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Classical Explanation for Photoelectric Effect

  1. Sep 27, 2008 #1
    Hello.

    I have a question regarding photoelectric effect. My textbook says that when photoelectric effect experiment was first performed, physicists could not explain the outcome with classical physics (e.g. maximum kinetic energy depends on frequency of light, not intensity). My question is, wasn't there even one thing that could be explained with classical physics about the results of this experiment? If anyone knows there was, please explain it to me.

    Thank you in advance.
     
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  3. Sep 27, 2008 #2

    mgb_phys

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    Classically you can explain light knocking out an electron,what you can't explain is how you need just one photon of high enough energy to do the job when billions of slightly lower energy put together can't.
     
  4. Sep 27, 2008 #3

    atyy

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    The original explanation was that light consisted of discrete particles called photons.

    This turns out not to be necessary, and it is possible that it is just the energy levels in the atoms that are discrete.

    So classical physics cannot explain the experiment, but you can choose to put the non-classicality in the atoms or the light.

    However, the photon concept turns out to be necessary for many other phenomena, so it is still useful to regard photons as a correct explanation of the photoelectric effect.
     
  5. Sep 27, 2008 #4
    I see. So only the thing they could explained with classical physics was that light could knock out electrons out of matter under some circumstances. Thank you very much for the very concise answer!

    I see, so without photon theory explanation is very tough. Thank you!
     
  6. Sep 27, 2008 #5
    Actually, would you mind explaining to me why "billions of slightly lower enegergy put together" cannot cause electrons to leave matter?
     
  7. Sep 27, 2008 #6

    ZapperZ

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    You may want to read the intro in J.J. Thorn et al. paper and references therein. You can get a copy of it here:

    http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

    The standard photoelectric effect strongly points in favor of the photon picture, but cannot rule out light being a classical field. It is other phenomena (photon antibunching, multiphoton photoemission, which-way experiments, etc.) that would make the convincing evidence for the photon picture.

    I'm a bit hesitant to point this out since it may complicate things beyond what you've asked for and what you wish to know, but what the hey... The Einstein photoelectric effect, at least the naive version of it, can actually be violated, even when using the photon picture. I'm mentioning this because of the point being made that "... billions of slightly lower energy put together..." cannot produce a photoelectron. This is only true for the single-photon photoemission, which is what the standard photoelectric effect is. We have plenty of observations of multiphoton photoemission using photons with energy LESS than the work function.

    Zz.
     
    Last edited: Sep 27, 2008
  8. Sep 27, 2008 #7

    atyy

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    I remember asking my high school teacher about 20 years ago about two-photon effects in the photoelectric effect, and was told that if the intensity is high enough you see a second peak, exactly as expected (she didn't tell me about selection rules, but still a pretty good answer I think). More recently I read that Einstein himself mentioned two-photon effects in his paper. I haven't read the paper myself, and I can't quite remember what the source is ...
     
  9. Sep 27, 2008 #8
    >ZapperZ, atyy
    I see. So if the light intensity is high enough, an electron can obtain enough energy from multiple photons with energy lower than work function before the energy decays. It is very clear now, thank you so much!
     
  10. Sep 27, 2008 #9

    mgb_phys

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    Not an expert but I thought you only have a 2photon photo electric effect in rather non-classical setups like nano-particles a surface plasmon or some metastable state.

    I don't know that you can hit a normal electron with two photons and eject it.
     
  11. Sep 27, 2008 #10

    Redbelly98

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  12. May 25, 2009 #11
    the photoelectric effect experiment has been done with extremely low light intensities. so low that it should have taken a single atom minutes to absorb enough energy to emit an electron. instead there is absolutely no noticable delay before electrons begin to be emitted by the metal. this does not rule out the possibility that light is a wave but of so would require that some of the atoms somehow start off with some random amount of energy already within them.
     
    Last edited: May 25, 2009
  13. May 25, 2009 #12

    malawi_glenn

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    isn't this a pretty old thread?
     
  14. May 25, 2009 #13
    you want me to start a new one?
     
  15. May 25, 2009 #14

    malawi_glenn

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    Well the discussion about it seemed quite dead, OP's last activity was december 2008.
     
  16. May 25, 2009 #15
    To clarify: Are you just arguing this proves the energy is concentrated at points that are randomly distributed over the beam's cross-section, rather than being uniformly spread over the beam cross-section (whilst making the assumption that the electron can only absorb power from its immediate vicinity and not the entire beam cross-section)? Or that it proves the energy is also longitudinally concentrated (in which case I take it the average delay matches a classical prediction but for some runs of the experiment the delay is longer or shorter)?

    Regards your concession: wouldn't the frequency dependence (the total non-response to any duration of redder light, and the increasing excess kinetic energy with bluer light), etc, disprove the hypothesis of the atoms having random pre-accumulated quantities of energy?

    Regards the competing model: is this behaviour truly not also predicted by the semi-classical approach of considering a quantum mechanical atom in a classical electromagnetic field?
     
  17. May 25, 2009 #16
    The energy of oxidation of a simple sugar molecule C6H12O6 is about 29 eV (electron volts). Visible light is about 2 to 3 eV. How does photosynthesis create a sugar molecule from CO2 and H2O when visible light photons have so little energy?
     
  18. May 25, 2009 #17

    mgb_phys

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    The photoelectric effect applies to promoting electrons in atoms, not breaking bonds in molecules.
    Photosynthesis involves a lot of intermediate states created by enzymes so the changes in energy at each step are fairly low. From what I remember of the Krebbs cycle the electrons are provided by the hydrogen, they aren't ejected directly from the atoms.
     
  19. May 25, 2009 #18

    alxm

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    Well, the reaction immediately catalyzed by Photosystem II is splitting water into molecular oxygen:
    2 H2O --> O2 + 4 H+ + 4 e-
    But the question still stands, since that reaction requires more energy than a single photon provides as well; it takes four.

    What happens in the first step, is that light is absorbed by the chlorophyll, which ejects an electron that goes off to reduce plastoquinone somewhere else*. So the photoelectric effect is in fact somewhat relevant here, in that the absorption of light leads to a charge separation:
    Chlorophyll + photon --> Chlorophyll+ + e-
    Although unlike the classical photoelectric effect, the electron isn't freed entirely - there's not enough energy for that - it moves to another part of the enzyme, but it remains in a bound state.

    This photo-oxidation of chlorophyll then leads to it being reduced back to its original state by a tyrosine amino-acid residue, which in turn is reduced by something else in a chain that ends over at another part of the enzyme, where there's a cluster of four manganese ions (and a calcium ion, although it probably doesn't take part). So one of the manganese ions gets reduced (either Mn(III) -> Mn(IV) or Mn(IV)->Mn(V), the exact details of the reaction are not entirely known). This is where the actual chemical reaction occurs.

    This whole procedure occurs four times, until the manganese cluster is oxidized 'enough' to form the energetically-expensive O-O bond, producing hydrogen ions and re-reducing the manganese ions in the process. (A simplification of course - the reaction doesn't occur in a single step, but the O-O bond formation is the 'biggest' one)

    The hydrogen ions (aka protons) form a concentration gradient across the chloroplast membrane, which is then used to power ATP-synthase, which powers most of everything else. Just the same as in our mitochondrial membrane.

    (* The energy of the reduced plastoquinone (=plastoquinol) is extracted elsewhere, which also ends up producing protons for the gradient.)
     
  20. May 26, 2009 #19

    Vanadium 50

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    Reference, please?
     
  21. May 26, 2009 #20

    malawi_glenn

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    pick up your chemistry book?
     
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