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Photoelectric effect

  1. Jan 28, 2008 #1
    Is in the photoelectric effect, the electrons are oppositing the electromagnetic field or they are excited from the energy?
     
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  3. Jan 28, 2008 #2

    mathman

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    I don't understand your question. However, the photoelectric effect involves an electron absorbing a photon and, being more energetic, leaving the atom.
     
  4. Jan 29, 2008 #3
    And do ultraviolet radiation pass to the metal, or just visible light?
     
  5. Jan 29, 2008 #4

    ZapperZ

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    You really need to put a bit more effort in asking your question here, expecially if you want others to put effort into responding. If not, you will continue to get responses from other people in the form of "HUH?", and you will have to keep on explaining yourself.

    UV radiation can cause photoemission if the work function is below the photon energy. Now what is it exactly that you want to know here in this thread?

    Zz.
     
  6. Jan 29, 2008 #5
    Ok, sorry. I just want to know, if UV radiation passes through the glass bulb, the glass will absorb that energy. So what is causing the electrons to flow? UV radiation or visible light, or maybe something else?
     
  7. Jan 29, 2008 #6

    ZapperZ

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    This is getting utterly confusing. You WERE asking about the photoelectric effect, weren't you? Why is it optical conductivity now?

    Glass bulb? When did that come in? Did you mention about glass anywhere till now? There's no "photoemission" using UV source on glass bulb, is there?

    Why don't you start from the very beginning and ask your question once again. But this time, please put in as much effort and information in your question so that we know (i) what you know (ii) what exact it is the picture that you have in your mind. If you see that you have to ADD new stuff as you go along, it means clearly that you omitted important information in your original question.

    Zz.
     
  8. Jan 29, 2008 #7
    Sorry for being rude, but have you ever heard about photoelectric tube or photoelectric cell? There is glass balloon or bulb or whatever... Inside there is vacuum, so it is called vacuum photocell. There are anode and cathode. So when I bring some kind of radiation to some metal, the electrons are excited and pulled off the cathode. What is that kind of radiation? UV or visible light? Also I want to know why the anode and cathode are put in vacuum photocell?
     
    Last edited: Jan 29, 2008
  9. Jan 29, 2008 #8

    mathman

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    The vacuum is used so the electrons can stream without interacting with anything. The radiation is always electromagnetic, as long as the frequency is high enough to excite the electrons of the material being used.
     
  10. Jan 29, 2008 #9

    ZapperZ

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    Not only have I heard of them, I've used them!

    What kind of radiation? Any radiation in which the photon energy is higher than the work function! I thought I mentioned this already.

    Electrons do not travel very far in air, and even if they do, a lot of them get scattered off and would not reach the anode. This is not what you want when you are using a photocell to detect EM radiation. If you are trying to detect UV, the "glass" has to be either quartz or fused silica, because ordinary glass absorbs UV.

    Zz.
     
  11. Jan 30, 2008 #10
    Ok, thanks. I understand now. I have two more questions. Why the cathode and anode are in vacuum balloon instead connected them with wire? Is this picture correct?
     
  12. Jan 30, 2008 #11

    ZapperZ

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    If they are connected to a wire, they would be the SAME thing, not two different things. The anode has to be at a higher potential then the cathode, the latter is usually grounded.

    Zz.
     
  13. Jan 30, 2008 #12
    And why in my text book says, that the kinetic energy of the electrons doesn't depends from the intensity of the radiation?
     
  14. Jan 30, 2008 #13

    ZapperZ

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    Because "intensity" only increases the number of photons, not the energy within each photon.

    Your textbook doesn't explain this? Please read for example, this:

    http://www.colorado.edu/physics/2000/quantumzone/photoelectric.html

    Zz.
     
  15. Jan 30, 2008 #14
    And what makes the current stronger? The more electrons ejected, or the speed of the electrons?
     
  16. Jan 30, 2008 #15

    ZapperZ

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    <scratching head>

    This is not clear?

    Zz.
     
  17. Jan 30, 2008 #16
    Yes, I understand that with the intensity. Thank you very much. There is one more thing in my text book. It says that the current is much stronger when I increase the intensity. So that means that more electrons are ejected and the current is getting stronger, right?
     
  18. Jan 30, 2008 #17

    Hootenanny

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    Correct.
     
  19. Feb 1, 2008 #18
    I want to ask you what happens with bar code readers, when the light is touching the bar code (but the white parts of the bar code), do they reflect in the photodiode of the bar code reader?
     
  20. Feb 1, 2008 #19

    dst

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    Yes. exactly so.
     
  21. Feb 1, 2008 #20
    Can you recognize this? Do you know what happens, when the light will touch the white parts? Will be sound produced? Shouldn't be black color instead of white?
     
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