In a photocell, with excitation wavelength $\lambda$, the faster electron has speed v. If the excitation wavelength is changed to $3\lambda /4$, the speed of the fastest electron will be

$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-W_0 .........\left(i\right)$

Let the speed of the fastest electron be V₁ when excitation wavelength is changed to $3\lambda /4$

$\begin{array}{l}\therefore \frac{1}{2}m{v}_1^2=\frac{4hc}{3\lambda }-W_0\\ \Rightarrow \frac{1}{2}m{v}_1^2=\frac{4}{3}\left(\frac{hc}{\lambda }-W_0\right)+\frac{W_0}{3}\\ \Rightarrow \frac{1}{2}m{v}_1^2=\frac{4}{3}\left(\frac{1}{2}m{v}^2\right)+\frac{W_0}{3} [u\sin g Eq.(i\left)\right]\\ \\ \Rightarrow v_1^2=\frac{4v^2}{3}+\frac{2W_0}{3m}\\ \therefore v_1>\sqrt{\frac{4}{3}}v\end{array}$