Question

$1.5mW$ of $400nm$ light is directed at a photoelectric cell. If $0.1\%$ of the incident photons produce photo electrons, find the current in the cell.

Answer 1

Power; $P=1.5mw$

$=1.5\times {10}^{-3}$ watt

Hence, Energy of each photon = $hc/\lambda$

$=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{400\times {10}^{-9}}$

$=4.96\times {10}^{-19}$ Joule

and Number of photo-electrons emitted per second

$=\frac{P}{Energytakenbyeach}$

$=\frac{1.5\times {10}^{-3}}{4.96\times {10}^{-19}}$

$n=3\times {10}^{15}$

No. of photon electrons that produced current

$=0.1\%$ of $n$

$=\frac{0.1}{100}\times 3\times {10}^{15}$

$N=3\times {10}^{12}$

Hence, current $I=Ne$

$=N\times 1.6\times {10}^{-16}$

$=3\times {10}^{12}\times 1.6\times {10}^{-19}$

$=4.8\times {10}^{-7}$

$=0.48\mu A$