Compton scattering is inelastic?

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  • #1
bobie
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I have a couple of simple questions:

- wiki says (http://en.wikipedia.org/wiki/Compton_scattering) that it " is an example of inelastic scattering". Is that true?, isn't all energy lost by the photon absorbed by the electron?

- is the electron really at rest? doesn't his energy/momentum affect the outcome?

- is the scattering angle θ (or the sum of the angles ) dependent on the energy of the photon?
This applet (http://www.ndt-ed.org/EducationResources/CommunityCollege/Radiography/Physics/comptonscattering.htm) gives the same curve for any value of energy.

- is there a lower threshold of energy under which the scattering cannot occur?

Thanks for your help
 

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  • #2
mathman
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I have a couple of simple questions:

- wiki says (http://en.wikipedia.org/wiki/Compton_scattering) that it " is an example of inelastic scattering". Is that true?, isn't all energy lost by the photon absorbed by the electron?

- is the electron really at rest? doesn't his energy/momentum affect the outcome?

- is the scattering angle θ (or the sum of the angles ) dependent on the energy of the photon?
This applet (http://www.ndt-ed.org/EducationResources/CommunityCollege/Radiography/Physics/comptonscattering.htm) gives the same curve for any value of energy.

- is there a lower threshold of energy under which the scattering cannot occur?

Thanks for your help
1. It is called inelastic because the photon loses energy.
2. Electron is a bound atomic electron. Energy of the electron is small effect.
3. The angular distribution is usually given by the Klein-Nishina formula, which doesn't take into account the binding energy of the electron. It is good enough for most analyses.
4. Minimum energy is that required to knock electron out of atom. If the photon is completely absorbed, it is the photoelectric effect.
 
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bobie
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3. The angular distribution is usually given by the Klein-Nishina formula, which doesn't take into account the binding energy of the electron.
Thanks mathman,
if I got it right we might consider the scattering an elastic collision between 2 particles, one with variable energy-momentum and one with 0 energy-momentum.
- what I meant is:
the link I quoted says that if a photon (.15 MeV) hits an electron and is scattered at 45° (θ) , then the electron recoils at 300° (-60°); if the photon had 2, 3 etc (* .15) energy, would angle θ have been still 45° and the sum of the angles 105° ?
- does the formula imply that the maximum energy a photon can absorb is 2*.511 MeV? (when θ is 180°)
 
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hello, mathman, when I read the post, I suddenly raise a problem. In quantum mechanics the particles have no trajectories. But in Compton scattering, how people can use the classical particle collision model to describe the effect ? I hope your answer, thank you !
 
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ZapperZ
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hello, mathman, when I read the post, I suddenly raise a problem. In quantum mechanics the particles have no trajectories. But in Compton scattering, how people can use the classical particle collision model to describe the effect ? I hope your answer, thank you !
First of all, scattering of light with other bodies can be described via the Born approximation.

Secondly, the scattering process is equivalent to the act of measurement, similar to you seeing a dot on a screen when an electron hits it. When you see that dot, you don't claim that particles have "no trajectories" anymore, do you? Same thing with Compton scattering. You are now already making a measurement of its momentum.

This is a bit off-topic from the original question. So if you want to deal into this further, you should start a new thread.

Zz.
 
  • #6
mathman
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hello, mathman, when I read the post, I suddenly raise a problem. In quantum mechanics the particles have no trajectories. But in Compton scattering, how people can use the classical particle collision model to describe the effect ? I hope your answer, thank you !
Compton scattering is usually described for high energy (gamma ray) photons, which act like particles in most interactions.
 
  • #7
bobie
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Compton scattering is usually described for high energy (gamma ray) photons, which act like particles in most interactions.
So, can we deal with scattering as we do with balls collision or there are some important differences?
- if so, does the scattering angle θ depend only on the contact angle (angle of recoil) or also on the photon energy?
 
  • #8
mathman
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The angular distribution is independent of the photon energy (Klein-Nishina formula). The angles after collision are related - conservation of momentum.
 
  • #9
bobie
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The angular distribution is independent of the photon energy (Klein-Nishina formula). The angles after collision are related - conservation of momentum.
Can someone, please, explain how the electron radius is related to the formula? (from wiki)
Note that this result may also be expressed in terms of the classical electron radius r_e=\alpha r_c. While this classical quantity is not particularly relevant in quantum electrodynamics, it is easy to appreciate: in the forward direction (for \theta ~ 0), photons scatter off electrons as if these were about r_e=\alpha r_c (~2.8179 fm) in linear dimension,
 

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