A ray of light with wavelength $\lambda$ is incident on three different photoelectric cells namely 1, 2 and 3. The threshold wavelength of these photoelectric cells are $\lambda_1$, $\lambda_2$, and $\lambda_3$, respectively and the magnitude of stopping potentials of these cells are $V_1$, $V_2$ and $V_3$, respectively. The relation between $\lambda$ and threshold wavelengths are $\lambda_1 < \lambda$, $\lambda_2 > \lambda$ and $\lambda_3 >> \lambda$. The correct option is :
Answer: (A) $V_1 = 0,\ V_2 < V_3$
Emission occurs only if $\lambda$ is less than the threshold wavelength.
Cell 1: $\lambda > \lambda_1$, so there is no emission and $V_1 = 0$.
Cells 2 and 3: $\lambda < \lambda_2$ and $\lambda < \lambda_3$, so both emit. The stopping potential is
$$eV = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)$$
A larger threshold wavelength $\lambda_0$ means a smaller work function and a larger $V$. Since $\lambda_3 > \lambda_2$, $V_3 > V_2$.
So $V_1 = 0$ and $V_2 < V_3$.
Solution by Sreeraj P, M.Sc Physics