Proton Scattering: Calculating Current Density

[madeinmsia]

Homework Statement

50uA beam of 1MeV proton
Target = Iron of 0.05um thick
Calculate scattered current density at distance 5 cm at 20 degrees angle

Homework Equations

$$\sigma(E,\theta)$$ = $$\pi*Z_{1}Z_{2}e^{4}(M_{1}/M_{2})/ET^{2}$$

The Attempt at a Solution

I figured the probability of scattering to the angle is $$\frac{\int^{20}_{20}\sigma(E,\theta)d\Omega}{\int^{\pi}_{0}\sigma(E,\theta)d\Omega}$$

Then how do i find the scattered current density with the info I have?

 

[Gokul43201]

The expression you've written down will give you an answer of 0 (zero) - since the probability of scattering exactly into a specified angle will be zero (as you can see from the limits of your integral)...but the probability density will not be zero. You need to find the current density, which is proportional to the probability density.

So, the correct term in the numberator is: $\sigma(E,\theta=20)$

In any case, I don't know what the terms in the given equation represent, and I don't see where the angular dependence (I expect there should be a $1-cos\theta$ factor somewhere) is embedded.