In the Geiger-Marsden scattering experiment, what is the distance of closest approach to the nucleus

Question

In the Geiger-Marsden scattering experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV $\alpha$-particle before it comes momentarily to rest? (Given Z for gold = 79)

Professor PC Thomas Classes · Answer

Using energy conservation, the initial kinetic energy equals the electrostatic potential energy at the distance of closest approach. $d = \frac{2Ze^2}{4\pi\epsilon_0 K} \approx 3.0 	imes 10^{-14}$ m.