If an alpha particle (Z = 2) of kinetic energy 7 MeV is incident on a gold nucleus (Z = 79), what is its closest distance of approach?

In Rutherford Scattering, an incident particle will reach the closest distance of approach when it is on a collision course head on with the target nucleus. At the closest distance of approach, the alpha particle comes to rest, hence it no longer has any kinetic energy. Because both the alpha particle and gold nucleus are positively charge, the initial kinetic energy is transformed into electric potential energy, and due to energy conservation, these must always sum to the initial kinetic energy. We can then equate the initial kinetic energy KEα to the final potential energy PEα at the instant the alpha particle is at rest with the equation PEα = KQ_alphaQ_Gold/r_min = KEα where K = 1/4πε₀ , Q_alpha= Z_alphae, Q_Gold = Z_Golde , e = 1.6x10^-19C and r_min is the closest distance of approach. Substituting these and equating gives PEα = kZ_alphaZ_Golde²/r_min = KEα. To convert KEα from MeV to Joules we must divide by e and multiply by x10⁶ hence kZ_alphaZ_Golde²/r_min = 7x10^6e  and rearranging for r_min gives r_min = kZ_alphaZ_Golde/7x10^6 = 3.25x10^-14m