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(C) Given: Work function $(W_0) = 4.80 \ eV$ and Kinetic energy $(KE) = 0.20 \ eV$. According to the photoelectric effect equation: Total energy $(E_T

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$1.21 	imes 10^{15}$

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(C) Given: Work function $(W_0) = 4.80 \ eV$ and Kinetic energy $(KE) = 0.20 \ eV$. According to the photoelectric effect equation: Total energy $(E_T) = W_0 + KE = 4.80 \ eV + 0.20 \ eV = 5.0 \ eV$. Convert energy to Joules: $E_T = 5.0 	imes 1.602 	imes 10^{-19} \ J = 8.01 	imes 10^{-19} \ J$. Using the relation $E_T = h
u$,where $h = 6.626 	imes 10^{-34} \ J \cdot s$: $
u = \frac{E_T}{h} = \frac{8.01 	imes 10^{-19} \ J}{6.626 	imes 10^{-34} \ J \cdot s} \approx 1.21 	imes 10^{15} \ Hz$.

When a metal surface is exposed to a certain frequency of electromagnetic radiation,the kinetic energy of the electron ejected from the metal surface is $0.20 \ eV$. If its work function $(W_0)$ is $4.80 \ eV$,the approximate frequency of the radiation falling on the metal surface in $Hz$ is:

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