1.5 mW of 400 nm light is directed at a photo electric cell. If 0.1% of the incident photons produce photo electrons, the current in the cell is-
Answer & explanation
Correct answer: option 1
$n=\frac{Pλ}{hc}$
$n_e=n×β\%=\frac{Pλ}{hc}×\frac{β}{100}$
$n_e=\frac{1.5×10^{-3}×400×10^{-9}}{6.6×10^{-34}×3×10^8}×\frac{0.1}{100}$
$I=n_e\,e=0.48μA$