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(C) From the given $I - V$ curve,the stopping potential $V_s = 5 \, V$ and the saturation current $I_s = 10 \, \mu A = 10^{-5} \, A$.The energy of inc

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$7$

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(C) From the given $I - V$ curve,the stopping potential $V_s = 5 \, V$ and the saturation current $I_s = 10 \, \mu A = 10^{-5} \, A$.The energy of incident photons is given by Einstein's photoelectric equation:$E = h
u = eV_s + \phi_0$$E = 5 \, eV + 2 \, eV = 7 \, eV$.The saturation current is related to the power $P$ of the incident light by:$I_s = \eta 	imes \left( \frac{P}{E} ight) 	imes e$where $\eta$ is the efficiency of photoemission $= 10^{-3} \% = 10^{-5}$.Substituting the values:$10^{-5} = 10^{-5} 	imes \left( \frac{P}{7 \, eV} ight) 	imes e$$1 = \frac{P}{7 \, eV} 	imes e$$1 = \frac{P}{7}$$P = 7 \, W$.

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