= Solution
With the proton fixed, $[-\hbar^2\nabla^2/(2m_\mu)-e^2/(4\pi\varepsilon_0r)]\psi=E\psi$: <kinetic energy> plus Coulomb attraction, assuming nonrelativistic point particles and neglecting spin, radiation, and proton motion (or replacing $m_\mu$ by the reduced mass). Substitution of the $2s$ form gives $E=-\hbar^2\alpha^2/(8m_\mu)$ and $\alpha=m_\mu e^2/(4\pi\varepsilon_0\hbar^2)$. Direct <differentiation> verifies it. The curve starts positive, crosses at $r=2/\alpha$, then approaches zero from below; ordinary hydrogen has the same shape but the smaller electron mass gives a larger radial scale.
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