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

T_max = (2m_e*c^2*β^2*γ^2)/(1 + (2γ*m_e/m_α) + (m_e/m_α)^2) ≈ 2 GeVδ = 0 (for α particles)Plugging these values into the Bethe Bloch formula, we get:-dE/dx = K*4*(Z/A)*1/(1)^2*[1/2*ln((2m_e*c^2*1*1*2 GeV)/(60 eV)^2) - 1
  • #1
jonmohajer1
3
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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:

Homework Statement


" 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"

Homework Equations



The Bethe Bloch formula is given by:
[itex]-\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 ][/itex]
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 [itex]T_{max}=\frac{2m_{e}c^2\beta^2{\gamma}^2}{1+\frac{2\gamma m_{e}}{M}+(\frac{m_{e}}{M})^2}[/itex]
which equals max kinetic energy transfer

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)
 
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  • #2


Hi Jon,

Welcome to the forum! It's great to see new members joining and seeking help with their studies. I'm a scientist and I'll do my best to help you understand this problem.

First of all, let's clarify the question. The total E_kin of the composite particle beam is equal to 2 GeV, which means that each particle in the beam has a kinetic energy of 2 GeV. This includes protons, α particles, and muons.

Now, let's look at the Bethe Bloch formula. This formula is used to calculate the average energy loss (dE/dx) of a charged particle (z, β, γ) in a medium (Z, A, I, δ). The terms in the formula represent:

-dE/dx: average energy loss (MeV/gcm^-2)
K: constant
z: charge of the particle
Z/A: atomic number/atomic mass of the medium
β: velocity of the particle
γ: Lorentz factor of the particle
m_e: rest mass of the electron
c: speed of light
T_max: maximum kinetic energy transfer
I: mean excitation energy of the medium
δ: density correction

Now, let's use this formula to calculate the ionisation energy loss of each particle in the beam. We will use the known masses, charges, and energies of the particles to calculate β and γ.

For protons:
z = 1
β = √(1 - (m_e/m_p)^2) ≈ 1
γ = 1/√(1 - β^2) ≈ 1
T_max = (2m_e*c^2*β^2*γ^2)/(1 + (2γ*m_e/m_p) + (m_e/m_p)^2) ≈ 2 GeV
δ = 0 (for protons)

Plugging these values into the Bethe Bloch formula, we get:
-dE/dx = K*1*(Z/A)*1/(1)^2*[1/2*ln((2m_e*c^2*1*1*2 GeV)/(60 eV)^2) - 1 - 0/2]
= K*(Z/A)*ln(7.4*10^7)

Similarly, for α particles:
z = 2
β = √(1 - (m
 

1. What is Bethe Bloch equation?

The Bethe Bloch equation is a mathematical formula used to calculate the energy losses of charged particles as they pass through a material. It takes into account the ionisation energy losses caused by interactions between the charged particle and the atoms in the material.

2. How is Bethe Bloch equation used in particle physics?

In particle physics, the Bethe Bloch equation is used to estimate the energy losses of protons, alpha particles, and muons as they pass through a material. This information is crucial in understanding the behavior of these particles and their interactions with matter.

3. What are the factors that affect energy losses in the Bethe Bloch equation?

The energy losses in the Bethe Bloch equation are affected by the charge of the particle, its velocity, and the properties of the material it is passing through. These factors determine the magnitude of ionisation energy losses and the range of the particle in the material.

4. How accurate is the Bethe Bloch equation?

The Bethe Bloch equation is a well-established and widely used formula in particle physics. It has been validated by numerous experiments and is considered to be accurate within a certain range of energies. However, at very high energies, the equation may not accurately predict the energy losses due to the effects of other factors such as multiple scattering.

5. Can the Bethe Bloch equation be applied to other particles besides protons, alpha particles, and muons?

Yes, the Bethe Bloch equation can be applied to other charged particles such as electrons, positrons, and heavy ions. However, the equation may need to be modified to take into account the different properties of these particles, such as their mass and charge.

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