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(A) According to Einstein's photoelectric equation, the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$, where

Vedclass · Accepted Answer

$V_1 = 0, V_2 < V_3$

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(A) According to Einstein's photoelectric equation, the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$, where $\lambda_0$ is the threshold wavelength.If $\lambda > \lambda_0$, the energy of the incident photon is less than the work function, so no photoelectric emission occurs, and the stopping potential $V_s = 0$.Given $\lambda_1  0$.Given $\lambda_2 > \lambda$, no photoelectric emission occurs, so $V_2 = 0$.Given $\lambda_3 >> \lambda$, no photoelectric emission occurs, so $V_3 = 0$.Wait, re-evaluating the condition: If $\lambda_1  \frac{1}{\lambda}$, so $V_1 > 0$. If $\lambda_2 > \lambda$, then $\frac{1}{\lambda_2}  0$. If $\lambda_3 >> \lambda$, then $\frac{1}{\lambda_3}  0$. Actually, the condition for emission is $\lambda \le \lambda_0$. For cell $1$: $\lambda_1 For cell $2$: $\lambda_2 > \lambda$ (Emission occurs, $V_2 > 0$).For cell $3$: $\lambda_3 >> \lambda$ (Emission occurs, $V_3 > 0$).Since $\lambda_3 > \lambda_2$, the work function $\Phi_3 = \frac{hc}{\lambda_3} Thus, $V_3 = \frac{hc}{e}(\frac{1}{\lambda} - \frac{1}{\lambda_3}) > V_2 = \frac{hc}{e}(\frac{1}{\lambda} - \frac{1}{\lambda_2})$.Therefore, $V_1 = 0$ and $V_2 < V_3$.

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