## ERQ · 12 marks · Topics: E.1 Structure of the atom + D.2 Electric and magnetic fields
**Stem.** A Rutherford-type scattering experiment is repeated using alpha particles directed head-on at a stationary gold-197 nucleus (Z = 79). The alpha particles are produced by a source and accelerated from rest through a potential difference V across parallel plates separated by 4.0 cm, before entering a vacuum chamber where the gold foil is mounted. A detector records the closest approach distance, d, of the alphas to a gold nucleus as 4.55 × 10⁻¹⁴ m. After scattering, some alpha particles enter a region of uniform magnetic field of flux density 0.85 T directed perpendicular to their velocity, and follow a circular arc of radius 0.123 m. Assume the gold nucleus remains stationary and treat all particles non-relativistically. (Mass of alpha = 6.64 × 10⁻²⁷ kg; charge of alpha = +2e.)
### Part (a) Outline [2 marks] · AO1 · Topic: E.1
Outline how the results of the Geiger–Marsden alpha scattering experiment support the nuclear model of the atom.
### Part (b)(i) Show that [3 marks] · AO2/AO3 · Topic: E.1
Show that the kinetic energy of an alpha particle at the moment of closest approach to the gold nucleus is approximately 5.0 MeV.
### Part (b)(ii) Determine [2 marks] · AO2/AO3 · Topic: E.1
Determine the accelerating potential difference V applied across the parallel plates.
### Part (c) Calculate [3 marks] · AO3 · Topic: E.1 + D.2
After scattering, an alpha particle enters the magnetic-field region and moves in a circular arc of radius 0.123 m. Calculate the kinetic energy of this scattered alpha particle, in MeV, and comment on whether the scattering was elastic.
### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
The kinetic energy obtained in (c) differs slightly from the value in (b)(i). Suggest two reasons, based on the assumptions of the model, why this discrepancy arises.
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## Mark Scheme
### Part (a) [2 marks]
- M1: Most alpha particles passed through with little/no deflection ⇒ atom is mostly empty space [ECF: no]
- M2: A very small fraction were deflected through large angles (>90°) ⇒ small, dense, positively charged nucleus containing most of the mass [ECF: no]
### Part (b)(i) [3 marks]
- M1: Recognise that at closest approach all KE → electrostatic PE: E_k = kQ₁Q₂/d with Q₁ = 2e, Q₂ = 79e [ECF: no]
- M2: Substitution: E_k = (8.99 × 10⁹)(2 × 1.60 × 10⁻¹⁹)(79 × 1.60 × 10⁻¹⁹) / (4.55 × 10⁻¹⁴) [ECF: yes]
- M3: E_k = 7.99 × 10⁻¹³ J ≈ 4.99 MeV ≈ 5.0 MeV (2 s.f.) — must show conversion 1 eV = 1.60 × 10⁻¹⁹ J [ECF: yes]
### Part (b)(ii) [2 marks]
- M1: Use qV = E_k ⇒ V = E_k / q = (7.99 × 10⁻¹³) / (2 × 1.60 × 10⁻¹⁹) [ECF: yes from b(i)]
- M2: V = 2.5 × 10⁶ V (2.50 MV, 2 s.f.) [ECF: yes]
### Part (c) [3 marks]
- M1: Equate magnetic force to centripetal: qvB = mv²/r ⇒ v = qBr/m = (2 × 1.60 × 10⁻¹⁹)(0.85)(0.123)/(6.64 × 10⁻²⁷) [ECF: no]
- M2: v = 5.04 × 10⁶ m s⁻¹; E_k = ½mv² = ½(6.64 × 10⁻²⁷)(5.04 × 10⁶)² = 8.43 × 10⁻¹⁴ J = 0.527 MeV (accept 0.5 MeV) [ECF: yes]
- M3: Comment: this is far less than 5.0 MeV, so scattering was NOT elastic — OR the scattered particle is not the original alpha / energy transferred to nucleus / recoil occurred. Must explicitly compare magnitudes. [ECF: yes]
### Part (d) [2 marks]
Award 1 mark each for any two distinct, physically valid critiques:
- The gold nucleus was assumed stationary, but in reality it recoils, removing kinetic energy from the alpha
- The model treats the nucleus as a point charge; at d ~ 4.5 × 10⁻¹⁴ m the alpha may interact via the strong nuclear force (not purely Coulombic)
- Non-relativistic treatment used, but at v ≈ 1.5 × 10⁷ m s⁻¹ for the incident alpha a small relativistic correction exists
- Electrons of the gold atom partially screen the nuclear charge, reducing the effective Coulomb potential
- The magnetic field may not be perfectly uniform, giving an inaccurate radius measurement
### Marker notes
- Alternative for (b)(i): students may quote E_k in J directly and then convert; full marks if all three steps visible.
- Alternative for (c): students who use p = qBr then E_k = p²/(2m) earn M1+M2 identically.
- (d) discriminator: reject vague answers like "experimental error" or "air resistance" (vacuum chamber). Must reference a stated model assumption. Accept any two from the bulleted list; one mark per distinct, physically valid critique.