In Rutherford's scattering experiments, alpha particles $($charge$=+2$e$)$ were fired at a gold foil. Consider an alpha particle with an initial kinetic energy KK heading directly for the nucleus of a gold atom $($charge$==+79$e$).$ The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy.

Find the distance of closest approach between the alpha particle and the gold nucleus for the case KK $=6\mathrm{.}5$ MeV

Express your answer using two significant figures. UNITS fm$.$