How Thick Should the Wall Be to Scatter Half the Particle Beam?

Click For Summary
SUMMARY

The discussion focuses on calculating the thickness of a wall required to scatter half of a particle beam, given a wall density of 2 x 1029 atoms per m3 and each atom having a radius of 3 x 10-15 m. The mean free path (λ) is defined as λ = 1/nσ, where σ is the microscopic cross-section calculated as σ = π*(3 x 10-15 m)2 = 2.83 x 10-29 m2. The macroscopic cross-section is determined using Σ = Nσ, and the exponential attenuation formula I = I0exp(-Σt) is applied to find the penetration depth (t).

PREREQUISITES
  • Understanding of particle physics concepts, specifically mean free path.
  • Familiarity with the calculation of microscopic and macroscopic cross-sections.
  • Knowledge of exponential attenuation in the context of particle interactions.
  • Basic proficiency in algebra and geometry for area calculations.
NEXT STEPS
  • Study the derivation and applications of the mean free path in particle physics.
  • Learn how to calculate macroscopic cross-sections for different materials.
  • Explore the implications of exponential attenuation in radiation physics.
  • Investigate real-world applications of particle scattering in nuclear physics.
USEFUL FOR

Students and professionals in physics, particularly those focusing on particle physics, nuclear engineering, and radiation safety, will benefit from this discussion.

leviathanX777
Messages
39
Reaction score
0

Homework Statement



A beam of particles strike a wall containing 2 x 10^29 atoms per m^3. each atom behaves like a sphere of radius 3 x 10^-15 m. Find the thickness of the wall that exactly half the particles will penetrate without scattering. What thickness would be needed to stop all but one particle in 10^6

Homework Equations



the mean free path (λ) = 1/nσ


The Attempt at a Solution



I'm not sure how to start the solution. I can't find an equation that will bring the thickness of the wall into the problem

Cheers.
 
Last edited:
Physics news on Phys.org
hmmm... you could consider the surface area effectivley covered by the spheres... what is that relative to the wall area in terms of the thickness?
 
I do not know the wall area, it didn't say.
 
Cool problem. I would approach it this way: The microscopic cross-section is ~ the area presented by a single nucleus, so in this case \sigma = \pi*(3x10-15 m)2 = 2.83 x 10-29 m2

Once you have that, you can calculate the macroscopic cross section using \Sigma=N\sigma and then you can use the exponential attenuation formula I = I0exp(-\Sigmat) to solve for the penetration depth t.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
6K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
16
Views
3K