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How does the Electron get knocked off by a photon

  1. Jul 14, 2009 #1
    Hi

    I was just thinking, but i am a little confused as to how the electron actually gets "knocked off " when it strikes the surface of a metal.

    Are we talking in the classical sense of billiard balls which are arranged in an ad hoc fashion or are we talking in the sense of some wave function embedded in the metal.

    Now why i ask this question is that, if we are talking about billiard ball arrangement then is it a random chance that the electrons are actually released, assuming that its the valence bad electrons we actually look at, there would not be that many electrons and at some point they would be depeleted, this obviously is a very classical sense theory, but i am looking for a quantum perspective.

    I thought it might be akin to resonance causing electrons of a particular amount of frequency actually getting released

    I am confused though
     
  2. jcsd
  3. Jul 14, 2009 #2

    diazona

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    Your resonance idea is closer to the truth. Basically an electron is bound to its atom (or in the valence band, bound to all atoms in the metal) and in order to be "knocked off," it needs to get a certain amount of energy. Enough energy to overcome the force binding it to the atom(s). When a photon with that amount of energy or more comes along, there is some chance that it will transfer its energy to the electron, thus freeing the electron. In quantum terms one would say that the electron gets promoted from a bound state to a scattering state (well maybe "free state" is a more intuitive term in this case).
     
  4. Jul 14, 2009 #3
    I havent quite understood thee transferrring part, how does it get transferred, that was my question.

    Whats a scattering state?
     
  5. Jul 14, 2009 #4
    The very simplest physical model which gives a realistic picture of the process is probably the one-dimensional potential well. The finite well, not the infinite well...a well just deep enough so that the first level is bound but the second level is free. The "tails" of the first level are exponentially decaying outside the well, but the "tails" of the second level oscillate to infinity outside the well. If you take the superposition of these two states and evalutate the charge distribution, you will see there is an oscillating dipole moment. It is this dipole moment which interacts with the incoming electromagnetic wave. At just the right frequency that wave will drive the electron from the ground state to the first excited state, where it has a chance of being detected outside the metal.
     
  6. Jul 14, 2009 #5

    Fredrik

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    There's some good stuff in this thread. See the posts by Civilized. My post #20 contains a link to the text he's talking about.
     
  7. Jul 14, 2009 #6

    ZapperZ

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    I don't think that helps, other than simply adding another layer of confusion to the issue. That issue has been discussed so many times before, especially within the context of "simple photoelectric effect" versus the more complex "photoemission" process involved with ARPES, RPES, etc, all of which have never been explained using "waves".

    Now, since we have decided to muddle this a bit rather than giving a straightforward answer to the OP, I feel that I must then step in and clarify the photoemission process in a metal. Using the standard Spicer's "3-step" process, the photoemission from a metallic surface can be grouped as such:

    1. absorption of photon
    2. excitation of a conduction band electron to an energy state above the vacuum energy
    3. diffusion of that electron until it escapes the bulk of the material.

    #3 implies that not all electrons that have been excited to above the vacuum energy will always escape the bulk. This is why we do not have a material with 100% quantum efficiency (yet). In many cases, the electron can collide with other conduction electrons and lose energy, or that the collisions will cause it to move even deeper into the bulk until it loses the energy it gained. This is why in semiconductors, and applying a band-bending technique resulting in a negative electron affinity semiconductor, we can get more electrons escaping the bulk.

    Having said that, it appears that the OP is more interested in how a system absorbs a photon and excite an electron. This issue is not confined to just the photoelectric effect, but also in atomic excitation. I believe this has been addressed already in at least a couple of posts. If that is the case, this isn't necessarily dealing with the photoelectric effect, which involves other processes as one can see beyond just photon absorption/excitation.

    Zz.
     
  8. Jul 14, 2009 #7
    I think the interesting thing to note here is how many things you can understand about a metal by treating it as a potential well.
     
  9. Jul 15, 2009 #8
    Ok let me get this question a little more organised

    1. Firstly how does the electron get excited at all?
    2. In the photoelectric phenom(which is the most simple i can think of) how do we rationalise an electron getting knocked out

    Essentially the 2nd part is the corollary to the first, why actually posted wrt photoelectric effect is to understand this question from an application perspective before actually plunging into the theory of it.


    Ok i have a little question here about the potential well, why do we call the metal electrons in the valence band to be a potential well, is it because the metal is an equipotential surface. Now we do obviously see that the metal is charged when connected a circuit and has photons bombarding it?

    Am i getting the fundae of potential well wrong here?
     
  10. Jul 15, 2009 #9

    ZapperZ

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    You have had several responses already addressing this. I'm not sure to what extent do you want your "how" to be answered. I also do not understand what you mean by "rationalise". If you are asking for the mechanism of photoemission, I believe that I've described it via the Spicer's 3-step model (or even the more "modern" one-step model). Those of us who do photoemission spectroscopy or study photocathodes apply these models extensively.

    Zz.
     
    Last edited: Jul 15, 2009
  11. Jul 15, 2009 #10

    ghc

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    The photon and the electron collide like billiard balls. If the photon doesn't have enough energy, it will "bounce" (it's called diffusion). If the photon has sufficient energy, the electron will be "extracted" from the metal. But this collision is described in terms of probabilities (wave functions) and we have to deal with these probabilities with the quantum mechanics rules. Why? Because in the real billiard balls case the balls are "visible" (can be detected without disturbing the collision) while in the photon-electron collision the particles can't be "seen" when moving or colliding without disturbing.
     
  12. Jul 15, 2009 #11
    Actually, the billiard ball model is especially inappropriate for the photo-electric effect. With billiard balls, the smaller, faster particle bounces of the larger heavier one without much transfer of energy at all. If they were billiard balls, the electron could never abosrb close to 100% of the photon energy.
     
  13. Jul 16, 2009 #12
    Rationalise i mean how do we countermand the classical billiard ball model.

    Also what exactly is absorption of energy is yet to be answered, as in what do we mean when we say energy is absorbed?
     
  14. Jul 16, 2009 #13

    ZapperZ

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    "countermand"? "classical billiard ball"?

    We rationalize it by pointing to the experimental evidence AND using quantum mechanics! This is now no longer a question of photoelectric effect, but rather a question on how QM describes our universe.

    We mean "photon comes in, no photon comes out". That's absorption.

    Zz.
     
  15. Jul 16, 2009 #14
     
  16. Jul 16, 2009 #15

    ZapperZ

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    That doesn't make any sense, does it? I mean, if you have a 5 eV photon, how do you "convert" that into an electron that has a "rest mass energy" of 0.511 MeV?

    Maybe you need to re-read all the posts in this thread from the beginning. There have been at least a couple of descriptions (albeit classical visualization) of the absorption process using the analogy of resonance. So please describe why you are still having a problem with that description.

    So why do we still hang on to this then?

    A QM picture is nothing like that. So which part of the QM picture of the physics do you have a problem with?

    Again, none of these are actually about the photoelectric effect. Rather, this is general Quantum Mechanics.

    Zz.
     
  17. Jul 16, 2009 #16

    jtbell

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    Also, the photon has no electric charge, whereas the electron does.
     
  18. Jul 23, 2009 #17
    does an antineutrino have a charge?
     
  19. Jul 23, 2009 #18

    Hootenanny

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    No. And what does that have to do with this thread?
     
  20. Jul 24, 2009 #19
    To Zapper Z

    Why are you suggesting that a photon has an electric charge and it transfers this charge over to the electron. This does not make sense. A photon has no charge and no mass as it is a bundle of energy. Shouldn't you mean energy (j) and not electron volts (eV).
     
  21. Jul 24, 2009 #20
    I find this very rude Hootenanny as it was me who initially created this thread. i was simply adding on to this thread by asking another question!!!!!
     
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