How to calculate thickness of the wall in a wire chamber

[Kyrios]

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 $m_{\pi}$ = 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

$$\theta = \frac{13.6}{\beta p} z \sqrt{\frac{x}{x_0}}$$

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?

 

[mfb]

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.