeV to 1.6 x 10-13 (J)

–10 Accept answer correctly rounded to at least 2 sf.

2.68 × 10 (J)

Calculator answer 2.6752 x 10-10 (J)

01.3 Idea that the rest energy of the products is greater than the MP1 allow: 2 AO2.1b

rest energy of the reactants. Rest energy of reactants = 1432.1 (MeV) and rest

AO3.1b

energy of products = 2663.6 (MeV) or 1231.5 MeV

seen.

MP2 allow:

Idea that kinetic energy of the reactants is greater than the The additional energy (1231.5 MeV) comes from the

kinetic energy of the products kinetic energy of the reactants.(Allow the idea that

products don’t have any kinetic energy).

Alternative: MP2 must relate to kinetic energy: speed / velocity /

momentum is insufficient (treat as neutral).

The rest energies of reactants + their (additional) kinetic

energy = rest energies of the products

Max 1 for the idea that the rest energies are not

equal and kinetic energy of the particles accounts

for the difference.

01.4 MP1 4 4 ×

Λ0 is a baryon or Λ0 consists of 3 quarks (condone any 3) AO3.1b

MP1: Applies conservation of baryon no. correctly to at least 0

or Λ has a baryon number = 1 OR

one decay

Ξ0 is a baryon or Ξ0 consists of 3 quarks (condone any 3)

or Ξ0 has a baryon number = 1

MP2: MP2:

Ω–(sss) 0 π– (d𝑢𝑢�)

Writes first or second decay in terms of quark compositions. Decay 1 Ξ (uss)

11 0 7

OR B

S –3 –2 0

Identifies decay is via weak interaction Q –1 0 –1

Ξ0 (uss) Λ0 (uds) π0 (u𝑢𝑢� or d𝑑𝑑̅)

OR Decay 2

B 1 1 0

Ξ0 has a strangeness = –2 or states quark structure as ssu

S –2 –1 0

Q 0 0 0

MP3: 0 –

Decay 3 Λ (uds) P(uud) π (d𝑢𝑢�)

The Λ0 has a strangeness = –1

B 1 1 0

S –1 0 0

Q 0 +1 –1

Writes first two decays in terms of quark compositions.

An answer of uds scores MP1 and MP4.

Must see MP2 and MP3 to award these marks.

MP4 : Award 1 mark if strangeness quoted as positive in both

MP2 and MP3 where MP2 and MP3 otherwise not

Λ0 =)

(Quark composition uds awarded.

OR Working can be shown on the equations above 01.4.

writes 3rd decay in terms of quark compositions

Writes all 3 decays in terms of quarks scores all 4 marks.

01.5 hc Condone POT error in any substituted values. 2 AO1.1a

Use of E = OR E = hf and 𝒄𝒄 =fλ

AO2.1b

Accept any answer correctly rounded to at least 2

–11 sf.

1.59 × 10 (J)

Max 1 mark for otherwise correct answer with POT

error.

Max 1 mark for an answer of 7.956 × 10-12 (J)

(Correct use of equation but divided energy by two.)

Max 1 mark for an answer of 2.55 × 10-30 (J)

(Assumes that –11

1.59 × 10 is in eV and

attempts to convert to J.)

Calculator display = 1.5912 × 10-11 (J)

01.6 − Tick in 2nd box only 1 AO3.1b

e + ve

Total 13

How to answer it

Particles, Conservation Laws, and Quark Structures

What this question tests

This comprehensive exam question assesses your core knowledge of particle physics from AQA AS Level. It tests particle conservation laws (baryon number, charge, strangeness), unit conversions (MeV to Joules), energy conservation in particle interactions involving kinetic energy, deducing multi-step decay quark structures, photon energy calculations using wavelength, and identifying allowed weak decay products.

Part 01.1 — Completing Table 1

Conservation Laws in Particle Interactions

✅ Correct Table Values for Particle Y

Baryon number: 0
Charge: +1e (or +1)
Strangeness: -1

💡 Key Knowledge

In any particle interaction, fundamental quantities must be conserved:

Baryon number (B): Sum on left = Sum on right. ( 0 + 1 = 1 + 0 + B_y → B_y = 0 )
Charge (Q): ( -1 + 1 = -1 + 0 + Q_y → Q_y = +1 )
Strangeness (S): Strong interactions conserve strangeness. ( -1 + 0 = -3 + 1 + S_y → -1 = -2 + S_y → S_y = -1 )

🎯 Mark scheme: 2 rows correct = 1 mark, 3 rows correct = 2 marks.

Part 01.2 — Rest Energy Calculation

Converting Rest Energy from MeV to Joules

📐 Step-by-Step Calculation

Identify given value: Rest energy of Omega-minus = 1672 MeV
Convert MeV to eV: Multiply by 10⁶ → 1672 × 10⁶ eV
Convert eV to Joules: Multiply by elementary charge ( 1.60 × 10⁻¹⁹ C ).
Calculation: 1672 × 10⁶ × 1.60 × 10⁻¹⁹ = 2.6752 × 10⁻¹⁰ J
Final Answer: 2.68 × 10⁻¹⁰ J (to 3 sf)

❌ Common Errors

Students frequently miss out the conversion factor from mega ( 10⁶ ) or incorrectly apply the electronvolt to joule multiplier. Always show your explicit substitution to pick up method marks even if your calculator trips up on powers of ten!

🎯 Mark scheme: 1 mark for correct conversion method (seeing 1.6 × 10⁻¹⁹ and 10⁶), 1 mark for final evaluated answer correctly rounded (at least 2 sf).

Part 01.3 — Energy Conservation

Explaining Rest Energy vs. Kinetic Energy

✅ Correct Answer Summary

The total rest energy of the products is greater than the total rest energy of the reactants. This 'extra' rest energy is provided by the kinetic energy of the incoming reactants (or kinetic energy is converted into rest mass energy).

🧠 Exam Technique & Guidance

Examiners look for two clear marking points here:

MP1: Acknowledge that product rest energy > reactant rest energy.
MP2: Explicitly state that reactant kinetic energy accounts for the difference (or is converted into mass). Avoid vague references to "speed" or "velocity".

🎯 Mark scheme: 2 marks total. 1 mark for rest energy comparison, 1 mark for linking the deficit to reactant kinetic energy.

Part 01.4 — Quark Structure Deduction

Deducing the Quark Structure of the Lambda-zero Particle

💡 Decay Chain Breakdown

Follow the decay sequence backwards or forwards using conservation laws:

Omega-minus (Ω⁻): sss (Strangeness = -3, Charge = -1, Baryon = 1)
Xi-zero (Ξ⁰): uss (Strangeness = -2, Charge = 0, Baryon = 1)
Lambda-zero (Λ⁰): uds (Strangeness = -1, Charge = 0, Baryon = 1)
Proton (p): uud (Strangeness = 0, Charge = +1, Baryon = 1)

✅ Final Answer

Quark structure of Λ⁰ = uds

🎯 Mark scheme: 4 marks. Awarded across applying conservation of baryon number, writing intermediate decay stages in terms of quarks, and correctly deducing strangeness and final quark composition uds .

Part 01.5 — Photon Energy Calculation

Calculating Gamma Photon Energy from Wavelength

📐 Step-by-Step Calculation

Equation: Use E = hc / λ
Constants:
Planck constant h = 6.63 × 10⁻³⁴ J s
Speed of light c = 3.00 × 10⁸ m s⁻¹
Substitution: E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (1.25 × 10⁻¹⁴)
Calculation: 1.989 × 10⁻²⁵ / 1.25 × 10⁻¹⁴
Final Answer: 1.59 × 10⁻¹¹ J

❌ Common Errors & Pitfalls

Forgetting to divide by wavelength or accidentally multiplying by it.
Using old or unrounded values for constants; always take values directly from your AQA data booklet.

🎯 Mark scheme: 2 marks. 1 mark for correct equation substitution, 1 mark for final evaluated answer ( 1.59 × 10⁻¹¹ J ).

Part 01.6 — Weak Decay Selection

Identifying Allowed Decay Products for a Negative Pion

✅ Correct Box to Tick

Tick the second box down: e⁻ + v̄ₑ (electron + electron antineutrino)

💡 Why this decay is valid

Pions decay via the weak interaction. Lepton number, charge, and family lepton numbers must be conserved:

Initial charge = -1 . Products charge = -1 + 0 = -1 .
Lepton number must be conserved (0 initially). An electron ( L = +1 ) requires an electron antineutrino ( L = -1 ) to sum to 0.

🎯 Mark scheme: 1 mark for ticking the second box exclusively.

Topics

Physics · 3.2 Particles and radiation

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