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Compton scattering problem

  1. Jun 23, 2008 #1
    Can Someone look over this and tell me if the work is correct.

    1. The problem statement, all variables and given/known data
    An object comes in with known velocity (v) and known mass (m) moving directly along the x-axis. It strikes another object of the same mass (m) which is stationary. Find the change in energy of the incoming object based on the angle at which it scattered ([tex]\theta[/tex]) and the velocity of the target after the collision (u). The collision only occurs in 2 dimensions. (Note: The angle at which the target object is moving can be removed algebraically.)

    [tex]E_{i}[/tex]=Energy of Incoming Object before collision
    [tex]E_{f}[/tex]=Energy of Incoming Object after collision
    [tex]E_{2}[/tex]=Energy of Target Object after collision

    [tex]p_{i}[/tex]= Momentum of incoming object before collision.
    [tex]p_{f}[/tex]=Momentum of Incoming object after collision.
    [tex]p_{2}[/tex]=Momentum of Target object after collion.

    [tex]\phi[/tex]= Arbitrary Angle of Target object scattering (should not matter based on the note.


    2. Relevant equations
    [tex]E=1/2*m*v^{2}[/tex]
    p=m*v
    [tex]p^{2}/(2*m)=E[/tex]

    3. The attempt at a solution

    Energy Balance:
    [tex]E_{i}=E_{f}+E_{2}[/tex]

    X-Momentum Balance:
    [tex]p_{i}=p_{f}*Cos(\theta) +p_{2}*Cos(\phi)[/tex]

    Y-Momentum Balance: (This should be a zero momentum system in y-direction)
    [tex]p_{f}*Sin(\theta)=p_{2}*Sin(\phi)
    [/tex]

    Squaring only the Momentum Equations and adding them together.

    Y-Balance:
    [tex]p_{f}^{2}*Sin^{2}(\theta)=p_{2}^{2}*Sin^{2}(\phi)[/tex]

    X-Balance: (after getting [tex]\phi[/tex] isolated on one side then squaring)
    [tex]p_{2}^{2}*Cos^{2}(\phi)=p_{i}^{2} - p_{i}*p_{f}*Cos(\theta)+p_{f}^{2}*Cos^{2}(\theta)[/tex]

    Adding the two momentum Equations and using [tex]Sin^{2}+Cos^{2}=1[/tex]
    [tex]p_{2}^{2}=p_{i}^{2}-p_{i}*p_{f}*Cos(\theta)+p_{f}^{2}[/tex]

    Relating back to energy, using the relationship defined in section 2.
    Since all masses are the same I divide the newly found momentum equation by 2m in order to get to energy:
    [tex]E_{2}=E_{i}+E_{f}-\sqrt{E_{i}}*\sqrt{E_{f}}*Cos(\theta)[/tex]

    (Note: [tex]\sqrt{2m}*\sqrt{2m}=2m[/tex] and [tex]p/\sqrt{2m}=\sqrt{E}[/tex] )

    Combining with the original Energy Balance Equation to Eliminate [tex]E_{2}[/tex]. This involves subtracting the equation I just solved for and the original equation.
    [tex]0=2*E_{f}-\sqrt{E_{i}}*\sqrt{E_{f}}*Cos(\theta)[/tex]

    Applying the Quadratic Equation
    [tex]\sqrt{E_{f}}=\sqrt{E_{i}}*Cos(\theta) +\- \frac{\sqrt{\sqrt{E_{i}}^{2}*Cos^{2}(\theta)-0}}{4}[/tex]


    Minus Sign Answer Leads to 0, so nontrivial answer is:
    [tex]E_{f}=\frac{E_{i}*Cos^{2}(\theta)}{4}[/tex]

    Is this the correct solution, or is there a step that I made a mistake or false assumption? It seems to me that this is wrong, because even a grazing trajectory decreases the initial energy by 3/4.
     
  2. jcsd
  3. Jun 23, 2008 #2
    Re: Scattering

    This is essentially compton scattering. Why do you have to angles? Also, are you assuming it is inelastic, or were you told that it is inelastic? I would probably assume an elastic collision.
     
  4. Jun 24, 2008 #3
    Re: Scattering

    Your right, this is basically compton scattering without the relativisitic effects. That is why the first 4 steps are the same as the Compton Scattering derivation. The difference is that the since this does not use relativistic effects:
    [tex]E=\frac{p^{2}}{2m}[/tex]

    For Compton Scattering, the conversion between energy and momentum for Photons and relativistic electron is different:

    For Relativistic Electrons: [tex]pc=\sqrt{E_{0}^{2}-m_{0}^{2}c^{4}}[/tex]

    For Photons: [tex]\frac{E}{c}[/tex]

    This is where the difference is. That is why I asked the question, because this answer does not seem intuitive that if I have a grazing trajectory (i.e. [tex]\theta[/tex] goes to 0) I still lose 3/4 of the energy.
     
  5. Jun 24, 2008 #4

    alphysicist

    User Avatar
    Homework Helper

    Re: Scattering

    Hi ajhunte,

    You are missing a factor of two here. The factor of two will keep the four from appearing in the denominator of the final answer.
     
  6. Jun 25, 2008 #5
    Re: Scattering

    Thank You, That is exactly what I was missing.
     
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