# Calculating range of muons in a detector?

1. May 2, 2015

### 1a2

1. The problem statement, all variables and given/known data
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?

2. Relevant 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)

3. 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?

Last edited: May 2, 2015
2. May 3, 2015

### Staff: Mentor

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.