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Compton scattering

  1. Feb 26, 2017 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    $$\lambda' - \lambda = h/mc (1-\cos\phi)$$

    3. 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.
     
  2. jcsd
  3. Feb 26, 2017 #2

    PeroK

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    Why momentum?
     
  4. Feb 26, 2017 #3

    kuruman

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    What do you think the statement below means?
     
  5. Feb 26, 2017 #4
    I assumed it meant the scattered wavelength but is it the original wavelength and I need to find the scattered wavelength?

    I can use it to get the wavelength I need because for the photons $$p=h/\lambda$$
     
  6. Feb 26, 2017 #5

    PeroK

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    I just thought there might be a quicker way without worrying about momentum.
     
  7. Feb 26, 2017 #6
    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)?
     
  8. Feb 26, 2017 #7

    PeroK

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    Using energy sounds better. If you need the wavelength to four significant figures, then is the electron's speed relativistic or not?
     
  9. Feb 26, 2017 #8
    Yea, it would be so for energy I need to use K = mc^2/sqrt(1-v^2/c^2) - mc^2 ?
     
  10. Feb 26, 2017 #9

    PeroK

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    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.
     
  11. Feb 26, 2017 #10
    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
     
  12. Feb 26, 2017 #11

    PeroK

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    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!
     
  13. Feb 26, 2017 #12
    I'm still not getting it to work. I'm now getting $$\lambda' = 1.715*10^-12$$
     
  14. Feb 26, 2017 #13

    PeroK

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    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}##?
     
  15. Feb 26, 2017 #14
    Thank you so much. I finally got it to work.
     
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