HSC Exam Practice

Sample response. The distance of closest approach (r_min) is the minimum distance between an incoming alpha particle and a nucleus during a head-on collision, at which point the alpha particle momentarily comes to rest before being repelled. It is calculated by applying conservation of energy: at closest approach, all the initial kinetic energy of the alpha has been converted into electrical potential energy between the alpha and the nucleus.

Marking notes. 1 mark for defining r_min as the minimum separation at which the alpha momentarily stops (or: where E_k = 0); 1 mark for correctly stating that conservation of energy (kinetic energy = electrical potential energy at closest approach) is the principle applied; 1 mark for identifying it as a head-on (direct) collision scenario.

Sample response. (1) Most alpha particles passed straight through with little or no deflection — indicates atoms are mostly empty space; the nucleus is tiny compared to the atomic radius. (2) Some alpha particles were deflected through small angles (<10°) — indicates the positive charge is concentrated rather than diffuse; near-miss trajectories feel a small Coulomb repulsion. (3) A very small fraction (~1 in 8,000) was deflected by more than 90°, some nearly straight back — indicates the existence of a tiny, very dense, positively charged nucleus capable of exerting enormous Coulomb repulsion at very short range.

Marking notes. 2 marks per observation (1 for the observation itself, 1 for the structural implication), for a total of 6. Award partial credit for partially correct inferences.

Sample response. The distance of closest approach is the point at which the alpha particle turns around due to Coulomb repulsion. At this point, the alpha is still outside the nuclear surface — it has been stopped by the electrostatic field before actually making contact with the nucleus. Therefore, the actual nuclear radius must be smaller than r_min. The value r_min represents the furthest from the nuclear centre that the alpha can be when stopped, so it sets an upper bound: the real nuclear surface is somewhere between zero and r_min.

Marking notes. 1 mark for stating that the alpha is turned around before reaching the nuclear surface; 1 mark for explaining that the real nucleus is smaller than r_min; 1 mark for clarifying that r_min is therefore an upper limit (not the radius itself).

Sample response. (i) Thomson’s model: positive charge is spread evenly throughout the atom (~10⁻¹⁰ m), like a diffuse cloud, with electrons embedded within it. Rutherford’s model: positive charge is concentrated in a tiny, dense nucleus (~10⁻¹⁵ m), with electrons orbiting at much larger distances. (ii) Thomson’s model predicts only small-angle scattering because the diffuse positive charge can only exert a weak repulsive force over a large volume. Rutherford’s model predicts mostly straight-through paths (most alphas miss the tiny nucleus) but occasional large-angle or back-scattering when an alpha passes close to the concentrated positive nucleus.

Marking notes. 1 mark for correctly contrasting charge distribution in Thomson’s model; 1 mark for correctly describing charge distribution in Rutherford’s model; 1 mark for Thomson’s scattering prediction (small angles only); 1 mark for Rutherford’s scattering prediction (mostly straight-through, rare large-angle).

Sample response. The key limitation of Rutherford’s nuclear model is that it cannot explain the stability of electron orbits. Classical electromagnetic theory predicts that any accelerating charge radiates energy. An electron in circular orbit around the nucleus is constantly accelerating (centripetal), so it should continuously emit electromagnetic radiation, lose kinetic energy, and spiral into the nucleus within approximately 10⁻¹² s — which would make all atoms instantly collapse. This was resolved by Bohr’s quantum model (1913), which introduced quantised energy levels and forbidden transitions that prevent electrons from spiralling in.

Marking notes. 1 mark for correctly identifying the limitation (accelerating electrons should radiate and spiral in); 1 mark for explaining the classical physics reasoning (centripetal acceleration → radiation → energy loss); 1 mark for identifying Bohr’s quantum model as the resolution.

Sample response (a) — r_min calculations (3 marks).

Common numerator: k × 2e² = 8.99×10⁹ × 2 × (1.6×10⁻¹⁹)² = 8.99×10⁹ × 2 × 2.56×10⁻³⁸ = 4.60×10⁻²⁸ N m² C⁻¹. E_k = 5.0 × 1.6×10⁻¹³ = 8.0×10⁻¹³ J.

Aluminium (Z = 13): r_min = 4.60×10⁻²⁸ × 13 / 8.0×10⁻¹³ = 7.5×10⁻¹⁵ m ≈ 7.5 fm [1].

Gold (Z = 79): r_min = 4.60×10⁻²⁸ × 79 / 8.0×10⁻¹³ = 4.54×10⁻¹⁴ m ≈ 45.4 fm [1].

Lead (Z = 82): r_min = 4.60×10⁻²⁸ × 82 / 8.0×10⁻¹³ = 4.72×10⁻¹⁴ m ≈ 47.2 fm [1].

Sample response (b) — nuclear radii (3 marks).

Aluminium (A = 27): R = 1.2× 27^1/3 = 1.2×3.0 = 3.6 fm [1].

Gold (A = 197): R = 1.2× 197^1/3 = 1.2×5.82 ≈ 6.98 fm ≈ 7.0 fm [1].

Lead (A = 208): R = 1.2× 208^1/3 = 1.2×5.93 ≈ 7.1 fm [1].

Comment: For all three elements, r_min ≫ R. E.g. for gold, r_min ≈ 45 fm vs R ≈ 7 fm — the alpha is turned around roughly 6× the nuclear radius away from the nuclear surface. This confirms that r_min is indeed an upper limit, not the actual nuclear radius.

Sample response (c) — student argument (2 marks). The student’s argument is correct [1]. A 5.0 MeV alpha approaching gold is turned around at ~45 fm, which is much larger than the nuclear radius (~7 fm). The alpha never gets close enough to the nuclear surface to probe it directly. To reduce r_min to the nuclear radius (~7 fm for gold), the required kinetic energy would need to be increased by a factor of ~45/7 ≈ 6.4, giving E_k ≈ 32 MeV — energies achievable only with a particle accelerator, not a natural alpha source [1].

Marking criteria. Part (a): 1 mark per correct r_min (accept ±5% rounding). Part (b): 1 mark per correct nuclear radius plus comment linking r_min ≫ R (accept ±0.1 fm). Part (c): 1 mark for correctly assessing the student’s argument as valid; 1 mark for quantitative or qualitative reasoning about the energy needed to reach the nuclear surface.

Sample response. The Geiger-Marsden experiment provided three key categories of evidence for Rutherford’s nuclear model. First, the observation that approximately 99.99% of alpha particles passed through the gold foil with minimal deflection is explained by the nuclear model because most of the atom is empty space; the nucleus (~10⁻¹⁵ m) is vastly smaller than the atom (~10⁻¹⁰ m), so most alphas miss it entirely. Second, the rare observation that ~1 in 8,000 alphas was deflected by more than 90° requires a concentrated positive nucleus. Using r_min = k 2Ze²/E_k for a 7.7 MeV alpha on gold (Z = 79), the closest approach is ~29.5 fm, already much larger than the nuclear radius (~7 fm). Only the intense Coulomb repulsion from a point-like concentrated charge could deflect an alpha by >90°; Thomson’s diffuse model explicitly predicts this is impossible and is therefore directly refuted. Third, small-angle deflections are consistent with near-miss trajectories past the nucleus, providing additional quantitative support for the model. A strength of Rutherford’s model is that it quantitatively predicts the scattering distribution using the Coulomb formula, which matches experimental scattering angle distributions extremely well — this predictive success is strong evidence in its favour. However, the model has a significant limitation: classical electromagnetism predicts that an electron in circular orbit is continuously accelerating, so it should emit radiation, lose energy, and spiral into the nucleus in ~10⁻¹² s. Rutherford’s model provides no mechanism to prevent this atomic collapse. This limitation was resolved by Bohr’s quantum model (1913), which is a later refinement of rather than a refutation of Rutherford’s nuclear model. In conclusion, the totality of the evidence from the Geiger-Marsden experiment strongly supports Rutherford’s nuclear model. The large-angle scattering data alone is sufficient to refute Thomson’s model, and the quantitative agreement between Rutherford’s predictions and all three categories of observation makes his model the best-supported account of atomic structure available from these experiments.

Marking criteria (7 marks). 1 = correctly explains observation 1 (straight-through) using empty space / tiny nucleus. 1 = correctly explains observation 2 (large-angle scattering) with reference to concentrated positive nucleus and Coulomb repulsion. 1 = uses formula or quantitative argument to support the explanation of closest approach. 1 = identifies and explains at least one strength of Rutherford’s model (quantitative prediction, explanatory power). 1 = identifies and explains the key limitation (orbital instability / electron radiation). 1 = explains why the limitation does not refute the nuclear model (resolved by Bohr; model is an improvement not a replacement). 1 = reaches an explicit, justified overall evaluative conclusion about the extent to which evidence supports Rutherford’s model.