In an experiment on photoelectric effect, light of wavelength 400nm is incident on a cesium plate at the rate of 5.0W. The potential of the collector plate is made sufficiently positive with respect to the emitter so that the current reaches its 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 circuit.

Answer

`(P lamda)/(hc xx 10^(6)) e A = 3.2 mu A`

Explanation

No of photons `= (P)/(E_(lamda)) = (P lamda)/(hc) s_(-1)`
no of photo electron s/s `= (P lamda)/(hc). (1)/(10^(6))`
`:.` Photo current `= (P lamda)/(hc xx 10^(6))`
`e =(5xx 800 xx10^(-9) xx(1.6 xx10^(-19)))/(6.63 xx 10^(-34) xx 3 xx 10^(8) xx 10^(6))A = 3.2 muA`.