Hard nuclear physics cross section problem

In summary, an aluminium foil of thickness L = 10E-6 m is placed in a beam of intensity of 6E12 particles per second. The differential scattering cross section is known to be of the form. 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.
  • #1
Max Eilerson
121
1
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.

[tex]\frac{d\sigma}{d\Omega} = A + Bcos^2 (\theta) [/tex]

where [tex]\theta [/tex] is the scattering angle and [tex]\Omega [/tex] 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 [itex]\theta [/itex] is 30 degrees and 31.5Hz when [itex]\theta [/itex] is 60 degrees. Find the values of A and B.


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

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

I looked on wiki and the differential cross section is
[tex]\frac{d\sigma}{d\Omega} = \frac{scattered flux / unit of solid angle
}{initial flux / unit of surface}[/tex] 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 news on Phys.org
  • #2
Max Eilerson said:
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.
 
  • #3
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
Ok, so the solid angle is just [tex] d\sigma = A / r^2 [/tex] 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.
 
Last edited:
  • #5
Max Eilerson said:
Ok, so the solid angle is just [tex] d\sigma = A / r [/tex] 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
 
  • #6
yup that's what I meant, just put it in wrong my latex seems to take an age to update.
 
  • #7
Think I've worked it out thanks to your hint Dan.
The scattering fraction
[tex] \frac{R_s}/{R_i}= \frac{N_{A}L\rho\sigma}{A_m}[/tex]
[tex] d\sigma = frac{A_mR_s}/{N_AL\rho R_i)} [/tex]
where R_s = scattered count rate, and R_i is the incident count rate.
 
Last edited:
  • #8
Max Eilerson said:
Think I've worked it out thanks to your hint Dan.
The scattering fraction
[tex] \frac{R_s}/{R_i}= \frac{N_{A}L\rho\sigma}{A_m}[/tex]
[tex] d\sigma = frac{A_mR_s}/{N_AL\rho R_i)} [/tex]
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.
 
  • #9
no, there isn't I can just say [tex] KR_s = A + Bcos^2 (\theta) [/tex]
where K is a constant. Not sure why he has including all the blurb.
 

1. What is a hard nuclear physics cross section problem?

A hard nuclear physics cross section problem refers to the calculation of the probability of a nuclear reaction occurring between two particles at a high energy. This problem is often encountered in particle accelerator experiments, where high energy collisions between particles can result in the production of new particles.

2. What factors affect the accuracy of cross section calculations?

Several factors can affect the accuracy of cross section calculations, including the energy and momentum of the colliding particles, the type of particles involved, and the nuclear forces at play. Theoretical models and experimental data are also important factors in determining the accuracy of cross section calculations.

3. How is the cross section of a nuclear reaction measured?

The cross section of a nuclear reaction is typically measured using a particle detector, which can detect and measure the particles produced in the reaction. This data is then used to calculate the cross section by comparing it to theoretical predictions or previous experimental results.

4. What is the significance of cross section calculations in nuclear physics?

Cross section calculations are important in nuclear physics because they provide valuable information about the fundamental interactions between particles and the structure of nuclei. They can also help to understand and predict the behavior of nuclear reactions, such as those that occur in the sun or in nuclear power plants.

5. How do scientists use cross section calculations in their research?

Scientists use cross section calculations in their research to study the properties of nuclei, the behavior of particles in high energy collisions, and to improve our understanding of the fundamental forces of nature. These calculations can also be used to design and optimize experiments in particle physics and nuclear engineering.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
831
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
728
  • Introductory Physics Homework Help
Replies
18
Views
3K
  • Introductory Physics Homework Help
Replies
15
Views
1K
Back
Top