Q: 10–3 W of 5000 A° light is directed on a photoelectric cell. If the current in the cell is 0.16 mA , the percentage of incident photons which produce Photoelectrons is
(A) 0.4 %
(B) 0.04 %
(C) 20 %
(D) 10
Solution : $\large E = \frac{h c}{\lambda } $
$\large E = \frac{12400}{5000} eV = \frac{12400}{5000} \times 1.6 \times 10^{-19} J$
No. of Photon $\large N = \frac{P}{E} = \frac{10^{-3}}{\frac{12400}{5000} \times 1.6 \times 10^{-19}}$
N = 0.25 x 1016
No. of Electron reaching $\large = \frac{0.16 \times 10^{-6}}{1.6 \times 10^{-19}} $
= 10+12
Percentage $\large = \frac{10^{12}}{0.25 \times 10^{16}} \times 100 $
= 0.04 %
Correct option is (B)