Suppose we take an example of $\text{Be}^{2+}$ and we add two muons to it. Will they go into L shell or there will be two K shells - one each for muons and electrons? If so, the system should have double noble gas ($\text{He}$) configuration?


5 Answers 5


Muons, due to their higher mass, have a much smaller Bohr radius in an atom than electrons do. Muons have two spin states, just like electrons, so two muons can occupy the same orbital.

A Be muonic atom with two muons will have a core with a helium-like structure, with two muons surrounding the nucleus. This will be much smaller than a helium atom, though.

Around this small core, you'll have two electrons, also in a helium-like structure. To an electron, the core is not much bigger than a helium nucleus and has the same charge.

Thus, the result will be like a heavy helium atom.

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    $\begingroup$ However, just dropping muons onto a $\text{Be}^{2+}$ ion will probably not result in that ground state (my guess would be that it is much more likely for at least one of the electrons to be ejected from). At least the resulting thing will be in an excited state, and then the next question is which lifetime is longer: the one of the excited electron state, or of the bound muon. $\endgroup$ Jun 1 at 19:36

A nucleus with both bound muons and bound electrons will have separate shell structures for the distinguishable leptons. A useful comparison is the nuclear shell model, in which nucleons are arranged into shells according to their total angular momentum, but there are separate shell structures for the distinguishable neutrons versus protons.

The spectroscopy of a $ \rm Be^{4+} + 2\mu^- + 2e^- $ system would be quite difficult to actually study. First there is the problem that getting two muons to capture on the same nucleus, on the microsecond timescale before the muons decay, would require an extremely large flux of cold muons. Second, it's probable that the x-rays released when the muon captures on the ion would be energetic enough to liberate the remaining electrons. If I were going to dig around in the literature for examples of the interaction between bound muons and bound electrons, I would expect to find data on muon capture by heavier targets.


Here is a hand wavy explanation. Electrons are very light. They must be treated quantum mechanically, while the nucleus can be treated classically.

The nucleus is treated as a point source of charge that generates an electric field. The potential energy of the electrons depends on their position in the field.

For electrons, the uncertainty principal plays a bigger role. If you confine an electron to a small orbital, the uncertainty of position is small. Therefore the uncertainty of momentum is big and the electron probably won't stay near the nucleus very long. These two competing uncertainties determine the size of the orbitals.

A muon has a mass between that of an electron and nucleus. If confined to the same size orbital as an electron, the uncertainty of momentum would be the same. But the velocity would be smaller. It would take a smaller orbital to balance the uncertainty of momentum.

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    $\begingroup$ So will there be two 1s shell or it will modify in some type of spectroscopic arrangements having 4 level $\endgroup$
    – Pradyuman
    Jun 1 at 14:19

The K,L,M...shells in atoms are a short hand way of describing the quantum mechanical complexity of the solutions for an atom. The K shell for electrons comes from solving the bound state for particles with the mass of the electrons. If instead of electrons you have particles with the mass of muons ( mind you the muons will decay very fast and get out of the bound state) it will be different orbitals in dimensions.

One must not confuse orbitals of the solutions for muons or for electrons with the shells which count by the closest to the nucleus K, next L..... The displacement of electrons will happen to the outer orbitals and only x-rays and harder radiation can hit out an electron from a K shell (the closest orbital to the nucleus) because they are very tightly bound.

You cannot just add two muons to an atom without changing the potentials that calculate the orbitals, so it is a no go problem you are discussing.

To the title question:

Do muons have different atomic shells than electrons?

The answer is yes, but this can happen on the same atom with the muons on their own orbital to the nucleus. . As the other answers explain because of its mass it will be practically in the nucleus( before decaying), and its shell might be be named as a K shell, but in a strict shell vocabulary the electrons will be in an L shell, as the next occupied orbital.

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    $\begingroup$ Assuming purely hydrogen-like solutions, the muons and electrons have different orbitals. Experimentally, the muon (captured in excited states for various reasons) rapidly relaxes down to the lowest muon level, which is not an electron level. Two muons in the lowest energy state do not preclude there being two electrons in an $s$ electron state. $\endgroup$
    – Jon Custer
    Jun 1 at 14:37
  • $\begingroup$ @j Yes, but have you read what the KL... means, K is the innermost orbital $\endgroup$
    – anna v
    Jun 1 at 16:04
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    $\begingroup$ Ok but the muons are about 200x bigger, so their radius will be 200x closer to the nucleus. Thus, from the view of the electron cloud, the muons will be essentially part of te nucleus. Thus, I think, we will have our K,L,M shells for the muons working independently for the electrons, and the chemical properties of that muonic beryllium atom will be very similar to helium (simply because the electron cloud will see essentially a 2+ charged "nucleus"). $\endgroup$
    – peterh
    Jun 1 at 16:21
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    $\begingroup$ Sad that munos decay so quickly, the world of this muonic chemistry would be so awesome. $\endgroup$
    – peterh
    Jun 1 at 16:23
  • $\begingroup$ @peterh Not only that, but what about muonics (the muon equivalent of electronics)? $\endgroup$
    – Michael
    Jun 3 at 0:43

Atomic shells arise from the Pauli exclusion principle which states that multiple identical fermions may not inhabit the same quantum state. Since muons are not identical to electrons, they will occupy states independently of the electron population. As other answers have already stated, the larger mass of muons will lead to their orbitals being much smaller than electrons.

As further example, consider a hypothetical fermion identical to the electron in every way except possessing a lepton number of 2. Adding two of these particles to Be$^{2+}$ will result in them both occupying $s$ orbitals identical to the inner electron orbitals.


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