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Compton vs Photoelectric effect?

  1. Sep 13, 2006 #1
    What is the main differences? Are they essentially the same thing?

    From what I have gathered, the two are fundalmentally the same. If a photon gives up all its energy after interaction with an electron than the photoelectric effect applies.
    On the other hand if the photon continues to travel and possess kinetic energy after collision with an electron than the Compton effect takes place.

    Also it seems the photoelectric effect happens mainly with electrons in metals whereas the Compton effect is with any material.
     
  2. jcsd
  3. Sep 13, 2006 #2

    Meir Achuz

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    They are completely different.

    In CE, a photon is elastically scattered by a charge which recoils.
    Conservation of energy and momentum results in a change in the energy and momentum of the photon (usually measured as a change in wave length).

    In PE, a photon is completely absorbed by a solid, and an electron is ejected in the process.
     
  4. Sep 13, 2006 #3
    What determines whether CE or PE occurs? Why is it that sometimes, an electron is able to absorb all of the photon's energy (PE) and sometimes, it just gets pushed around like a "ball collision" (CE)?

    I'll have a guess. Is it because PE occurs only if the photon matches a transition state in the atom? Hence the electron is able to completely absorb the photon's energy. When it doesn't occur (i.e. energies don't match) than you get CE?
     
  5. Sep 13, 2006 #4

    ZapperZ

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    There are no more "atoms" in a solid. The energy bands that you get in a solid such as a metal is the collective effect of all the atoms in such a way that you no longer get the individual atomic behavior. That is why you get band of energy, rather than discrete states of atoms.

    Photoelectric effect will only occur in solids due to the requirement of longitudinal momentum conservatoin. Compton effect occurs when a photon encounter a "free" particle or atoms. It can occur, in a limited sense, in solids, but the effect isn't dominant when compared with other phenomena that can occur.

    Zz.
     
  6. Sep 13, 2006 #5

    Astronuc

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    If one looks at the cross-sections of PE and CE, one finds that PE dominates at lower energies. As gamma energy increases, the probability of Compton scattering increases.

    PE and CE occur in solid, liquid or gaseous states of any element. The reaction rate increases with electron density, which is a function of the element and state (solid, liquid or gas).
     
  7. Sep 13, 2006 #6

    ZapperZ

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    The standard Photoelectric effect doesn't actually occur in gasses. That is more commonly called photoionization. This is different than photoelectric effect because you don't have a continuous energy bands, and you also don't have a "work function".

    Zz.
     
  8. Sep 13, 2006 #7

    Astronuc

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    It maybe that nuclear engineers are abusing the term 'photoelectric'. I learned PE means the electron absorbs the photon completely, which is also photoionziation. I think we (nuclear engineers and radiation physicists) misappopriated the term. :rofl:

    I was wondering about the distinction between the classic 'photoelectric' effect where if the photon (on order of a few eV's) strikes a metal surface resulting in the ejection of a photo-electron, and the interaction of a keV X-ray or gamma-ray which strikes any atomic electron and is completely absorbed. For nuclear engineers and health physicists, there is not distinction. If a photon is completely absorbed, then we called it 'photoelectric effect' and the ejected electron is a photo-electron.
     
    Last edited: Sep 13, 2006
  9. Sep 13, 2006 #8

    jtbell

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    That depends of course on the energy of the photon. In one of our undergraduate labs, students send 662-keV gammas from Cs-137 into a NaI scintilator and do a pulse-height analysis of the spectrum that comes out of the scintillator's photodetector. I don't have a spectrum handy here at home, but the numbers of pulses from photoelectric and Compton interactions are in the same general ballpark, with probably more Comptons.

    For higher-energy gammas, say the approx. 1.2 MeV ones from Co-60, the number of Comptons is definitely a lot larger than the number of photoelectrics.
     
    Last edited: Sep 13, 2006
  10. Sep 13, 2006 #9
    Yup. Compton scattering spectroscopy is usually done in the 100 keV range so that's where CS is a strong enough process to produce decent statistical certainty. It's a pretty handy technique to probe certain things. One can map fermi surfaces, probe coordination and bond lengths etc. with CS. It's pretty painful with lab x-ray equipment but at 3rd gen synchrotrons have increased interest in it.
     
  11. Sep 14, 2006 #10
    This looks exactly the experiment I am doing and is what led me to ask this question while thinking about the theory behind it. A convenient reason why the photoelectric effect occur at lower energy levels is because the incoming photons doesn't have enough energy to give energy to the electron and go off on its own (as in Compton). In this regard, the photoelectric and compton effects are much closer related.

    Although as Zapper suggested, the type of atom is also important as to which effect takes place.
     
  12. Dec 31, 2009 #11
    So, which comparison between CS and PE is correct? I am confuse.
     
  13. Dec 31, 2009 #12
    Compton scattering can occur on either free electrons or lightly bound electrons. The defining characteristic is that there is always a recoil photon. For low energies, Compton scattering is sometimes called Thomson scattering. The differential Compton scattering is sometimes referred to as Klein-Nishina scattering. The low energy (classical Thomson) total cross section is about
    σ = (8 pi/3) re2 = 0.666 barns, where re is the classical electron radius.
    For photon energies over ~0.5 MeV, the Compton cross section drops off due to relativistic effects. It is independent of the nuclear charge Z.
    Below ~60 KeV in air (~600 KeV in lead), the dominant photon attenuation cross section is where the photon energy is completely absorbed by a bound electron. For photon energies in the KeV region, this is often referred to as deep core photo-ejection. Very roughly, the cross section is crudely proportional to Z4/Ephoton3. The photo-ejection cross section can suddently jump by an order of magnitude as the photon energy crosses the binding energy (absorption edge) of the K-shell electron (~33 KeV in sodium iodide, ~90 KeV in lead). Complete photon absorption by a bound electron in the visible and UV region is usually called the photoelectric effect or photoionization.
    Bob S
     
    Last edited: Dec 31, 2009
  14. Feb 12, 2010 #13
    Why does not this relativistic effect come into account when photoelectric effect takes place?
     
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