Exercise 80 discusses the idea of reduced mass. When two objects move under the influence of their mutual force alone, we can treat the relative motion as a one particle system of mass $\mathbf{\mu }\mathbf{=}{\mathbf{m}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{2}}\mathbf{/}\left(m_1+m_2\right)$. Among other things, this allows us to account for the fact that the nucleus in a hydrogen like atom isn’t perfectly stationary, but in fact also orbits the centre of mass. Suppose that due to Coulomb attraction, an object of mass ${\mathbf{m}}_{\mathbf{2}}$and charge $\mathbf{-}\mathbf{e}$orbits an object of mass ${\mathbf{m}}_{\mathbf{1}}$ and charge +Ze . By appropriate substitution into formulas given in the chapter, show that (a) the allowed energies are $\frac{{\mathbf{Z}}^{\mathbf{2}}\mathbf{\mu }}{\mathbf{m}\mathbf{}}\frac{{\mathbf{E}}_{\mathbf{1}}}{{\mathbf{n}}^{\mathbf{2}}}$, where is the hydrogen ground state, and (b) the “Bohr Radius” for this system is $\frac{\mathbf{m}}{\mathbf{z\mu }}{\mathbf{a}}_{\mathbf{0}}$ ,where ${\mathbf{a}}_{\mathbf{0}}$is the hydrogen Bohr radius.

From Section 7.8, you know that, all formulas for hydrogen apply to hydrogen like atoms if you simply replace ${\mathbf{e}}^{\mathbf{2}}$to${\mathbf{Ze}}^{\mathbf{2}}$.

Hence,

Energy levels of hydrogen like atoms, ${\mathbf{E}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{m}{\left({\mathrm{Ze}}^2\right)}^{\mathbf{2}}}{\mathbf{2}{\left(4{\mathrm{\pi \epsilon }}_0\right)}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{2}}{\mathbf{n}}^{\mathbf{2}}}\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{2}}}$ …………(1)

Where, z is the atomic mass, h is Plank’s constant, n = principal quantum number,${\mathbf{\epsilon }}_{\mathbf{0}}$is the permittivity of free space.

And the Bohr’s Radius is,

${\mathbf{r}}_{\mathbf{n}}\mathbf{=}{\mathbf{n}}^{\mathbf{2}}\frac{\left(4{\mathrm{\pi \epsilon }}_0\right){\mathbf{h}}^{\mathbf{2}}}{\mathbf{m}\left({\mathrm{ze}}^2\right)}$ ….. (2)

Where, m is the mass.

Now, using eq. (1) and replacing mass (m) with Effective mass $\left(\mathrm{\mu }\right)$, you get,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{n}}=\frac{\mathrm{m}{\left({\mathrm{Ze}}^2\right)}^2}{2{\left(4{\mathrm{\pi \epsilon }}_0\right)}^2{\mathrm{h}}^2}\frac{1}{{\mathrm{n}}^2} \\ =\frac{{\mathrm{Z}}^2\mathrm{\mu }}{\mathrm{m}}\frac{-{\mathrm{me}}^4}{2{\left(4{\mathrm{\pi \epsilon }}_0\right)}^2{\mathrm{h}}^2}\frac{1}{{\mathrm{n}}^2} \\ =\frac{{\mathrm{Z}}^2\mathrm{\mu }}{\mathrm{m}}\frac{{\mathrm{E}}_1}{{\mathrm{n}}^2} \end{gathered}$$

Hence, the allowed energies are $\frac{{\mathrm{Z}}^2\mathrm{\mu }}{\mathrm{m}}\frac{{\mathrm{E}}_1}{{\mathrm{n}}^2}$.

Where, ${\mathrm{E}}_1$is the Energy of hydrogen ground state.

Again, using eq. (2) and replacing mass (m) with Effective mass $\left(\mathrm{\mu }\right)$, you get,

$$\begin{gathered} {\mathrm{r}}_{\mathrm{n}}=1^2\frac{\left(4{\mathrm{\pi \epsilon }}_0\right){\mathrm{h}}^2}{\mathrm{\mu }\left({\mathrm{Ze}}^2\right)} \\ =\frac{\mathrm{m}\left(4{\mathrm{\pi \epsilon }}_0\right){\mathrm{h}}^2}{\mathrm{Z\mu }\left({\mathrm{me}}^2\right)} \\ =\frac{\mathrm{m}}{\mathrm{Z\mu }}{\mathrm{a}}_0 \end{gathered}$$

Hence, the Bohr’s Radius is $\frac{\mathrm{m}}{\mathrm{Z\mu }}{\mathrm{a}}_0$.

Where, $a_0$is the hydrogen Bohr radius.