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Photoelectric wavefunction

  1. Jun 24, 2005 #1
    When a photon with the right frequency hits the electron of an atom. At what portion does the wave function of the atom collapse. Is it when the photon enters the atomic space or when the photon hit the potential electron probability location?
  2. jcsd
  3. Jun 25, 2005 #2


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    The problem with your question is that it is based on a lot of inaccuracy of what a "photoelectric effect" is and quite a bit on quantum mechanics.

    I'll say this again. The standard photoelectric effect is a process of light hitting on a SOLID surface, typically a metal. In metal, the majority of the properties of the material is due to the CONGLOMERATION of ALL the atoms in such a way that, because they are so close together, they lose their individuality. This then forms various energy bands, and the conduction band is one example. The electrons in the conduction band does NOT belong to any particular atom. So you do not have an "electron stuck to the atom" scenario anymore when you deal with, for example, the typical photoelectric effect. So if you don't learn anything else from this posting, you should at least make sure you understand this. Photoelectric effect is NOT equal to photoionization of atoms.

    Secondly, when you deal with "one photon being absorbed by the metal conduction band", one no longer deals with 'wavefunction'. It is just too cumbersome. When you study QM a little bit more, you will come across what is known as Second Quantization. This is a different way to do QM, and it is more efficient in dealing with processes in physics, especially in condensed matter physics (where photoelectric effect/photoemission belongs). You may want to look into this first (if you haven't) and then see how it treats the photoelectric effect.

    Thirdly, and this is something a typical photoelectric effect process doesn't cover, the photoelectric effect, as we have found out, has MORE to do with the material properties. So in fact, you have to also know the QM description of the material (i.e. those "bands") to be able to consistently explain this phenomenon. So you may have to deal with "wavefunctions", but it's the wavefunction of the material, rather than the photon, that is more important.

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