A ray of light with wavelength λ is incident on three different...

According to Einstein's photoelectric equation, the magnitude of the stopping potential $V$ is given by:

$eV = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_{th}}$

For photoelectric emission to take place, the incident wavelength $\lambda$ must be less than the threshold wavelength $\lambda_{th}$.

For cell 1, $\lambda > \lambda_1$. No photoelectric emission occurs, so $V_1 = 0$.

For cell 2, $\lambda < \lambda_2$. Photoelectric emission occurs, and $V_2 = \dfrac{hc}{e} \left( \dfrac{1}{\lambda} - \dfrac{1}{\lambda_2} \right) > 0$.

For cell 3, $\lambda \ll \lambda_3$. Photoelectric emission occurs, and $V_3 = \dfrac{hc}{e} \left( \dfrac{1}{\lambda} - \dfrac{1}{\lambda_3} \right) > 0$.

Since $\lambda_3 \gg \lambda$ and $\lambda_2 > \lambda$, it follows that $\lambda_3 > \lambda_2$, which gives $\dfrac{1}{\lambda_3} < \dfrac{1}{\lambda_2}$.

Therefore, $\dfrac{1}{\lambda} - \dfrac{1}{\lambda_3} > \dfrac{1}{\lambda} - \dfrac{1}{\lambda_2}$, which implies $V_3 > V_2$.

Thus, $V_1 = 0$ and $V_2 < V_3$.

Answer: $V_1=0,\,V_2 < V_3$