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.


  • $\begingroup$ Intensity = Energy/(Time * Area) = (No. of photons)(Energy per photon)/(Time * Area) .. Yes, it depends on the no. of photons, but that's not the only thing to consider. You can increase the energy of individual photons to increase intensity. This way, you do not need to increase the number of photons. $\endgroup$ Commented Apr 23, 2018 at 9:32
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    $\begingroup$ Please consider writing more useful question titles, see How do we write good question titles? $\endgroup$
    – ACuriousMind
    Commented Apr 23, 2018 at 16:49

1 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.

  • $\begingroup$ How so? Since the no. of electrons being ejected per unit time remains constant and their KE is increasing, ie, they are covering the distance between the electrodes in lesser and lesser time, shouldn't the current increase? $\endgroup$ Commented Apr 23, 2018 at 6:39
  • $\begingroup$ The quantum yield is not one. Everything varies with frequency: both the reflectivity and the absorption coefficient (the depth at which photons are absorbed). $\endgroup$
    – user137289
    Commented Apr 23, 2018 at 6:44
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    $\begingroup$ Further Density of States of the electrons will come into play. $\endgroup$
    – Tomi
    Commented Apr 23, 2018 at 8:09
  • $\begingroup$ Can't say that you're wrong, but the photoelectric effect is typically only discussed over a relative narrow range of photon frequencies. Of course if you want to discuss all interactions of electromagnetic radiation with matter then that opens up a huge discussion. $\endgroup$
    – MaxW
    Commented Apr 23, 2018 at 12:47
  • $\begingroup$ @MaxW The variations in yield with photon energy are quite strong also when one limits oneself to visible and near-UV light. This means electron kinetic energies up to a few eV, the region of plasmon energies, of large changes in the joint densities of states, etcetera. $\endgroup$
    – user137289
    Commented Apr 23, 2018 at 14:36

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