When a certain metallic surface is illuminated with monochromatic light wavelength $\lambda$, the stopping…
When a certain metallic surface is illuminated with monochromatic light wavelength $\lambda$, the stopping potential for photoelectric current is $4 V_0$. When the same surface is illuminated with light of wavelength $3 \lambda$, the stopping potential is $\mathrm{V}_0$. The threshold wavelength for this surface for photoelectric effect is
$9 \lambda$
$\frac{\lambda}{9}$
$3 \lambda$
$\frac{\lambda}{3}$
Solution
Using Einstein's photoelectric equation, $\mathrm{h} v=\phi_0+\mathrm{KE}_{\text {max }}$
$\therefore \quad \frac{\mathrm{hc}}{\lambda}=\phi_0+\mathrm{e}\left(4 \mathrm{~V}_0\right)$
....(i) $\left(\because \mathrm{KE}_{\max }=\mathrm{eV}_{\mathrm{s}}\right)$
Also, $\frac{\mathrm{hc}}{3 \lambda}=\phi_0+\mathrm{eV}_0$...(ii)
Subtracting equation (i) from $4 \times$ equation (ii) we get,
$\left(\frac{4}{3}-1\right) \frac{\mathrm{hc}}{\lambda}=4 \phi_0-\phi_0$ or $\phi_0=\frac{\mathrm{hc}}{9 \lambda}$
But $\phi_0=\frac{\mathrm{hc}}{\lambda_0}$, where $\lambda_0$ is the threshold wavelength, hence $\lambda_0=9 \lambda$.
Hence, option (A) is correct.