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(C) The energy of a single incident photon is given by $E = \frac{hc}{\lambda} = \frac{1240\,eV\cdot nm}{400\,nm} = 3.1\,eV$.Converting this energy to

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$50$

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(C) The energy of a single incident photon is given by $E = \frac{hc}{\lambda} = \frac{1240\,eV\cdot nm}{400\,nm} = 3.1\,eV$.Converting this energy to Joules: $E = 3.1 	imes 1.6 	imes 10^{-19}\,J = 4.96 	imes 10^{-19}\,J$.The number of photons incident per second is $n = \frac{P}{E} = \frac{1.55 	imes 10^{-3}\,W}{4.96 	imes 10^{-19}\,J} = 3.125 	imes 10^{15}\,	ext{photons/s}$.Since only $10\%$ of photons produce photoelectrons,the number of photoelectrons emitted per second is $N_e = 0.10 	imes n = 0.10 	imes 3.125 	imes 10^{15} = 3.125 	imes 10^{14}\,	ext{electrons/s}$.The photoelectric current is $I = N_e 	imes e = 3.125 	imes 10^{14} 	imes 1.6 	imes 10^{-19}\,A = 5.0 	imes 10^{-5}\,A$.Converting to $\mu A$: $I = 50\,\mu A$.

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