1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.1 % of the incident photons produce photoelectrons, find the current (in $\mu$A) in the cell.

Number of photons incident per second

                $\begin{array}{l}=\frac{\text{ Power }}{\text{ Energy of one photon }}\\ =\frac{\mathrm{P}}{(\mathrm{hc}/\mathrm{\lambda })}=\frac{\mathrm{P\lambda }}{\mathrm{hc}}\end{array}$

Number of electrons emitted per second = 0.1% of

                 $\frac{\mathrm{P\lambda }}{\mathrm{hc}}=\frac{\mathrm{P\lambda }}{1000\mathrm{hc}}$

$\therefore$Current = Charge (on photoelectrons per second)

                $\begin{array}{l}=\frac{\mathrm{P\lambda e}}{1000\mathrm{hc}}\\ =\frac{\left(1.5\times {10}^{-3}\right)\left(400\times {10}^{-9}\right)\left(1.6\times {10}^{-19}\right)}{\left(1000\right)\left(6.63\times {10}^{-34}\right)\left(3\times {10}^8\right)}\\ =0.48\times {10}^{-6}\mathrm{A}=0.48\mathrm{\mu A}\end{array}$