When radiation of wavelength ' $\lambda$ ' is incident on a metallic surface, the stopping potential is $4.8…
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
$2 \lambda$
$4 \lambda$
$6 \lambda$
$8 \lambda$
Solution
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 $4 \lambda$.