Chemistry • Year 11 • Module 1 • Lesson 13
Atomic Models – Historical Development
Apply your understanding of subatomic particle calculations, Rutherford’s gold foil experiment, and model development to real data, scenario reasoning, and comparative tasks.
1. Interpret data — subatomic particle table
The table below lists six nuclides. Complete all empty cells. Show working in the space below the table where required. 12 marks (1 per cell, 2 cells per row)
| Nuclide | Z (protons) | A (mass number) | Neutrons | Electrons (neutral) | Electrons (as ion) |
|---|---|---|---|---|---|
| 126C | 6 | 12 | n/a | ||
| 3517Cl− | 18 | 17 | |||
| 2311Na+ | |||||
| 5626Fe2+ | |||||
| 168O2− | |||||
| 19779Au | n/a |
1.1 For the Fe2+ ion, explain in one sentence why the electron count differs from the neutral iron atom. 1 mark
1.2 Which two nuclides in the table are not ions? State how you know. 1 mark
2. Interpret graph — alpha particle scattering in Rutherford’s experiment
The bar graph below shows the approximate percentage of alpha particles deflected at different angle ranges in Rutherford’s gold foil experiment. Use it to answer the questions. 7 marks
Figure 2. Approximate distribution of alpha particle deflection angles in Rutherford’s gold foil experiment. Illustrative data adapted from Geiger & Marsden (1913), Philosophical Magazine 25: 604–623.
2.1 Describe the overall pattern shown in the graph. What proportion of alpha particles were deflected at large angles? 2 marks
2.2 Explain why the 98.5% of particles deflected at 0–5° is consistent with Rutherford’s nuclear model but inconsistent with Thomson’s plum pudding model. 2 marks
2.3 Rutherford said the back-scatter result was “almost as incredible as if you fired 15-inch shells at tissue paper and they came back and hit you.” Using data from the graph, explain what structural feature of the atom Rutherford concluded from the ~0.05% back-scatter result. 3 marks
3. Compare four atomic models across five criteria
Complete the table below. For each model and criterion, write a concise description. 10 marks (1 per shaded cell)
| Criterion | Thomson (1904) | Rutherford (1911) | Bohr (1913) |
|---|---|---|---|
| Location of positive charge | |||
| Electron arrangement | |||
| Key experimental evidence | |||
| Critical limitation | |||
| What superseded it |
4. Predict and justify — a spectrum scenario
A scientist heats a sample of pure sodium vapour and observes that it emits only two very close, bright yellow lines at wavelengths of 589 nm and 589.6 nm, rather than a continuous rainbow spectrum.
5 marks
4.1 Using Bohr’s model, explain why sodium emits only these specific wavelengths rather than a continuous spectrum. 3 marks
4.2 Predict what would happen to the emission spectrum if the atoms were cooled back to ground state. Would the lines still be observed? Justify your prediction. 2 marks
Q1 — Subatomic particle table (completed)
12C: Z=6, A=12, neutrons=6, electrons(neutral)=6, ion n/a.
35Cl−: Z=17, A=35, neutrons=18, electrons(neutral)=17, electrons(ion)=18 (gains 1).
23Na+: Z=11, A=23, neutrons=12, electrons(neutral)=11, electrons(ion)=10 (loses 1).
56Fe2+: Z=26, A=56, neutrons=30, electrons(neutral)=26, electrons(ion)=24 (loses 2).
16O2−: Z=8, A=16, neutrons=8, electrons(neutral)=8, electrons(ion)=10 (gains 2).
197Au: Z=79, A=197, neutrons=118, electrons(neutral)=79, ion n/a.
Q1.1 — Fe2+ electron count
Fe2+ is a cation that has lost 2 electrons to form a 2+ charge, so it has 26 − 2 = 24 electrons instead of the 26 in a neutral iron atom.
Q1.2 — Non-ions
12C and 197Au are the neutral atoms (no ionic charge symbol); their electron count equals their atomic number (6 and 79 respectively).
Q2.1 — Pattern in graph (2 marks)
The vast majority (98.5%) of alpha particles were deflected at only 0–5° (i.e. passed through with negligible deflection) [1]. Only a very small fraction — approximately 0.2% combined — were deflected at angles greater than 45°, and fewer than 0.05% were deflected at greater than 90° (back-scatter) [1].
Q2.2 — Consistency with nuclear model, inconsistency with plum pudding (2 marks)
Rutherford’s nuclear model predicts that the atom is mostly empty space, so most alpha particles will encounter nothing and pass straight through with minimal deflection — consistent with the 98.5% result [1]. Thomson’s plum pudding model predicted diffuse positive charge spread throughout the atom, which would cause small but roughly uniform deflections for all alpha particles; it cannot account for the overwhelming majority passing through essentially undeflected, as the charge would not be concentrated enough to let particles pass freely [1].
Q2.3 — Back-scatter conclusion (3 marks)
Only ~0.05% of alpha particles (roughly 1 in 20,000) were reflected back at angles >90° [1]. This tiny fraction indicated that almost all of the atom is open space; only on the very rare occasion that an alpha particle travels on a near-direct collision path with the concentrated region does it get deflected strongly [1]. Rutherford concluded that nearly all the positive charge and mass of the atom is concentrated in a tiny, extremely dense central nucleus — because only a highly localised and very intense electrostatic repulsion could reverse the trajectory of a fast, massive alpha particle [1].
Q3 — Compare four atomic models
Location of positive charge — Thomson: uniform sphere throughout atom. Rutherford: tiny central nucleus. Bohr: tiny central nucleus (same as Rutherford).
Electron arrangement — Thomson: embedded randomly in positive sphere. Rutherford: orbit nucleus at large distances in empty space. Bohr: fixed circular orbits (shells) at discrete, quantised energy levels.
Key experimental evidence — Thomson: cathode ray tube (electrons deflected by E & B fields). Rutherford: gold foil experiment (large-angle alpha particle deflections). Bohr: hydrogen emission spectrum (discrete spectral lines match calculated energy level differences).
Critical limitation — Thomson: cannot explain large-angle deflections in Rutherford’s experiment. Rutherford: cannot explain why electrons don’t spiral inward (classical radiation) or why emission spectra are discrete. Bohr: only accurate for hydrogen; fails for multi-electron atoms; no explanation of fine spectral structure.
What superseded it — Thomson: Rutherford’s nuclear model (1911). Rutherford: Bohr’s quantised shell model (1913). Bohr: quantum mechanical model / Schrödinger’s orbital model (1926).
Q4.1 — Bohr’s model and sodium emission (3 marks)
According to Bohr’s model, electrons in sodium occupy fixed, discrete energy levels [1]. When the sodium vapour is heated, electrons absorb energy and jump to higher energy levels (excited state). When they fall back to lower energy levels, they emit photons whose energy equals the energy difference between the two levels (E = hf) [1]. Because only specific energy jumps are allowed (only fixed levels exist), only specific photon frequencies (specific wavelengths) are emitted — in this case the two yellow lines at 589 nm and 589.6 nm correspond to two slightly different electron transitions. No other wavelengths are emitted because no other energy transitions occur, giving a discrete line spectrum rather than a continuous one [1].
Q4.2 — Prediction when cooled (2 marks)
When cooled, electrons return to ground state (lowest available energy level) and stop emitting photons [1]. No further emission would be observed: the yellow spectral lines would disappear because no electron transitions are occurring. The same wavelengths would reappear if the atoms were re-excited (heated again), but at room temperature the atoms emit no visible light [1].