Example

Question:

In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV \(\alpha\)-particle before it comes momentarily to rest and reverses its direction?

Solution:

Throughout the scattering, the total mechanical energy is conserved. Initial energy \(E_i\) (kinetic) = final energy \(E_f\) (potential) at closest approach. Let \( d \) be the center-to-center distance between \(\alpha\)-particle and gold nucleus at this point.
The conservation gives: \[ K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{d} \] \[ K = \frac{2Ze^2}{4\pi\epsilon_0 d} \implies d = \frac{2Ze^2}{4\pi\epsilon_0 K} \] With \( K = 7.7~\text{MeV} = 1.2 \times 10^{-12}~\text{J} \), \( 1/4\pi\epsilon_0 = 9.0 \times 10^9~\text{N m}^2/\text{C}^2 \), \( e = 1.6 \times 10^{-19}~\text{C} \), \( Z = 79 \) for gold: \[ d = \frac{2(9.0 \times 10^9)(1.6 \times 10^{-19})^2 Z}{1.2 \times 10^{-12}} \] \[ = 3.84 \times 10^{-16}~\text{Z m} \] For gold (\( Z = 79 \)): \[ d(\text{Au}) = 3.0 \times 10^{-14}~\text{m} = 30~\text{fm} \] The actual radius of gold nucleus is about 6 fm (\( 6.0 \times 10^{-15}~\text{m} \)), which is much less than the distance of closest approach. Thus, the \(\alpha\)-particle reverses motion without touching the nucleus.