Rutherford scattering and nuclear radiusAQA A-Level Physics: Revision notes

In the Geiger–Marsden experiment, a narrow beam of alpha particles was fired at very thin gold foil in a vacuum, and the number scattered through different angles was counted. The foil is thin so that each alpha particle is scattered by a single nucleus, and a vacuum prevents absorption by air.

Observations and conclusions:

• Most alpha particles pass straight through, so most of an atom is empty space.
• A small fraction (about 1 in 8000) is deflected through more than 90°, so the positive charge and almost all the mass are concentrated in a tiny nucleus, which repels the positive alpha particles.
• The scattering matches the inverse-square electric repulsion between nucleus and alpha particle.

Ideas about the atom have developed as new evidence appeared:

• Plum pudding model (Thomson): a sphere of positive charge with electrons embedded in it. It predicts only small deflections of alpha particles.
• Nuclear model (Rutherford): a small, dense, positive nucleus, with electrons orbiting at a much larger distance.
• Neutron discovered by Chadwick, explaining the extra mass of nuclei.
• Quarks: high-energy electron scattering showed that protons and neutrons are not fundamental but have internal structure.

A theory is accepted until new evidence cannot be explained by it.

A head-on alpha particle slows as it approaches a nucleus, because of the electric repulsion. At the distance of closest approach it is momentarily at rest, so all its kinetic energy has become electric potential energy:

$E_k = \frac{Qq}{4\pi\varepsilon_0 r}$ so $r = \frac{Qq}{4\pi\varepsilon_0 E_k}$

where $Q$ and $q$ are the charges of the nucleus and the alpha particle. Because the alpha particle never touches the nucleus, $r$ is an upper limit to the nuclear radius.

Worked example: 7.7 MeV alpha particle on gold ($Z = 79$): $E_k = 7.7\times10^6 \times 1.60\times10^{-19} = 1.23\times10^{-12}$ J, so $r = \frac{(2e)(79e)}{4\pi\varepsilon_0 E_k} = 3.0\times10^{-14}$ m.

High-energy electrons (hundreds of MeV) have a de Broglie wavelength $\lambda = h/p$ about the size of a nucleus (around $10^{-15}$ m). For such electrons, $p \approx E/c$.

When the beam is scattered by a nucleus, a diffraction pattern appears: the intensity of scattered electrons plotted against angle falls to a first minimum and then rises to a smaller maximum. The first minimum is at

$\sin\theta \approx \frac{1.22\lambda}{D}$

where $D$ is the nuclear diameter, so the radius is $D/2$. Electron diffraction gives a more accurate radius than closest approach, because electrons do not feel the strong force.

Typical nuclear radii are between about 1 fm and 8 fm ($10^{-15}$ m), around 10⁵ times smaller than an atom.

Experiments show that the radius depends on nucleon number $A$:

$R = R_0A^{1/3}$ with $R_0 \approx 1.2$ fm

Since volume $\propto R^3 \propto A$ and mass $\propto A$, the density is the same for all nuclei: the equation is evidence that nuclear matter has constant density.

$\rho = \frac{Au}{\frac{4}{3}\pi R_0^3 A} = \frac{3u}{4\pi R_0^3} \approx 2.3\times10^{17}$ kg m⁻³

For ⁵⁶Fe: $R = R_0A^{1/3} = 1.2\times10^{-15}\times56^{1/3} = 1.2\times10^{-15}\times3.83 = 4.6\times10^{-15}$ m.

The nuclear density of $2.3\times10^{17}$ kg m⁻³ is about 3 × 10¹³ times the density of ordinary solids such as iron ($7.9\times10^3$ kg m⁻³), because an atom is almost entirely empty space around a tiny nucleus.