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Why do we have a saturation current in photoelectric effect?

  1. Dec 23, 2015 #1
    I had this question popped into my mind when I was reading this topic one day.
    In the photoelectric experiment, when light, having frequency greater than the threshold frequency, falls on a metal, electrons are emitted. Since electrons emitted are of different energies (I presume it's because they're coming from different energy levels in the lattice of the metal), not all of them are able to reach the anode per second.

    Now, in the curve between applied potential and photocurrent measured, on increasing potential, number of electrons reaching cathode per second increases. So naturally, the photocurrent also increases. But apparently, when "all" the electrons emitted are able to reach the cathode, the current reaches a saturation. Why?

    Current here is basically number of electrons reaching the cathode reaching per second. On increasing the potential, electrons move faster, right? And since the photoelectric effect is continuously happening, (electrons getting emitted continuously by photons), more number of electrons should reach the cathode per second.

    I KNOW THIS MAY SOUND WRONG, BECAUSE NUMBER OF ELECTRONS EMITTED DEPENDS ON INTENSITY OF LIGHT U....

    Wait, so the number of electrons emitted per second is fixed because the intensity is fixed. Therefore yes, if we are increasing the potential, we are allowing those electrons (emitted at that time frame) to reach the cathode quicker so as to allow more electrons to reach cathode into that one second time period. But there aren't any more emitted electrons left to reach in that time period, are there? The number of electrons emitted per second is fixed.

    SO, SATURATION CURRENT IS THAT CURRENT WHEN THE RATE OF FLOW OF CHARGE IN THE CIRCUIT BECOMES EQUAL TO THE RATE OF EMISSION OF ELECTRONS .

    So I just answered my question while writing it. Is my understanding correct? Have I reached the right conclusion, or am I greatly flawed? I would really like to hear your views on this topic!
     
    Last edited: Dec 23, 2015
  2. jcsd
  3. Dec 24, 2015 #2

    BvU

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    Hello Shubadh, :welcome:

    Yes. Light intensity determines the current when the frquency of the light is high enough. Google "photoelectric effect current vs voltage" to see pictures.
     
  4. Dec 24, 2015 #3

    ZapperZ

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    Yes, you found the answer to your question. The rate of emission depends on the intensity (number of photons per second) of light. So by the time you reach saturation, you are capturing all the electrons emitted.

    When I measure the quantum efficiency of photocathodes, I always make sure that this is done within the saturation voltage. However, I also make sure that the gradient isn't so high that the Schottky effect becomes no longer negligible that it significantly alters the effective work function.

    Zz.
     
    Last edited: Dec 24, 2015
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