Explanation

(d) $hv - W_{0} = \frac{1}{2}mv_{\max}^{2} \Rightarrow \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}} = \frac{1}{2}mv_{\max}^{2}$
$\Rightarrow hc\left( \frac{\lambda_{0} - \lambda}{\lambda\lambda_{0}} \right) = \frac{1}{2}mv_{\max}^{2} \Rightarrow v_{\max} = \sqrt{\frac{2hc}{m}\left( \frac{\lambda_{0} - \lambda}{\lambda\lambda_{0}} \right)\ }$
When wavelength is $\lambda$ and velocity is $v,$ then
$v = \sqrt{\frac{2hc}{m}\left( \frac{\lambda_{0} - \lambda}{\lambda\lambda_{0}} \right)}\ \ \ \ \ \ \ldots(i)$
When wavelength is $\frac{3\lambda}{4}$ and velocity is $v'$ then
$v^{'} = \sqrt{\frac{2hc}{m}\left\lbrack \frac{\lambda_{0} - (3\lambda/4)}{(3\lambda/4) \times \lambda_{0}} \right\rbrack}\ \ \ \ \ \ \ldots(ii)$
Divide equation (ii) by (i), we get
$\frac{v^{'}}{v} = \sqrt{\frac{\lbrack\lambda_{0} - (3\lambda/4)\rbrack}{\frac{3}{4}\lambda\lambda_{0}} \times \frac{{\lambda\lambda}_{0}}{\lambda_{0} - \lambda}}$
$v^{'} = v\left( \frac{4}{3} \right)^{1/2}\sqrt{\frac{\lbrack\lambda_{0} - (3\lambda/4)\rbrack}{\lambda_{0} - \lambda}}$
$i.e.\ v^{'} > v\left( \frac{4}{3} \right)^{1/2}$