2
$\begingroup$

Why do Rydberg atoms have electrons that can transition above the first ionization energy?

This can only happen in a multi-electron system but I don't see how the Rydberg formula would even work for n = infinity + 1. I thought it might have to do with the second ionization energy but the definition of that is just another electron being ionized not anything above the first ionization energy.

$\endgroup$
4
  • 2
    $\begingroup$ Why do Rydberg atoms have electrons that can transition above the first ionization energy? Any refererence to it ? As it is literally asking how $a < b$ can be at the same time $a > b$. $\endgroup$ – Poutnik Dec 2 '20 at 10:51
  • $\begingroup$ Pretty sure only three-component particles can have bound states above zero energy. The theoretical Rydberg atom certainly cannot. I guess one spin-state of an actual Rydberg-like atom has a slightly larger disscociation limit than the other. Within an electromagnetic field, things might be a bit different, too. $\endgroup$ – Karl Dec 2 '20 at 11:22
  • 1
    $\begingroup$ @Poutnik this is something my professor tried to explain. This is translated so it might not be the best working. "there are also internal electronic levels in multi-electron atoms that can be excited. These occur at higher excitation energies than those Rydberg transitions that converge to the first ionization energy (corresponds to the lowest energy to ionize an atom). They are therefore inevitably above the first ionization energy. Note: This is not possible only in atomic hydrogen and other one-electron atoms, since there is only one electron to be excited." $\endgroup$ – bobsburger Dec 2 '20 at 11:50
  • 2
    $\begingroup$ Remember that too laconic question elaboration is figurally very expensive, leading to clarifications much longer than the saved words in both characters and time. $\endgroup$ – Poutnik Dec 2 '20 at 11:52
3
$\begingroup$

A useful concept in Rydberg atoms is the notion of a channel. This concept is borrowed from scattering theory and describes all solutions of the Schrodinger equation for the scattering of the Rydberg electron of specified angular momentum with the ionic core in a certain energetic state. The bound Rydberg states correspond to scattering at negative energy and the scattering at positive energy corresponds to the ion-electron continuum of the channel.

If the ion has multiple electrons, it can have several excited states. Each of these excited states can form a channel. The energy of a Rydberg electron is given by

$E=E_\text{IP} + E' - \frac{hcR}{(n-\delta_\ell)^2}$

Where $E_\text{IP}$ is the first ionization energy, $E'$ is the excitation energy of the ionic core, $R$ is the mass-corrected Rydberg constant, $n$ is the principal quantum number and $\delta_\ell$ is the quantum defect for a Rydberg electron with orbital angular momentum $\ell$.

Each excited state of the ionic core thus gives a new Rydberg series with each own set of principal quantum numbers. So, Rydberg states do exist above the first ionization energy, but their principal quantum number is finite; it just belongs to a different channel.

enter image description here

Image source

$\endgroup$
2
  • $\begingroup$ if I wanted to know more about this could you reference a book or website. $\endgroup$ – bobsburger Dec 2 '20 at 19:05
  • $\begingroup$ The classic book on Rydberg atoms is from Tom Gallagher. But this is quite an advanced text. You could also try to google for quantum defect theory. $\endgroup$ – Paul Dec 2 '20 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.