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Kyrios
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Homework Statement
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?
Homework Equations
[tex] \theta = \frac{13.6}{\beta p} z \sqrt{\frac{x}{x_0}} [/tex]
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?
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