Explain why the saturation current in photoelectric effect experiment with a light of one frequency and intensity is independent of the anode potential.

Explanation

(i) When the frequency of the incident light is doubled, the kinetic energy of emitted photoelectron becomes more than double as explained below. Let `E_(1)` and `E_(2)` be the K.E. of photoelectrons for incident light of frequency v and 2v respectively.
Then `hv=E_(1)+phi_(0)` and `2hv=E_(2)+phi_(0)`.
So `2=(E_(2)+phi_(0))/(E_(1)+phi_(0))` or `2E_(1)+2phi_(0)=E_(2)+phi_(0) or E_(2)=2E_(1)+phi_(0)` or `E_(2)gt2E_(1)`
(ii) Photoelectric current remains unchanged as it depends upon the intensity of incident light.
(iii) The stopping potential becomes more than double as
`eV_(0)=(K.E.)_(max)=hv-phi_(0)` or `hv=eV_(0)+phi_(0)`
and `eV_(0)=h2v-phi_(0)=2(eV_(0)+phi_(0))-phi_(0)=2eV_(0)+phi_(0)`. So `V'_(0) gt 2V_(0)`