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Experiment involving photoelectric effect

  1. Jan 29, 2009 #1
    Reading through the lecture notes, I had a weird idea which came in the form of an experiment that could be done.

    Imagine you shine light through two slits. Obviously you will get an interference pattern with bright and dark lines (constructive and destructive interference). Then on the other side of the slit, you have a shiny metal surface where you view the interference pattern. Not sure about this but would you be able to see an interference pattern on a shiny metal surface? The question is if you could, would you still be able to say that the particle theory of light (photoelectric effect), which states that light is composed of individual packets called photons which each carry a certain energy E= hf, holds true?


    When you try and detect the current flowing through the metal, i am guessing that current still flows otherwise by now quantum theory would be in a big mess haha. But I fail to understand how they could get out of this experiment. Anyone care to explain?

    Cheers
     
  2. jcsd
  3. Jan 29, 2009 #2
    Current in metal no good when interfere because lines too close. Need heat to get electrons off metal.
     
  4. Jan 29, 2009 #3

    ZapperZ

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    I am not sure what you are trying to get at here. Why would you want to "view" the interference effect using the photoelectric effect? Just to show that light can still have particle properties? Just because this light came from a 2-slit experiment makes no difference. It is still a light, and you can do to it the same thing you can do to any light. How were you to know the intricate history of the light that was used to go through the 2-slit in the first place? Would that make any difference if I used a synchrotron source, or an arc lamp? Those two used completely different principles to generate light.

    Please note that the spectrum of photoelectrons emitted from a photoemission process isn't trivial, and the emitted electrons can go in many different directions. You cannot "view" this the same way you view those interference pattern.

    Zz.
     
  5. Jan 29, 2009 #4
    I am suggesting that if the wave theory of light explains the interference pattern and you do what I have told above, why does the classical theory of light not work. In other words, an increasing intensity of light should increase the current flow but it does not. I used the double slit in forcing light to be a wave and hit the metal as a wave or does it somehow change back into particle form?

    I thought light interacts as a wave in some situations and as a particle in others?
     
  6. Jan 29, 2009 #5

    ZapperZ

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    Er... there is only ONE description of light within the QM formalism. Read the FAQ in the General Physics forum. Using the particle picture, one can also arrive at the interference pattern.

    "wave theory" IS the classical theory of light. And increasing the intensity DOES increase the current flow in a photoelectric effect - you get more electrons out since there are more photons per unit time per unit area.

    I think there's a serious misunderstanding of not only the photoelectric effect, but basic quantum physics here.

    Zz.
     
  7. Jan 29, 2009 #6
    Oh so there is only one unified description of light according to quantum mechanics. I get that but how does quantum mechanics achieve that at least through the Schrodinger equation. How do you unify something that displays both wave like nature and particle like behaviour?
     
  8. Jan 30, 2009 #7

    ZapperZ

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    Read the Marcella paper that I've mentioned several times on here. He derived all the interferences effects not using waves, but purely using QM.

    http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

    Zz.
     
  9. Jan 31, 2009 #8
  10. Jan 31, 2009 #9

    Redbelly98

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    That's just what standard quantum mechanics does. However, a lot of physicists do object to the term "particle" for light, and prefer instead to talk about energy quanta or packets of energy.

    There is a difference between saying that the amount of energy is present in discrete units (integer multiples of hf), vs. calling something a particle.

    My understanding is that it behaves as both, all the time. (But see my comment above on using the term "particle".)
     
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