Published by Patrick Mutisya · 14 days ago
To infer from the results of the α‑particle scattering experiment the existence and extremely small size of the atomic nucleus.
Before 1911 the prevailing model of the atom was the Thomson “plum‑pudding” model, in which positive charge was spread uniformly throughout the atom with electrons embedded like raisins in a pudding.
If positive charge were spread uniformly, the electric field inside the atom would be weak and the α‑particles would experience only small, gradual deflections. The observation of large‑angle scattering is incompatible with this picture.
Rutherford proposed that:
The differential cross‑section for scattering of a particle of charge \$Z1e\$ by a nucleus of charge \$Z2e\$ is given by the Rutherford formula:
\$\frac{d\sigma}{d\Omega} = \left(\frac{Z1 Z2 e^2}{16\pi\varepsilon_0 E}\right)^2 \frac{1}{\sin^4(\theta/2)}\$
where \$E\$ is the kinetic energy of the α‑particle and \$\theta\$ is the scattering angle.
The impact parameter \$b\$ is the perpendicular distance between the initial trajectory of the α‑particle and the centre of the nucleus. Large‑angle scattering requires \$b\$ to be comparable to the nuclear radius \$R\$.
For a back‑scattered α‑particle (\$\theta \approx 180^\circ\$) the closest approach \$r_{\min}\$ satisfies the energy‑conservation condition:
\$\frac{1}{4\pi\varepsilon0}\frac{Z1 Z2 e^2}{r{\min}} = E\$
Solving for \$r_{\min}\$ gives an upper bound on the nuclear radius:
\$R \lesssim r{\min} = \frac{Z1 Z2 e^2}{4\pi\varepsilon0 E}\$
For typical α‑particles (\$E \approx 5\,\$MeV, \$Z1=2\$, \$Z2=79\$ for gold):
\$R \lesssim \frac{(2)(79)(1.60\times10^{-19}\,\text{C})^2}{4\pi(8.85\times10^{-12}\,\text{F m}^{-1})(5\times10^{6}\,\text{eV})(1.60\times10^{-19}\,\text{J eV}^{-1})} \approx 1.5\times10^{-14}\,\text{m}\$
This is roughly two orders of magnitude smaller than the atomic radius (\$\sim10^{-10}\,\$m), confirming that the nucleus is extremely compact.
Experimental data across many nuclei lead to the empirical relationship:
\$R = r_0 A^{1/3}\$
where \$A\$ is the mass number and \$r_0 \approx 1.2\,\$fm (\$1\,\$fm = \$10^{-15}\,\$m). This agrees with the order‑of‑magnitude estimate from Rutherford scattering.
| Quantity | Symbol | Typical \cdot alue (Gold foil) | Units |
|---|---|---|---|
| α‑particle kinetic energy | \$E\$ | 5 MeV | eV |
| Charge of α‑particle | \$Z_1\$ | 2 | – |
| Charge of gold nucleus | \$Z_2\$ | 79 | – |
| Estimated nuclear radius | \$R\$ | \$\sim1.5\times10^{-14}\$ | m |
| Atomic radius (order of magnitude) | \$a\$ | \$\sim1\times10^{-10}\$ | m |