For ${\mathrm{v}}_{\mathrm{m}\mathrm{i}\mathrm{n}}$ collision should be completely inelastic.

According to energy conservation principle,

$\therefore \frac{1}{2}\mathrm{m}{\mathrm{v}}_{\mathrm{m}\mathrm{i}\mathrm{n}}^2=\frac{1}{2}\mathrm{m}{\mathrm{v}}^2+\frac{1}{2}\mathrm{m}{\mathrm{v}}^2+∆\mathrm{E}$ ...(i))

According to momentum conservation principle.

$\mathrm{m}{\mathrm{v}}_{\mathrm{m}\mathrm{i}\mathrm{n}}=\mathrm{m}\mathrm{v}+\mathrm{m}\mathrm{v}$

$\therefore \mathrm{v}=\frac{{\mathrm{m}}_{\mathrm{m}\mathrm{i}\mathrm{n}}}{2}$ ...(ii)

After solving eq. (i) an d(ii), we get

$\frac{1}{2}\mathrm{m}{\mathrm{v}}_{\mathrm{m}\mathrm{i}\mathrm{n}}^2=2∆\mathrm{E}$

$\therefore {\mathrm{v}}_{\mathrm{m}\mathrm{i}\mathrm{n}}=\sqrt{\left(\frac{4∆\mathrm{E}}{\mathrm{m}}\right)}$

Here, $∆\mathrm{E}=$ minimum excitation energy

$=10.2\mathrm{ }\mathrm{e}\mathrm{V}$

$\mathrm{m}=1.67\times {10}^{-27}\mathrm{ }\mathrm{k}\mathrm{g}$

$\therefore {\mathrm{v}}_{\mathrm{m}\mathrm{i}\mathrm{n}}=\sqrt{\left(\frac{4\times 10.2\times 1.6\times {10}^{-19}}{1.67\times {10}^{-27}}\right)}$

$=\text{6.25}\times {10}^4 \mathrm{m} {\mathrm{s}}^{-1}$