CUET Preparation Today
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- |
0.48μA 0.42mA 0.48 mA 0.42μA |
0.48μA |
$n=\frac{Pλ}{hc}$ $n_e=n×β\%=\frac{Pλ}{hc}×\frac{β}{100}$ $n_e=\frac{1.5×10^{-3}×400×10^{-9}}{6.6×10^{-34}×3×10^8}×\frac{0.1}{100}$ $I=n_e\,e=0.48μA$ |
