Find the number of photons emitted per second by a 40 W source of monochromatic light of wavelength 6000Å . What is the photoelectric current assuming 5% effeciency of photoelectric effect? Given `h=6.62xx10^(-34)Js and c=3xx10^(8) ms^(-1),e=1.6xx10^(-19)C`

Answer

`0.97A`

Explanation

No. of photons emitted per second by a source is
`=("power")/("energy of each photon")=P/(hc//lambda)=(Plambda)/(hc)`
`=(40xx(6000xx10^(-10)))/((6.63xx10^(-34))xx(3xx10^(8))) = 12.12xx10^(19)`
As one photon can eject one photoelectron only, and efficiency of photoelectric emission is 5% so
`eta=("no. of photoelectrons emitted per sec"(n))/("no. of photons falling per sec" (12.12xx10^(19))`
`:. 5/100=n/(12.12xx10^(19))`
or `n=(5xx12.12xx10^(19))/100=60.60xx10^(17)s^(-1)`
Photoelectric current `I=("ne")/(t)=("ne")/(1)`
`=60.6xx10^(17)xx1.6xx10^(-19)=96.96xx10^(-2)A =0.97 A`