In photo-emissive cell, with exciting wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3 λ /4, the speed of the fastest emitted electron will be:

Einstein's photoelectric equation is given by

${\mathrm{E}}_{\mathrm{k}}=\mathrm{E}-\mathrm{w}$

$\text{ but }{\mathrm{E}}_{\mathrm{k}}=\frac{1}{2}{\mathrm{mv}}^2\text{ and }\mathrm{E}=\frac{\mathrm{ch}}{\mathrm{\lambda }}$

$\therefore \frac{1}{2}{\mathrm{mv}}^2=\frac{\mathrm{ch}}{(3\lambda /4)}-\mathrm{w}$

$\text{ or }\frac{1}{2}m{v}^2=\frac{4}{3}\frac{hc}{\lambda }-w\dots \text{ (2) }$

Dividing Eq. (ii) by Eq. (i), we get

$\frac{{\mathrm{v}}^2}{{\mathrm{v}}^2}=\frac{\frac{4}{3}\frac{\mathrm{ch}}{\mathrm{\lambda }}-\mathrm{w}}{\frac{\mathrm{ch}}{\mathrm{\lambda }}-\mathrm{w}}$

$=\frac{\frac{4}{3}\frac{\mathrm{ch}}{\mathrm{\lambda }}-\frac{4}{3}\mathrm{w}+\frac{1}{3}w}{\frac{\mathrm{ch}}{\mathrm{\lambda }}-\mathrm{w}}$

$=\frac{4}{3}+\frac{\mathrm{w}}{3\left(\frac{\mathrm{ch}}{\mathrm{\lambda }}-\mathrm{w}\right)}>\frac{4}{3}$

$\therefore \frac{{\mathrm{v}}^2}{{\mathrm{v}}^2}>\sqrt{\frac{4}{3}}\text{ or }{\mathrm{v}}^{\mathrm{\prime }}>\sqrt{\frac{4}{3}}\mathrm{v}$