Derivation using reduced mass for electron orbitals in the Bohr model

Question

Recently in my class, we used the Bohr model to consider the motion of the nucleus about the center of mass of the electron-nucleus system. I understand that this also happens in the Earth-moon system, where the Earth is kind of wobbling along a point, i.e. the COM is actually inside the surface of the Earth. Our teacher told us to quickly calculate the radius and speed of the electron, and that we can simply substitute the reduced mass of the electron in the electron-nucleus system in the original equation (which didn't consider nuclear motion). I have tried my best to derive it using four equations (centripetal force on electron and nucleus, angular momentum conservation, and com) but I can't get the expression.

Answer 1

Let the proton mass be $m_p$ and its position $\mathbf{r}_p$. Let the electron mass be $m_e$ and its position $\mathbf{r}_e$. Then the equations of motion are

$$m_e\frac{d^2\mathbf{r}_e}{dt^2}=\mathbf{F}$$

$$m_p\frac{d^2\mathbf{r}_p}{dt^2}=-\mathbf{F}$$

where their electrostatic attraction is

$$\mathbf{F}=-\frac{e^2}{4\pi\epsilon_0}\frac{\mathbf{r}_e-\mathbf{r}_p}{|\mathbf{r}_e-\mathbf{r}_p|^3}.$$

Now consider the separation vector between the two particles,

$$\mathbf{r}=\mathbf{r}_e-\mathbf{r}_p.$$

We have

$$\begin{align} \frac{d^2\mathbf{r}}{dt^2}&=\frac{d^2\mathbf{r}_e}{dt^2}-\frac{d^2\mathbf{r}_p}{dt^2}\\ &=\frac{1}{m_e}\mathbf{F}-\frac{1}{m_p}(-\mathbf{F})\\ &=\left(\frac{1}{m_e}+\frac{1}{m_p}\right)\mathbf{F}\\ &=\frac{m_p+m_e}{m_pm_e}\mathbf{F} \end{align}$$

and thus we find an equation of motion for $\mathbf{r}$,

$$\mu\frac{d^2\mathbf{r}}{dt^2}=-\frac{e^2}{4\pi\epsilon_0}\frac{\mathbf{r}}{r^3},$$

where

$$\mu=\frac{m_pm_e}{m_p+m_e}$$

is called the reduced mass. Instead of two equations, we have just one to solve, for a particle of effective mass $\mu$! The whole standard Bohr analysis applies, but with the substitution

$$m_e\to\mu$$

to take into account the motion of the proton. For example, instead of the energy levels being

$$E_n=-\frac{m_ee^4}{8\epsilon_0h^2}\frac{1}{n^2},$$

they are

$$E_n=-\frac{\mu e^4}{8\epsilon_0h^2}\frac{1}{n^2}.$$

We have

$$\mu= m_e\left(1+\frac{m_e}{m_p}\right)^{-1}\approx m_e\left(1-\frac{m_e}{m_p}\right)$$

so the motion of the proton introduces small corrections of fractional size

$$\frac{m_e}{m_p}\approx \frac{1}{1836}.$$