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I'm studying some chemistry on my own in anticipation for the new school year and in my book, I came across the Rydberg equation for the first time. I worked through some examples and everything was fine until I came across this comment on the question, "Calculate the wavelength of the radiation released when an electron moves from n=5 to n=2":

For future reference: the Rydberg formula only works for hydrogen-like atoms.

What is meant by "hydrogen-like"? I've heard that solving for multi-electron systems is (near) impossible, so I understand why hydrogen is used here, but I don't understand what "hydrogen-like" is.

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A hydrogen-like atom (or ion) is simply any particle with a nucleus and one electron.


That should be sufficient to answer the question at hand, but I thought I should say a bit more, as some of these answers are potentially confusing.

The historical reason why the Rydberg formula only works for hydrogen-like atoms is because it was originally formulated to explain the spectral lines of hydrogen. It was never intended to explain the spectra of multi-electron atoms.

The physical reason, though, is because the Rydberg formula uses energy levels that only depend on the principal quantum number $n$, which has to be a positive integer:

$$\bar{\nu} = Z^2\mathcal{R}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \qquad n_1,n_2 \in \mathbb{Z}^+$$

and nowadays we know that this is only true for hydrogen-like atoms;$^*$ the energy levels of multi-electron atoms depend on both $n$ and $l$.$^\dagger$


The $n$-dependency was later successfully rationalised by the Bohr model, but saying that "the Rydberg formula only works for hydrogen-like atoms because the Bohr model only works for them" is misleading and misses the point, as:

  1. This implies that the Rydberg formula was derived from the Bohr model, which is not true; it was merely empirically determined, and the formula predated the Bohr model by 25 years.
  2. The Bohr model simply does not work for hydrogen-like atoms. The fact that it reproduces the Rydberg formula should merely be considered a serendipity; Bohr arrived at the correct result by the wrong method.
  3. It does not lend any real insight into the proper reason why the Rydberg formula does not apply for helium, etc. (which I briefly mentioned above).

$^*$ In fact, the energy levels of hydrogen are not only dependent on $n$ (due to various small effects such as – but not limited to – spin-orbit coupling, and hyperfine splitting). Wikipedia has a good overview of the topic here and most QM textbooks have a chapter on the hydrogen atom, where they discuss these perturbations to the Hamiltonian and their effects on the energies. Not surprisingly, the inability to explain this was one of the failures of the Bohr model.

$^\dagger$ Of course, there is also a series of approximations here. The energy levels of multi-electron atoms are only approximately described by sums of orbital energies, so the transition energies are only approximately equal to a difference in energy between two orbitals.

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  • $\begingroup$ I like your answer, but I think you confuse Balmer's formula with Rydberg's formula. Rydberg was aware of Balmer's equation predicting the spectral positions of the hydrogen lines and Rydberg successfully(!) tried to find a similar expression for non-hydrogenic elements (see Z. Phys. Chem. 5, 227 (1890)). He did this by introducing an $\ell$ dependent quantum defect, which for H is equal to zero. $\endgroup$ – Paul Aug 15 '17 at 15:49
  • $\begingroup$ @Paul Thanks for your comment and sorry for my late reply. I didn't know that! Wikipedia isn't entirely clear on it, and the different form of the equations used back then don't help, but from what I can tell: (1) Balmer found the formula in my answer for the specific case of $n_1 = 2$; (2) Rydberg extended it to alkali metals introducing a quantum defect; (3) the formula above is a specific case of Rydberg's formula with quantum defects set to zero. Am I understanding it right? I just wanted to make sure before I edit. $\endgroup$ – orthocresol Aug 27 '17 at 17:08
  • $\begingroup$ sorry for my late reply as well, I was on holiday the last weeks... You're absolutely right. $\endgroup$ – Paul Sep 18 '17 at 9:08
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Hydrogen-like atoms are atoms with a single electron "orbiting" a nucleus which has more than one nucleon. As @Xerxes pointed out in a comment, you can in principle have a nucleus made up of particles other than protons and neutrons (nucleons). Positronium might be an extreme example of this.

Wikipedia actually has an entry about Hydrogen-like atoms which goes somewhat beyond what you asked.

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    $\begingroup$ There are technical issues with this answer: The core of a hydrogen-like atom need not be made of nucleons. Also, deuterium has more than one nucleon and it is not hydrogen-like, it is hydrogen. $\endgroup$ – Xerxes Jul 22 '17 at 16:55
  • $\begingroup$ And now it is no surprise what a helium-like ion is. $\endgroup$ – Jeppe Stig Nielsen Jul 22 '17 at 20:35
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    $\begingroup$ I'd argue that hydrogen counts as a hydrogen-like atom. But that's splitting hairs. $\endgroup$ – orthocresol Jul 24 '17 at 5:09
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Hydrogen-like ions are ions that possess only one electron, just like a hydrogen atom.

The fact that these ions have only one ion in the outermost shell makes it simpler to analyze their radii and energies, as a simple electrostatic model can be used to describe them. Species having multiple electrons are difficult to study, and are beyond the scope of the Bohr model. This is because the inter-electronic replusions are hard to account for in electrodynamic interactions that makes up the atom's bound system.

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    $\begingroup$ You don't actually need electrodynamics for multi-electron atoms more than you do for hydrogen atom. It's just that Schrödinger's equation for multi-electron atoms is vastly more difficult to solve — even in its purely electrostatic form. $\endgroup$ – Ruslan Jul 22 '17 at 6:28
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    $\begingroup$ The Bohr model is just a historical viewpoint, it really shouldn't be used to rationalise these things anymore. Electrodynamic interactions don't need to enter the Hamiltonian, as Wildcat explained (the full Hamiltonian is here). The difficulty arises from the electron-electron repulsions. $\endgroup$ – orthocresol Jul 22 '17 at 8:39
  • $\begingroup$ @orthocresol Yes, I am very much aware of that. However this is a question about something regarding Bohr's model and that the reason why Bohr's model doesn't work with multielectronic atoms is because Bohr's model uses electrostatic interactions. $\endgroup$ – Pritt Balagopal Jul 22 '17 at 9:59
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    $\begingroup$ I don't see a single mentioning of Bohr's model neither in the OP nor in the post linked there. And as Bohr's model doesn't work correctly even for hydrogen atom (e.g. ground-state angular momentum), I see no reason to use it at all, at least in the context of current question. $\endgroup$ – Ruslan Jul 22 '17 at 10:17
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    $\begingroup$ On top of the previous comment, the Rydberg formula was only empirically determined, and is therefore not tied to any particular theory. It is not derived from the Bohr model; in fact, the Bohr model was developed to rationalise it. However, the use of the Bohr model is no longer necessary, as the QM model accounts for the Rydberg formula perfectly well. So this is not "a question about something regarding the Bohr model". $\endgroup$ – orthocresol Jul 22 '17 at 10:38
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In the spirit of notable University Of Nottingham chemist Sir Martyn Poliakoff's comment in the Periodic videos episode Helium in Disguise (also YouTube):

Again and again, for scientists, if they hear something surprising that makes them think in a different way, it’s really good.

I'll add to the other excellent answers by listing some less common hydrogen-like "atoms".

The video is about Muonic Heavy Hydrogen or so called ${}^{4.1}H$ is discussed at the end of this list. Yes, that's pronounced "hydrogen four-point-one." See the New Scientist article Atomic disguise makes helium look like hydrogen, and the primary article published in Science Kinetic Isotope Effects for the Reactions of Muonic Helium and Muonium with H2 Fleming, D. G. et al. Science 28 Jan 2011: Vol. 331, Issue 6016, pp. 448-450, DOI: 10.1126/science.1199421


Positronium: (e+e−)

Positronium (Ps) is a system consisting of an electron and its anti-particle, a positron, bound together into an exotic atom, specifically an onium.

It has Rydberg states with $E_n \sim -(6.8eV)/n^2$ or half that of normal hydrogen, because the reduced mass is half that for an electron bound to a much heavier item. The longer-lived triplet ${}^3S_1$ state has a mean lifetime of about 142ns, and decays by electron positron annihilation into three gamma-ray photons. It is sometimes studied by slowing and stopping positrons in powdered MgO where they capture an electron and tend to remain relatively unperturbed by the other atoms.


Muonium: (μ+e−)

Muonium is an exotic atom made up of an antimuon and an electron, which was discovered in 1960 and is given the chemical symbol Mu. During the muon's 2.2 µs lifetime, muonium can enter into compounds such as muonium chloride (MuCl) or sodium muonide (NaMu). Due to the mass difference between the antimuon and the electron, muonium (μ+e−) is more similar to atomic hydrogen (p+e−) than positronium (e+e−). Its Bohr radius and ionization energy are within 0.5% of hydrogen, deuterium, and tritium, and thus it can usefully be considered as an exotic light isotope of hydrogen.


True Muonium: (μ+μ−)

True muonium or muononium is an exotic atom made up of an antimuon and a muon. It is yet to be observed, but it may have been generated in the collision of electron and positron beams.


Muonic hydrogen:

Negative muons can, however, form muonic atoms (previously called mu-mesic atoms), by replacing an electron in ordinary atoms. Muonic hydrogen atoms are much smaller than typical hydrogen atoms because the much larger mass of the muon gives it a much more localized ground-state wavefunction than is observed for the electron.


Muonic helium:

Muonic helium is created by substituting a muon for one of the electrons in helium-4. The muon orbits much closer to the nucleus, so muonic helium can therefore be regarded like an isotope of helium whose nucleus consists of two neutrons, two protons and a muon, with a single electron outside. Colloquially, it could be called "helium 4.1", since the mass of the muon is slightly greater than 0.1 amu. Chemically, muonic helium, possessing an unpaired valence electron, can bond with other atoms, and behaves more like a hydrogen atom than an inert helium atom.

Muonic heavy hydrogen atoms with a negative muon may undergo nuclear fusion in the process of muon-catalyzed fusion, after the muon may leave the new atom to induce fusion in another hydrogen molecule. This process continues until the negative muon is trapped by a helium atom, and cannot leave until it decays.

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above: From New Scientist.

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I think one key aspect that is overlooked in these replies is the Born-Oppenheimer approximation. The nucleus of any atom can be approximated as a point with very few relativistic correction due to its mass. An electron on the other hand is essentially massless (comparatively speaking. I know it is still a fermion and has mass). Once you put more than one electron in any orbital those relativistic corrections have consequences. So ''hydrogen-like'' means any atom that does not need Pauli or Dirac's insight to explain the deviations in the spectra i.e. the bohr model.

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    $\begingroup$ I am not sure if you don't confuse some things. For example you can straight forward analytically solve the hydrogen atom considering the electrons and protons mass. You simply switch to center of mass coordinates. Levels change by roughly 1+(1/(1+1836)). Thats actually beyond BO. Relativity you need for high "nuclear" effective charges $Z$. Thats quite different things. $\endgroup$ – Rudi_Birnbaum Jul 22 '17 at 20:25
  • $\begingroup$ That is a good point. I guess I thought that Dirac applied relativity to explain the Pauli exclusion principle. In the case of the hydrogen atom you can simply switch to the center of mass. Is that the same as normal coordinates? I thought that is what the BO approximation did to limit the 3N degrees of freedom thus reducing the amount of parameters needed to approximate more complicated systems i.e anything past a two body problem. $\endgroup$ – Shawn CoteBurk Jul 23 '17 at 18:32
  • $\begingroup$ When you solve the H-atom problem the "conventional way" you first use the BO approximation and then fix the proton to the origin. When you solve it in the way I have scetched you dont apply the BO and you don't fix the proton. Rather you decompose the problem in relative motion of the two wrt centre of mass and the motion of the centre of mass (which you don't care about since thats just plane wave). the relative motion problem is completely identical to the conventinal H-atom except you obtain a different effective mass, hence energy levels. $\endgroup$ – Rudi_Birnbaum Jul 24 '17 at 7:48
  • $\begingroup$ That makes sense. Thanks for the clarification. $\endgroup$ – Shawn CoteBurk Jul 24 '17 at 19:49
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Single electron systems are known as hydrogenic / hydrogen like species like He+, Li2+, Be3+ etc.

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