When radiations of wavelength ' $\lambda$ ' is incident on a metallic surface the stopping potential…
When radiations of wavelength ' $\lambda$ ' is incident on a metallic surface the stopping potential required is 4.8 volt. If same surface is illuminated with radiations of double the wavelength, then required stopping potential becomes 1.6 volt. Then the value of threshold wavelength for the surface is
$2 \lambda$
$4 \lambda$
$6 \lambda$
$8 \lambda$
Solution
We know, $\mathrm{V}=\frac{\mathrm{hc}}{\lambda}-\phi_0$
$\Rightarrow 4.8=\frac{\mathrm{hc}}{\lambda}-\phi$ ....(i)
Given $\lambda=2 \lambda$
$\Rightarrow 1.6=\frac{h c}{2 \lambda}-\phi$ ....(ii)
$\therefore \quad$ Dividing (i) by (ii),
$3\left(\frac{\mathrm{hc}}{2 \lambda}-\phi\right)=\frac{\mathrm{hc}}{\lambda}-\phi$
$3 \frac{\mathrm{hc}}{2 \lambda}-3 \phi=\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)$
$\frac{\mathrm{hc}}{2 \lambda}=2 \phi$
$\therefore \quad$ Threshold wavelength is $4 \lambda$
.