In an experiment on photoelectric effect of light wavelength 400 nm is incident on a metal plate at rate of 5W. The potential of the collector plate is made sufficiently positive with respect to emitter so that the current reaches the saturation value. Assuming that on the average one out of every `10^6` photons is able to eject a photoelectron, find the photocurrent in the cirucuit.

Answer

`(Plambda)/(hcxx10^(6))e A=3.2 mu A`

Explanation

No. of photons `(P)/(E_(lambda))=(P_(lambda))/(hc).s^(-1)`
no of photo electron `s//s=(Plambda)/(hc).(1)/(10^(6))`
`:.` photo current `=(Plambda)/(hcxx10^(6)).e=(5xx800xx10^(-9)xx(1.6xx10^(-19)))/(6.63xx10^(-34)xx3xx10^(8)xx10^(6))A=3.2 mu A.`