# Modifying the Bohr Model for Muonic Hydrogen

I am trying to find the energy between the $n=2\leftrightarrow3$ transition for Muonic Hydrogen. My approach was to modify the Bohr model for standard hydrogen but taking the mass of the Muon $m_\mu\approx 207m_e$ instead of the mass of the electron $m_e$ and then substituting in the effective mass $m=\frac{m_\mu m_p}{m_\mu+m_p}\approx186.03m_e$ (I took $m_p\approx1836m_e$).

Now taking $R=\frac{m_ee^4}{8\varepsilon_o^2h^3c}\approx10.97\times 10^6$ it gives $\Delta E\approx340.12\times 10^6\text{ J}$

Is this correct? Because it's quite a large value (assuming I got the units right)

• Since mass enters linearly in the Rydberg constant, shouldn't your "new" Rydberg constant be just $R_{new}= R_{old}\times m/m_e$? You could then simply continue with $R_{new}$. – ZeroTheHero May 16 '18 at 13:55
• Yes it is $186.03R_{old}$ – Jepsilon May 16 '18 at 13:57
• Given that $\hbar c R_{old}\sim 13.6$eV (see en.wikipedia.org/wiki/…) you must have a conversion error somewhere. – ZeroTheHero May 16 '18 at 14:01
• I will double check my calculation then – Jepsilon May 16 '18 at 14:14

$E_n = \frac{Z \alpha^2 \mu} {2 n^2}$ for hydrogen like atoms in the nonrelativistic approximation. For the hydrogen ground state ($Z=1$, $n=1$, $\mu=\mu_e$) this gives 1 Ry. Your answer should therefore be about $\mu_{\mu} /\mu_e$ times $1/4-1/9$ Ry, that is about 26 Ry ~ 400 eV ~ 6.4 $10^{-17}$ J.