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(C) The intensity of the radiation is $I = 6.4 	imes 10^{-5} \; W/cm^{2}$ and the area is $A = 1 \; cm^{2}$.The power incident on the surface is $P =

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$11$

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(C) The intensity of the radiation is $I = 6.4 	imes 10^{-5} \; W/cm^{2}$ and the area is $A = 1 \; cm^{2}$.The power incident on the surface is $P = I 	imes A = 6.4 	imes 10^{-5} \; W$.The energy of a single photon is $E_{ph} = \frac{hc}{\lambda} = \frac{1240 \; eV \cdot nm}{310 \; nm} = 4 \; eV$.Converting this energy to Joules: $E_{ph} = 4 	imes 1.6 	imes 10^{-19} \; J = 6.4 	imes 10^{-19} \; J$.The number of photons incident per second $(n)$ is given by $n = \frac{P}{E_{ph}} = \frac{6.4 	imes 10^{-5} \; J/s}{6.4 	imes 10^{-19} \; J} = 10^{14} \; 	ext{photons/s}$.Given that one in $10^{3}$ photons ejects an electron, the number of electrons emitted per second is $N_e = n 	imes 10^{-3} = 10^{14} 	imes 10^{-3} = 10^{11}$.Comparing this with $10^{x}$, we get $x = 11$.

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