 Solution

The alpha particle has electric charge $2e$ and acquires kinetic energy $K=2e(20\mathrm{kV})=40\mathrm{keV}$. For Coulomb head-on closest approach with target recoil, the initially stationary beryllium nucleus is free to recoil. Approximating the nuclear masses by $4u$ and $9u$, the initial relative-motion kinetic energy isAt closest separation the relative radial velocity vanishes, although the centre of mass continues moving. Conservation of energy for the relative motion givesUsing $e^2/(4\pi {\epsilon }_0)=1.43996\mathrm{MeV}\mathrm{fm}$ gives $r_{\min }\simeq 416\mathrm{fm}$. If the nucleus were externally held fixed, the relevant energy would be $K$, yielding $288\mathrm{fm}$ instead. Initially stationary is not the same as constrained stationary; the given proton and neutron counts supply the recoil correction. Nuclear sizes and strong interactions are negligible at this separation.