The atomic emission spectrum for sodium ($\ce{Na}$) is completely dominated by a line in the range of yellow, about $590~\mathrm{nm}$ (to be more precise, it's a doublet). Here is how it looks like:

a) shows the emission spectrum and b) the absorption

This line is due to the transition $3\mathrm{p} \to 3\mathrm{s}$. I know the Rydberg equation which can be used to predict the transitions in hydrogen. I would like to know if there is a way to predict the transitions for other elements, such as the alkali metal $\ce{Na}$.

  • $\begingroup$ When the Schrödinger equation is able to be solved for multi-electron systems, then it should be possible to predict the transition energies. $\endgroup$
    – LDC3
    May 3, 2015 at 16:13
  • $\begingroup$ The answer is no: the spectra are way too complex! Even the neon spectrum is too complex. So we measure the spectra empirically. $\endgroup$
    – Ed V
    Jul 19, 2023 at 23:16

1 Answer 1


The origin of the Rydberg equation is the Bohr model's equation for the energy of a hydrogenic state: $E_\text{B} = -\dfrac{R_{H}hc}{n^2}$

This is accurate for other hydrogenic atoms, with only one electron too (bearing in mind that $R_{X}$ is proportional to the atomic number squared).

When you get to non-hydrogenic atoms you need to model inter-electron repulsion as well. This can be done by introducing an empirical parameter $\delta$, the quantum defect: $E_\text{B} = -\dfrac{R_{X}hc}{(n-\delta_X)^2}$. For sodium the quantum defect is 1.37, allowing good approximation of the transition energies between n-states in the electronic spectrum.

I don't think there is a similar back-of-the-envelope method to calculate the differences in energies between multi-electron atom l-energy levels (like the 3s-3p). The theory of angular momentum tends to mathematical complexity. There are likely good computationally derived approximations though.


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