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?

(D) The key idea is that throughout the scattering process,the total mechanical energy of the system consisting of an $\alpha$-particle and a gold nucleus is conserved.
At the distance of closest approach $d$,the $\alpha$-particle momentarily comes to rest,meaning its kinetic energy $K$ is entirely converted into electric potential energy $U$.
Using the conservation of energy,$K = U = \frac{1}{4 \pi \varepsilon_{0}} \frac{(2e)(Ze)}{d} = \frac{2Ze^{2}}{4 \pi \varepsilon_{0} d}$.
Rearranging for $d$,we get $d = \frac{2Ze^{2}}{4 \pi \varepsilon_{0} K}$.
Given $K = 7.7 \; MeV = 7.7 \times 10^{6} \times 1.6 \times 10^{-19} \; J \approx 1.232 \times 10^{-12} \; J$,$Z = 79$ for gold,and $\frac{1}{4 \pi \varepsilon_{0}} = 9.0 \times 10^{9} \; Nm^{2}/C^{2}$.
Substituting these values: $d = \frac{2 \times 9.0 \times 10^{9} \times 79 \times (1.6 \times 10^{-19})^{2}}{1.232 \times 10^{-12}} \approx 2.95 \times 10^{-14} \; m$,which is approximately $30 \; fm$.