Bohr's atomic model when nucleus is in motion The radius of the orbit of an electron, Considering Bohr's postulate of angular momentum quantization is given by:
$$r_n=\frac{\epsilon_0n^2h^2}{\pi m_eZe^2}$$
Now, I came across a problem that asked to derive the radius of the orbit, energy in the ground state, et cetera considering the nucleus to be non-stationary. 
The solution to the problem simply substituted the reduced mass of the nucleus and electron in the place of the mass of electron to arrive at the answer.
But when I tried to derive the result using basic principles, I got a slightly different answer. Moreover, I am confused about the angular momentum quantization- do we take the angular momentum about the nucleus or the center of mass (which is also the center of rotation) here?
Considering angular momentum about the center of mass, because that seems more logical, here is my solution:
Let the distance between the center of mass and the electron be $r_n$. Then, we can say that the distance between the center of mass and nucleus is $r_n\cdot \frac{m_e}{m_n}$ where $m_n$ is the mass of the nucleus. Total distance between nucleus and electron is $r_n\frac{m_e+m_n}{m_n}$.
Angular momentum quantization: $m_er_nv=n\cdot\frac{h}{2\pi}$
Balancing electrostatic force with centripetal force:$\frac{1}{4\pi\epsilon_0}\cdot\frac{Ze^2m_n^2}{r_n^2(m_e+m_n)^2}=\frac{m_ev^2}{r_n}$Substituting and solving for $r_n$ gives:$$r_n=\frac{(m_e+m_n)^2}{m_em_n^2}\cdot\frac{n^2h^2\epsilon_0}{\pi Ze^2}$$But according to the solution with direct substitution of reduced mass, it should be:$$r_n=\frac{(m_e+m_n)}{m_em_n}\frac{n^2h^2\epsilon_0}{\pi Ze^2}$$So, where am I going wrong here?Edit 1:
And moreover, why is the use of reduced mass appropriate here? When we use the reduced mass in the bigger world, we don't make a postulate which quantizes the angular momentum of an object to add a constraint to the problem.

Edit 2:
As said in the comment, I equated the value of $\frac{nh}{2\pi}$ with the angular momentum of the system, and that worked well to arrive at the answer. The value of $r_n$ matches exactly with the value of $r_n$ if the nucleus was stationary. But, the value of the distance between the electron and nucleus, which is $r_n\cdot\frac{m_e+m_n}{m_n}$ matches with the reduced mass substituted formula. It still seems a bit unintuitive, although logical to use the angular momentum of the entire system.
 A: 
Let the distance between the center of mass and the electron be $r_n$.

This isn’t a great choice for your independent variable, because the electrical force depends on the distance between the two charges.
Instead, let’s use

*

*$a = r_e + r_p$ for the distance relevant for electromagnetism.

*reduced mass $\mu$ obeying $\frac1\mu = \frac1{m_p} + \frac1{m_e}$

*$r_e = \frac{\mu a}{m_e}$ for the distance between the center of mass and the electron

*$r_p = \frac{\mu a}{m_p}$ for the distance between the center of mass and the proton(s)

Note that we have $\mu a = m_p r_p = m_e r_e$.
Your force-balance equation for the electron becomes
\begin{align}
\frac{\beta}{a^2}
& = \frac{m_e v_e^2}{r_e}
=
m_e r_e \omega_e^2
= \mu a \omega_e^2
\end{align}
where $\beta = \frac{e^2 Z}{4\pi\epsilon_0}$ hides all the details about the electric charge, and $\omega_e = v_e/r_e$ is the electron’s orbital angular frequency.  The corresponding relationship for the nucleus is $\beta/a^2 = \mu a \omega_p^2$, which should confirm your intuition that the electron and the proton have the same orbital frequency $\omega_e = \omega_p = \omega$.
To find the angular momentum, convince yourself that the moment of inertia for this two-body rotor is
\begin{align}
I &= I_e + I_p 
\\ &= m_e r_e^2 + m_p r_p^2 = \mu a^2
\end{align}
and find the particle separations $a_n$ which give quantized angular momenta $L_n = I \omega_n = n\hbar$.
The stationary-nucleus approximation works because $m_p \gg m_e$, so $\mu\approx m_e$ and $r_p/r_e \approx 0$.
