Physics • Year 12 • Module 8 • Lesson 11
The Nuclear Model of the Atom
Apply your understanding of Rutherford scattering, the closest approach formula, and model comparison to real data, calculations, and a diagram critique.
1. Interpret experimental data — alpha scattering across different target elements
A student fires 5.0 MeV alpha particles at thin foils of different elements and records the fraction deflected by more than 90° and the calculated distance of closest approach. The table shows the data. 8 marks
| Target element | Atomic number (Z) | Fraction deflected >90° | Closest approach rmin (calculated) | Consistent with nuclear model? |
|---|---|---|---|---|
| Aluminium (Al) | 13 | Very rare | 7.1 × 10−15 m | |
| Silver (Ag) | 47 | Rare, but more than Al | 2.7 × 10−14 m | |
| Gold (Au) | 79 | Rare, but more than Ag | 4.5 × 10−14 m | |
| Uranium (U) | 92 | Rare, but most of the four | 5.3 × 10−14 m |
Closest approach values calculated using rmin = k 2Ze2 / Ek. Illustrative data.
1.1 Complete the “Consistent with nuclear model?” column for all four rows. Briefly justify each entry. 4 marks (1 per row)
1.2 Describe the trend in closest approach distance as Z increases and explain this trend using the formula rmin = k 2Ze2 / Ek. 2 marks
1.3 All four closest approach values are larger than the actual nuclear radii (which are on the order of 10−15 m). What does this confirm about the relationship between rmin and nuclear radius? 2 marks
2. Interpret graph — closest approach vs kinetic energy
A researcher calculates the distance of closest approach for alpha particles of different kinetic energies approaching a gold nucleus (Z = 79). The results are plotted below. 7 marks
Figure 2. Distance of closest approach (rmin) versus kinetic energy (Ek) for alpha particles approaching gold (Z = 79). Calculated using rmin = k 2Ze2 / Ek.
2.1 Describe the shape of the graph and identify what type of mathematical relationship it represents between rmin and Ek. 2 marks
2.2 Use the graph to estimate rmin for a 5.0 MeV alpha and then verify this using rmin = k 2Ze2 / Ek. Show full working. (k = 8.99×109 N m2 C−2; e = 1.6×10−19 C; 1 MeV = 1.6×10−13 J) 3 marks
2.3 A student concludes: “Doubling the kinetic energy of the alpha particle doubles the closest approach distance.” Is this correct? Use the formula to justify your answer. 2 marks
3. Compare Thomson’s and Rutherford’s atomic models
Complete the two-column table below. For each feature, write a concise description that contrasts the two models. 10 marks (1 per cell)
| Feature | Thomson’s plum pudding model | Rutherford’s nuclear model |
|---|---|---|
| Distribution of positive charge | ||
| Location of electrons | ||
| Prediction for alpha scattering angles | ||
| Explains large-angle scattering? | ||
| Key limitation / failure |
4. Predict and justify — effect of changing the target nucleus
Consider a scattering experiment where 7.7 MeV alpha particles are fired at a thin foil. 5 marks
4.1 A researcher switches from a gold target (Z = 79) to an aluminium target (Z = 13). Predict whether the fraction of alphas deflected by more than 90° will increase, decrease, or stay the same. Justify your prediction using the closest approach formula. 3 marks
4.2 The same researcher argues that using a target with a very high Z makes the gold foil experiment a more sensitive probe of nuclear size. Evaluate whether this claim is correct, linking your answer to the formula for rmin. 2 marks
5. Diagram critique — what’s wrong with this student’s diagram?
A Year 12 student drew the diagram below to explain the Geiger-Marsden experiment. There are three errors in the student’s description beneath the diagram. Identify each error and write the correction. 6 marks (2 per error: 1 identify, 1 correct)
Student’s description:
“Rutherford fired positively-charged beta particles at a thick sheet of aluminium. Nearly all of the particles bounced back, with only a few passing straight through. Those that bounced back showed that atoms are mostly empty space. The small number that passed straight through showed that there must be a tiny, concentrated, negatively charged nucleus at the centre of each atom.”
5.1 Error 1: What is wrong?
Correction:
5.2 Error 2: What is wrong?
Correction:
5.3 Error 3: What is wrong?
Correction:
Q1.1 — Consistent with nuclear model?
All four rows: Yes. The nuclear model predicts that large-angle scattering is possible but rare, because the nucleus is tiny. Higher-Z targets have a larger, more positively charged nucleus, exerting greater Coulomb repulsion at a given distance — so larger-Z targets produce more large-angle deflections. Each result is consistent with this prediction.
1 mark per row for “Yes” plus a valid justification linking Z to scattering frequency or Coulomb force.
Q1.2 — Trend in closest approach vs Z
As Z increases, rmin increases (i.e. alphas are turned around at a greater distance from the nucleus) [1]. From rmin = k 2Ze2 / Ek, since k, e, and Ek are constant, rmin ∝ Z; doubling Z doubles the closest approach distance [1].
Q1.3 — Upper limit confirmed
All calculated rmin values (e.g. 4.5×10−14 m for gold) are larger than actual nuclear radii (~10−15 m) [1]. This confirms that the alpha is stopped by Coulomb repulsion before reaching the nuclear surface — the distance of closest approach is therefore an upper limit on, not equal to, the nuclear radius [1].
Q2.1 — Graph shape and relationship
The graph shows a curve that decreases steeply from high values at low energy and levels off at higher energies — a hyperbolic (inverse) shape [1]. The relationship is an inverse proportion: rmin ∝ 1/Ek [1].
Q2.2 — Estimate and calculation (5 MeV, gold)
From the graph: rmin ≈ 45 fm (accept 40–50 fm) [1].
Calculation [1 for substitution, 1 for correct answer]:
rmin = (8.99×109) × 2 × 79 × (1.6×10−19)2 / (5.0 × 1.6×10−13)
= (8.99×109 × 2 × 79 × 2.56×10−38) / (8.0×10−13)
= (3.64×10−26) / (8.0×10−13)
≈ 4.55×10−14 m (≈ 45.5 fm)
Q2.3 — Doubling kinetic energy
The student is incorrect [1]. From rmin = k 2Ze2/Ek, doubling Ek halves rmin (inverse relationship). For example, doubling from 5 MeV to 10 MeV reduces rmin from ~45 fm to ~23 fm [1].
Q3 — Compare and contrast table
Distribution of positive charge: Thomson — spread evenly through the whole atom (~10−10 m), like a diffuse cloud. Rutherford — concentrated in a tiny nucleus (~10−15 m).
Location of electrons: Thomson — embedded throughout the diffuse positive cloud like fruit in a pudding. Rutherford — orbit the nucleus at relatively large distances (~10−10 m).
Prediction for alpha scattering angles: Thomson — only small-angle deflections (weak, diffuse repulsion). Rutherford — mostly undeflected, occasional large-angle scattering (concentrated repulsion from dense nucleus).
Explains large-angle scattering? Thomson — No; diffuse positive charge cannot produce sufficient repulsive force for large deflections. Rutherford — Yes; concentrated nucleus provides intense Coulomb repulsion at short range.
Key limitation / failure: Thomson — completely fails to explain large-angle scattering (refuted by Geiger-Marsden data). Rutherford — cannot explain why orbiting electrons (accelerating charges) do not radiate energy and spiral into the nucleus (required Bohr's quantum model).
Q4.1 — Switching to aluminium target (3 marks)
The fraction deflected by more than 90° will decrease [1]. From rmin = k 2Ze2/Ek, the closest approach distance is proportional to Z [1]. With Z = 13 (Al) vs Z = 79 (Au), the closest approach is much smaller, meaning alphas must come much closer to the smaller aluminium nucleus to be deflected by >90° — which is less likely because the target is also smaller. The lower Coulomb repulsion from Al means fewer alphas experience the intense force needed for large-angle deflection [1].
Q4.2 — High-Z target as sensitive probe (2 marks)
The claim is partially correct but requires qualification [1]. Higher Z increases rmin ∝ Z, so the alpha is turned around further from the nuclear surface. This actually makes it harder to probe the nuclear surface itself because the alpha never gets close enough. A higher-energy beam with the same target is a better strategy for probing nuclear size (smaller rmin at higher energy). Using high-Z targets does increase the frequency of large-angle events (improving statistics), but it gives a larger (less accurate) upper limit for the nuclear radius [1].
Q5 — Diagram critique (6 marks)
5.1 Error 1 (wrong particle type): The student describes “beta particles”. Beta particles are electrons (negatively charged) and would not be repelled by the nucleus [1]. Correction: Rutherford used alpha particles (helium nuclei, +2e), which are positively charged and are electrostatically repelled by the positive nucleus, causing the large-angle deflections observed [1].
5.2 Error 2 (outcomes reversed): The student says “nearly all bounced back” and “a few passed straight through.” This is the wrong way around [1]. Correction: In reality, most alpha particles passed straight through with little or no deflection; only a very small fraction (~1 in 8,000) was deflected by more than 90° [1].
5.3 Error 3 (wrong charge of nucleus): The student describes the nucleus as “negatively charged.” This is incorrect [1]. Correction: The nucleus is positively charged (it contains protons). The large-angle scattering results from the electrostatic repulsion between the positive alpha particle and the positive nucleus. A negatively charged nucleus would attract the alpha particles, not repel them [1].