Energy of scattered photon

  • Thread starter nelufar
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  • #1
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Main Question or Discussion Point

How to calculate energy of scattered photon when the energy of incident photon is equal to the rest energy of an electron and the angle between the direction of the recoiling electron and the incident photon is 40 degrees?
In compton effect the energy of scattered photon is related to angle
between the scattered photon and the incident photon. How to relate the angle between the direction of the recoiling electron and the incident photon and angle between the scattered photon and the incident photon?
 

Answers and Replies

  • #2
Derive this from conservation of energy and momentum in relativistic mechanics. Or look it up. I know it is on HyperPhysics.
 
  • #3
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Can't we use derivation for Compton effect?
 
  • #5
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I tried solving the problem using relativistic approach but there also the relation is for angle of the scattered photon not for angle of recoiling electron.So,can there be any other way to solve it.
 
  • #6
Momentum is conserved. The transverse component was zero before the interaction, so theta and phi are related by
[tex]0 = \frac{h}{\lambda '} \sin \theta - p_e \sin \phi[/tex]
 
  • #7
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But we only know the energy of the incident photon and rest mass of electron. Recoiling electron will also have energy which is not given.So how to calculate P_e and what about /theta?
 
  • #8
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Is there no solution to the problem? Can there be any other approach to solve it?
 
  • #9
nelufar said:
Is there no solution to the problem? Can there be any other approach to solve it?
This looks like a homework problem, so of course there is a solution.
You just need to work on it a bit.
 
  • #10
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I have tried a lot but the equation which you have provided, I got stuck there. So, if you can help , it would be really helpful. Atleast tell me can there be some other approach which probably I am not thinking of.
 
  • #11
I referred you to HyperPhysics. Momentum is conserved so
[tex]\vec{p_i} = \vec{p}_f + \vec{p_e} \Leftrightarrow \vec{p}_f= \vec{p_i}-\vec{p_e}[/tex]

[tex]p_f^2 = p_i^2 + p_e^2 - 2\ p_i\ p_e \cos{\phi}. [/tex]
That is how you can get it as a function of the electron angle, I think.
If you now proceed along similar lines as Nave, I think you should get there.
 
Last edited:
  • #12
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Pieter Kuiper said:
I referred you to HyperPhysics. Momentum is conserved so
[tex]\vec{p_i} = \vec{p}_f + \vec{p_e} \Leftrightarrow \vec{p}_f= \vec{p_i}-\vec{p_e}[/tex]

[tex]p_f^2 = p_i^2 + p_e^2 - 2\ p_i\ p_e \cos{\phi}. [/tex]
That is how you can get it as a function of the electron angle, I think.
If you now proceed along similar lines as Nave, I think you should get there.
Thanks. I will try with this.
 

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