= 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> is
$$
K_* = \frac{m_{\rm Be}}{m_\alpha+m_{\rm Be}}K=\frac9{13}(40\,\mathrm{keV})=27.6923\,\mathrm{keV}.
$$
At closest separation the relative radial <velocity> vanishes, although the <centre of mass> continues moving. <Conservation of energy> for the relative motion gives
$$
\frac{(2e)(4e)}{4\pi\varepsilon_0r_{\min}}=K_*,\qquad
\boxed{r_{\min}=\frac{8e^2}{4\pi\varepsilon_0K_*}\simeq4.16\times10^{-13}\,\mathrm m.}
$$
Using $e^2/(4\pi\varepsilon_0)=1.43996\,\mathrm{MeV\,fm}$ gives $r_{\min}\simeq416\,\mathrm{fm}$. 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.
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