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Bethe-Bloch Equation different materials

  1. Oct 12, 2008 #1
    I am trying to use the Bethe-Bloch equation to investigate how a Muon loses its energy as it penetrates through different materials in the muon lifetime experiment. In order to do so I need 2 determine both the total energy and kinetic energy.
    For kinetic energy maximum I have been using

    Tmx = (2*m*p^2*c^4) / (m^2*c^4 + M^2*c^4 + 2*m*c^2*(p^2*c^2 + M^2*c^4)^0.5 )

    Typical values for Tmax I should be getting I have been informed are around 645MeV however I keep getting values in the region of 45MeV so I have obviously went wrong somewhere. Am I correct in saying:
    m = electron mass (0.511MeV)
    p = momentum of muon (achieved via p= gamma*mass of muon*c)
    M = mass of the muon (105.6584 MeV)

    If it is correct is there a chanae I am wrong with units of c somewhere? Or is it even density realted?


    Also the velocity of a muon, is it correct that it is 0.9901c (i.e Beta = v/c for special relativity) before striking the material?

    Much help would be greatly appreciated.
     
  2. jcsd
  3. Oct 12, 2008 #2

    clem

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    I don't understand your equation, but
    if all masses, energies, and momenta are in MeV, you don't need any factors of c.
     
  4. Oct 13, 2008 #3
    An equivalent of the equation is on wikipedia. It is slightly different but it can be simplified to what I have above for Tmax (N.B. this Tmax is the one used in the bethe-bloch equation.
     
  5. Oct 13, 2008 #4
    To get 645 MeV with either equation, you need gamma ~ 25. The velocity you're quoting (0.9901c) corresponds to gamma ~ 7. That could be the problem. I don't know what the correct velocity is, it depends on your specific experiment.
     
  6. Oct 13, 2008 #5

    Vanadium 50

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    There aren't many muon sources with the velocity known to 4 digits of accuracy.
     
  7. Oct 13, 2008 #6
    Many muons produced in LHC collisions have energies greater than ~8 GeV and therefore have velocities of 0.9999c (to 4 digits of accuracy, rounded down).
     
  8. Oct 14, 2008 #7

    Vanadium 50

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    Fine. You got me there.

    Nonetheless, there are very few monoenergetic muon sources at the energies under discussion, which leads me to suspect something may be wrong here.
     
  9. Oct 14, 2008 #8

    malawi_glenn

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    Have LHC started to colliding yet?

    ...trying to save Vanadium 50 ;-) ...
     
  10. Oct 14, 2008 #9

    Vanadium 50

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    The LHC collided 450 GeV protons on steel. That gives a maximum muon energy of 30 GeV and an average muon energy of ~1.5 GeV.
     
  11. Oct 14, 2008 #10

    malawi_glenn

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    Ahh yeah that is true, during detector performance tests
     
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