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Energy transfer from photon to an electron - Compton vs Photoelectric

  1. Jun 4, 2012 #1
    Hello,

    I have a problem with the two seemingly conflicting descriptions of the energy transfer from a photon to an electron I found in my textbook.

    The first one appears in the description of the photoelectric effect:
    "In Einstein's picture, an individual photon arriving at the surface [of the material] is absorbed by a single electron. This energy transfer is an all-or-nothing process, in contrast to the continuous transfer of energy in the wave theory of light; the electron gets all of the photon's energy or none at all."

    I cannot understand how this isn't inconsistent with Compton scattering:
    "The incident photon would give up part of its energy and momentum to the electron [...].The scattered photon can fly off at a variety of angles with respect to the incident direction, but it has less energy and less momentum than the incident photon."

    Could anyone explain how these two are not in contradiction?
    Thanks!
     
  2. jcsd
  3. Jun 4, 2012 #2
    There are three primary gamma ray interactions:
    1. Photoelectric absorption - The target atoms completely absorbs the gamma ray and emits and electron.
    2. Compton Scattering - The elastic scattering of a gamma photon from an electron.
    3. Pair production - An energetic gamma ray produces an electron-positron pair.

    The predominant mechanisms vary with the atomic number Z of the target material and energy of the incident gamma photon.
    At a given Z, the probability of photoelectric absorption is highest for lower energies. Likewise, the probability of pair production is highest for high energy photons. A gamma photon must possess more than 1.02 MeV of energy for pair production to occur since the rest mass of an electron is 0.511 MeV. Compton scattering occurs for the in-between energies.

    In all, they are two different interactions so there is no contradiction.
     
  4. Jun 4, 2012 #3

    jtbell

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    Staff: Mentor

    In Compton scattering, the electron is considered as basically "free" and not interacting with anything else but the scattering photon. This is actually an idealization, because the electron is usually part of an atom. Nevertheless, it's a good approximation because the energy of an X-ray (on the order of 104 eV) or a gamma-ray (on the order of 105 or 106eV) is much larger than the binding energy of an "outer" electron (a few eV).

    In the photoelectric effect with light, the photon energy is very small (a few eV). The interaction of the electron with the atom that it's part of, and the crystal or whatever that the atom is part of, comes into play.

    So generally, for low-energy photons you get the photoelectric effect, and for higher-energy photons you get the Compton effect. But there's not a sharp boundary, because these are quantum-mechanical processes. You have probabilities instead. If you have gamma-rays with an energy of say 662 keV impinging on a certain material, e.g. a NaI(Tl) scintillator crystal, you have a certain probability for each process, and you can actually observe the effects of both processes in the output from the scintillator.

    The higher the energy, the smaller the probability of getting the photoelectric effect, and the higher the probability of getting the Compton effect.

    [added: and with high enough energy as EulersFormula noted, pair-production also comes into play.]
     
  5. Jun 4, 2012 #4
    Now I get it, they didn't make that distinction in my textbook. Thanks for your explanation!
     
  6. Jun 4, 2012 #5
    Is that somehow related to limited values of energy an electron can absorb? (I know that it's true when its bounded, but when it escapes its energy stored as kinetic doesn't have to be discrete, right?)
     
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