When radiation of wavelength $\lambda$ is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength $3 \lambda$, the
stopping potential is $\frac{\mathrm{V}}{4}$. If the threshold
wavelength for the metallic surface is $n \lambda$ then
value of $\mathrm{n}$ will be
$\frac{\mathrm{hc}}{\lambda}=\frac{\mathrm{hc}}{\lambda_{0}}+\mathrm{eV}$ ............(I)
$\frac{\mathrm{hc}}{3 \lambda}=\frac{\mathrm{hc}}{\lambda_{0}}+\frac{\mathrm{e} \cdot \mathrm{V}}{4}$ ..................(II)
(multiply by 4)
$\frac{4 \mathrm{hc}}{3 \lambda}=\frac{4 \mathrm{hc}}{\lambda_{0}}+\mathrm{eV}$ .................(III)
From (i) \& (iii)
$\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{0}}=\frac{4 \mathrm{hc}}{3 \lambda}-\frac{4 \mathrm{hc}}{\lambda_{0}}$
$-\frac{\mathrm{hc}}{3 \lambda}=-\frac{3 \mathrm{hc}}{\lambda_{0}}$
$9 \lambda=\lambda_{0}$
$\mathrm{n}=9$