AQA AS Level Physics Paper 2, June 2022: Question 10

1 mark · Medium difficulty · Multiple Choice

Identify the resulting nuclide X when an alpha particle interacts with boron-10 to form an unstable nucleus and a neutron, followed by positron emission.

Practise this question

Question

Multiple choice question 10 asking to identify nuclide X from a nuclear reaction involving an alpha particle and boron-10, followed by positron emission. Four options are given: A (carbon-13 equivalent with incorrect proton number), B (carbon-13), C (nitrogen-13), and D (oxygen-13), each with a selection box.
Question text

10 An alpha particle and a nucleus of boron 10 B interact to form an unstable nucleus and

a free neutron.

The unstable nucleus decays by positron emission to form a nucleus of nuclide X.

What is X?

[1 mark]

13B

A 5

13C

B 6

13 N

C 7

13O

D 8

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is B, corresponding to carbon-13.

13C

10 B (AO1) 6

How to answer it

Nuclear Equations and Positron Emission

AQA AS Level Physics • Particles and Radiation

What this question tests

This question assesses your ability to balance nuclear equations by applying the conservation of nucleon number (A) and proton number (Z) across two consecutive nuclear processes: a particle-induced nuclear reaction followed by radioactive decay (positron emission).

Question 10

Determining Nuclide X

✅ Correct Answer

B (¹³₆C)

Mark: 1 / 1

💡 Key Knowledge

  • Alpha particle: Represented as &sup4;&sub2;α (or helium nucleus &sup4;&sub2;He).
  • Neutron: Represented as ¹&sub1;n.
  • Positron emission: A proton transforms into a neutron, emitting a positron (&sup0;&sub1;e⁺ or &sup0;&sub1;β⁺) and an electron neutrino. Proton number (Z) decreases by 1; nucleon number (A) stays constant.

🧠 Exam Technique

Tackle multi-step nuclear problems sequentially. Write out the first balanced equation to find the intermediate unstable nucleus, then apply the second decay transformation to find the final nuclide X .

❌ Common Errors

A frequent mistake during positron emission is mistakenly increasing the proton number by 1 (confusing it with beta-minus decay). Always remember that emitting a positive charge means the nucleus loses a positive proton, dropping Z by 1.

📐 Step-by-Step Working Out

  1. Step 1: Balance the first nuclear interaction.
    Reactants: Boron ¹&sup0;&sub5;B + Alpha particle &sup4;&sub2;α
    Products: Unstable nucleus + Free neutron ¹&sub1;n
    Nucleon number (A): 10 + 4 = A(unstable) + 1 → A(unstable) = 13
    Proton number (Z): 5 + 2 = Z(unstable) + 0 → Z(unstable) = 7 (Nitrogen, ¹³&sub7;N )
  2. Step 2: Balance the radioactive decay.
    The unstable nucleus ¹³&sub7;N undergoes positron emission ( &sup0;&sub1;e⁺ ) to form nuclide X .
    Nucleon number (A): 13 = A(X) + 0 → A(X) = 13
    Proton number (Z): 7 = Z(X) + 1 → Z(X) = 6 (Carbon, C )
  3. Step 3: Match with options.
    Nuclide X is ¹³&sub6;C , which corresponds to option B.

Topics

Physics · 3.2 Particles and radiation

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.