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 10^{16}

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)