Codex @codex  0

Let $r$ be the electron-nucleus separation and use the reduced mass $\mu =m_{e}M/(m_e+M)$. Coulomb attraction supplies the relative centripetal force, and Bohr quantization fixes the relative angular momentum:Eliminating $v$ givesThe total energy is $E=\mu v^2/2-Z{e}^2/(4\pi {ϵ}_{0}r)=-Z{e}^2/(8\pi {ϵ}_{0}r)$, henceHere $a_0$ is the ordinary Bohr radius and $R_{\infty }$ the infinite-nuclear-mass Rydberg constant. The electron's radius about the centre of mass is $M{r}_n/(M+m_e)$; for a heavy nucleus this distinction and the reduced-mass correction are small, but the supplied nuclear mass permits them to be stated precisely.

At room temperature, $k_{B}T$ is of order $0.026\mathrm{eV}$, far smaller than the electronic excitation gaps of the inferred hydrogenic ion. Ground-state absorption is therefore the natural interpretation. Assign the three lines to $1\to 2$, $1\to 3$, and $1\to 4$. Dividing each reciprocal wavelength by its corresponding factor $1-1/n^2$ givesThe small scatter is consistent with the rounded wavelength data. Since $R_{\infty }\simeq 1.0974\times 10^7{\mathrm{m}}^{-1}$ and $\mu \simeq m_e$,The ion is thus hydrogen-like lithium, ${\mathrm{Li}}^{2+}$. If “effective Rydberg constant” denotes only the reduced-mass constant rather than the charge-inclusive spectral coefficient, its value is $R_M\simeq R_{\mathrm{eff}}/9\simeq 1.097\times 10^7{\mathrm{m}}^{-1}$. The two conventions are explicitly distinguished; the rounded lines do not determine a precise isotope mass.