Effect Of Frequency on Photocurrent

Question

The situation is this.. I am performing a photoelectric effect experiment - just the usual - evacuated glass tube containing two metal electrodes connected to an external source of DC voltage (say an alkaline cell) and ammeter in series. I shine a light on one of the electrodes - the frequency is more than enough to cause eject photoelectrons, and the voltage of the cell opposes the electrons which are trying to get to the other electrode. The question is this - what happens to the current on increasing the intensity of light while keeping no. of photons falling on the electrode constant (in other words increasing the frequency)?

1. Does the current remain the same because the no. of photons falling on the electrode is constant and so no. of electrons being ejected is constant?
2. Or does it increase because the electrons have more kinetic energy, ie, they are moving faster towards the other electrode so that the $t$ (time) term in $I$ = $q$/$t$ is less ($q$ remains constant because no. of electrons being ejected is still the same)?

This question appeared in one of my tests and the answer key says that the current remains constant but think that case 2 is correct and current would increase, which is why I asked this question.

Thanks.

Answer 1

The photoyield varies with frequency, in different ways for different materials. There is no way of answering such a question.

Edit to expand a bit on this: The quantum yield (number of electrons emitted/number of incident photons) is always less than unity, often much less. The art of making photocathodes is to increase this ratio, at least for the range of wavelengths of interest.

Reflected light does not contribute to the photoyield, so the variation of the reflectivity with frequency is an important factor. For alkali metals etc reflectivity is low above the plasma frequency.

Photoemission can be treated in a three-step model: 1) absorption, 2) transport of the electron to the surface, and 3) transmission through the interface.

In the first step, the photon is absorbed by excitation of an electron from an occupied state to an unoccupied state. So this depends on the density of states, both occupied and unoccupied. The probability is largest for vertical transitions where the wavevector $k$ of the electron is conserved. This joint density of states is strongly dependent on photon energy.

The second step depends on the kinetic energy of the excited electron. It does not depend that much on the material, there is the "universal curve".

The third step involves the work function. At threshold, only electrons at normal incidence can make it into the vacuum.