# Hard nuclear physics cross section problem

1. Nov 13, 2006

### Max Eilerson

In a scattering experiment, an aluminium foil of thickness L = 10E-6 m is placed in a beam of intensity of 6E12 particles per second. The differential scattering crosss section is known to be of the form.

$$\frac{d\sigma}{d\Omega} = A + Bcos^2 (\theta)$$

where $$\theta$$ is the scattering angle and $$\Omega$$ the solid angle.

With a detector area of 0.1x0.1 m^2, placed at a distance of 5m from the foil, it is found that the mean counting rate is 40 Hz when $\theta$ is 30 degrees and 31.5Hz when $\theta$ is 60 degrees. Find the values of A and B.

Clearly I had to set up two simulataneous equations using the different $\theta$ values.
I've found the number of targets which is just
$$\frac{N_{A}L\rho}{A}$$

where A is the mass number (27 for Aluminium).

I looked on wiki and the differential cross section is
$$\frac{d\sigma}{d\Omega} = \frac{scattered flux / unit of solid angle }{initial flux / unit of surface}$$ but I don't really understand how to relate this to the information I have (I guess the incident flux is just the number of targets x multiplied by the beam intensity per unit area)

2. Nov 13, 2006

### physics girl phd

Watch out -- The flux is going to be the number of particles that hit the target PER unit area.

3. Nov 13, 2006

### OlderDan

To add a bit more to the other reply, I think dσ is proportional to the count rate (counts per second per unit area of the detector), and dΩ is a solid angle that you can calculate from the geometry of the apparatus. I don't think the number of targets enters into this calculation.

4. Nov 13, 2006

### Max Eilerson

Ok, so the solid angle is just $$d\sigma = A / r^2$$ where A is the detector area, and r is the distance between the foil and the detector. I'm still unsure about dσ.

Just realised the density of aluminium is also quoted in the question, so probably needed.

Last edited: Nov 13, 2006
5. Nov 13, 2006

### OlderDan

The solid angle is dΩ. The total solid angle of a sphere is 4π steradians and the surface area of a sphere is S=4πr² so dΩ = A/S = A/r²

6. Nov 13, 2006

### Max Eilerson

yup that's what I meant, just put it in wrong my latex seems to take an age to update.

7. Nov 13, 2006

### Max Eilerson

Think I've worked it out thanks to your hint Dan.
The scattering fraction
$$\frac{R_s}/{R_i}= \frac{N_{A}L\rho\sigma}{A_m}$$
$$d\sigma = frac{A_mR_s}/{N_AL\rho R_i)}$$
where R_s = scattered count rate, and R_i is the incident count rate.

Last edited: Nov 13, 2006
8. Nov 13, 2006

### OlderDan

I assume there was more to your problem than just finding the A and B you first asked about.

9. Nov 13, 2006

### Max Eilerson

no, there isn't I can just say $$KR_s = A + Bcos^2 (\theta)$$
where K is a constant. Not sure why he has including all the blurb.