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I'm looking at a question which starts off as:

A proton is travelling through a material and scattering the electrons in the material. Express the scattering angle in terms of the impact parameter, $b$, ther reduced mass $\mu$, the relative speed $v$, and the scattering angle in the ZMF. Hence show that the momentum transfer is

$$q=\frac{2\mu{v}}{\sqrt{1+z^2}}\quad\text{where}\quad z=b\mu{v^2}/\alpha.$$

My attempt: Somethings are confusing me straight off, I'm not sure whether this is inelastic scattering or elastic scattering and I'm not sure whether $v$ is the relative speed in the LAB or ZMF frame. Also, assuming that $\alpha$ is meant to be the fine structure constant, this means that $z$ has units which seems a bit of an issue to me... I've also assumed that this is all none relativistic.

By looking at Rutherford scattering, we have that (in the electron rest/LAB frame) the scattering angle $\chi$ is given by

$$\cot\left(\frac{\chi}{2}\right)=\frac{mv^2b}{\alpha\hbar{c}}$$

I have assume that $v$ is the relative speed in the LAB frame (not sure how correct that is). From here I'm really not sure where I am meant to be going, I am pretty sure that this is pretty basic mechanics but I can't see how I should be going about this so any tips/ advice would be much appreciated!

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    $\begingroup$ "I'm not sure whether this is inelastic scattering or elastic scattering" Electrons are structureless, point particles at attainable energies and protons get a new name if they are excited. There is no place for the energy to go in a inelastic case which wouldn't be reflected in the description of the interaction. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Apr 17, 2013 at 12:29
  • $\begingroup$ @dmckee sorry I don't quite follow your second sentence, doesn't deep inelastic scattering contradict what you're saying (I may well be confusing myself here) $\endgroup$
    – Dmist
    Commented Apr 17, 2013 at 20:45
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    $\begingroup$ Yes, but in DIS you don't get a proton out. You get a haronic spray. You can also scatter off a proton and create a $\Delta^+$, but again, you don't have a proton in the final state. Or produce pions or kaons, but then you have more than two particles in the final state. If you have an electron and proton in and exactly an electron and a proton out there is no freedom to be inelastic. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Apr 17, 2013 at 21:19
  • $\begingroup$ What about scattered radiation? At some point, if the speeds are high enough, isn't radiation generated by the collision and consequent acceleration of the two particles? Then the energy of the outgoing particles must be less than the incoming energy. $\endgroup$
    – Marty Green
    Commented Jun 11, 2016 at 10:56

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First, I shall refer you to this problem of elastic Rutherford scattering (your problem is identical to it because of you use the same force equation, and the angles in both repulsive and attractive are identical), where the question states two results:

$$|\Delta {\bf{p}}| = 2 p \sin\left(\frac{\Theta}{ 2}\right)$$ and $$|\Delta {\bf{p}}| = \frac{2 \alpha}{v_0 b} \cos \Big(\frac{\Theta}{2} \Big)$$

The first can be derived from triangular cosine rule, while the second has been worked out in my response to that same problem in that link.

Using $\sin^2(\Theta/2 )+\cos^2(\Theta/2 )=1$ and $p=\mu v_0$, you should be able to obtain $$|\Delta {\bf{p}}| = \frac{2 \mu v_0}{\sqrt{1+z^2}} $$

where $z=b\mu v^2_0/\alpha$. In your case, replacing $q=|\Delta {\bf{p}}|$ and $v=v_0$ will give your desired answer.

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