# Rutherford scattering differential probability function

1. Sep 29, 2011

### Uniquebum

1. The problem statement, all variables and given/known data
Spray of alpha particles hit a gold membrane whose density is $\rho$ and thickness d, with a kinetic energy of $E_k$. The spray's intensity is I. Alpha particles are measured from an area of A which is set R away from the membrane. If $\theta$ is the angle between the spray hitting the membrane and the center of the area A, what is the amount of alpha particles hitting the area A in one hour when $\theta$ = 10 degrees.

2. Relevant equations
The propability equation for alpha particles going into an angle of $\theta$ after hitting the membrane.

$\frac{d\sigma}{d\Omega}$ = $(\frac{z*Z*e^2}{8*\pi*\epsilon*M*v^2})^2*\frac{1}{sin(\theta/2)^4}$
Where z = alpha particle's charge, Z = gold's atomic number, M = alpha particle's mass and \epsilon = vacuum permittivity

$\rho = 19.3(g/cm^3), M_{au} = 197u, Z_{au} = 79, d = 0.1*10^{-6}m, I = 10^4(Particles/Second), A = 1.0cm^2, R = 10cm$
3. The attempt at a solution
I would assume the straight forward way of doing this would be to calculate the probability for the particles to hit the angle and then multiply that with the intensity and then multiply that with 60s*60min to get the particles per one hour.

First i changed the equation to this form. ( z = 2e and i added 2*0.5 = 1 to the denominator)
$\frac{d\sigma}{d\Omega}$ = $(\frac{2e*79*e^2}{8*\pi*\epsilon*0.5*2*M*v^2})^2 * \frac{1}{sin(\theta/2)^4}$

Thus i can just throw in the kinetic energy to the denominator and make it look like
$\frac{d\sigma}{d\Omega}$ = $(\frac{79*e^3}{8*\pi*\epsilon*E_k})^2*\frac{1}{sin(\theta/2)^4}$

And after putting the numbers in i get $\approx 5.1216 \dots * 10^{-62}$
That's most likely not correct. Anyone care to help?