result for this state is [12] and the eigenenergy becomes complex beyond that point. The resolution of this paradox is, as has been well-known for a long time, the creation of electron-positron pairs from the Dirac-Fermi sea when Z > 1 /α .[29] When a negatively-charged muon slows down in matter, sometimes it is captured by the Coulomb potential of an atomic nucleus before it decays. If so, it then quickly cascades down through its hydrogen-like levels to the 1 s ground state, emitting x-rays along the way. From there it then either decays or can be be captured by the nucleus, both of which are weak interaction processes. The capture process depends sensitively upon the value of the 1 s wave function at the origin. The numerical coding for muonic atom ground states for values of Z of interest goes pretty much as for the electronic atom case for such a Z . The differences come from replacing the mass of the electron (0.511 MeV) with that of the muon (105.66 MeV). This must be done for the “muonic Bohr radius”, a$_{Bµ}$ = a$_{B}$ ( m$_{e}$/m$_{µ}$ ) = 255.8 fm, and consequently in the choices of x match and x max which are now proportional to a$_{Bµ}$/Z . Also, the non-relativistic binding energy B is now Z 2 ( m$_{µ}$/m$_{e}$ ) Ry. Calculating, as above, the ground state for muonic $^{40}$Ca ( Z = 20), starting with B = 1 . 1 MeV and a 0 = 0.1, the iterations converge to final values of B = 1 . 13136 MeV and a 0 = 0 . 239462. However, this initial calculation used a value of ϵ = 10 − 4 (dimensionless), or a cut-off at the origin of 0.000187 fm, rather smaller than the size of the $^{40}$Ca nucleus,[30] The nuclear charge is not a point charge, of course, but has a charge density distribution, ρ ( r ), which is spread out over the volume of the nuclear sphere. In fact, one of the major reasons for having studied muonic atoms in the past, through the last 2 p → 1 s x-ray, is for determining this charge density.[11] As a simple example, suppose the nucleus is a sphere of radius R and charge eZ with a charge density ρ which is constant out to its surface and zero for r > R : We need to 19 R Ca ≈ r 0 A 1 / 3 = 1 . 2 × 40 1 / 3 = 4 . 1 fm . (46) E (1 s ) = m (1 − Z $^{2}$α$^{2}$)1 / 2 (45) F. Muonic Atoms