When radiation of wavelength ' \(\lambda\) ' is incident on a metallic surface, the stopping potential is \(4.8 \mathrm{~V}\). If the surface is illuminated with radiation of double the wavelength then the stopping potential becomes \(1.6 \mathrm{~V}\).' The threshold wavelength for the surface is

Stopping potential:
\(\begin{aligned}
& \mathrm{eV}_0=\mathrm{h} v-\phi_0 \\
& \mathrm{eV}_0=\frac{\mathrm{hc}}{\lambda}-\phi_0
\end{aligned}\)
For first case: \(\mathrm{e}(4.8)=\frac{\mathrm{hc}}{\lambda}-\phi_0... (i)\)
For Second case: \(e(1.6)=\frac{h c}{2 \lambda}-\phi_0... (ii)\)
Dividing equation (i) by equation (ii),
\(\begin{aligned}
& 3\left(\frac{\mathrm{hc}}{2 \lambda}-\phi_0\right)=\frac{\mathrm{hc}}{\lambda}-\phi_0 \\
\therefore \quad & \frac{3 \mathrm{hc}}{2 \lambda}-3 \phi_0=\frac{\mathrm{hc}}{\lambda}-\phi_0 \\
& \frac{\mathrm{hc}}{2 \lambda}=2 \phi_0 \\
& \frac{\mathrm{hc}}{4 \lambda}=\phi_0
\end{aligned}\)
\(\therefore \quad\) Threshold wavelength is \(4 \lambda\).