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I Photoelectric absorption versus Compton scattering

  1. Feb 15, 2017 #1
    During the photoelectric process, photons are absorbed and electrons are ejected. In Compton scattering, photons are scattered rather than absorbed. My textbook (Quantum Physics of Atoms,Molecules,Solids,Nuclei, and Particles) explains that absorption occurs in the photoelectric process because the electrons are bound to atoms and a truly free electron cannot absorb a photon and conserve energy. So far so good. No questions. However, during compton scattering, there are two peaks of intensity. One peak (shifted wavelength) corresponds to the photons which struck free electrons and lost a bit of energy. The other peak (unshifted wavelength) corresponds to photons which actually hit more tightly bound inner electrons. So these photons are hitting BOUND electrons just as in the photoelectric effect and yet they are being scattered. So to summarize, I was told that photons in the photoelectric process are absorbed not scattered because the electrons are bound not free and yet here is this example of a photon hitting a bound electron and being scattered not absorbed?
     
  2. jcsd
  3. Feb 15, 2017 #2
    If the photon is of the right wavelength it can excite the bound electron to a higher energy level and be absorbed. If not, it can still be scattered by it.
     
  4. Feb 16, 2017 #3
    I understand this. So what can be said about the atoms of the target material used in the photoelectric effect versus the atoms of the target material during x-ray scattering? Is it just a coincidence that the atoms of the target material in the photoelectric effect had just the right electron configuration so as to absorb the incident photons and the atoms in the target material in x-ray scattering experiments had just the right electron configuration so that when the x-ray photons hit the tightly bound electrons they are scattered? Why is it being stressed to me that the big difference is that in x-ray scattering the electrons are more or less free and that is why the photons are scattered in contrast to the bound electrons of the photoelectric process?
     
  5. Feb 16, 2017 #4
    I would say it is because the X-Ray's have higher energy. Photoelectric effect transition energy frequencies are more towards the visible range, but with X-Ray's if electrons were not free to start with they would be afterwards!
     
  6. Feb 16, 2017 #5

    ZapperZ

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    There is a whole lot of confusion in this thread from what I have read.

    1. The standard photoelectric effect is done typically on METAL and using photons up to UV range. Thus, the photoelectrons are emitted from the conduction band electrons, not from core-level electrons. These conduction band electrons are bound to the solid but NOT to any individual atoms. So it does not exhibit the discrete atomic energy spectrum, but rather the continuous conduction band spectrum.

    2. x-ray sources CAN induce photoemission. This technique is what is used in x-ray photoemission spectroscopy (XPS). In this case, the photoelectrons may come from core level electrons, i.e. not electrons from the conduction band or valence band of the solid, but rather from the ones that are tied to the atoms/molecules.

    3. The "crossover" between dominant photoemission versus dominant compton scattering depends very much on the material involved. It is why light detection in PET scans can involve more than two different ranges of photon energies. Both types of phenomena are possible in any photon-matter interaction. It is a matter of which one dominates at any given range. To dive into the details of this, one has to look at the cross-section for each process, and this may not be that trivial.

    Zz.
     
  7. Feb 16, 2017 #6
    Thank you. These processes were explained to me in the context of an introductory quantum class and many of the details of the experiments were brushed over because our primary focus is on the necessity of the quantum nature of light to explain apparent paradoxes in the results. However, I would really like to have a deeper understanding of the details of the experiments.
     
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