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(C) The power of the light source is $P = 5 	imes 10^{-3} \, W$ and the number of photons emitted per second is $n = 8 	imes 10^{15} \, s^{-1}$.The

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(C) The power of the light source is $P = 5 	imes 10^{-3} \, W$ and the number of photons emitted per second is $n = 8 	imes 10^{15} \, s^{-1}$.The energy of a single photon $E$ is given by $E = \frac{P}{n}$.$E = \frac{5 	imes 10^{-3}}{8 	imes 10^{15}} \, J = 6.25 	imes 10^{-19} \, J$.To convert this energy into electron-volts $(eV)$,divide by the charge of an electron $e = 1.6 	imes 10^{-19} \, C$:$E = \frac{6.25 	imes 10^{-19}}{1.6 	imes 10^{-19}} \, eV = 3.90625 \, eV \approx 3.9 \, eV$.According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E)_{\max}$ is related to the stopping potential $V_s$ as $(K.E)_{\max} = e V_s = 2.0 \, eV$.The photoelectric equation is $E = \phi + (K.E)_{\max}$,where $\phi$ is the work function.$\phi = E - (K.E)_{\max} = 3.9 \, eV - 2.0 \, eV = 1.9 \, eV$.Therefore,the work function of the metal is $1.9 \, eV$.

$A$ monochromatic light source of intensity $5 \, mW$ emits $8 \times 10^{15}$ photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is $2.0 \, V$. The work function of the metal will be ............ $eV$. (in $.9$)

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