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Photon Electron Collision

  1. Aug 29, 2012 #1
    1. The problem statement, all variables and given/known data
    A 100KeV collides with an electron at rest. It is scattered through 90 degrees. What is the energy after the collision? What is the kinetic energy in eV of the electron after the collision and what is the direction of its recoil.


    2. Relevant equations
    Compton Scattering
    [tex]λ_{f}-λ_{i}=\frac{h}{m_{e}c}(1-cos(θ))[/tex]

    Energy of a Photon
    [tex]E=hf[/tex]


    3. The attempt at a solution

    What confuses me the most is the wording. What does "Scattered THROUGH 90 degrees mean?" Does that mean that the scattering angle of the photon is 90 degrees? Or does it mean that the scattering angle is between 0 and 90? I cannot confidently do this problem unless I fully understand the wording.

    I took it as a scattering angle of 90 degrees.

    So first I found the energy of the photon before the collision in Joules, and plugged this and theta=90 degrees into the Compton Scattering formula to get the final wave length, (from which I can get the energy of the photon.) I got about 84 KeV. Seems reasonable. Then I did an energy balance to get the energy of the electron.

    [tex]E_{pi}+E_{ei}=E_{pf}+E_{ef}[/tex]
    [tex]100keV+\frac{m_{e}c^{2}}{1.602×10^{-19}}=84keV+E_{ef}[/tex]
    Solving this I get the final energy of the electron to be 527.89keV. Again seems reasonable considering its rest energy is on the order of 500 keV.

    The last part is what I am having confusion with. If the scattering angle of the photon was indeed 90 degrees, that would mean the angle of the electron scatter would be 0 degrees. I did a momentum balance and I got an angle that was extremely close to zero. My concern is that I misunderstood the meaning of the words "through 90 degrees" and did the problem incorrectly.
     
  2. jcsd
  3. Aug 30, 2012 #2

    vela

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    That's correct.

    Why are you dividing by the elementary charge?

    You need to show your work here. I'm not sure where you got the idea that the electron's scattering angle would be close to 0.
     
  4. Aug 30, 2012 #3
    Thank you for your reply

    I divided by the charge because I was converting to KeV (I should have also divided by 1000) since the rest of the terms were already in KeV.

    Also, for some reason I had it in my head that the sum of both the scattering angles should add up to 90 degrees. (Isn't there a rule like this? Is it 180 degrees?)

    In posting the work for the momentum balance I found an error.

    It it too late at the moment for me to get back into the problem so I will post back tomorrow.
     
  5. Aug 30, 2012 #4

    vela

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    You'll find it much more convenient to work directly with electron-volts instead of converting to and from joules. A couple of useful combinations to memorize are:
    \begin{align*}
    hc &= 1240~\text{nm eV} \\
    \hbar c &= 197~\text{nm eV} = 197~\text{MeV fm} \\
    m_ec^2 &= 511000~\text{eV} = 0.511~\text{MeV}
    \end{align*} Those combinations show up often. For example, in the Compton scattering formula, you have
    $$\lambda_f - \lambda_i = \frac{h}{m_ec}(1-\cos\theta) = \frac{hc}{m_ec^2}(1-\cos \theta) = \frac{1240~\text{nm eV}}{511000~\text{eV}}(1-\cos\theta) = 0.00242~\text{nm }(1-\cos\theta)$$ The energy of a photon is related to its wavelength via
    $$ E = hf = \frac{hc}{\lambda} = \frac{1240~\text{nm eV}}{\lambda} $$
     
  6. Aug 30, 2012 #5

    vela

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    You're thinking of the case where you have two objects of equal mass that collide elastically.
     
  7. Aug 30, 2012 #6
    I cannot figure out how I can get the direction of the electron after the collision. I already used the Compton Formula in the beginning of the problem and that comes from conservation of momentum and energy so I don't know what else I can do. :[

    Plus that formula uses the scattering angle (which is the angle between photon and electron is it not?), wouldn't I need to get the electrons angle with respect to the horizontal to state its direction?

    EDIT:
    I am going to try conservation of momentum, treating it as a two dimensional problem.

    EDIT:
    This is turning out really messy, I don't think its the correct approach.
     
    Last edited: Aug 30, 2012
  8. Aug 30, 2012 #7
    Hint (1): The photon has 100 keV before being scattered, and 84 keV afterwards.

    What is the kinetic energy of the electron before and after the collision?

    Do you need to treat the electron relativistically?

    Hint (2): Write the incident and scattered photon momenta as 2D vectors. Given that momentum is conserved, what is the 2D momentum vector of the electron after the collision?
     
  9. Aug 30, 2012 #8
    The kinetic Energy of the electron is zero before the collision. After the collision the kinetic energy of the electron would be 16 keV. This seems like a lot of energy, doesn't that mean I WOULD need to treat the electron relitavistically? So wouldn't that mean..

    [tex]E=\sqrt{(m_{e}c^{2})^{2}+(pc)^{2}}[/tex]

    And E would be 16 keV? I don't know if this tells me anything I don't already have at this point.

    As for the vectors, in the X direction it would be.. (Making a coordinate system where the photon comes in completely horizontally..)
    [tex]p_{photon-xi}=p_{photon-xf}+p_{electron-xf}[/tex]
    And for the y direction..
    [tex]0=p_{photon-yf}-p_{electron-yf}[/tex]

    So my instinct is to say that [itex]θ_1[/itex] is the angle between the horizontal and the photon and [itex]θ_2[/itex] is the angle between the horizontal and the electron.

    So for the x direction..
    [tex]p_{photon-i}=p_{photon-f}cosθ_1+p_{electron-f}cosθ_2[/tex]
    And for the y direction..
    [tex]0=p_{photon-f}sinθ_1-p_{electron-f}sinθ_2[/tex]

    This is where it was getting messy. Do I now solve these equations for [itex]θ_2[/itex]?
     
  10. Aug 30, 2012 #9

    vela

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    No, it's not.
     
  11. Aug 30, 2012 #10

    vela

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    No, E is not 16 KeV. Your difficulties seem to arise because you don't know exactly what the various variables represent. You should consult your textbook and notes to clear that up.

    Also, you can't say 16 KeV is a lot of energy without having some scale to determine what qualifies as "a lot." If you were talking about a thrown baseball, for example, 16 KeV is nothing. The quantity that sets the scale here is the rest energy of the electron. Is 16 KeV a significant fraction of that?
     
  12. Aug 30, 2012 #11
    I went through the math of the Compton scattering formula and discovered that theta is the angle between the initial and final momentum vectors (or the angle between the horizontal and final momentum vectors) of the photon. For some reason I thought it was the angle between the electron and the photon.

    For the rest energy I got 511.8 keV so having that I can say that 16 keV is not a significant fraction of the rest mass and therefore I do not have to treat the electron relitavistically.

    So wouldnt that mean the electrons kinetic energy is
    [tex]K_e=\frac{p_e}{2m_e}[/tex]

    From here I can find the electrons momentum. Now that I know what the scattering angle truly is, I can plug it into the equation for the momentum vectors in the x direction or y direction. (it would be θ_1) and solve for θ_2

    [tex]p_{photon-i}=p_{photon-f}cosθ_1+p_{electron-f}cosθ_2[/tex]
    [tex]0=p_{photon-f}sinθ_1-p_{electron-f}sinθ_2[/tex]

    Am I getting closer? :confused:
     
  13. Aug 30, 2012 #12

    vela

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    You need to square the momentum in your expression for the kinetic energy. You don't really need that formula here though.

    Forge ahead. You're almost done.
     
  14. Aug 30, 2012 #13
    I squared the momentum and plugged everything in and got about 41 degrees. Seems reasonable!

    I am curious though, what other way could I have done it where I didn't have to use the kinetic energy formula?

    Thank you!!
     
  15. Aug 30, 2012 #14
    You know that total energy and momentum are conserved in Compton scattering.

    So the first step for me would have been to calculate the energy of the scattered photon, 84 keV.

    The energy difference has to go to the electron, 16 keV.

    The momentum of a photon is proportional to its energy.

    For the incident photon we have a*(100, 0), for the scattered photon a*(0,84),
    where "a" is a proportionality factor that does not matter. The angle between these two vectors is 90 deg, the scattering angle. For a different angle you would have to use sin and cos.

    The difference in momentum has again to go to the electron. That is a*(100,-84). If you are only interested in the angle of the photon, not it's speed, then you just have to calculate the angle this vector makes with the incident vector, i.e. about -40 deg. The proportionality factor "a" conveniently drops out of the angle calculation.
    (note that I use the minus sign to indicate that the direction is opposite that of the scattered photon)

    So the kinetic energy formula is not really necessary. Neither is the question if the electron is relativistic or not. If you wanted the speed of the electron, then you would have to worry about both of these questions.
     
  16. Aug 30, 2012 #15
    Your method is much better than the way I did it! Thank you for this insight. :]
     
  17. Aug 30, 2012 #16
    The first half is exactly what you did.

    For the second half I took a shortcut :-)
     
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