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Compton effect problem - Find angle and energy of scattered electron

  1. Mar 28, 2014 #1
    1. The problem statement, all variables and given/known data

    http://puu.sh/7Nw8T.png [Broken]

    2. Relevant equations
    From what I've read in my text book I believe I should be using the equations:
    Photon:
    p=E/c

    Electron:
    E(kinetic) = (Gamma-1)*m*c^2
    p= (Gamma)*mass(electron)*velocity(electron)


    3. The attempt at a solution
    For the first part:
    Ok, so first of all I wasn't entirely sure what energy the 0.7M electron volts was referring to, but I assumed it meant the incident photon. So I took that 0.7M electron volts and worked out the initial momentum of the system, in the x direction. I also noted that the initial y momentum of the system was zero.

    Then, using trig and the angles given, seperated this momentum into x and y components for both the photon and the electron. The expressions were in E', v etc.. Basically unknown quantities. I assume I must equate these knowing that.

    Px = P(photon)+P(electron) = 0.7x10^-6eV /c
    Simularly, Py = P(Photon)-P(Electron) = 0
    Given these both have the angles in them I assume there must be some trick to work out the angle. I'm stuck at that point, assuming I haven't made some mistake... Which is quite possible as I'm struggling with quantum mechanics lol.

    For the second part, I assume that knowing the angles the particles are scattered I can find the velocity of the electron and hence it's kinetic energy.



    Py = P(photon)-P(electron) = 0
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 28, 2014 #2

    Simon Bridge

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    ... the pic says "A 0.700MeV photon scatters off a free electron..." so certainly the energy refers to the incident photon.

    Good start, it is basically a conservation of energy problem.

    Shouldn't that be a +6 in the exponent? Why not work in MeV/c?

    You are told that the scattering angle of the photon is twice the scattering angle of the electron. You are also given a very strong hint at the bottom that you have yet to use.

    There's also a special relationship that applies to the Compton effect.
     
    Last edited: Mar 28, 2014
  4. Mar 29, 2014 #3
    I had another read of my text book and I think I found the "special relationship". I assume it's the relationship between incident wavelength - scattered wavelength and (1-cos(theta)). I put in hc/E in place of the incident wavelength and got a messy expression for the scattered one in terms of theta.

    Here I'm stuck again. Should I use that scattered wavelength to get an expression for the scattered photon momentum?

    I apologize for the lack of intuition on my part. This is the first physics course I've taken in 2 years, and I'm really struggling to recall some of the basic stuff. Thanks for taking the time to help.
     
  5. Mar 29, 2014 #4

    Simon Bridge

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    Conservation of momentum gets you two expressions, the Compton scattering formula (yes, the delta-lambda one) gives you a third.

    Have you used conservation of energy yet?

    Did you remember that ##\theta=2\phi## is the photon angle?

    When you have the same number of equations as you have unknowns, you can start solving for things.
     
  6. Mar 29, 2014 #5
    Alright thanks I'll see if I can finish it from here.
     
  7. Mar 29, 2014 #6
    I've given it a huge go and still no luck unfortunately. Am I using the right formular for the momentum of an electron, p= gamma*m*v?

    Also, could I get a hint as to what order I should solve the equations?
    :)
     
  8. Mar 30, 2014 #7

    Simon Bridge

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    Note: $$p_ec=\sqrt{E_{tot}^2-(mc^2)^2}$$
    You have to get clever about it or you'll get a mess of trig functions.
    Show me the 4 equations you are starting with.

    You should have something like:
    >>> photon enters with ##E_0=p_0c = hc/\lambda## given (700keV),
    ... and is deflected through unknown angle ##2\phi## from it's initial direction,
    ... departing with unknown energy ##E_\gamma = p_\gamma c##,
    ... causing an electron recoil at angle ##\phi##

    * conservation of energy: $$p_0c+mc^2 = \gamma mc^2 + p_\gamma c$$ ... mc^2=511keV is the electron rest-mass energy E.

    * conservation of momentum: $$p_0c=p_\gamma c\cos 2\phi + p_e c\cos\phi\\
    p_\gamma c\sin 2\phi = p_e c\sin\phi$$
    * Compton effect tells us:
    $$\frac{1}{p_0c}-\frac{1}{p_\gamma c } = \frac{1}{mc^2}(1-\cos 2\phi)$$
    Notice how you can express everything in terms of energy just by multiplying through by c?
     
  9. Mar 30, 2014 #8
    Thanks a lot!
    The 4 equations I started with were

    The wavelength relationship I posted about above, and the following

    c(Gamma*mass(electron)*velocity(electron))cos(electron angle) + c*(h/lambda')cos(theta) = E0
    (h/Lambda')sin(theta) = (Gamma*mass(electron)*velocity(electron))sin(electron angle)
    (Gamma - 1)*mass(electron)*c^2 + hc/lamda' = E0

    lamda' = scattered wavelength of photon

    I just realized I wasn't including the electrons initial mass energy in the conservation of energy.
     
    Last edited: Mar 30, 2014
  10. Mar 30, 2014 #9

    Simon Bridge

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    Gah! Probably a good idea to learn LaTeX :/
    https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

    That list of equations is actually the end of the physics - I would not normally have listed them like that but I decided you had made a credible effort and you'd learn quicker by comparing yours with mine.
    The rest is algebra ...

    I would work in terms of energy rather than momentum.
    Your unknowns are the final energies for the photon and the electron, gamma, and the angle.

    ##E_{tot}^2=m^2c^4+p^2c^2=\gamma^2m^2c^4## ... can get you out of some trouble.
    ... you'll want a list of trig identities too.

    ... well it subtracts from the other side so you get the kinetic energy term left over - but it is good practice to include it at the start: it is total energy that is conserved.
     
  11. Apr 28, 2015 #10
    I am stuck on this same question - mainly with the algebra, also isn't the Compton formula meant to have the LHS back to front as below?

    $$\frac{1}{p_\gamma c }-\frac{1}{p_0c} = \frac{1}{mc^2}(1-\cos 2\phi)$$
    $$p_0c+mc^2 = \gamma mc^2 + p_\gamma c$$
    $$p_0c=p_\gamma c\cos 2\phi + p_e c\cos\phi\\
    p_\gamma c\sin 2\phi = p_e c\sin\phi$$

    Process is as follows : identify the unknowns - $$p_\gamma\ phi\ p_e\ gamma$$

    I need to rearrange equations but I keep getting convoluted formulas that I am sure I have made an error in.

    To get rid of some variables I can make an uknown eg $$p_e$$ the subject in two equations and then set them equal to each other right?

    Eventually continue on this process until there is only one unknown by substituting the new formula for unknowns into a formula with just two unknowns eg comptons equation

    Is this on the right track?
     
    Last edited: Apr 28, 2015
  12. Apr 28, 2015 #11

    Simon Bridge

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    Yep... you can also make an unknown the subject in one equation, then sub the result into all the others.
    Did you try drawing a diagram?

    Aside: notice that the lhs of the compton equation has to be positive.
    This can be used to check which way around the terms have to be.
     
    Last edited: Apr 28, 2015
  13. Apr 28, 2015 #12
    Your a good man Simon, I'll have another go at it tonight.

    Do you use any programs to check your math when playing with formula like this?

    Also - I've just clicked regarding the LHS, I made my error when going from the standard formula to your one by just converting the wavelength to momentum*c
     
  14. Apr 28, 2015 #13

    jtbell

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    If you need to work on this further, please feel free to start your own thread, showing your own work. We prefer that to piggybacking on someone else's year-old thread.
     
  15. Apr 28, 2015 #14
    Ok I'll have another go now and then post up where I get stuck.

    Is using latex mandatory too?
     
  16. Apr 28, 2015 #15

    jtbell

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    You can write many equations fairly decently using the subscript and superscript buttons and the Greek letter / math symbol palette (the big ∑ button) in the toolbar of the editing window. That would be OK. On the other hand, using LaTeX for such equations isn't too complicated, either. See for example

    https://www.physicsforums.com/help/latexhelp/

    and skip over the more complicated parts that you don't need. Your choice. :oldsmile:
     
  17. Apr 29, 2015 #16

    Simon Bridge

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    I would consder LaTeX as "best practise" rather than compulsory.
    So long as we can read it.
     
  18. Apr 29, 2015 #17
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