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The way I understand PES is that you shine a light on a bunch of atoms and measure the kinetic energy of photons to determine ionisation energies (or energies of electrons... is that different?).

enter image description here

I expected to see two peaks, one for the first IE, another for the 2nd ionization. Of course IE1 and IE2 are different. Why can't we see that?

Is this some sort of averaged-out data between the two spreads of electrons being ejected?

I know that in the He atom, both electrons are at the same level. But I can't see how this method would shed light (no pun included) on the very energy (orbital+pairing) and not individual ionization energies.

Any help/explanation would be much appreciated!

Edit: Found this on Libretext:

enter image description here Figure 10.4.6 : Photoelectron spectrum HCl . (CC BY-NC; Ümit Kaya via LibreTexts)

This is really close to, if not precisely what I expected. Are the other PES graphs oversimplified representations of the true data? Or am I missing something??

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  • $\begingroup$ Have you looked up the ionisation energies to see if either matched your data? $\endgroup$
    – porphyrin
    Commented May 5 at 11:16
  • $\begingroup$ I conducted no experiments myself. I was just learning about this stuff and looked at these diagrams. They (the majority of the diagrams online with "PES" as a search - Figure 1.) weren't as intuitively pleasing, given that IE1 and IE2 are different for atoms. Figure 2 however is close to my understanding of how ionisation and orbitals work. $\endgroup$
    – Maddy
    Commented May 5 at 11:43
  • $\begingroup$ Also, just looked up the data... the peak corresponds to Helium's 1st IE... around 24eV The 2nd which can also be determined by Bohr's postulates (13.6*4) is around 54.4 eV. So I would have expected another pick at around that energy. But I see none. Are all the diagrams online oversimplified or am I missing something? $\endgroup$
    – Maddy
    Commented May 5 at 11:46
  • $\begingroup$ I do realise how the peak for He is gonna be at 2e... either one of the 2 electrons can be ejected at this energy. So the probability of ejection is twice that of Hydrogen. But I don't understand why I don't see the other peak that I would have expected otherwise. $\endgroup$
    – Maddy
    Commented May 5 at 11:49

2 Answers 2

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It is very hard to study light elements like H and He with PES. I could find one experimental paper: 1 (does not seem to be behind a paywall). They did not look at the energy range where the 2nd ionization would be. That paper used a very advanced setup to be able to get the results.

I think the second ionization is theoretically present but very, very weak. You would have to extract an electron twice from the same He atom. However, at any given time in the sample, only a small fraction of the atom experience the loss of an electron. It would be very rare to measure the extraction of the second electron.

As for the HCl spectra from libretext you show, it is the PES spectrum of an HCl molecule so you have peaks for each molecular orbital in the molecule, it is not related to a 2nd ionization.

Note on PES: if you are learning about the technique, it is mostly used to study (the surface of) solids. So the results are generally presented with the notion of energy bands (s,p,f...) of elements in a solid, not molecular orbitals or ionization energy of single atoms (I am sure there are some exceptions).

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  • $\begingroup$ So how does the method tell you about the energies of different subshells when it won't necessarily eject all the electrons out? As far as I've looked into it, it measures the number of electrons and their kinetic energies that have been ejected out due to some incident light ray... Is that correct? $\endgroup$
    – Maddy
    Commented May 6 at 14:37
  • $\begingroup$ Yes, the binding energy of the electron in the atom is equal to the difference between the energy of the light and the kinetic energy of the ejected electron. In general, each electron you measure comes from a different atom, or at least the atom goes back to the ground state before being excited again. We are dealing with macroscopic samples, you are not stripping a single atom from its core electrons to get the spectra. $\endgroup$
    – Guillaume
    Commented May 6 at 15:39
  • $\begingroup$ So you mean that IE2 is rarely achieved say in the case of lithium... then how is it that we get data regarding the binding energy in the inner orbital (1s) in lithium? $\endgroup$
    – Maddy
    Commented May 6 at 16:28
  • $\begingroup$ You get an electron from the 1s shell ejected and its energy is measured. There is no need to remove other electrons from the atom. In PES, binding energies are measured, not successive ionization energies. You should read a little bit more on the technique and the way it works, I do not think you have a correct mental picture of how it works. $\endgroup$
    – Guillaume
    Commented May 6 at 19:06
  • $\begingroup$ I'm sorry, I am unable to intuitively explain this entire thing. I don't know how it would make sense to know about the binding energy and hence kinetic energy of an electron without knocking it off. Say, I'm calculating BE for 1s in Berrilium. Then I would have to first knock off electrons with higher energy, i.e. 2s electrons... isn't that it? So I would first knock off electron#1 with some IE (binding energy of 2s). The next electron would have a different IE and hence different KE, hence a different peak. It is only after that that we would be able to say anything about core electrons. $\endgroup$
    – Maddy
    Commented May 6 at 20:12
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@Guillaume is right... but I would really like to have a very clear and concrete explanation for anyone thinking about this in the future...

The key is in the fact that successive ionisations are very unlikely. An atom would lose the outermost electron and would ionise. This ion would be highly unstable. And we would rarely have enough time to even reach IE2 let alone other successive ionisations.

So to be clear, that is NOT how PES works. It's not a step-wise electron removal process where you kind of peel away layers of the electron cloud (that is what I thought it was but I was wrong).

Instead, given enough energy, core electrons can be ejected off even before the outermost electrons. Yes, it's a little confusing to think higher energy electrons are ejected off before lower energy electrons but that's how thermodynamics works in conjugation with Quantum mechanics. Its quite intuitive... the resonant energy of core electrons matches with the energy of the photons and thus they have a larger probability of being ejected. The relative number of ejected electrons at each binding energy then depends on the number of electrons there are in a particular degenerate set of orbitals. It's just a matter of probability.

You don't see the atom's layers of electrons peeling off... instead, you ground the entire thing. One layer gets off, and almost immediately is replaced with another layer of an electron from the ground. What is amazing is that you can have inner electrons being ejected too, without disturbing the outer electrons much.

It took me a while to realise how this works. Thank you for helping me.

https://physics.stackexchange.com/questions/41775/why-possibility-for-x-ray-to-excite-inner-electrons-higher-than-outer-electrons#:~:text=Yes%2C%20what%20the%20original%20poster,binding%20energy%20of%20the%20electron.

Here's another useful discussion that helped me get to this conclusion with a lot more clarity.

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