When a muon decays from rest, typically what fraction of the energy is carried off by the electron? I tried looking into some papers, but I wasn't sure how to interpret the graphs they displayed.
[Physics] Electron energy from muon decay
particle-physics
Related Solutions
If you consider a muon decay in vacuum then there is no difference between the lifetimes of muons and antimuons.
However a muon can interact with a proton via the weak force to form a neutron, while an antimuon cannot. Since the air is full of protons this means the muon lifetime in air is slightly shorter than the antimuon lifetime. If the air were full of antiprotons instead of protons the effect would be the other way round.
To explore the contributions of electronic (ion-electron scattering) and nuclear (ion-target atom) stopping, you can look at SRIM, available for download free from srim.org. This program was create by Ziegler, Ziegler, and Biersack ages ago and has been update by them since (yes, the same Ziegler as on a label on your graph above). Note that SRIM does not deal with the radiative effects at very high particle energy - it is/was mainly concerned with modeling ion implantation effects, not relativistic high energy particle physics.
Calculating the stopping power/ranges for protons in water, it is readily seen that the electronic stopping dominates nuclear interactions along almost all of the proton path length (again, I'm ignoring radiative contributions). A few examples are given here (stopping units of MeV/(mg/cm2)):
Proton energy dE/dx (elec) dE/dx (nucl) Range +/- straggle (um)
10 MeV 4.657E-2 2.700E-5 1200 +/- 54
1 MeV 2.402E-1 2.090E-4 26.55 +/- 1.16
100 keV 8.252E-1 1.476E-3 1.43 +/- 0.11
10 keV 4.283E-1 8.66E-3 0.26 +/- 0.07
1 keV 1.447E-1 3.364E-2 0.03 +/- 0.02
100 eV 4.577E-2 6.354E-2 0.004 +/- 0.004
So, for protons into water, the peak of the overall stopping is near 100keV, and almost all of that stopping is electronic. Only at very low energies does the nuclear stopping finally exceed the electronic. This is why protons have a pretty sharp end of range - the electronic stopping is a very smooth stopping as the proton scatters off of lots of electrons, losing very little energy each time. The nuclear stopping is a more violent event (larger momentum transfer), but really only happens at the end of range. So, the proton transfers energy to target atoms only at the end of range. Energy is being continually lost to electron scattering along the whole trajectory (in varying amounts depending on the proton energy at that point).
That last bit is important - you can get the energy loss at a given energy, but as the proton loses energy, you have to re-evaluate what the energy loss is at that new energy. If you are above the energy loss peak, a little more proton energy gives you a much longer range until it finally slows closer to the peak energy loss. (Note that a 10MeV proton has a range of 1.2 millimeters, while a 1 MeV proton goes only 26 microns, and a 100keV proton about 1.5 microns.)
Best Answer
A (somewhat idealized plot) is shown on page 6 of this paper. In a three body decay, the energy peaks close to the maximum due to phase space. The maximum is $53$ MeV and the peak looks to be around $45-48$ MeV