In an experiment on photoelctric effect, light of wavelength `800 nm` ( less than threshold wavelength) is incident on a cessium 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 of every `10^(6)` photons is able to eject a photoelectron, find photo current in the circuit.

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.`