In a photoelectric experiment set up, photons of energy `5 eV` falls on the cathode having work function `3 eV` (a) if the saturation current is ` i= 4 mu A` for intensity `10^(-5) W//m^(2)`, then plot a graph between anode potential and current (b) Also draw a graph for intensity of incident radiation `2 xx 10^(-5) W//m^(2)`

Explanation

Maximum kinetic energy of the photoelectrons would be `K_("max")=E-W=(5-3) eV=2eV`
Therefore, the stopping potential is `2V`. Saturation current depends on the intensity of light incident. When the intensity is doubled the saturation current will also become two fold. the corresponding graphs are shown in figure.