Rutherford's Alpha Scattering Experiment

In 1909, under Ernest Rutherford's direction at Manchester, Hans Geiger and Ernest Marsden directed a beam of energetic alpha particles ($\sim 5.5$ MeV from a radium source) at a thin gold foil and counted the scattered alphas at various angles using a zinc-sulphide scintillation screen. The experiment was a direct test of Thomson's plum-pudding model — but the results forced a revolutionary picture: almost all of the atom's mass and positive charge sits in a tiny nucleus.

Concept

Setup. A collimated alpha beam from a $^{226}$Ra source struck a $\sim 4\times 10^{-7}$ m thick gold foil. A movable detector (a fluorescent screen viewed through a microscope) recorded scattered alphas at any chosen angle $\theta$ measured from the forward direction.

Observations.

• Most alphas (more than 99 percent) passed nearly straight through, deflected by less than $1^\circ$.
• A small fraction were scattered through large angles, with about 1 in 8000 deflected by more than $90^\circ$.
• A few (rare) were almost back-scattered ($\theta \approx 180^\circ$).

In Rutherford's own words: "It was about as credible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

Conclusion.

• The atom is mostly empty space (most alphas undeflected).
• All the positive charge and almost all the mass is concentrated in a region of size $\sim 10^{-15}$ m — the nucleus.
• Electrons orbit the nucleus at distances $\sim 10^{-10}$ m.

Derivation

Distance of closest approach $r_0$. For an alpha aimed head-on at a nucleus (impact parameter $b=0$), it slows, stops momentarily at $r_0$, and bounces back. All initial kinetic energy converts to electric potential energy at $r_0$:

$\frac{1}{2}m_\alpha v^2 = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r_0}$

Solving,

$r_0 = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K}$

where $K = \tfrac12 m_\alpha v^2$ is the alpha's kinetic energy. This $r_0$ gives an upper bound on the nuclear radius.

Rutherford scattering formula (stated). The number of alphas scattered per unit area at angle $\theta$ is

$N(\theta) \propto \frac{1}{\sin^4(\theta/2)}$

This sharp $\sin^{-4}(\theta/2)$ dependence (verified by Geiger and Marsden) is the signature of a Coulomb scattering centre, i.e. a point-like nucleus.

Worked Example

A 5.5 MeV alpha is aimed head-on at a gold nucleus ($Z=79$). Find the distance of closest approach.

$K = 5.5\,\text{MeV} = 5.5\times 10^6 \times 1.6\times 10^{-19}\,\text{J} = 8.8\times 10^{-13}\,\text{J}$

$r_0 = \frac{(9\times 10^9)\, (2)(79)(1.6\times 10^{-19})^2}{8.8\times 10^{-13}}$

Numerator: $9\times 10^9 \cdot 2\cdot 79\cdot 2.56\times 10^{-38} = 3.64\times 10^{-26}$.

$r_0 = \frac{3.64\times 10^{-26}}{8.8\times 10^{-13}} \approx 4.1\times 10^{-14}\,\text{m}$

So $r_0 \approx 41$ fm — well outside the actual gold nuclear radius (about 7 fm). The alpha never touches the nucleus; Coulomb repulsion alone is enough to reverse it.

Common Confusions

• The distance of closest approach is not the nuclear radius. It is only an upper bound — at MeV energies, the alpha is reversed long before reaching the nucleus.
• Rutherford's scattering formula assumes a point charge. Deviations from $1/\sin^4(\theta/2)$ at very high energy revealed finite nuclear size in later experiments.
• Most alphas are undeflected because the nucleus is small ($10^{-15}$ m) compared with the atomic spacing ($10^{-10}$ m) — the atom is mostly vacuum.
• The fraction back-scattered tells you the nuclear size, not the atomic size.

Key Takeaways

• Geiger-Marsden 1909: gold foil scattering of alpha particles.
• Most alphas pass through; a small fraction scatter by more than 90 degrees; very few back-scatter.
• Hence the atom has a tiny, dense, positive nucleus ($\sim 10^{-15}$ m) surrounded by mostly empty space.
• Distance of closest approach $r_0 = \dfrac{2Ze^2}{4\pi\varepsilon_0 K}$ gives an upper bound on the nuclear radius.
• Rutherford scattering: $N(\theta) \propto 1/\sin^4(\theta/2)$ confirms a Coulombic, point-like scattering centre.