Physics • Year 12 • Module 8 • Lesson 11
The Nuclear Model of the Atom
Build HSC Band 5–6 extended-response technique by analysing scattering data, evaluating atomic models, and designing investigations using the closest approach calculation.
1. Data + scenario: Rutherford’s reanalysis of the Geiger-Marsden data (Band 5–6)
8 marks Band 5–6
Scenario. In the original Geiger-Marsden experiment (1909), alpha particles of kinetic energy approximately 7.7 MeV were fired at a gold foil (~200 atoms thick). Geiger and Marsden observed that approximately 1 in 8,000 alpha particles was deflected by more than 90°. Before Rutherford’s 1911 analysis, Thomson’s plum pudding model predicted that the positive charge was distributed evenly throughout the atom (~10−10 m radius). The table below gives a summary of observations and model predictions.
| Observation | Thomson’s prediction | Rutherford’s explanation |
|---|---|---|
| ~99.99% of alphas pass straight through with <1° deflection | Consistent (diffuse positive charge causes minimal deflection) | |
| ~1 in 8,000 alphas deflected by >90° | Impossible (diffuse charge too weak for large deflection) | |
| Some alphas deflected by small angles (1°–10°) | Consistent (weak deflection from diffuse charge) | |
| Calculated rmin for 7.7 MeV alpha on gold: ~29 fm | Not applicable (Thomson model has no nucleus) |
(k = 8.99×109 N m2 C−2; e = 1.6×10−19 C; 1 MeV = 1.6×10−13 J; ZAu = 79)
Q1. Analyse and evaluate the experimental data above to assess the validity of Thomson’s and Rutherford’s models of atomic structure. In your response you must:
- Complete Rutherford’s explanation column for rows 2 and 4 in the table, incorporating a calculation to verify the rmin value of ~29 fm.
- Explain why the observation in row 2 is sufficient on its own to refute Thomson’s model.
- Assess the extent to which the observation in row 1 supports Thomson’s model, and explain why Rutherford’s model also accounts for it.
- State and explain one limitation of Rutherford’s nuclear model that Thomson’s model does not face.
- Reach a justified conclusion about which model is better supported by the totality of the evidence.
2. Experimental design — testing nuclear radius scaling (Band 5–6)
7 marks Band 5–6
Research question. A Year 12 student argues that if the nuclear radius formula R = R0 A1/3 is correct, then the distance of closest approach for alpha particles of the same energy should increase with Z in a predictable way. The student has access to alpha sources of known energy (5.0 MeV and 7.7 MeV), thin foils of aluminium, silver, gold, and lead, and detection equipment that can measure the angle and approximate energy of deflected alphas.
Claim to test: “The distance of closest approach is directly proportional to Z, consistent with the formula rmin = k 2Ze2/Ek.”
Q2. Design the investigation and present it in the format below.
- State a testable hypothesis in the form “If… then… because…”, naming the independent and dependent variables.
- Identify at least two controlled variables and explain why each must be controlled.
- Describe the procedure in at least four numbered steps including how you would calculate rmin from experimental data.
- Explain what result would falsify your hypothesis.
- State two limitations of your design and one specific improvement.
Q1 — Sample Band 6 response (8 marks), annotated
Table completion (rows 2 and 4):
Row 2 — Rutherford’s explanation: The existence of a tiny, dense, positively charged nucleus means that a very small number of alpha particles pass close enough to experience an intense Coulomb repulsion that deflects them by >90°. Because the nucleus occupies only ~10−15 m of the ~10−10 m atom, close encounters are extremely rare (~1 in 8,000), consistent with observation [1 — clear mechanism linked to nuclear model].
Row 4 — Rutherford’s explanation with calculation [1]: At closest approach all kinetic energy is converted to electrical potential energy:
rmin = k 2Ze2 / Ek = (8.99×109 × 2 × 79 × (1.6×10−19)2) / (7.7 × 1.6×10−13)
= (8.99×109 × 158 × 2.56×10−38) / (1.232×10−12)
= (3.637×10−26) / (1.232×10−12)
≈ 2.95×10−14 m ≈ 29.5 fm. This is an upper limit on the nuclear radius, since Coulomb repulsion stops the alpha before it reaches the actual nuclear surface.
Why row 2 alone refutes Thomson’s model [1]: Thomson’s model explicitly predicts that large-angle scattering is impossible, because the diffuse positive charge (~10−10 m spread) can only exert a small, distributed force on passing alphas, which could only cause small-angle deflections. The observation that ~1 in 8,000 alphas is deflected by >90° is a direct, quantitative contradiction of this prediction. A single well-verified contradicting observation is sufficient to refute a model in science.
Row 1 and model support [1]: Row 1 (most alphas straight through) is consistent with Thomson’s model. However, consistency is not the same as support — Rutherford’s nuclear model also predicts that most alphas pass straight through (because the nucleus is tiny and the atom is mostly empty). Thomson’s model fails to distinguish itself from Rutherford’s on this observation. A model that cannot account for all the data (specifically row 2) is refuted regardless of partial agreement with other data.
Limitation of Rutherford’s model [1]: Classical electromagnetism requires that accelerating charges emit electromagnetic radiation. An electron in circular orbit around the nucleus is continuously accelerating (centripetal acceleration), so it should constantly radiate energy, lose speed, and spiral into the nucleus within ~10−12 s. Thomson’s model does not have electrons in circular orbits, so it does not face this problem. Rutherford’s nuclear model was subsequently improved by Bohr’s quantum model to resolve this issue.
Justified conclusion [1]: Rutherford’s nuclear model is better supported by the totality of the evidence. It explains all four rows of observations quantitatively, including the critical large-angle scattering that Thomson’s model explicitly predicts cannot occur. Thomson’s model is consistent with two observations but fails definitively on the key row 2 result. Although Rutherford’s model has a significant limitation (orbital stability), this was later resolved by quantum mechanics. Rutherford’s model represents a greater advance in explanatory power.
Marking criteria summary (8 marks): 1 = row 2 correctly explained using concentrated nucleus and Coulomb repulsion; 1 = row 4 calculation correct and ≈29.5 fm with correct formula use; 1 = clear explanation of why row 2 alone refutes Thomson (specific prediction vs observation); 1 = accurate assessment of row 1 (consistent with both; not diagnostic); 1 = Rutherford’s limitation stated correctly (radiating electrons / orbital instability); 1 = limitation linked to Thomson (Thomson does not face this); 1 = explicit evaluative conclusion comparing models using at least two pieces of evidence; 1 = precise physics terminology throughout (Coulomb repulsion, electrostatic potential energy, upper limit, nuclear model).
Q2 — Sample Band 6 response (7 marks), annotated
Hypothesis [1]: If the distance of closest approach is directly proportional to atomic number Z (as predicted by rmin = k 2Ze2/Ek), then plotting calculated rmin values against Z for four target elements at a fixed alpha energy will produce a straight line through the origin, because the formula shows rmin ∝ Z when Ek is constant. Independent variable: target element (Z = 13, 47, 79, 82 for Al, Ag, Au, Pb). Dependent variable: calculated rmin.
Controlled variables [1 for two with reasons]: (1) Alpha particle kinetic energy: must be kept constant (use the same 5.0 MeV source for all foils) because rmin ∝ 1/Ek; different energies would change the calculated rmin independent of Z, confounding the results. (2) Foil thickness: same thickness for all target foils to ensure alphas undergo the same number of potential interactions and to minimise multiple-scattering effects that could distort results.
Procedure [1 for four clear steps with calculation method]: (1) Set up a collimated 5.0 MeV alpha source directed at a thin aluminium foil in an evacuated chamber; position a movable scintillation detector to measure the intensity of scattered alphas at angles from 0° to 180°. (2) Record the energy of the incoming alpha beam and confirm it using the known source specification. (3) Calculate rmin for aluminium using the formula rmin = k 2Ze2/Ek, substituting Z = 13 and Ek = 5.0 MeV = 8.0×10−13 J. (4) Repeat steps 1–3 for silver (Z = 47), gold (Z = 79), and lead (Z = 82), keeping all experimental conditions identical. Plot calculated rmin vs Z and test for linearity.
Falsification [1]: The hypothesis would be falsified if the graph of rmin vs Z is not linear (e.g. shows a curve, a different power law, or no clear trend), indicating that rmin does not scale proportionally with Z, contradicting the Coulomb formula.
Limitations [1 each, 2 total]: (1) The closest approach is calculated theoretically from the formula using known Z and Ek, not directly measured; any uncertainty in alpha energy or slight energy loss in the foil would introduce error. (2) The experiment only measures an upper limit on nuclear radius; to actually probe the nuclear surface, much higher energy alphas would be needed (requiring a particle accelerator), which are not available with a simple radioactive source.
Improvement [1]: Repeat each measurement at two different energies (5.0 MeV and 7.7 MeV) for each target element. This provides an internal consistency check: if rmin × Ek is constant for the same element at both energies, it confirms the inverse relationship and validates the formula.
Marking criteria summary (7 marks): 1 = hypothesis with IV and DV and “because” justification from formula; 1 = two controlled variables each with a reason; 1 = four steps including calculation method for rmin; 1 = states what would falsify the hypothesis (non-linear graph); 1 = one valid limitation; 1 = second valid limitation; 1 = specific improvement.