c
For minimum potential difference electron has to make transition from \(n=3\) to \(n=2\) state but first electron has to reach to \(\mathrm{n}=3\) state from ground state. So, energy of bombarding electron should be equal to energy difference of \(n=3\) and \(n=1\) state.
\( \Delta \mathrm{E}=13.6\left[1-\frac{1}{3^2}\right] \mathrm{e}=\mathrm{eV} \)
\( \frac{13.6 \times 8}{9}=\mathrm{V} \)
\( \mathrm{V}=12.09 \mathrm{~V} \approx 12.1 \mathrm{~V}\)
So, \(\alpha=121\)