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) $V_1 = 0,\ V_2 < V_3$
Concept. Emission needs $\lambda < \lambda_0$; then $eV = hc\left(\dfrac1\lambda - \dfrac{1}{\lambda_0}\right)$.
Step 1. Cell 1. $\lambda > \lambda_1$: no emission, $V_1 = 0$.
Step 2. Cells 2 and 3. Both emit; a larger $\lambda_0$ gives a larger $V$, and $\lambda_3 > \lambda_2$, so $V_3 > V_2$.
Answer. $V_1 = 0,\ V_2 < V_3$ (Option A).
Solution by Sreeraj P, M.Sc Physics