Geometry of Rutherford head-on scattering: An alpha particle ($q = +2e$) is fired directly towards a gold nucleus ($Z = 79$). At the distance of closest approach $r_0$, all initial kinetic energy is converted into electrostatic potential energy.
An $\alpha$-particle with an initial kinetic energy of 5.5 MeV approaches a gold nucleus ($Z=79$) head-on. Calculate the distance of closest approach $r_0$. (Take $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2/\text{C}^2$, $e = 1.6 \times 10^{-19}\text{ C}$)