Rutherford nuclear model

The experiment. In the Geiger–Marsden setup, a narrow beam of alpha particles ($^4\text{He}^{2+}$, charge $+2e$) from a radioactive source was aimed at a thin gold foil (about $10^{-7}$ m thick), and a rotatable zinc-sulphide screen counted scintillation flashes at every scattering angle around the foil. Three observations had to be explained together: most alpha particles passed through with little or no deflection; a small fraction deflected through large angles, well beyond $90°$; and a very small fraction (roughly 1 in 8000) bounced back at angles close to $180°$.

Why Thomson's model failed. In Thomson's earlier "plum-pudding" picture, positive charge was spread smoothly through the whole atom, like a diffuse cloud with electrons studded inside it. A charge spread that thinly can only push an incoming alpha particle a little — the repulsion at any point is weak because the charge nearby is small. Such a model can account for the small deflections but has no mechanism at all for throwing a fast, heavy alpha particle almost straight back. The back-scattering data alone ruled it out.

Rutherford's model. To produce occasional sharp, large-angle deflections, Rutherford proposed that an atom's positive charge and nearly all its mass are concentrated in an extremely small region at the centre — the nucleus — with electrons orbiting it at comparatively enormous distances, held by electrostatic attraction so the atom is neutral overall. Most of the atom's volume is then empty space, which is why the great majority of alpha particles pass through undisturbed; only the rare particle that happens to aim almost directly at a nucleus feels a strong enough repulsive Coulomb force to swing through a large angle or rebound.

Estimating nuclear size. For a head-on collision, the alpha particle slows down as it approaches the nucleus and momentarily stops at the point of closest approach $r_0$, where all of its initial kinetic energy has converted into electrostatic potential energy:

where: $\tfrac12mv^2$ is the alpha particle's initial kinetic energy far from the nucleus, $2e$ is the alpha particle's charge, $Ze$ is the nuclear charge for atomic number $Z$, $\varepsilon_0$ is the permittivity of free space (folded into Coulomb's constant $k = 1/4\pi\varepsilon_0 = 9\times10^9\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$), and $r_0$ is the distance of closest approach. Solving for $r_0$ gives an upper bound on the nuclear radius, since a real alpha particle only reaches the nuclear surface if it happens to have enough energy and is aimed exactly head-on. Experiments of this kind consistently give

— about ten thousand times smaller than the size of the atom as a whole ($\sim10^{-10}\ \text{m}$), confirming that the nucleus is an extremely small, dense core carrying essentially all the atom's mass. Within this model, the atomic number $Z$ is the number of protons (equal to the number of electrons in a neutral atom), the neutron number is $N$, and the mass number $A = Z + N$ gives the total number of nucleons packed into that tiny core.

An alpha particle with kinetic energy $K = 5.0\ \text{MeV}$ is fired head-on at a stationary gold nucleus ($Z = 79$). Find the distance of closest approach.

Step 1. Convert the energy to joules: $K = 5.0\ \text{MeV} = 5.0\times10^6\times1.6\times10^{-19}\ \text{J} = 8.0\times10^{-13}\ \text{J}$.

Step 2. Write the closest-approach relation from the derivation above: $r_0 = \dfrac{k(2e)(Ze)}{K}$, with $k = 9\times10^9\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$, $e = 1.6\times10^{-19}\ \text{C}$ and $Z = 79$.

Step 3. Compute the numerator: $k(2e)(79e) = 9\times10^9\times2\times79\times(1.6\times10^{-19})^2 = 9\times10^9\times158\times2.56\times10^{-38} = 3.64\times10^{-26}\ \text{J}\,\cdot\,\text{m}$.

Step 4. Divide by $K$: $r_0 = \dfrac{3.64\times10^{-26}}{8.0\times10^{-13}} = 4.55\times10^{-14}\ \text{m}$, about $45.5\ \text{fm}$.

This is larger than the true nuclear radius of gold, because a 5 MeV alpha particle simply does not carry enough energy to force its way in to the actual nuclear surface — it stops well outside and gets deflected back before ever "feeling" the nucleus directly. Higher-energy alpha particles push $r_0$ down toward the true nuclear scale of a few femtometres.

Competitive stretch

For the same gold nucleus, what closest-approach distance would a $50\ \text{MeV}$ alpha particle reach? Since $r_0 \propto 1/K$, scaling the previous answer by the energy ratio gives $r_0 = 4.55\times10^{-14}\times\dfrac{5.0}{50} = 4.55\times10^{-15}\ \text{m} \approx 4.6\ \text{fm}$ — now within a factor of a few of gold's actual nuclear radius, which is why physicists used progressively higher-energy beams to pin down real nuclear sizes.

• Treating the distance of closest approach as the nuclear radius rather than an upper bound on it — the formula assumes a purely electrostatic, head-on interaction and stops applying once the particle would otherwise penetrate the nucleus.
• Using charge $e$ instead of $2e$ for the alpha particle, or $e$ instead of $Ze$ for the target nucleus — a very common arithmetic slip that changes the answer by a large factor.
• Assuming the Thomson model was ruled out simply because particles were deflected at all — small deflections are expected from any charge distribution; it is specifically the rare large-angle and back-scattering events that only a concentrated central charge can produce.
• Attempting to invoke the full Rutherford scattering formula with impact parameter and differential cross-section — this is outside the ISC syllabus, which requires only the qualitative model and the energy-conservation estimate of $r_0$.

Ask the tutor: want to work through how the distance of closest approach changes for a different target nucleus or a different beam energy, step by step?