# Scattering cross sections

1. Oct 23, 2007

### strangequark

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

A beam of $$\alpha$$-particles, of kinetic energy 5.3 MeV and intensity $$10^{4}$$ particle/sec, is incident normally on gold foil with thinckness $$1 x 10^{-5} cm$$. (The density, atomic weight and atomic number of gold are 19.3 g/cm, 197 and 79 respectively.) A particle counter of area $$1 cm^{2}$$ is placed on the opposite side of the foil from the incoming beam at a distance of 10 cm. Suppose $$\theta$$ is the angle between the center of the foil and the center of the detector. How many counts should we expect per second if the detector registers all particles passing through it at $$\theta=10 degrees$$ and $$\theta=45 degrees$$?

2. Relevant equations

$$\frac{d\sigma}{d\Omega}=(\frac{1}{4\pi\epsilon_{0}})^{2} (\frac{zZe^{2}}{2Mv^{2}})^{2} I n \frac{1}{sin^{4}(\frac{\theta}{2})}$$

where n is the number of nuclei per unit area, and I is the incident intensity.

also told that:

$$2 \pi sin(\theta) d\theta=d\Omega$$

and

$$dN=\frac{d\sigma}{d\Omega} d\Omega$$

3. The attempt at a solution

First I calculated the constant:

$$(\frac{1}{4\pi\epsilon_{0}})^{2} (\frac{zZe^{2}}{2Mv^{2}})^{2}$$

where I had $$2Mv^{2}=4(5.3 MeV)$$, $$z=2, Z=79$$

Then I found n:

$$n=\frac{19.3g}{cm^{3}}\frac{1}{197amu}1 x 10^{-5}cm$$

now's where I have a problem... I'm trying to integrate:

$$2\pi(\frac{sin(\theta)}{sin^{4}(\frac{\theta}{2})})d\theta$$

but I don't know what angles to integrate thru... when I draw a diagram, I get a pretty small angular range and my answer comes out too small.

for $$45 degrees = 11.4 counts/sec.$$
and for $$10 degrees = 4020 counts/sec.$$

I could really use some help on seeing how these answers are obtained... thanks!

Last edited: Oct 23, 2007
2. Oct 23, 2007

### javierR

Haven't looked in detail at your attempt. First, make sure you're using the correct units, and consistently. Usually safest to work in SI units unless an equation specifies to use some non-standard units. Next, your integration region is indeed the small angular width of the detector (10deg-small amount to 10deg+small amount and similarly for 45deg case). For the integration, you can define $\theta'=\theta/2$ and use the trig identity $\sin(2\theta')=2\sin\theta' \cos\theta'$. Then $d\theta=2d\theta'$ and the cosine can be "brought into the differential": $\cos\theta' d\theta'=d \sin\theta'$.

3. Oct 23, 2007

### 3uc1id

i am doing that exact integral now. i just posted it in cal.analysis. i am trying to use the substitution u=cos(theta/2) but it hasnt produced anything of value yet.

4. Oct 24, 2007

### 3uc1id

ok i finally finished this problem. you have to find d(omega), not by integrating the sin stuff u have above. use d(omega) = INTEGRAL of dA/r^2 so you pull out the r^s terms using 10^2 cm^2, and use 1 cm for the area A. youll get 1/100.second thing is you have to correct your n value. i converted the density to kg/m^2. then i divided it 197, and multiplied it by avagad's number to arrive at 5.897x10^28. crunching all this then youll get close. you should get 6....x10^-5 x 1/sin^4(theta/2). the last trick ill leave for you to do. remember you have to find the counts per hr. but the equation will be in s^-1.
good luck.