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Bethe-Bloch equation for high-energy muons

  1. Sep 7, 2007 #1
    Can someone help me? I want to use the Bethe-Bloch equation for stopping power of muons. I have put it all together but it is coming out with stupid answers (like 10^-64). I have kept the units consistently SI. Any ideas?
  2. jcsd
  3. Sep 8, 2007 #2


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    SI units?... *shudder*

    Anyways, what's the incoming velocity of the muons? the bethe formula fails badly if the energies get too low...

    other than that there are a lot of little side calculations, density, etc, that go into your calculation that should be re-checked (by you) because getting those factors wrong can mess up your answer...

    also, what mean excitation energy are you using?... I know that the rule of thumb is 10*Z eV... but then you have to convert that to SI units if you stick with those...
  4. Sep 8, 2007 #3
    Thanks for your reply.

    They are cosmic-ray muons so I'm not worried about energies below ~0.1GeV (which I think is well above the lower limit of application). I will be looking at energies as high as I can go without having to worry too much about radiation losses (which I know is Z-dependent and which I will find out from working out the muon critical energy - I think it's just over 110GeV for uranium).

    I'm getting the mean excitation energy from table (which /i believe is from the formula you mentioned) but am converting from eV to J.

    I'm using SI units to try and keep everything consistent. I have gone through all the side equations and as far as I can tell they are all consistent. The next thing I will try to do is convert the answer to MeV/g cm^2 and see if it gets me any nearer.

    Just one side point, ze is the charge of the incident particle. Since it is muons I am looking at, when z^2 appears, should this just equal 1?
  5. Sep 8, 2007 #4


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    Yes, if what you call 'e' is the charge of the electron, then z^2=1 for the muon.
  6. Nov 8, 2007 #5
    How can I drow the beth-bloch formula
    please explane 2 me step by step
  7. Nov 13, 2007 #6


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    step 1: consider a probe electron traversing a solid.
    step 2: pretend the response of the solid to the probe is linear.
    step 3: calculate the force on the probe.
    step 4: The stopping power (not a power, actually) is the magnitude of that force.
  8. Dec 21, 2010 #7
    At high energies, typically 100 GeV, radiative processes (bremsstrahlung and pair production) become import and dominate losses described by the Bethe equation. A description is given at pdg.lbl.gov --> reviews, tables, ... -->passage of particles through matter. Click on
    Atomic and Nuclear Properties on the home page to access the crossover point (muon critical energy) for > 300 elements, compounds, and mixtures. A longer and more technical discussion of all this can be found at pdg.lbl.gov/2008/AtomicNuclearProperties/adndt.pdf
  9. Dec 23, 2010 #8
    The Bethe equation does not describe energy loss by electrons and positrons, just heavier particles. I don't understand olgranpappy's steps very well, but stopping power (= - dE/dx) is not a force; it is e.g. MeV / (g/cm^2). The electron/positron case is described very well in NIST ICRU-37. An interactive version is at
  10. Jan 30, 2011 #9
    I am searching for the stopping power of positron.Could you give me an idea about calculation of stopping power of positron except bethe....
  11. Feb 1, 2011 #10
    Charged particles lose energy in passing through matter almost entirely through collisions with
    atomic and/or molecular electrons. The electrons are excited to a higher state or simply ejected (ionization). The interaction is fundamentally different if the incident particle is an electron or positron for a couple of reasons. In the first place, it has exactly the same mass
    as the electrons it will interact with. This means that in principle it can give ALL its kinetic
    energy to the target electron (try it with billiard balls). Secondly, the electron is identical and indistinguishable from the target electron, and the rules (Pauli principle) says that no two electrons can be in the same state. Not true for positrons, so they lose energy slightly differently than electrons. Both of these complications result in energy loss almost but not quite as described by the Bethe equation; I find it more complicated.

    All of this is explained in ICRU Report 37, which also gives tables for several elements and compounds. (International Commission on Radiation Units and Measurements). It's pretty hard to get, but the same thing, by the same people, is on the NIST pages:
    (You can find this and related links on my
    pdg.lbl.gov/2010/AtomicNuclearProperties/index.html ) dE/dx tables for muons are provided, which scale for other heavy particles.

    I don't "do" electrons and positrons there for another reason: I am mainly interested in high-energy particles. At energies above a few tens of MeV they lose energy mainly by scattering from nuclei with the production of photons, which in turn scatter from nuclei
    with the production of electron-positron pairs. And so on, making for a vast shower of photons, electrons, and positrons. This is discussed lots of places, among them in our Review of Particle Physics, pdg.lbl.gov/2010/reviews/contents_sports.html --> click on
    "Passage of particles through matter." (It's fairly technical.)
  12. Feb 7, 2011 #11
    Thank you very much...
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