Compton scattering

  • #1

Homework Statement


A photon with wavelength 0.1385 nm scatters from an electron that is initially at rest. What must be the angle between the direction of propagation of the incident and scattered photons if the speed of the electron immediately after the collision is 9.30×106 m/s?

Homework Equations


$$\lambda' - \lambda = h/mc (1-\cos\phi)$$

The Attempt at a Solution


I have $$\lambda' = 0.1385 nm$$ but I'm unsure as to how to get the original wavelength so that I can use the equation stated above. I think I need to use conservation of momentum $$\overrightarrow{Pe} = \overrightarrow{p} - \overrightarrow{p'}$$ but I'm not really sure how to apply it.
 

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  • #2
PeroK
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Homework Statement


A photon with wavelength 0.1385 nm scatters from an electron that is initially at rest. What must be the angle between the direction of propagation of the incident and scattered photons if the speed of the electron immediately after the collision is 9.30×106 m/s?

Homework Equations


$$\lambda' - \lambda = h/mc (1-\cos\phi)$$

The Attempt at a Solution


I have $$\lambda' = 0.1385 nm$$ but I'm unsure as to how to get the original wavelength so that I can use the equation stated above. I think I need to use conservation of momentum $$\overrightarrow{Pe} = \overrightarrow{p} - \overrightarrow{p'}$$ but I'm not really sure how to apply it.

Why momentum?
 
  • #3
kuruman
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but I'm unsure as to how to get the original wavelength
What do you think the statement below means?
A photon with wavelength 0.1385 nm scatters from ...
 
  • #4
What do you think the statement below means?

I assumed it meant the scattered wavelength but is it the original wavelength and I need to find the scattered wavelength?

Why momentum?

I can use it to get the wavelength I need because for the photons $$p=h/\lambda$$
 
  • #5
PeroK
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I can use it to get the wavelength I need because for the photons $$p=h/\lambda$$

I just thought there might be a quicker way without worrying about momentum.
 
  • #6
I just thought there might be a quicker way without worrying about momentum.

Can I use that the kinetic energy is 1/2mv^2 with m = mass of an electron(?) and the v = 9.3 * 10 ^6 m/s. Then the wavelength is just h/sqrt(2*E*m)?
 
  • #7
PeroK
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Can I use that the kinetic energy is 1/2mv^2 with m = mass of an electron(?) and the v = 9.3 * 10 ^6 m/s. Then the wavelength is just h/sqrt(2*E*m)?

Using energy sounds better. If you need the wavelength to four significant figures, then is the electron's speed relativistic or not?
 
  • #8
Using energy sounds better. If you need the wavelength to four significant figures, then is the electron's speed relativistic or not?

Yea, it would be so for energy I need to use K = mc^2/sqrt(1-v^2/c^2) - mc^2 ?
 
  • #9
PeroK
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Yea, it would be so for energy I need to use K = mc^2/sqrt(1-v^2/c^2) - mc^2 ?

This is perhaps a good example where you should solve the problem algebraically and then, once you see how ##K## appears in the relationship for ##\cos \phi## you can make a decision on how to calculate ##K##. You could do it for both and see what difference it makes.
 
  • #10
This is perhaps a good example where you should solve the problem algebraically and then, once you see how ##K## appears in the relationship for ##\cos \phi## you can make a decision on how to calculate ##K##. You could do it for both and see what difference it makes.

I did it for both and the difference is small so I guess either works. I got $$\lambda = 7.79*10^-11$$ and when I used this in the equation I can't get it to work. It worked out to be $$\cos\phi = -23.99$$, which doesn't work
 
  • #11
PeroK
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I did it for both and the difference is small so I guess either works. I got $$\lambda = 7.79*10^-11$$ and when I used this in the equation I can't get it to work. It worked out to be $$\cos\phi = -23.99$$, which doesn't work

That value for ##\lambda'## might be wrong. Notice also that ##\lambda## is the original wavelength.

PS I make it that the difference between classical are relativistic KE makes a difference of 0.1 degrees!
 
  • #12
That value for ##\lambda'## might be wrong. Notice also that ##\lambda## is the original wavelength.

PS I make it that the difference between classical are relativistic KE makes a difference of 0.1 degrees!

I'm still not getting it to work. I'm now getting $$\lambda' = 1.715*10^-12$$
 
  • #13
PeroK
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I'm still not getting it to work. I'm now getting $$\lambda' = 1.715*10^-12$$

I'm going to bed now, but perhaps someone else can pick this up.

How did you get ##\lambda'##?

To help you a bit, did you get:

##\frac{1}{\lambda'} = \frac{1}{\lambda} - \frac{K}{hc}##?
 
  • #14
I'm going to bed now, but perhaps someone else can pick this up.

How did you get ##\lambda'##?

To help you a bit, did you get:

##\frac{1}{\lambda'} = \frac{1}{\lambda} - \frac{K}{hc}##?

Thank you so much. I finally got it to work.
 

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