When a surface is illuminated with light of wavelength $\lambda$, the stopping potential is $V$. Whe

Question

When a surface is illuminated with light of wavelength $\lambda$, the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2 \lambda$, the stopping potential is $\frac{V}{3}$. Threshold wavelength for the metallic surface is

Professor PC Thomas Classes · Answer

$$
\begin{aligned}
& e V=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}ight) \ldots \ldots ....(1) \
& e \frac{V}{3}=h c\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}ight) \ldots \ldots . (2)\
& \frac{(1)}{(2)} \Rightarrow \lambda_0=4 \lambda
\end{aligned}
$$