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B Effect of frequency and intensity on PE

  1. Mar 26, 2017 #1
    if two light with different frequency have the same intensity then the PE current is the same. But that doesn't make sense, and some graphs show that if you increase frequency while intensity is constant, the current increases up to a point then drops.

    A couple of old threads, people mentioned that as you increase frequency if intensity is constant then the current goes down (Less photons)....

    but don't you have a high probability of a photon succeeding in ejecting an electron?

    Which one is right?

    Thank you in advance.
  2. jcsd
  3. Mar 26, 2017 #2


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    There are a number of different things going on here in those different threads that you may be referring to (it would have been clearer if you made specific links and references here).

    Let's make sure we clear up certain terminologies here.

    Each photon carries an energy hf.

    We can define "intensity" for this case as the number of photons per second. This may not be true in all applications, because there are other instances where "intensity" includes the energy as well. But for this purpose, let's stick with number of photons per second.

    Now, if you have a light source with a fixed power (which is usually the case), then if you vary the frequency, the intensity also changes because the power has to remain constant. So if the frequency goes up (so the energy per photon goes up), then the intensity has to drop due to the constant power from the light source. So this is the origin for a lower number of photocurrent being emitted, because the light source is producing less photons per second.

    Let's now have the situation where the intensity remains the same, no matter the frequency. In real life situation, you will not get the same number of photoelectrons being emitted. This is because there is something called the quantum efficiency (QE) of a material that is dependent on the energy of the photons. QE is the number of photoelectrons emitted per photon. It is a statistical concept. Most metals have very low QE, as in 10-4%-10-2%, meaning that it takes hundreds to thousands of photons to emit one photoelectrons.

    Your original guess is correct, in that the higher the energy, the more likely is the possibility of an emission. So in general, QE goes higher with a higher photon energy. However, this is not true in all cases, as in for all materials. Because of the mechanism of photoemission, it depends very much of not only the density of states of the occupied states, but also the density of states of the unoccupied states. So the band structure of the material plays a role in the QE of the material. This leads to in some material having a drop in QE at some point as the frequency of the photon increases.

    BTW, you need to also pay close attention to how these photocurrent versus frequency data are done. In many instances, the drop is often due to the transmission cut-off of the window that the photon passes through.

  4. Mar 27, 2017 #3
    Thank you for your reply.

    I thought that the drop happened because when you increase the frequency, the number of photons decreases and the total result(QE + decreasing number of photons) will be decreasing the current. Sorry I don't know the band structure but I will look for it

    The thing is that in my book and exams they use this graph
    Where you have different frequency but same intensity(Not sure if they mean power or number of photons). So If they mean the same number of photons then the highest frequency will have the highest current.. If they mean as power then it depends on how high the frequency is.

    Here is the graph of Current vs wavelength ( The only one I found):

    And this is one of the threads I was talking about:

    The last answer was that intensity means power per area (which is the definition we took) and if you increase frequency the number of photons decreases so current decreases but that assumes 100% efficiency and that every photons removes exactly one electron

    And if you consider that intensity means the # of photons then if you assume that the efficiency is 100% then regardless of the frequency the current will be the same, Is that is what they are implying in my book and exams? It is really confusing.
  5. Mar 27, 2017 #4


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    I wish you would have cited the exact problem that you're having with your book. It would have been a lot more direct for me to address the specific issue, rather than wasted my time delving into something you don't understand or don't need.

    But I've explained this already when I talked about the constant power of the light source. Did you miss it?

    First of all, there is a difference between asking about the photoelectric effect, versus the EXPERIMENT of doing the photoelectric effect.

    When a photoelectron is emitted, it is usually emitted in many different directions, often spanning the whole available angles. This means that there will be those that will not get to the anode, and thus, will be be counted as part of the current being measured. So what you end up doing is applying a positive potential bias to the anode. As the potential increases, the current that you detect increases until you get to the point where you are collecting ALL of the emitted electrons. This is the start of the plateau in the current. Beyond this, the current won't increase anymore.

    That graph has been normalized to the power for each frequency. That is why they all have the same saturation current.

    Again, for that second graph, unless there is a proper source for it, the explanation that I've given in my previous post applies. So I am not sure why you're asking it again here, unless you either missed it, or you didn't understand it.

  6. Mar 27, 2017 #5

    My exact problem is the first graph with the stopping potentials.

    I absolutely didn't, Not sure why always my wording seems to imply that. I was talking about something else. I didn't know that QE dropped as you increase the frequency. I thought that it increases ( which is wrong) but the number of photons decreases so it results in negative changeso nvm.

    How come what you said above doesn't apply? Why do they have the same current in the end if they have the same power but different frequencies?
    They must have meant same number of photons and 1 photon - 1 electron This way frequency won't matter

    I didn't ask about it. I brought it because it contradicts the first one.
  7. Mar 27, 2017 #6


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    I have no idea what's going on here.

  8. Mar 27, 2017 #7


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    I think we're dealing with an experiment with more than one input variable here - plus the definitions are a bit shaky. That's a recipe for misunderstandings.
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