The maximum potential energy due to electrostatic repulsion between two hydrogen nucleus is nearly (radius…

The separation between two hydrogen nucleus should be equal to the diameter of nucleus,

$\Rightarrow r=2r_0=2\times 1.1\times {10}^{-15} \mathrm{m}$

so potential energy of the system,

$U=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r}=9\times {10}^9\times \frac{{\left(1.6\times {10}^{-19}\right)}^2}{2\times 1.1\times {10}^{-15}}=0.65 \mathrm{MeV}$