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How to calculate thickness of the wall in a wire chamber

  1. Nov 28, 2014 #1
    1. The problem statement, all variables and given/known data
    This is a question on multiple coulomb scattering in a wire chamber

    momentum p = 500MeV/c
    wire resolution = 120 microns
    distance from wall to wire = 0.01m
    radiation length of wall material X_0 = 2E-3 m
    mass of charged particle [itex]m_{\pi} [/itex] = 139.6 MeV/c^2
    charge z = 1

    How thick is the wall if 68% of the particles are scattered less than the wire resolution?

    2. Relevant equations

    [tex] \theta = \frac{13.6}{\beta p} z \sqrt{\frac{x}{x_0}} [/tex]

    3. The attempt at a solution

    What I am trying to find is x, but am uncertain of where exactly the 120 microns and 68% comes in.
    I looked up a distribution table and for z=0.68, the area below z = 0.7517. (I think this is what you have to do for 68% scatter less than the resolution?)
    The gaussian distribution forms a triangle to work out theta, with the side opposite to the angle. I assume the mean would be zero, so would this side just be 0.7517x120microns?
     
    Last edited: Nov 28, 2014
  2. jcsd
  3. Nov 29, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    I don't understand your attempt.

    The formula gives the average deflection angle.

    I would first see which angle gives the "maximal" scattering (a deviation by one times the wire resolution within the wire chamber). Then you can find x/x0 to have 68% of the particles below this angle.
     
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