Define the'distance of closest approach'. An $\displaystyle \alpha$-particle of kinetic energy K is bombarded on a thin gold foil. Derive an expression for the'distance of closest approach'.

The distance between the \(\displaystyle \alpha\) particle and the target nucleus when whole kinetic energy of an \(\displaystyle \alpha\) particle gets converted into potential energy. Alternatively: It is the distance from the nucleus at which alpha particle stops momentarily and then begins to retrace its path. At distance of closest approach \[\begin{aligned} \mathrm{K} & =\mathrm{U} \\ \mathrm{~K} & =\frac{(2 e)(Z e)}{4 \pi \varepsilon_{0} d} \\ \therefore d & =\frac{2 Z e^{2}}{4 \pi \varepsilon_{0} \mathrm{~K}} \end{aligned} \]