When a metallic surface is illuminated with radiation of wavelength ' $\lambda$ ', the stopping potential is…
When a metallic surface is illuminated with radiation of wavelength ' $\lambda$ ', the stopping potential is ' $\mathrm{V}$ '. If the same surface is illuminated with radiation of wavelength ' $2 \lambda$ ', the stopping potential is ' $\left(\frac{\mathrm{v}}{4}\right)$,. The threshold wavelength for the metallic surface is
$\frac{5}{2} \lambda$
$3 \lambda$
$4 \lambda$
$5 \lambda$
Solution
For stopping potential $\mathrm{V}, \mathrm{eV}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0}$
For stopping potential $\frac{\mathrm{V}}{4}, \frac{\mathrm{eV}}{4}=\frac{\mathrm{hc}}{2 \lambda}-\frac{\mathrm{hc}}{\lambda_0}$
Taking the ratio, we get
$\begin{aligned}
& \quad 4=\frac{\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)}{\frac{1}{2 \lambda}-\frac{1}{\lambda_0}} \\
& 4\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right)=\frac{1}{\lambda}-\frac{1}{\lambda_0} \\
& \left(\frac{2}{\lambda}-\frac{4}{\lambda_0}\right)=\frac{1}{\lambda}-\frac{1}{\lambda_0} \\
& \left(\frac{2}{\lambda}-\frac{1}{\lambda}\right)=\frac{1}{\lambda_0}+\frac{4}{\lambda_0} \\
& \quad\left(\frac{1}{\lambda}\right)=\frac{3}{\lambda_0} \\
& \therefore \quad \lambda_0=3 \lambda
\end{aligned}$