-2
$\begingroup$

I posted a question on Chemistry Stack Exchange a while back. It was related to the naming of atomic orbitals. One of the answers to it mentioned a research experiment. The link to the question is given below.

Size of Orbitals, Making Intuitive Sense of Quantum Model, Nomenclature of Subshells in the Quantum Model

As mentioned in the answer, alkali metals show various different bands of spectrum, on the basis of which we named the S, P, D and F orbitals.

But I didn't get why we don't have these seperate spectra in Hydrogen. (I've studied Hydrogen Spectrum in my classes, and it doesn't talk about anything like this. It only talks about Bohr using them to justify the quantisation of energy.) So what makes Hydrogen different?

Also, do elements other than alkali metals show a similar spectrum pattern or is it a single spectrum like Hydrogen? Which ones do? What's the defining criteria for a single spectrum and multiple spectrum?

$\endgroup$
3
  • 1
    $\begingroup$ The fact a hydrogen atom has just a single electron is a deal breaker. Similarly as for the classical gravitational analogy, the general analytical solution of the system state exists only for systems with 2 (point-like or spherically symmetric) objects. $\endgroup$
    – Poutnik
    Mar 25, 2022 at 9:34
  • 3
    $\begingroup$ The basis for orbital naming was the hydrogen spectrum not the alkali metal spectrum. And I have no idea what you mean by "we don't have these separate spectra in hydrogen". The biggest difference between the hydrogen spectrum and others is that hydrogen has only one electron making it far simpler (and analytically soluble). $\endgroup$
    – matt_black
    Mar 25, 2022 at 9:45
  • $\begingroup$ The H and alkali atomic spectra are very similar in structure. The main difference being that the upper levels of the alkalis are different in energy between the s, p, d orbitals because the inner electrons shield the nucleus. For H the levels with the same n are of equal energies. Herzberg, Atomic Spectra and Atomic Structure, Explains this nicely and in excruciating detail $\endgroup$
    – jimchmst
    Mar 25, 2022 at 21:50

1 Answer 1

4
$\begingroup$

As mentioned in my previous answer to your first query, sharp, principal, diffuse, and fundamental notation for modern orbital "labels" comes from alkali and alkaline earth metal spectra. This is very well established in historical spectroscopy. It did not come from analyzing hydrogen spectrum. Why? The reason is that these visual "labels" for spectrum lines come from the time when the electron was not discovered and it was very easy to study alkali and alkaline earth metals by electric arcs or sparks. It was just pure mathematical analysis (hence the name series spectra...the series is the empirical mathematical series).

Hydrogen atom is the simplest atom with only one electron. How did they find out why hydrogen has only one electron is another story. This was the first atom that was tackled with sound and rigorous physics principles known to man at that time. Those wonderful years were the mid-1920s for physics but perhaps not for mankind. In fact, hydrogen atom spectrum lines also showed "mathematical series" behavior, their names were Lyman, Balmer, Paschen, Pfund series etc, named after European spectroscopists. They did not call them sharp, principal, diffuse, fundamental series.

Today these s,p,d,f labels just have a notational meaning related to orbital angular momentum $l$ values in Schrodinger's equation. If $l$ =0, you label it as $s$, $l$ =1, it is $p$ etc. This is pure algebraic use. Do not associate any other historical meaning to them now.

$\endgroup$
2
  • $\begingroup$ Okay right, being a multielectron system makes a lot of difference over here. That is exactly why the Bohr model failed and we had to look up to Quantum Menchanics right, because it couldn't explain the spectra of multielectronic species. Thanks a lot, @M. Farooq! $\endgroup$
    – user122625
    Mar 25, 2022 at 14:37
  • 1
    $\begingroup$ Yes, multielectron atoms are still a challenge. $\endgroup$
    – AChem
    Mar 25, 2022 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.