Calculating range of muons in a detector?

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SUMMARY

The discussion focuses on calculating the range of muons in a NOvA detector, which consists of PVC plastic and oil layers. The user applied the formula for energy loss per unit length, using specific densities for Butane and C-552 air-equivalent plastic. The calculated range for a 200 MeV muon is approximately 39.1 cm, based on the derived values of energy loss and density. The user expressed uncertainty about the appropriateness of the materials used in the calculations, particularly regarding the assumption of minimal ionization for lower energy muons.

PREREQUISITES
  • Understanding of muon physics and particle interactions
  • Familiarity with energy loss calculations in materials
  • Knowledge of density and composition of materials like Butane and C-552 plastic
  • Basic proficiency in using equations for particle range calculations
NEXT STEPS
  • Research "energy loss of charged particles in matter" for deeper insights
  • Study "Bethe-Bloch formula" for a comprehensive understanding of ionization energy loss
  • Explore "material properties of Butane and C-552 plastic" for accurate density values
  • Investigate "muon range calculations at varying energies" to compare results
USEFUL FOR

Physicists, students in particle physics, and engineers involved in detector design and analysis will benefit from this discussion.

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Homework Statement


If one can approximate NOvA detectors being built out of hundreds of "sandwiches" or PVC plastic of 1cm thickness and a "plane" of oil of 5.6 cm, what is the range of muon of energy 200MeV, 500MeV, 1GeV, and 1.8GeV in the detector?

Homework Equations



\frac{dE}{dρ*x} = .15*(1.76) + .85*(2.2) = 2.14 MeV*cm2 / g
rho = .15*(1.76) + .85*(2.5) = 2.42 g/cm3
(using Butane as "oil" and "C-552 air-equivalent plastic" as CVC plastic)

The Attempt at a Solution


I got the values above from: http://pdg.lbl.gov/2014/AtomicNuclearProperties/adndt.pdf
I also use .15 and .85 because the detector is 15% plastic and 85% oil, if I use 1cm/(1cm + 5.6cm) and 5.6cm/(1cm+5.6cm)

From the equations above, I get

x = E / (ρ)*(2.14) = 200MeV / (2.4)(2.14) = 39.1 cm

Though I'm not sure its correct to use Butane as "oil" and "C-552 air-equivalent plastic" as CVC plastic

Am I doing this correctly?
 
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I have no idea what "C-552 air-equivalent plastic" is, but that should be fine.
The assumption that the particles are always minimally ionizing is more problematic, especially at 200 MeV.