${10}^{–3}$ 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 

Energy of each photon,
$E=\frac{124\cancel{00}}{50\cancel{00}}$  eV
$E=2.48$  eV
E = e(2.48) joule
Number of photons striking the target per second = $\frac{P}{\left(2.48\right)e}$   $=\frac{{10}^{-3}}{\left(2.48\right)e}$ 
   
 No. of electrons from surface per second  =  $\frac{i}{e}$$\frac{0.16\text{×}{10}^{-6}/\cancel{e}}{({10}^{-3}/2.48)\cancel{e}}\text{×100\%}$  

$\therefore n=\frac{{\text{0.16×10}}^{\text{-6}}\text{×2.48}}{{\text{10}}^{\text{-3}}}\text{×100\%}$   

 $n=0.04\%$