${10}^{-3}$ W of 5000 Å light is directed on a photoelectric cell. If the current in the cell is 0.16 $\mu$A, the percentage of incident photons which produce photoelectrons, is

Number of photons falling per second:

$N_p=\frac{{10}^{-3}}{{\displaystyle \frac{6.6\times {10}^{-34}\times 3\times {10}^8}{5000\times {10}^{-10}}}}$

No of electrons per unit time $=\frac{I}{e}$

Percentage of photons producing photoelectrons = $\frac{N_e}{N_p}\times 100=\frac{{10}^{12}}{2.5\times {10}^{15}}\times 100=0.04\%$