What happens after Compton scattering event?

[mitch_1211]

Hi All,

So the photoelectric effect is the phenomenon where an orbital electron fully absorbs an incoming photon (assuming the energy of the photon is greater than the binding energy of the electron) and is ejected from its shell. The electron can then undergo its own interactions in the material. Meanwhile there is a 'hole' left in the atoms shell where the electron was so if no free electron fills is the shell energy levels are reshuffled so that the hole is filled resulting in a release of another photon (characteristic x-ray) to compensate for the energy difference of the shells involved in the reshuffle.

Now my question is, when a photon is Compton scattered from an electron and it deposits enough energy to release the electron from the shell does a reshuffle occur here also resulting in characteristic x-ray emission? Surely the 'hole' can't just remain there?

Thanks in advance

Mitch

 

[sophiecentaur]

It isn't "orbital electrons" that supply the photo-electrons. The electrons that are involved in the photoelectric effect are dissociated electrons and are not bound to any particular positive ion core and are the ones which contribute to an electric current (one per atom).
The result of the photoelectric effect is to leave the bulk of the metal positively charged and there is no hole formation. If the target is insulated then it will gradually go more and more positive until the escaping electrons are held in a space charge around it. They are constantly falling back down and dissipating the photon energies as heat in the target. (That's where your energy goes.) There is not enough energy involved to produce any X Ray emissions as the only potentials involved are in the order of a very few volts. (Google "Stopping Volts" in the context of the PE effect; it tells you the surplus KE of the electrons that are released.)

 

[mitch_1211]

It isn't "orbital electrons" that supply the photo-electrons. The electrons that are involved in the photoelectric effect are dissociated electrons and are not bound to any particular positive ion core and are the ones which contribute to an electric current (one per atom).
The result of the photoelectric effect is to leave the bulk of the metal positively charged and there is no hole formation. If the target is insulated then it will gradually go more and more positive until the escaping electrons are held in a space charge around it. They are constantly falling back down and dissipating the photon energies as heat in the target. (That's where your energy goes.) There is not enough energy involved to produce any X Ray emissions as the only potentials involved are in the order of a very few volts. (Google "Stopping Volts" in the context of the PE effect; it tells you the surplus KE of the electrons that are released.)

Thanks for getting back to me :)

I should have been more specific, I wasnt referring to the photoelectric effect that describes the photocurrent produced from metals that you explained. I should have said photoelectric absorption as in one of the interaction methods that gamma rays and x-rays have with matter.

I was just curious if Compton scattering (another interaction method for gamma and x-rays) had similar consequences to photoelectric absorption

:)

 

[Simon Bridge]

Now my question is, when a photon is Compton scattered from an electron and it deposits enough energy to release the electron from the shell does a reshuffle occur ... Surely the 'hole' can't just remain there?

... physics continues as normal after the Compton scattering event, yes.

 

[gadong]

In brief, core holes disappear via two mechanisms:

(1) X-ray emission: electron from a higher shell drops into the hole, and an X-ray is emitted;
(2) Auger electron emission: electron from a higher shell drops into the hole, and an ["Auger"] electron is emitted.

Note that #2 is a radiationless decay process.

These processes are studied using X-ray photoelectron spectroscopy (or XPS). The core hole state exists for long enough (1 fs) to be useful for thermodynamic estimations and even as a "clock" in some XPS experiments.

 

[mitch_1211]

thanks for the replies...

I figured out that photoelectric absorption occurs with *inner* shell electrons hence resulting in a re-shuffle involving significant energies and so the Auger electrons or characteristic x-rays are produced.

However Compton scattering events predominantly occur with *outer* shell electrons, those that could be considered basically 'free'. So even though the electron is still ejected after a Compton scattering even there is no reshuffle due the position of the hole and hence no characteristic x-rays or Auger electrons are produced.

Thanks again