In high school chemistry, students are taught that the convergence limit in the emission spectrum of an element can be used to determine the first ionization energy.

However, what about the second, third or nth ionization energy? Wouldn't that be a higher energy and therefore appear as a higher energy / smaller wavelength spectral line? Shouldn't the convergence limit be the ionization energy for the 1s1 electron?

  • 2
    $\begingroup$ They have smaller wavelength - waaaay smaller. I guess they also have some "convergence limits", but what makes you think you'll even see them in the spectrum? $\endgroup$
    – Mithoron
    Sep 1, 2023 at 18:05
  • 3
    $\begingroup$ As you state, X-ray absorption spectrsocopy can determine ionization potential for inner electrons: en.wikipedia.org/wiki/X-ray_absorption_spectroscopy . It's elementary, just looking at things in a different light. $\endgroup$ Sep 1, 2023 at 19:03
  • $\begingroup$ @Mithoron, The 2nd ionization energy of sodium is 4563 kJ/mol. This corresponds to a wavelength of 26 nm, which is UV. So it should be detectable with UV spectrophotometry. $\endgroup$
    – Rafael
    Sep 1, 2023 at 19:40
  • $\begingroup$ Therefore, on a UV spectrum, I would think the line observed for the 2nd I.E. should be closer to the convergence limit than the line observed for the 3s electron. Wouldn't this make any calculations performed from the convergence limit inaccurate, because it's not necessarily corresponding to the 1st ionization energy, but for some n>1 I.E. ? $\endgroup$
    – Rafael
    Sep 1, 2023 at 19:46

1 Answer 1


The "Convergence limit" is for a single excitation like in the Balmer series for a hydrogen atom. For the Bohr model, the Rydberg formula gives the wavelength of the emission line:

$$\frac{1}{\lambda} = Z^2 * R_\infty * (\frac{1}{n'^2} - \frac{1}{n^2})$$

where n is the principal atomic numb from which the electron is decaying. So as $n -> \infty$ then $\frac{1}{n^2}$ converges to 0 and the wavelength converges to a limit.

So the same notion would apply to other excitations of electrons with multiple electrons, but of course there is no such "nice" mathematical solution for multi-electron atoms.

To get to the point, there is just no way to compare the excitation of a single electron with the excitation of multiple electrons.

  • $\begingroup$ There actually is a way to compare the excitation of a single-electron atom with that of a multi-electron atom. In quantum defect theory, the excited electron is considered to collide (at negative, i.e. bound energy) with the positively charged ion core. At long distance, the ion core acts like a point charge and hydrogen-like wave functions are expected and at short distance the scattering electron is disturbed by the field of the other electrons. Matching these regimes results in a phase shift of the wave function which manifests itself as a quantum defect ($n$ is no longer integer). $\endgroup$
    – Paul
    Sep 2, 2023 at 3:21
  • $\begingroup$ Note that the collision and therefore quantum defect depends on the orbital angular momentum $\ell$ and s, p, d, f electrons have different Rydberg series converging on the same ionization energy. Higher $\ell$ states have less core penetration, a smaller quantum defect and are more hydrogen-like. $\endgroup$
    – Paul
    Sep 2, 2023 at 3:24
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    $\begingroup$ @Paul - You correctly point out some of the nuances of multi-electron atoms. Hopefully it will suffice to say that the mathematics is very much more complicated than the simple Rydberg formula. I also should have also pointed out that multi-electron atoms require quantum mechanical calculations and the selection rules for electron transitions yield much more complicated spectra. $\endgroup$
    – MaxW
    Sep 2, 2023 at 5:56

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