## ERQ · 12 marks · Topics: E.1 Structure of the atom + D.2 Electric and magnetic fields
**Stem.** In a modern re-enactment of Rutherford's scattering experiment, a thin gold foil of thickness 8.0 × 10⁻⁷ m is bombarded by alpha particles of kinetic energy 5.5 MeV emitted from an americium-241 source. A particular alpha particle is observed to undergo a head-on collision with a single gold nucleus (Z = 79) and is scattered directly back along its incoming path. After scattering, the alpha particle is steered into a velocity selector for identification, consisting of crossed uniform electric and magnetic fields. The electric field has magnitude 1.20 × 10⁵ V m⁻¹ directed downward, and a magnetic field is applied perpendicular to both the electric field and the particle's velocity so that the alpha particle passes through undeflected. Take the mass of an alpha particle as 6.64 × 10⁻²⁷ kg and its charge as +2e.
### Part (a) State [2 marks] · AO1 · Topic: E.1
State two experimental observations from Rutherford's original scattering experiment, and state the conclusion about atomic structure that each observation supports.
### Part (b)(i) Show that [3 marks] · AO2/AO3 · Topic: E.1
Show that the distance of closest approach of the alpha particle to the gold nucleus during the head-on collision is approximately 4.1 × 10⁻¹⁴ m.
### Part (b)(ii) Determine [3 marks] · AO2/AO3 · Topic: E.1+D.2
After scattering, the alpha particle retains essentially all of its initial kinetic energy. Determine the magnitude of the magnetic field that must be applied in the velocity selector so that the alpha particle passes through undeflected.
### Part (c) Show that [2 marks] · AO3 · Topic: E.1
Using your answer to (b)(i), show that the calculated upper limit of the radius of the gold nucleus is significantly smaller than the empirical nuclear radius predicted by R = R₀A^(1/3) with R₀ = 1.20 × 10⁻¹⁵ m for gold (A = 197).
### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
Suggest two reasons why the distance of closest approach calculated in (b)(i) overestimates the actual radius of the gold nucleus.
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## Mark Scheme
### Part (a) [2 marks]
- Mark 1: Any one valid observation–conclusion pair, e.g. "most alpha particles passed straight through with little/no deflection" → "atom is mostly empty space" [ECF: no]
- Mark 2: A second distinct pair, e.g. "a small fraction were deflected through large angles (>90°)" → "atom contains a small, dense, positively charged nucleus" [ECF: no]
- Note: Award zero if observation given without matching conclusion (or vice versa).
### Part (b)(i) [3 marks]
- Mark 1: Equate initial KE to electric PE at closest approach: ½mv² = kQq/d, i.e. E_k = (2e)(79e)/(4πε₀ d) [ECF: no]
- Mark 2: Correct conversion 5.5 MeV = 5.5 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.80 × 10⁻¹³ J AND correct substitution: d = (8.99 × 10⁹)(2)(79)(1.60 × 10⁻¹⁹)² / (8.80 × 10⁻¹³) [ECF: yes]
- Mark 3: d = 4.13 × 10⁻¹⁴ m ≈ 4.1 × 10⁻¹⁴ m, shown to at least 2 s.f. [ECF: yes]
### Part (b)(ii) [3 marks]
- Mark 1: Condition for undeflected passage: qE = qvB, so B = E/v [ECF: no]
- Mark 2: Find v from KE: v = √(2E_k/m) = √(2 × 8.80 × 10⁻¹³ / 6.64 × 10⁻²⁷) = 1.63 × 10⁷ m s⁻¹ [ECF: yes from b(i) conversion]
- Mark 3: B = (1.20 × 10⁵)/(1.63 × 10⁷) = 7.4 × 10⁻³ T (accept 7.3–7.5 × 10⁻³ T) [ECF: yes]
### Part (c) [2 marks]
- Mark 1: Empirical radius R = (1.20 × 10⁻¹⁵)(197)^(1/3) = (1.20 × 10⁻¹⁵)(5.819) = 6.98 × 10⁻¹⁵ m ≈ 7.0 × 10⁻¹⁵ m [ECF: no]
- Mark 2: Comparison statement: d (≈ 4.1 × 10⁻¹⁴ m) is roughly 6× larger than the empirical R (≈ 7.0 × 10⁻¹⁵ m), confirming d is an upper bound only [ECF: yes]
### Part (d) [2 marks]
- Mark 1: Any one valid reason, e.g. "the alpha particle is repelled by Coulomb force and decelerates to rest before physically contacting the nucleus — so d > R" [ECF: no]
- Mark 2: A second distinct reason, e.g. "the model assumes the nucleus is a point charge / ignores finite nuclear size", OR "the strong nuclear force is not modelled — only electrostatic repulsion is considered", OR "5.5 MeV is too low an energy for the alpha to reach the nuclear surface (penetration would require higher KE)" [ECF: no]
### Marker notes
- (b)(i) alternative: working in eV throughout — d = k(2)(79)e/E_k(in eV) with k·e = 1.44 × 10⁻⁹ eV·m gives d = (1.44 × 10⁻⁹ × 158)/(5.5 × 10⁶) = 4.14 × 10⁻¹⁴ m. Full credit.
- (b)(ii) alternative: relativistic check — v/c ≈ 0.054, so non-relativistic treatment justified; no penalty either way.
- (d) discriminator: accept any two of {alpha never touches nucleus / point-charge assumption / strong force neglected / insufficient KE to overcome Coulomb barrier / recoil of gold nucleus neglected (treats nucleus as infinitely massive)}. Do NOT accept vague answers like "experimental error" or "air resistance".