# Rutherford scattering calculation

1. Oct 20, 2007

### strangequark

1. The problem statement, all variables and given/known data

The fraction of $$6.0 MeV$$ protons scattered by thin gold foil, of density $$\rho=19.3 g/cm^{3}$$, from the incident beam into a region where scattering angles exceed $$60 degrees$$ is equal to $$2.0 x 10^{-5}$$. Calculate the thickness of the gold foil using the result of the previous problem.

2. Relevant equations
The result I got for the previous problem was a formula for the number of particles scattered thru and angle of $$\theta$$ or greater and is:

$$N(\theta)= (\frac{1}{4\pi\epsilon_{0}})^{2} \pi I \rho t (\frac{zZe^{2}}{Mv^{2}})^{2} cot^{2}(\frac{\theta}{2})$$

3. The attempt at a solution

So I took $$\frac{N(\theta)}{I}$$ which is equivalent to removing the I from the above equation and attempted to solve for t, which is the thickness...

$$t=\frac{2x10^{-5}}{(\frac{1}{4\pi\epsilon_{0}})^{2} \pi (19.3 g/cm^{3}) (\frac{79(1.602x10^{-19}coul)^{2}}{2(6 MeV)})^{2} cot^{2}(\frac{\pi}{6})}$$

problem is that I keep getting an answer in m/kg, so I must be missing some term, though I can't figure out what it is...

any help is very much appreciated.

2. Oct 20, 2007

### Staff: Mentor

Absorber thickness is sometime measured in kg/m2 or gm/cm2 such that dividing by density in kg/m3 or gm/cm3 yields a thickness in m or cm, respectively.

Make sure the relationship between N and I is correct, and check one's units on the various terms.