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## ERQ · 12 marks · Topics: E.1 Structure of the atom + D.2 Electric and magnetic fields · Archetype: theory_application
**Integration:** primary=E.1 Structure of the atom, secondary=D.2 Electric and magnetic fields (strength: supporting)
**Stem.** In a modern repeat of the Geiger–Marsden experiment, a thin gold foil (Z = 79) is bombarded with alpha particles of kinetic energy 5.40 MeV from a ²¹⁰Po source. The alpha particles are treated as point charges that interact with each gold nucleus only through the Coulomb force; the gold nucleus is taken to be stationary. The empirical nuclear-radius formula R = R₀A^(1/3) with R₀ = 1.20 fm and A = 197 may be used. In a second run, the source is replaced by one producing 30.0 MeV alpha particles from an accelerator; the same foil is used and the detector array measures the angular distribution of scattered particles.
### Part (a) Calculate [2 marks] · AO2 · Topic: E.1
Calculate the radius R of a gold nucleus, in fm.
### Part (b)(i) Show that [3 marks] · AO2 · Topic: E.1+D.2
For a head-on collision in the 5.40 MeV run, show that the distance of closest approach d between the alpha particle and the gold nucleus is approximately 4.21 × 10⁻¹⁴ m.
### Part (b)(ii) Comment on [2 marks] · AO3 · Topic: E.1
By comparing your answer to (a) with d from (b)(i), comment on whether the 5.40 MeV data are consistent with the Rutherford model treating the nucleus as a point charge.
### Part (c) Explain [3 marks] · AO3 · Topic: E.1+D.2
In the 30.0 MeV run, the measured angular distribution of scattered alpha particles deviates significantly from the 1/sin⁴(θ/2) prediction at large scattering angles. Explain this deviation.
### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR · Topic: E.1
The derivation of d in (b)(i) assumes the gold nucleus remains stationary throughout the collision. Suggest how relaxing this assumption would alter the value of d obtained for the 5.40 MeV alpha particles, justifying your reasoning.
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## Mark Scheme
### Part (a) [2 marks] — Calculate
- M1: substitution R = 1.20 × 10⁻¹⁵ × 197^(1/3) (= 1.20 × 10⁻¹⁵ × 5.819) [no ECF]
- M2: R ≈ 6.98 × 10⁻¹⁵ m ≈ 7.0 fm [no ECF]
### Part (b)(i) [3 marks] — Show that
- M1: equates initial KE to Coulomb PE at closest approach: 5.40 × 10⁶ × 1.60 × 10⁻¹⁹ = kQq/d, with Q = 79e and q = 2e [no ECF]
- M2: rearranges d = (2 × 79 × (1.60 × 10⁻¹⁹)² × 8.99 × 10⁹) / (5.40 × 10⁶ × 1.60 × 10⁻¹⁹) [no ECF]
- M3: d ≈ 4.213 × 10⁻¹⁴ m, demonstrating ≥ 3 sig fig agreement with 4.21 × 10⁻¹⁴ m [no ECF]
### Part (b)(ii) [2 marks] — Comment on (interpretation + data per §4.4.1)
- M1: identifies that d (≈ 42 fm) is much larger than R (≈ 7 fm), so the alpha particle never reaches the nuclear surface [ECF from (a)]
- M2: concludes the data are consistent with Rutherford's point-charge / pure-Coulomb assumption, since only the external Coulomb field is sampled
### Part (c) [3 marks] — Explain (causal chain per §4.4.1)
- M1: at 30.0 MeV, d scales as 1/E so d ≈ 4.21 × 10⁻¹⁴ × (5.40/30.0) ≈ 7.6 × 10⁻¹⁵ m, which is comparable to R ≈ 7.0 fm, **therefore** the alpha particle penetrates the nuclear charge distribution rather than scattering from an external point charge, **which means** the inverse-square Coulomb law underpinning the 1/sin⁴(θ/2) prediction no longer holds and the angular dependence deviates [ECF from (a), (b)(i)]
- M2: at such separations the alpha particle enters the range of the strong nuclear force, **therefore** the interaction is no longer purely electrostatic and additional (attractive, short-range) scattering contributes, producing further departure from the Rutherford formula
- M3: large-angle scattering corresponds to small impact parameter, **therefore** it is precisely these trajectories that probe r ≲ R where the above effects dominate, explaining why the deviation appears specifically at large θ
### Part (d) [2 marks] — Suggest (proposal + reasoning per §4.4.1)
- M1: proposes that the true d would be larger than the value in (b)(i) [or equivalently: the alpha particle does not reach as close to the nucleus]
- M2: justifies via conservation of momentum: the recoiling nucleus carries away kinetic energy, **therefore** less of the initial KE is available to do work against the Coulomb potential, so equating KE to kQq/d gives a larger d
### Marker notes
- (a): accept 6.98–7.05 fm.
- (b)(i): alternative method via d = kQq/E in eV form acceptable; answers in range 4.20–4.22 × 10⁻¹⁴ m accept full marks; show-that target 4.213 × 10⁻¹⁴ m given to 3 sig figs so student working must demonstrate intermediate substitution.
- (b)(ii): student must reference both numerical values to score M1; ECF from (a) and the given show-that value of (b)(i) per §4.5.
- (c): award M1 OR M2 in full if the candidate constructs a complete causal chain from one mechanism (charge-distribution breakdown OR strong nuclear force); M3 is awarded independently for linking small impact parameter to large θ. Maximum 3 marks. Accept use of the given 4.21 × 10⁻¹⁴ m from (b)(i) per §4.5; ECF accepted from (a).
- (d): accept equivalent reasoning via reduced-mass argument (effective mass μ = m_α m_Au/(m_α+m_Au) replaces m_α, so KE available in the centre-of-mass frame is reduced by factor m_Au/(m_α+m_Au) ≈ 0.98, giving a fractional increase in d of ≈ 2%); a numerical estimate is not required for M2.
- Show-that target in (b)(i): 4.21 × 10⁻¹⁴ m given to 3 sig figs; student-derived 4.20–4.22 × 10⁻¹⁴ m acceptable.
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