8
$\begingroup$

This question is closely related to my previous question Bound states in QED.

Muonium is a system of electron and anti-muon. This article in wikipedia claims that muonium is unstable.

QUESTION: Why it is unstable? Is it due to presence of weak interactions, or it can be explained only in the framework of QED (involving both electrons and muons and their anti-particles).

To compare with, the positronium, consisting from electron and positron, in an unstable particle (see e.g. this article in wikipedia). It can annihilate to two photons. This effect can be explained by the usual QED. This explanation does not seem to work for the muonium.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ @BebopButUnsteady: Is there a literature where this is discussed in detail? $\endgroup$
    – MKO
    Jul 21, 2011 at 17:17
  • 2
    $\begingroup$ @MKO: I'm sure there's a literature on muonium but I don't really know a great place to point you. I can tell why what I said is true though. With only QED, not only is the total charge conserved, but the number of muons minus antimuons is conserved and the number of electrons minus positrons is conserved. So if you start with an electron and a antimuon, you can't get rid of them, and so their lowest energy state (which is muonium by definition) must be stable. Whereas turning an electron and a positron into two photons is still allowed, so that decay happens. $\endgroup$
    – BebopButUnsteady
    Jul 21, 2011 at 17:32

3 Answers 3

Reset to default
9
$\begingroup$

Let's start with nuclear beta decay as a similar example. 12C, with 6 protons and 6 neutrons, is stable. As Marek has pointed out, it's not enough just to say that neutrons are unstable. If all neutrons were unstable with respect to beta decay, then 12C could decay into 12N. The reason this can't happen is that the mass of a 12N nucleus is higher than the mass of a 12C. That means that if the decay were to proceed, the total mass-energy of the products (12N, electron, and antineutrino) would be greater than the initial mass-energy of the 12C.

In the muonium example, the muon wants to decay into a positron plus some neutrinos. The difference between the mass of a muon and the mass of a positron is 105.147 MeV. (The masses of the neutrinos are negligible.) This decay would be forbidden if muonium's binding energy was -105.147 MeV or stronger. But it's not. The binding energy of muonium is basically the same as the binding energy of hydrogen, about 14 eV.

Improve this answer
$\endgroup$
7
$\begingroup$

Here is a rough estimate why the muonium atom is unstable against the muon beta decay (in contrast to a neutron in a deutron). The binding energy of the electron in a muonium atom neglecting the reduced mass effect is approximately equal to its binding energy in a Hydrogen atom according to the Bohr's model, i.e., 13.6 ev. The muon performs a beta decay to a positron and a neutrino. The rest mass difference between the muon and the positron is about 105.6 - 0.5 = 105.1 Mev. Thus the binding energy is so small compared to the mass difference, thus the muon decay mode is much favorable. In contrast the mass of the deutron is 1875.6 MeV while the rest mass of the outputs of a neutron beta decay + a proton is 2(938.27 MeV) + 0.511 MeV = 1877.05. Thus the deutron favors the stable state.

Improve this answer
$\endgroup$
3
$\begingroup$

As Marek said, it is as unstable as any other radioactive atom. It is a weak decay of muon or antimuon that destroys the muonium.

Improve this answer
$\endgroup$

Your Answer

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

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