AQA A-Level Physics Paper 2, June 2025: Question 1

📐 Step-by-Step Calculation

1. Identify Charges:
   Alpha particle: Q_α = 2e = 2 × 1.60 × 10⁻¹⁹ C
   Gold nucleus (Z = 79): Q_Au = 79e = 79 × 1.60 × 10⁻¹⁹ C
2. Convert Energy from MeV to Joules:
   E_k = 5.59 × 10⁶ × 1.60 × 10⁻¹⁹ J = 8.944 × 10⁻¹³ J
3. Conservation of Energy Equation:
   At distance of closest approach r , all initial E_k becomes electrostatic E_p :
   E_k = (Q_α × Q_Au) / (4πε₀r)
4. Rearrange & Solve for r:
   r = (2 × 1.60 × 10⁻¹⁹ × 79 × 1.60 × 10⁻¹⁹) / (4π × 8.85 × 10⁻¹² × 8.944 × 10⁻¹³)
   r = 4.07 × 10⁻¹⁴ m

✅ Correct Answer & Guidance

radius of nucleus = 4.1 × 10⁻¹⁴ m (or 4.07 × 10⁻¹⁴ m)

❌ Common Errors

• Forgetting to convert MeV to Joules: Leaving energy in MeV or eV results in a power-of-ten error of at least 10¹³.
• Using Mass Number (197) instead of Atomic Number (79): Charge depends solely on protons ( Z = 79 ), not total nucleons.
• Using wrong alpha charge: Setting Q_α = 1e instead of 2e .

🧠 Exam Technique

Remember that the distance of closest approach is an upper limit for the nuclear radius. The alpha particle stops before it actually touches the nucleus due to electrostatic repulsion, so r_closest > R_nucleus .

✅ Acceptable Points (Any Two)

• The alpha particle does not touch the nucleus / it stops at a distance from the nucleus (gives an upper limit to the nuclear radius; the estimate is larger than the true radius).
• The distance of closest approach depends on the initial kinetic energy of the alpha particle (does not yield a single, definitive value).
• The recoil of the target gold nucleus is ignored.
• The alpha particle itself has a finite size (it is not a point charge).

❌ What Examiners Rejected

• Simply stating "it is the distance of closest approach" without explaining that it gives an upper limit or is larger than the actual radius.
• Referring to the shape of the nucleus (e.g. "the nucleus is not spherical").
• Mentioning standard experimental uncertainty or human error.

🎨 What to Sketch on Figure 1

Examiner Visual Description:

• Start on the y-axis: Maximum intensity at angle = 0 . The curve must start high and curve downward with a negative gradient.
• First Minimum: Curves down to a distinct dip (minimum), but the intensity must NOT reach zero (it must remain above the horizontal axis).
• Subsequent Peaks: Rises to at least one subsidiary maximum whose height is much lower than the central peak, before decreasing again.

✅ Mark Scheme Breakdown

• Mark 1 (Graph shape): Maximum at angle = 0 , first minimum is non-zero, followed by at least one subsidiary smaller maximum.
• Mark 2 (Feature stated): The angle of the first diffraction minimum (first minimum) is measured and used to calculate radius via sin θ ≈ 0.61 λ / R (or 1.22 λ / 2R ).

❌ Common Mistakes on the Graph

• Touching the horizontal axis (zero intensity) at the minimum — in electron diffraction by nuclei, the minima never drop to zero intensity.
• Drawing symmetrical double-slit interference fringes or light-diffraction grating patterns (equal spacing/equal intensity).
• Failing to identify the first minimum as the critical feature.

📐 Step-by-Step Derivation

1. Express Nuclear Volume:
   Assuming the nucleus is spherical:
   V = (4/3) π R³
   Substitute R = R₀A¹ᐟ³ :
   V = (4/3) π (R₀A¹ᐟ³)³ = (4/3) π R₀³ A
   Hence, V ∝ A .
2. Express Nuclear Mass:
   If m_n is the average mass of a nucleon:
   m = A × m_n (mass is directly proportional to A ).
3. Calculate Density (ρ = m / V):
   ρ = (A × m_n) / ((4/3) π R₀³ A) = m_n / ((4/3) π R₀³)
4. Conclude:
   Because A cancels out completely, ρ depends only on constants ( m_n, R₀, π ). Therefore, nuclear density is constant across all nuclei.

💡 Key Knowledge & Marking Conditions

• Mark 1: Stating either that volume is spherical and proportional to A ( V = (4/3)πR₀³A ) OR that nuclear mass is proportional to A ( m = A × m_n ).
• Mark 2: Showing that in the density expression ρ = m / V , A cancels out completely, leaving only constant terms (independent of A ).