= Circular-orbit derivation of the Bohr model
{title2=$r_n=n^2\hbar^2/(\mu\kappa),\quad E_n=-\mu\kappa^2/(2\hbar^2n^2)$}
Consider a single <Electron> of <mass> $m_e$ and a point <atomic nucleus> of <mass> $M$ and positive <electric charge> $Ze$, with $e>0$ the <elementary charge>. Use nonrelativistic <classical mechanics>, omit external fields and interactions with other <Electrons>, and write
$$
\mu=\frac{m_eM}{m_e+M},\qquad
\kappa=\frac{Ze^2}{4\pi\epsilon_0},\qquad
\hbar=\frac{h}{2\pi}.
$$
Here $\mu$ is the <Coulomb two-body reduced mass>, $\epsilon_0$ the <vacuum permittivity>, and $h$ the <Planck constant>. The separation $r$ and relative speed $v$ describe the <two-particle relative coordinate>; neither body is silently pinned in place. The <Electron>'s own radius about the <centre of mass> is $Mr/(M+m_e)$.
For the assumed circular <orbit>, <Coulomb's law> provides the required <centripetal acceleration>, and the imposed <orbital angular momentum> condition is
$$
\frac{\mu v^2}{r}=\frac{\kappa}{r^2},\qquad
L=\mu vr=n\hbar,\qquad n=1,2,3,\ldots.
$$
Substitute $v=n\hbar/(\mu r)$ into the first equation and cancel $r^{-2}$ to obtain
$$
r_n=\frac{n^2\hbar^2}{\mu\kappa},\qquad
v_n=\frac{\kappa}{n\hbar}.
$$
The <kinetic energy> is $K=\mu v^2/2=\kappa/(2r)$ and the <Coulomb potential energy> is $U=-\kappa/r$, with zero <energy> at infinite separation. Hence
$$
E_n=K+U=-\frac{\kappa}{2r_n}
=-\frac{\mu Z^2e^4}{2(4\pi\epsilon_0)^2\hbar^2n^2}.
$$
This derives the <energy> sequence from the stated circular model; the modern <Schrödinger equation> derivation is the distinct <series-termination quantization of the Coulomb radial equation>. The radius above scales as $n^2/Z$, whereas the <binding energy> scales as $\mu Z^2/n^2$. The <Bohr radius> article supplies the corresponding <Coulomb potential> length scale. The nonrelativistic assumption requires $v_n\ll c$, with $c$ the <speed of light>.
For an original numerical example choose $Z=1$, $M=1.67262192369\times10^{-27}\,\mathrm{kg}$, $m_e=9.1093837015\times10^{-31}\,\mathrm{kg}$, $e=1.602176634\times10^{-19}\,\mathrm C$, $h=6.62607015\times10^{-34}\,\mathrm{J\,s}$, $\epsilon_0=8.8541878128\times10^{-12}\,\mathrm{F\,m^{-1}}$, and $c=299792458\,\mathrm{m\,s^{-1}}$. Treat these as stipulated constants for the calculation. They give $\mu/m_e=0.99945568$, $r_1=0.0529465\,\mathrm{nm}$, $E_1=-13.598287\,\mathrm{eV}$ and $v_1/c\simeq0.00730$, where $\mathrm{eV}$ denotes an <electronvolt>. At $n=2$, $r_2=4r_1$ and $E_2=E_1/4$; the ground-state <ionization energy> is $-E_1$.
Taking $M\to\infty$ gives $\mu\to m_e$. A finite nuclear <mass> reduces every <binding energy> by $\mu/m_e$ and increases each separation by $m_e/\mu$. This follows from relative motion, not a fitted correction to individual <spectral lines>. Replacing the <atomic nucleus> by a heavier <isotope> changes the same factor, providing another check of the model.
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