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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)

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    $\begingroup$ 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 $R_{new}$. $\endgroup$ – ZeroTheHero May 16 '18 at 13:55
  • $\begingroup$ Yes it is $186.03R_{old}$ $\endgroup$ – Jepsilon May 16 '18 at 13:57
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    $\begingroup$ Given that $\hbar c R_{old}\sim 13.6$eV (see en.wikipedia.org/wiki/…) you must have a conversion error somewhere. $\endgroup$ – ZeroTheHero May 16 '18 at 14:01
  • $\begingroup$ I will double check my calculation then $\endgroup$ – Jepsilon May 16 '18 at 14:14
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$E_n = \frac{Z \alpha^2 \mu} {2 n^2}$ 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.

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In order to obtain the energy levels of muonic hydrogen you can simply replace the reduced mass corresponding to the proton and electron with the reduced mass corresponding to the proton and the muon particle as you have already indicated.This should give you the correct Rydberg constant for the spectrum of this atom.

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