When a metallic surface is illuminated with radiation of wavelength λ , the stopping potential is V . If the…

Given,

Stopping potential is $V$ when the metal is illuminated with $\lambda$.

Stopping potential is $\frac{V}{4}$ when the metal is illuminated with $2\lambda$.

In photo electric effects
$\mathrm{e}{\mathrm{V}}_0=\mathrm{h}\nu -\mathrm{W}$
$\mathrm{e}{\mathrm{V}}_0=\frac{\mathrm{h}\mathrm{c}}{\mathrm{\lambda }}-\mathrm{W}$
$\mathrm{e}\mathrm{V}=\frac{\mathrm{h}\mathrm{c}}{\mathrm{\lambda }}-\mathrm{W}$ ...(i)
$\mathrm{e}\frac{\mathrm{V}}{4}=\frac{\mathrm{h}\mathrm{c}}{2\mathrm{\lambda }}-\mathrm{W}$ ....(ii)
From (i) and (ii)
$\frac{\mathrm{h}\mathrm{c}}{\mathrm{\lambda }}-\mathrm{W}=4\left(\frac{\mathrm{h}\mathrm{c}}{2\mathrm{\lambda }}-\mathrm{W}\right)$
$\frac{\mathrm{h}\mathrm{c}}{\mathrm{\lambda }}-\mathrm{W}=\frac{2\mathrm{h}\mathrm{c}}{\mathrm{\lambda }}-4\mathrm{W}$
$3\mathrm{W}=\frac{\mathrm{h}\mathrm{c}}{\mathrm{\lambda }}\Rightarrow \mathrm{W}=\frac{\mathrm{h}\mathrm{c}}{3\mathrm{\lambda }}$
$\frac{\mathrm{h}\mathrm{c}}{{\mathrm{\lambda }}_{\mathrm{m}\mathrm{a}\mathrm{x}}}=\frac{\mathrm{h}\mathrm{c}}{3\mathrm{\lambda }}\Rightarrow {\mathrm{\lambda }}_{\mathrm{m}\mathrm{a}\mathrm{x}}=$ threshold wavelength $=3\mathrm{\lambda }$