# Scattering cross sections

## Homework Statement

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 $$\theta=45 degrees$$?

## Homework 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

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

## 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:

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": $\cos\theta' d\theta'=d \sin\theta'$.