Hydrogen atom is bombarded with electrons accelerated through a...

For minimum potential difference, electron has to make transition from $n = 3$ to $n = 2$  state but first electron has to reach to $n = 3$ state from ground state. So, energy of bombarding electron should be equal to energy difference of $n = 3$ and $n = 1$ state.

$\DeltaE = 13 . 6\left[1 - \dfrac{1}{{3}^{2}}\right]e = eV$

$⇒ \dfrac{13 . 6 \times 8}{9} = V$

$⇒ V = 12 . 09V \approx 12 . 1V$

So, $\alpha = 121$.