In photoelectric effect, the intensity of light is directly related to the number of photons incident on metal. This is indicated by the photocurrent vs. voltage applied for different frequencies having the same saturation current.

Intensity= (No. of photons per unit time) x (Energy of one photon =hv)

So if frequency is increased keeping intensity constant, shouldn't the no of photons incident per unit time decrease and thereby saturation current should decrease?

  • $\begingroup$ You are correct. See physics.stackexchange.com/questions/128964/… $\endgroup$
    – Andrei
    Commented Aug 2, 2016 at 4:15
  • $\begingroup$ What about the actual experimentall graph that shows saturation current is independent of frequency? $\endgroup$ Commented Aug 2, 2016 at 5:33
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    $\begingroup$ Do you have such a graph? If so, its source should describe the conditions which produced the data. Include a link if you would like our help to analyze it. $\endgroup$
    – rob
    Commented Aug 2, 2016 at 6:00
  • $\begingroup$ You should look at G.W. Gobeli and F.G. Allen, Direct and Indirect Excitation Processes in Photoelectric Emission from Silicon, Phys. Rev. 127(1) 141-149 (1962). While your conclusion may be true for photon energies well above threshold, it certainly is not true near threshold where the photoelectric yield (electrons/quantum as they put it) is increasing rapidly, by orders of magnitude over perhaps 0.2eV of photon energy. $\endgroup$
    – Jon Custer
    Commented Aug 2, 2016 at 13:42
  • $\begingroup$ Also many introductory textbooks have graphs that show that with constant intensity, saturation current in independent of frequency. $\endgroup$ Commented Aug 3, 2016 at 3:38

1 Answer 1


for a given photosensitive material and frequency of incident radiation,saturation current is found to be proportional to the intensity of incident radiation whereas the stopping potential is independent of its intensity.


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