If $I = naevd$ where $v$ is the drift velocity then why in photoelectric effect increase potential across plates only increase current for a while?

Question

Just studying photoelectric effect and got this confusion.

Answer 1

The equation you give for the drift velocity applies to a conductor where the density of the conduction electrons is constant.

However in the PE experiment if we keep the light intensity constant it is the number of electrons released per second that is constant. That means if we increase the (average) velocity of the electrons by increasing the bias voltage the density of the electrons in the space between the metal surface and the detector decreases. Since the density is changing we cannot use the regular drift velocity equation.

At equilibrium the number of electrons per second received at the detector has to be equal to the number of photoelectrons per second emitted by the metal, and that depends only on the light intensity i.e. the number of photons per second hitting the metal. Changing the bias voltage cannot affect this so it cannot change the equilibrium current.