An antimuon and electron may bind together via Coulombic attraction and then decay, but is the following process possible? $$(\mu^++e^-) → \gamma + \gamma$$

$\mu^+$ : antimuon

$\gamma$ : photon

$e^-$ : electron

Which of the following claims are correct?

a) This decay mode can proceed naturally.

b) This decay would violate electron-lepton number. $$0 + 1 (e) \rightarrow 0 + 0$$

c) This decay would violate muon-lepton number. $$-1(\mu^+) + 0 \rightarrow 0 + 0$$

d) This decay would violate tau-lepton number. $$0 + 0 \rightarrow 0 + 0$$

e) This decay would violate charge conservation. $$+1 - 1 \rightarrow 0 + 0$$

The antimuon has a lepton number of -1. The electron has a lepton number of +1. So lepton number isn't violated - that takes out b) c) d).

I thought annihilation had to occur with a particles respective antiparticle. i.e. electron-positron, etc. Meaning a) is also not correct.

Charge of the electron is -1, and antimuon is +1. therefore charge would be conserved - ruling out e)

Of all my assumptions here a) is the only one I'm not too sure about.

  • $\begingroup$ Lepton numbers of each kind are conserved separately. Also, I'm not very sure what those 5 points are... Are they general claims? Options in an MCQ? But the decay mode won't proceed, and points (d) and (e) are false. $\endgroup$
    – user191954
    Jul 18 '18 at 11:53
  • $\begingroup$ Yes, sorry it's a MCQ. $\endgroup$
    – says
    Jul 18 '18 at 13:20
  • $\begingroup$ Are you supposed to isolate 1 of those 5 claims which is wrong, or 1 which is right, or something else? Because unless I'm severely confused, both c and d are correct, but a, d, and e are incorrect propositions... $\endgroup$
    – user191954
    Jul 18 '18 at 13:24
  • $\begingroup$ No, there's more than one correct answer. So in this case I'd say it's b) and c) $\endgroup$
    – says
    Jul 18 '18 at 13:25
  • $\begingroup$ See physics.stackexchange.com/questions/403834/… $\endgroup$ Jul 18 '18 at 18:58

There isn’t one type of lepton number; every generation has one and all of them have to be conserved seperately.

  • 2
    $\begingroup$ Not quite correct. There is "lepton number" and there is "lepton family number". Lepton family number applies to each generation separately, while lepton number applies to all leptons in general. Neutrino oscillation does not preserve lepton family number, but does conserve lepton number. $\endgroup$
    – ohwilleke
    Jul 18 '18 at 13:32
  • $\begingroup$ Could the antimuon and electron decay to an electron neutrino and muon neutrino? (µ+e-) → νe + νµ . Charge would be conserved, lepton numbers conserved. $\endgroup$
    – says
    Jul 18 '18 at 14:36
  • $\begingroup$ @says what you are writing is an interaction, not a decay, to start with. Single particles decay, two interact. The interaction has to conserve muon number and electron number and to be ok it should be a mu+ and an antimuon on the right sid , charge has to be conserved as well as muon lepton number and electron lepton number, on the right. $\endgroup$
    – anna v
    Jul 18 '18 at 17:26
  • $\begingroup$ i of course meant antimunneutrino :( on the right side $\endgroup$
    – anna v
    Jul 19 '18 at 8:54

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