What is probability of scattering of muons with lead?

For relativistic muons, the RMS multiple-scattering angle in lead is approximately:

theta0 = (13.6 MeV / beta pc) * sqrt(x/X0) * [1 + 0.038 ln(x/X0)]

where:

  • theta0 = RMS scattering angle (radians)
  • x = thickness of lead
  • X0 = radiation length of lead = 0.56 cm
  • p = muon momentum
  • beta ~= 1 for cosmic-ray muons

Example: a 3 GeV muon passing through 1 cm of lead:

x/X0 = 1/0.56 = 1.79

The RMS scattering angle is about:

theta0 ~= 0.006 rad = 6 mrad = 0.34 degrees

Therefore:

  • Nearly all 3 GeV cosmic-ray muons pass through 1 cm of lead.
  • Typical deflections are only a few tenths of a degree.
  • Large-angle scattering (>10 degrees) is very rare.
  • Energy loss is about 25 MeV per cm of lead, so 1 cm removes less than 1% of the muon’s energy.

By Admin

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