Hard nuclear physics cross section problem

[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)

 

[physics girl phd]

I guess the incident flux is just the number of targets x multiplied by the beam intensity per unit area

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

 

[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.

 

[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 realized the density of aluminium is also quoted in the question, so probably needed.

 

[OlderDan]

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

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

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

 

[Max Eilerson]

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

 

[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.

 

[OlderDan]

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.

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

 

[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.