AQA A-Level Physics Paper 2, June 2023: Question 19
1 mark · Medium difficulty · Multiple Choice
Calculate the closest distance of approach of an alpha particle with known kinetic energy towards a gold nucleus.
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Question text
19 An alpha particle is moving towards a stationary gold nucleus. The alpha particle has a
kinetic energy of 9.0 × 10−13 J when it is a large distance from the gold nucleus.
The gold nucleus contains 79 protons.
What is the closest possible distance of approach of the alpha particle to the gold nucleus?
[1 mark]
A 2.5 × 10−16 m
B 2.0 × 10−14 m
C 4.0 × 10−14 m
D 2.0 × 10−7 m
Mark scheme
Show the mark scheme
19 C 4.0 × 10−14 m
How to answer it
Closest Approach of an Alpha Particle to a Gold Nucleus
This question assesses your ability to apply conservation of energy in electric fields to calculate the distance of closest approach during Rutherford alpha scattering. Specifically, it tests:
- Equating initial kinetic energy ( Ek ) to electric potential energy ( Ep ) at the point of momentary rest.
- Correctly identifying the charges of both the alpha particle ( q = +2e ) and the target nucleus ( Q = +79e ).
- Manipulating Coulomb's Law for electric potential energy to solve for distance ( r ).
Question 19 Walkthrough
Determining Distance of Closest Approach
✅ Correct Answer
C: 4.0 × 10⁻¹⁴ m
At the point of closest approach, all initial kinetic energy has converted entirely into electrostatic potential energy. Solving for r yields 4.0 × 10⁻¹⁴ m .
💡 Key Knowledge
- Conservation of Energy:
Ek = Ep (at closest approach, velocity is momentarily zero). - Electric Potential Energy Formula:
Ep = (1 / 4πε₀) × (q₁q₂ / r) - Alpha particle charge: Contains 2 protons, so q₁ = +2e = 2 × 1.60 × 10⁻¹⁹ C .
- Gold nucleus charge: 79 protons, so q₂ = +79e = 79 × 1.60 × 10⁻¹⁹ C .
- Coulomb constant:
1 / (4πε₀) ≈ 8.99 × 10⁹ N m² C⁻²
📐 Step-by-Step Calculation
- Set up the energy conservation relation:
Ek = (1 / 4πε₀) × (q₁q₂ / r) - Rearrange to make the closest approach distance (r) the subject:
r = (1 / 4πε₀) × (q₁q₂) / Ek - Substitute the known physical constants and values:
q₁ = 2 × (1.60 × 10⁻¹⁹ C) = 3.20 × 10⁻¹⁹ C
q₂ = 79 × (1.60 × 10⁻¹⁹ C) = 1.264 × 10⁻¹⁷ C
Ek = 9.0 × 10⁻¹³ J
1 / (4πε₀) = 8.99 × 10⁹ N m² C⁻² - Compute the product in the numerator:
Numerator = 8.99 × 10⁹ × (2 × 79) × (1.60 × 10⁻¹⁹)²
Numerator = 8.99 × 10⁹ × 158 × 2.56 × 10⁻³⁸ = 3.636 × 10⁻²⁶ J m - Divide by kinetic energy:
r = (3.636 × 10⁻²⁶) / (9.0 × 10⁻¹³) = 4.04 × 10⁻¹⁴ m
To 2 significant figures: 4.0 × 10⁻¹⁴ m.
❌ Common Traps & Distractors
- Selecting B (2.0 × 10⁻¹⁴ m): This is the most common distractor! Students mistakenly set the charge of the alpha particle to +1e (proton charge) instead of +2e , halving the numerator and resulting in exactly half the correct distance.
- Selecting A (2.5 × 10⁻¹⁶ m): Arises from squaring the distance r² by confusing potential energy with Coulomb's electrostatic force formula ( F = k q₁q₂ / r² ).
- Power of 10 Errors (Option D): Forgetting to square the elementary charge e (using e only once instead of e² ).
🧠 Exam Technique & Insight
- Physics Sense Check: Nuclear radii are of the order ~10⁻¹⁵ m to ~10⁻¹⁴ m . A closest approach distance of ~10⁻¹⁴ m makes physical sense as an upper limit on the size of the gold nucleus.
- Speed Tip for MCQs: Write the numerator as (8.99 × 10⁹) × (2 × 79) × (1.60 × 10⁻¹⁹)² directly into the calculator with brackets around the denominator to avoid order-of-operation errors.
- Particle Recognition: An alpha particle is a helium-4 nucleus ( ₂⁴He²⁺ ). Always double-check you've used Z = 2 for the alpha charge.
Topics
Physics · 3.7 Fields and their consequences (A-level only) · 3.8 Nuclear physics (A-level only)
Question and mark scheme from the AQA A-Level Physics examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.