Thomson Rutherford Models

12.1 Introduction — What is Inside an Atom?

Greek philosophers (Democritus, ~400 BCE) speculated that all matter is made of indivisible particles called atomos. Two thousand years later, Dalton (1808) made the same idea quantitative. But it was the discovery of the electron by J. J. Thomson in 1897 — using cathode rays in a discharge tube — that proved atoms have internal structure. Since the atom as a whole is electrically neutral, an equal amount of positive charge must also live somewhere inside. The race to map that interior was on.

12.2 Thomson's Plum-Pudding Model (1898)

J. J. Thomson proposed the first concrete model of the atom in 1898. He visualised it as a uniform sphere of positive charge, about \(10^{-10}\) m in radius, with the electrons embedded inside it like plums in a pudding (or seeds in a watermelon). The arrangement of electrons was to be determined by their mutual repulsion balanced by attraction toward the positive jelly.

Thomson's model could explain the neutrality of atoms and the existence of electrons. It even gave reasonable estimates for the atom's size. But two stubborn problems remained:

• It predicted the wrong line spectra (only a few frequencies of vibration, not the rich Rydberg-pattern lines actually observed).
• It made a definite, testable prediction about how alpha particles should pass through metal foils — a prediction that turned out to be spectacularly wrong.

12.3 Geiger–Marsden α-Particle Scattering (1909)

Ernest Rutherford (Nobel Laureate 1908 for his work on radioactivity) directed his students Hans Geiger and Ernest Marsden to test Thomson's model directly. Their experiment used:

• A polonium-214 source (radioactive) that emits alpha particles (He²⁺ nuclei) of about 5.5 MeV kinetic energy.
• A series of pinholes that produced a narrow, parallel beam.
• A very thin gold foil (~100 nm thick — only ~400 atoms across), used because gold can be hammered into extremely thin sheets.
• A movable detector consisting of a zinc sulphide screen on which each α-particle hit produced a tiny scintillation, observed through a microscope.

The Astonishing Result

The experimental observations were:

1. The vast majority of α-particles passed through the foil almost undeflected.
2. About one in 8000 was scattered through angles larger than 90°.
3. A handful were scattered straight back toward the source.

12.4 Rutherford's Nuclear Model (1911)

To explain the back-scattering, Rutherford concluded that almost all the positive charge and almost all the mass of the atom must be concentrated in an unimaginably tiny central region — the nucleus. The electrons orbit around this nucleus at relatively huge distances, leaving the atom almost entirely empty space.

From the fraction of large-angle scatterings he could estimate the nuclear size:

• Atomic radius ≈ \(10^{-10}\) m
• Nuclear radius ≈ \(10^{-15}\) m (one hundred-thousandth of atomic radius)
• Volume ratio nucleus/atom ≈ \(10^{-15}\)

If a hydrogen atom were the size of a football stadium, the nucleus would be a grain of sand at the centre — and the rest of the stadium would be empty.

12.4.1 The Trajectory Equation — Coulomb Repulsion

Each α-particle (charge +2e) and each gold nucleus (charge +Ze, with Z=79 for gold) repel each other through the Coulomb force. Conservation of energy and angular momentum gives a hyperbolic trajectory characterised by the impact parameter b — the perpendicular distance from the centre that the α would pass at if there were no force. Geometric analysis (which we will not derive here) gives:

\[b = \frac{1}{4\pi\varepsilon_0}\,\frac{Ze^2\cot(\theta/2)}{(1/2)mv^2}\]

Small b → close approach → large scattering angle θ. Most α-particles have large b (they miss the nucleus by lots) and are deflected hardly at all. Only the rare head-on encounters produce backscattering.

12.4.2 Distance of Closest Approach r₀

For the special case of a head-on collision (b = 0, θ = 180°), the α-particle slows, stops momentarily at distance r₀ from the nucleus, and then bounces straight back. By energy conservation:

\[\frac{1}{2}m v^2 = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{r_0}\]

Solving for r₀:

For 5.5 MeV α-particles on gold (Z=79): r₀ ≈ 4.1 × 10⁻¹⁴ m. So the gold nucleus must have a radius smaller than this — perhaps 10⁻¹⁵ m.

12.5 Electron Orbits — A Classical Crisis

If the electrons orbit the nucleus, the Coulomb force provides the centripetal force:

\[\frac{1}{4\pi\varepsilon_0}\,\frac{e^2}{r^2} = \frac{m_e v^2}{r}\]

Multiplying both sides by r/2 gives the kinetic energy of the orbiting electron:

\[K = \frac{1}{2}m_e v^2 = \frac{1}{8\pi\varepsilon_0}\,\frac{e^2}{r}\]

The potential energy of the electron-nucleus pair is:

\[U = -\frac{1}{4\pi\varepsilon_0}\,\frac{e^2}{r}\]

Total mechanical energy:

\[E = K + U = -\frac{1}{8\pi\varepsilon_0}\,\frac{e^2}{r}\]

Negative E confirms that the electron is bound to the nucleus.

Worked Examples