Ans : (a)
$E_{max} = eV_{\circ}$
So, $ \lambda_e = \large\frac{h}{\sqrt{2mE_{max}}} = \large\frac{h}{\sqrt{2meV_{\circ}}}$
$ \Rightarrow V_{\circ} =\large\frac{ h^2} { 2me \lambda_e^2}$
$ \Rightarrow V_{\circ} =\large\frac{ (6.6 \times 10^{-34} )^2}{ [2 \times (9.1 \times 10^{-31} ) \times 1.6 \times 10^{-19} \times (10^{-10} ) ^2}$
$ \Rightarrow V_{\circ} = 151\: V$