The photoelectric equation is given by: $$eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$$ where $V$ is the stopping potential, $\lambda$ is the incident wavelength, and $\lambda_0$ is the threshold wavelength.
Photoelectric emission occurs only if $\lambda < \lambda_0$. If $\lambda \ge \lambda_0$, the kinetic energy is zero, and thus the stopping potential $V = 0$.
Given $\lambda_1 < \lambda$, emission is not possible, so $V_1 = 0$. Given $\lambda_3 > \lambda$, emission occurs. Since $\lambda_2$ is between $\lambda_1$ and $\lambda_3$, we compare the potentials. As $\lambda_0$ increases, the work function $\phi = \frac{hc}{\lambda_0}$ decreases, leading to a higher stopping potential for a fixed $\lambda$. Since $\lambda_2 < \lambda_3$, the work function for cell 2 is higher than cell 3, hence $V_2 < V_3$.
Final Answer: B
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