Energy loss of muons in lead plate

I want to determine how much energy $100$ GeV-Muons loose in a Pb-Absorber of length $l=1$m using the following graphic:

In a solution I was given it just says $dE=-\frac{dE}{dx}\cdot\rho\cdot l=2$GeV where $-\frac{dE}{dx}=1.8$MeVcm²/g, but I can't see how to obtain this result. How does one read the term $dE/dx$ off the graphic and why is $dE=\dots$?

• Myons? Did you mean muons? – John Rennie Feb 19 '18 at 19:13
• Yes, sorry. Thought the name was the same in english. – Buh Feb 19 '18 at 20:11

How does one read the term $dE/dx$ off the graphic
So what you need is a quantity to match the horizontal scale. The $\beta$ that appears there is the particle velocity over the speed of light $\beta = v/c$, and the gamma that is there is the Lorentz factor $\gamma = [1 -\beta^2]^{-1/2}$.
Your muons are ultra-relativistic (because $E \gg m_\mu c^2$), so it is fairly easy to compute a a good approximation to $\beta \gamma$.
Now, as you note the line for lead drops a little below $2\,\mathrm{eV \cdot cm^2/g}$ meaning that the detailed result could be a bit different from that rule-of-thumb value, but the fact is that the energy involved is so high that it just doesn't make much differences so people stick with the rule of thumb.