Neutrinos are detected when they 'hit' a nucleus and are re-directed (z-boson) or 'create' a leptons that matches the type of neutrino....

What happens when the neutrino(s) in a neutrinoless double beta decay are (re)absorbed by the nucleus? Does the nucleus have to emit a third electron?

  • $\begingroup$ There is actually one neutrino emitted by one nucleon and then absorbed in the nucleus (which would violate lepton number since two electron decay). I think this process is mediated by the $W^-$ boson and not the neutral Z boson. Why do you think the nucleus will then have to re-emit another third electron? $\endgroup$
    – joseph h
    Sep 4, 2020 at 23:26
  • 1
    $\begingroup$ WP. $\endgroup$ Sep 5, 2020 at 0:49

1 Answer 1


In neutrinoless double beta decay, two electrons are released. So, using your model language: the first nucleus emits one electron and one antineutrino. This antineutrino is then re-absorbed as a neutrino by the other nucleus, which in turn too emits one electron. So there's only two electrons. From wikipedia comes this plot that says it all:

enter image description here

  • $\begingroup$ Or it could be two positrons. $\endgroup$
    – J.G.
    Sep 22, 2021 at 19:45
  • $\begingroup$ well, yes, but then it's called neutrinoless double positron decay $\endgroup$
    – rfl
    Sep 22, 2021 at 19:52

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