Photoeffect and compton scattering

In summary, the photoelectric effect is when electrons are ejected from an atom because of a high energy photon, and the Compton effect is when electrons are emitted from the nucleus because of a low energy photon.
  • #1
vabamyyr
66
0
I was asked question: "Why in case of photoelectric effect electrons closer to the atom are ejected and in case of Compton effect electrons in the outer shells are emitted from nucleus by x-ray photon?".

Well I know that this topic is very deep and one has to carefully select words to explain this but I think that in case of CE photon has far too much energy and momentum to eject outer shell electron which is not too tightly bound to the atom. And in case of PE the photon is absorbed completely and for that to happen it has to encounter electron which is closer to the atom and therefore has higher binding energy.

Am I thinking wrong? If so then smarter opinions are welcome.
 
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  • #2
vabamyyr said:
in case of CE photon has far too much energy and momentum to eject outer shell electron

I don't understand this. Please explain
 
  • #3
sorry, i meant in case of CE it is not possible for photon to disappear (absorbed by electron) so it only gives some part of energy and momentum to electron which is sufficient to eject it from the atom.
 
  • #4
The more I read this forum, especially topics concerning CE and PE I'm getting more and more confused.
 
  • #5
vabamyyr said:
The more I read this forum, especially topics concerning CE and PE I'm getting more and more confused.

Well, I tend to oversimplify things intentionally because I think there is a natural human tendency to obfuscate.

That said, I would suggest that the photoelectric effect and compton scattering are the same thing - photons hitting matter.
 
  • #6
yes, but still I don't understand why Compton effect involves free or loosely bound electrons and not electrons from inner shells. Can u enlighten me on that one?
 
  • #7
When one does compton scattering the photon energy is far beyond the absorption edges of whatever elements you're shooting the photons at. The probability that the photon interacts with atom via the photoelectric effect is much lower then.
 
  • #8
Do I understand this? If the target is tightly bound, then and interaction is more likely with a low energy photon. If the target is loosely bound, then the interaction is more likely with a high energy photon.
 
  • #9
Not really. It's more about how far off resonance the photon energy is from the absorption edges. I've seen a nice schematic picture of the interaction probabilities vs energy in the x-ray regime, I'll try to find it for you. You could also look up the Kramers-Heiseberg formula which describes the physics involved well.

edit: the pic seems to be very evasive. too evasive considering that I've seen it atleast a gazillion times.
 
Last edited:
  • #10
>>interaction probabilities vs energy in the x-ray regime <<

How about this: High energy photons are more like little bullets and interact more rarely, but low energy photons are more spread out and therefore more likely to interact?
 
  • #11
in case of compton effect,elastic collision between x-ray photon and target electron takes place. The electron close to the nucleus of the atoms are tightly bound. When such bound electrons scatter x-rays,the collision is considered to be taking place between the photon and the whole atom. the whole atom recoils instead of the individual electron.The compton shift in this case is given by same expression except that the mass of electron is replaced with that of the atom. The mass of the atom is about ten thousand times greater than mass of electron. Therefore the compton shift is so small as to be unobservable. this process gives unmodified line and we say that compton effect involves outer electrons only.
 

Related to Photoeffect and compton scattering

1. What is the photoelectric effect?

The photoelectric effect is a phenomenon in which electrons are emitted from a material when it is exposed to electromagnetic radiation, such as light. This can occur when the energy of the radiation is high enough to overcome the binding energy of the electrons in the material.

2. What is the role of the work function in the photoelectric effect?

The work function is the minimum amount of energy required to remove an electron from a material. In the photoelectric effect, the energy of the incoming radiation must be equal to or greater than the work function in order for electrons to be emitted from the material.

3. How does the intensity of light affect the photoelectric effect?

The intensity of light does not have a direct effect on the photoelectric effect. However, increasing the intensity of light will increase the number of photons hitting the material, which in turn increases the likelihood of electrons being emitted.

4. What is Compton scattering?

Compton scattering is a phenomenon in which an incoming photon collides with a free electron, transferring a portion of its energy to the electron and causing it to scatter. This results in a decrease in the energy and an increase in the wavelength of the scattered photon.

5. How does the energy of the scattered photon in Compton scattering relate to the scattering angle?

The energy of the scattered photon in Compton scattering is related to the scattering angle through the Compton formula: Δλ = h/mc(1-cosθ), where Δλ is the change in wavelength, h is Planck's constant, m is the mass of the electron, c is the speed of light, and θ is the scattering angle.

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