AQA A-Level Physics Paper 1, June 2025: Question 1

6 marks · Medium difficulty · Short Answer

Complete a Feynman diagram for electron capture at the quark level, explain the stability of an atom versus a bare nucleus that only decays by electron capture, and calculate the number of neutrons from specific charge.

Practise this question

Question

Question 01 consists of three parts about an isotope of titanium that only decays by electron capture. Part 01.1 shows Figure 1, an unlabelled Feynman diagram with incoming and outgoing fermion lines on the left and right connected by a wavy line in the centre with an arrow pointing left, asking to complete the diagram to show the quark change and label incoming, outgoing, and exchange particles. Part 01.2 asks to explain why a neutral atom of this isotope is less stable than a nucleus that has no orbiting electrons. Part 01.3 gives a proton number of 22 and specific charge of 4.8 × 10^7 C kg^-1, asking to determine the number of neutrons in the nucleus.
Question text

01 This question is about an isotope of titanium that only decays by electron capture.

01.1 Complete Figure 1 to show the quark change that occurs during electron capture.

Label the incoming and outgoing particles and the exchange particle.

[3 marks]

Figure 1

01.2 In some circumstances, a nucleus of this isotope can exist with no orbiting electrons.

Explain why a neutral atom of this isotope is less stable than a nucleus that has no

orbiting electrons.

[1 mark]

01.3 A nucleus of this isotope of titanium has a proton number of 22 and a specific charge

of 4.8 × 107 C kg−1.

Determine the number of neutrons in this nucleus.

[2 marks]

number of neutrons =

Mark scheme

Show the mark scheme Mark scheme for Question 01 detailing marks for parts 01.1, 01.2, and 01.3. For 01.1 (3 marks): up becoming down quark; electron becoming electron neutrino; W boson with consistent charge connecting the two vertices. For 01.2 (1 mark): in order to decay it needs orbiting electrons because it only decays by electron capture. For 01.3 (2 marks): calculation showing total charge divided by specific charge gives mass 7.3 × 10^-26 kg, dividing by nucleon mass gives 44 nucleons, and subtracting 22 protons gives 22 neutrons.

Question Answers Additional comments/Guidance Mark AO

01.1 Diagram with: Accept uud changes to udd. 3 AO2

Up becoming down 1 Do not accept capital D for d.

Electron becoming (electron) neutrino 2 Accept e or e- for electron but reject β-.

Condone ve for νe (ie same size e on same

line)

Accept v (roman) for ν (greek).

W connecting the two, with a boson charge consistent with Boson charge must be consistent bottom left

their diagram to top right AND bottom right to top left.

Do not accept a change in boson arrow

direction in 3 .

01.2 Treat irrelevant right or wrong comments as 1 AO3

In order for the nucleus to decay it needs (orbiting) neutral.

electrons because it (only) decays by electron capture Accept reverse argument.

Condone suggestion that probability of

electron capture is decreased.

Mention of ‘ionisation’ is neutral.

01.3 Evidence of Expect to see 2 1 × AO2

22 × 1.6 × 10−19 or 3.52 ×10-18 (to determine charge)

22 ×1.6 ×10−19 1 × AO3

mass = = 7.3×10−26 (kg)

4.8×107

OR

7.3×10−26

their charge number of nucleons = −27 = 44

(to determine mass) 1 1.67×10

4.8 ×107

Accept alternatives for the nucleon mass

Accept for 1 the idea that specific charge is

0.5 × specific charge of proton

Evidence of calculation leading to 22 neutrons, e.g.number of

neutrons = their nucleon number (44) – 22 = 22 (neutrons) Do not accept ‘22’ without some evidence.

2 Evidence to support 2 can be seen earlier in

the calculation e.g. subtraction of masses.

Total 6

How to answer it

Titanium Isotope: Electron Capture & Specific Charge

📋 What this question tests

This question assesses fundamental particle physics and nuclear properties from the AQA A-Level specification:

  • Feynman Diagrams: Quark transformations during weak interactions, exchange boson identification, and conservation laws (charge, baryon number, lepton number).
  • Decay Mechanisms: Evaluating radioactive stability based on physical conditions required for decay (specifically orbital electron capture).
  • Specific Charge Calculations: Applying the definition of specific charge ( charge / mass ) to a multi-nucleon system to deduce neutron number.

Question 01.1

Quark Transformation & Feynman Diagram for Electron Capture

[3 Marks]

✅ Correct Answer Breakdown

  • Mark 1: Up quark ( u ) changing into a down quark ( d ) on the nucleon side. (Also accepted: uud → udd ).
  • Mark 2: Incoming electron ( e or e⁻ ) interacting to become an outgoing electron neutrino ( νe ).
  • Mark 3: Correct exchange boson W⁺ linking the two vertices, consistent with the arrow pointing left-to-right.
Diagram Layout:
• Left branch: incoming u (bottom-left) → outgoing d (top-left).
• Right branch: incoming e⁻ (bottom-right) → outgoing νe (top-right).
• Boson line: Wavy line with arrow pointing from left to right must be labelled W⁺ .

💡 Key Knowledge

Electron capture overall equation:

p + e⁻ → n + νe

  • Proton quark structure: uud
  • Neutron quark structure: udd
  • Quark transformation: u → d
  • Charge conservation at left vertex:
    Charge in = +2/3 e (up quark)
    Charge out = -1/3 e (down quark) + +1 e (W⁺ boson) = +2/3 e ✅
  • Charge conservation at right vertex:
    Charge in = +1 e (W⁺) + -1 e (e⁻) = 0
    Charge out = 0 (νe) ✅

🧠 Exam Technique: Reading the Boson Arrow

The question diagram pre-prints an arrow on the wavy boson line pointing from left to right. You must match the sign of the W boson to this direction:

  • A positive charge transferred from left to right balances the vertex: u (+2/3) → d (-1/3) + W⁺ (+1) .
  • Do not redraw or alter the pre-printed direction arrow. If the arrow points left-to-right, the boson must be W⁺ .

❌ Common Errors

  • Writing a capital D for the down quark (strict mark scheme penalty: "Do not accept capital D for d").
  • Using β⁻ instead of e or e⁻ for the incoming atomic electron.
  • Labelling the neutrino as an antineutrino ( ν̄e ). Lepton number must be conserved: Le = +1 → +1 .
  • Labeling the boson as W⁻ while leaving the arrow pointing towards the lepton vertex (violates charge conservation).

Question 01.2

Nuclear Stability of a Bare Nucleus vs. Neutral Atom

[1 Mark]

✅ Correct Answer

The nucleus can only decay by electron capture, which requires orbiting (inner-shell) electrons to capture. A bare nucleus has no electrons available, so it cannot decay (making it more stable than the neutral atom).

Mark Scheme: In order for the nucleus to decay it needs (orbiting) electrons because it (only) decays by electron capture. [1 mark]

🧠 Exam Technique: Focus on the Given Premise

  • Notice the bold word in the question prompt: "only decays by electron capture".
  • "Stability" in radioactivity refers directly to a nucleus's likelihood or ability to undergo radioactive decay.
  • Avoid discussing electrostatic repulsion, binding energy, or ionization energy; the question specifically links stability to the decay mode.

Question 01.3

Determining the Neutron Number from Specific Charge

[2 Marks]

📐 Step-by-Step Calculation

Step 1: Calculate total charge of the nucleus ( Q )
Proton number Z = 22
Elementary charge e = 1.60 × 10⁻¹⁹ C
Q = 22 × (1.60 × 10⁻¹⁹ C) = 3.52 × 10⁻¹⁸ C
Step 2: Calculate the total mass of the nucleus ( m )
Definition: Specific Charge = Q / m
Rearranging for mass:
m = Q / Specific Charge = (3.52 × 10⁻¹⁸ C) / (4.8 × 10⁷ C kg⁻¹) = 7.333... × 10⁻²⁶ kg
(Award Mark 1 for evidence of calculating total charge OR setting up mass = Q / Specific Charge)
Step 3: Determine the nucleon number ( A )
Mass of one nucleon ( mnuc ≈ 1.67 × 10⁻²⁷ kg ):
A = m / mnuc = (7.333... × 10⁻²⁶ kg) / (1.67 × 10⁻²⁷ kg) = 43.91 ≈ 44 nucleons
Step 4: Calculate the number of neutrons ( N )
N = A - Z = 44 - 22 = 22 neutrons
(Award Mark 2 for correct method showing nucleon number minus 22 resulting in 22)

💡 Alternative Shortcut Method

Compare the specific charge of the nucleus directly to a proton:

  • Specific charge of a single proton = (1.60 × 10⁻¹⁹) / (1.67 × 10⁻²⁷) ≈ 9.58 × 10⁷ C kg⁻¹ .
  • Given specific charge = 4.8 × 10⁷ C kg⁻¹ , which is exactly half ( 0.5× ) the proton's specific charge!
  • Since specific charge is proportional to Z / A , a value equal to half of a proton means Z / A = 1 / 2 .
  • Therefore, A = 2Z = 44 , giving N = 44 - 22 = 22 .

❌ Common Errors & Mark Traps

  • Bald Answers: Writing just 22 without showing working receives 0 marks. The mark scheme explicitly states: "Do not accept '22' without some evidence."
  • Including Electron Mass: The question asks about the nucleus, not the atom. Including orbital electrons in charge or mass will give incorrect values.
  • Incorrect Nucleon Mass: Forgetting to use the data booklet value for nucleon mass ( 1.67 × 10⁻²⁷ kg ).

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.