In a Geiger-Marsden experiment. Find 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. ($Z$ for gold nucleus $= 79$)

Let $d$ be the distance of closest approach then by the conservation of energy,

Initial kinetic energy of incoming $\alpha$-particle $K$ = Final electric potential energy $U$ of the system

 $\Rightarrow K=\frac{1}{4\pi {\epsilon }_0}\times \frac{\left(2e\right)\left(Ze\right)}{d} \therefore d=\frac{1}{4\pi {\epsilon }_0}\frac{2Z{e}^2}{K} ....\left(i\right)$

Here, $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9 N{m}^2 C^{-2}, Z=79, e=1.6\times {10}^{-19} C.$

$K=7.7 MeV=7.7\times {10}^6\times 1.6\times {10}^{-19} J= 1.2\times {10}^{-12} J$

Substituting these values in (i)

$$\begin{gathered} d=\frac{2\times 9\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2\times 79}{1.2\times {10}^{-12}} \\ d= 3\times {10}^{-14} m= 30 fm \left(\because 1 fm={10}^{-15 }m\right) \end{gathered}$$