# Homework Help: Bethe Bloch: Calculating ionisation energy losses for protons, alpha and muons

1. Jan 15, 2012

### jonmohajer1

Hi, this is my first post here. The forum seems like a great idea! I've got finals coming up in a few days and can't seem to get my head around this example and my tutor seems unreachable at the moment. Anyway, here it is:
1. The problem statement, all variables and given/known data
" A beam of particles contains protons, α particles and muons with E_kin = 2 GeV.
Calculate their respective ionisation energy loss dE/dx in a gas with Z/A = 0.5 and
I = 60 eV.
How can the dE/dx of a particle be used to determine its electrical charge z"

2. Relevant equations

The Bethe Bloch formula is given by:
$-\frac{dE}{dx} = Kz^2\frac{Z}{A}\frac{1}{\beta^2}\left [\frac{1}{2}\ln\frac{2m_{e}c^{2}\beta^{2}\gamma^{2}T_{max}}{I^{2}}-\beta^{2}-\frac{\delta }{2} \right ]$
where-dE/dx is average energy loss (MeV/gcm^-2) of a charged particle (z, β, γ) in a medium (Z, A, I, δ)
also I've got $T_{max}=\frac{2m_{e}c^2\beta^2{\gamma}^2}{1+\frac{2\gamma m_{e}}{M}+(\frac{m_{e}}{M})^2}$
which equals max kinetic energy transfer

3. The attempt at a solution
Like I said, I just can't seem to visualise it, I think partly because of the ambiguity of the question wording. Is the total Ekin of the composite particle beam equal the stated value, or is that only the kinetic energy of the muons? My guess is that I calculate -dE/dx three times using known masses, charges etc of each particle type, perhaps obtaining β from the stated E_kin, treating it as the relativistic kinetic energy of a single particle? However, it seems there's still information missing, such as a value for delta (density correction).Should the formula simplify in anyway?
I also have in my notes "Applications of Bethe-Bloch: if E_kin is known, dE/dx~z^2 is a sensible measure for z" which I feel probably answers the final part of the question, but I can't say I understand how that was obtained.

Any help would be greatly appreciated, and I'll do my best to help others with questions in future. Cheers, Jon (leeds, UK)

Last edited: Jan 15, 2012