**Estimate** the kinetic energy of the following particles after their material interactions:

- a) a muon with \(E_{\text{kin}}\) = 400MeV traverses a 10 cm iron target,
- b) a proton with \(E_{\text{kin}}\) = 200MeV traverses a 20 cm iron target,
- c) a charged pion with \(E_{\text{kin}}\) = 1GeV traverses a 20 cm iron target.

The density of iron is \(\rho=7.9\text{gcm}^{-3}\).

## Background & Theory

The main way heavy charged particles (i.e., anything but \(e^-\)) lose energy when interacting with matter is through *ionization*. Electrons lose their energy mainly through *Bremsstrahlung* in the material.^{1}

To evaluate the energy of the other charged particles after their respective journeys through the iron target, we must use the **Bethe-Bloch** formula (sometimes just called the Bethe formula):

\begin{equation}

- \frac{dE}{dx} = \frac{4 \pi}{m_e c^2} \cdot \frac{nz^2}{\beta^2} \cdot \left(\frac{e^2}{4\pi\varepsilon_0}\right)^2 \cdot \left[\ln \left(\frac{2m_e c^2 \beta^2}{I \cdot (1-\beta^2)}\right) - \beta^2\right]

\end{equation}

Looking at the form of the Bethe-Bloch equation, we can see there is a region where ionization is at a minimum around \(\beta\gamma \sim [4, 5]\); we call particles with energies falling into this region “Minimally ionizing particles”, or MIPs, and it’s often useful when estimating particle energy loss to exploit this part of the graph as we will now do.

So, we need only consider the behaviour of the Bethe-Bloch formula at the respective incoming energies of the particles. Indeed, if you ever find yourself plugging numbers into the Bethe-Bloch formula, *you’re probably doing it wrong!*

## Solution

So for **part a)**, we have a muon with \(E_{\text{kin}}\) = 400MeV, \(m_\mu\) = 105MeV. By my calculation, estimating \(p\sim E_\text{kin}\), that gives a \(\beta\gamma = \frac{1}{m_\mu}\cdot400\text{MeV} \sim 4 \rightarrow\) MIP.^{3} So, taking the y-value for a minimally ionizing particle as 1.5MeVcm²/g, \[\Delta E = 1.5\text{MeVg}^{-1}\text{cm}^2\cdot7.9\text{gcm}^{-3}\cdot10\text{cm} = 120\text{MeV}\]\[E_\text{after} = 280\text{MeV}\]

For **part b)**, we cannot simply say E~p because the proton mass is too high at ~1GeV. The proton has kinetic energy 200MeV and therefore momentum 644MeV. So, \(\beta\gamma = p/m \approx 0.6 \rightarrow\) not a MIP, rather the proton finds itself in the exponential region of the Bethe-Bloch, which suggests it’s going to get stuck in the iron as it passes through. Even if the proton were in the MIP region, ΔE would be \(\approx\) 240MeV (we’re now traversing a 20cm target), larger than its original energy.

For **part c)**, m_{π}=140MeV, E=1GeV. Again saying \(E \sim p\) because E >> m_{π}, we see \(\beta\gamma \sim 7\), which is slightly more than a MIP. \(\Delta E_\text{MIP}\) = 240MeV, but the pion loses slightly more than if it were a MIP, so its energy loss is somewhere in the region \[\Delta E = 240\rightarrow400\text{MeV}\]\[E_\text{after} \sim 600\rightarrow700\text{MeV}\]

- Bremsstrahlung, a purely electromagnetic process, allows us to evaluate the energy loss of electrons with the formula \(E=E_0e^{-x/x_0}\) where \(x_0\) is the
*radiation length*, often given in g/cm². In the context of an electromagnetic calorimeter, we can also think of \(x_0\) as the average distance a photon travels through a medium before pair production occurs. [↩] - http://mxp.physics.umn.edu/s06/Projects/S06_ParticleShowerEnergy/theory.htm [↩]
- Without the approximation P~E, we would have, from E²=p²+m² and E=E
_{kin}+m, p=493MeV => p/m=4.7, so it turns out the approximation wasn’t so bad. [↩]