Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is a beta decay possible with out the emission of an electron in the process ?

Beta decay involves the formation of a proton and an electron from a neutron.

share|cite|improve this question

migrated from Apr 3 '13 at 13:59

This question came from our site for scientists, academics, teachers and students.

Since the definition of beta decay contains the emission of a $\beta^-$ particle (another word for an electron), the question is meaningless.

However, if you define beta decay as "neutron becomes a proton", then yes, it is possible.

This is "normal" beta decay:

enter link description here

A down quark becomes an up quark, in the process releasing a $W^-$ boson. The boson then decays into a $e,\bar \nu_e$ pair

However, the $W^-$ boson can also decay into a:

  • $\tau,\bar\nu_\tau$ pair
  • $\mu,\bar \nu_\mu$ pair
  • A quark pair consisting of one of $\{d,s,b\}$ and one of $\{\bar u, \bar c, \bar t\}$. ($\pi^-,K^-$, etc)

in which case you will have a phenomenon similar to beta decay with different particle(s) coming out.

However, these are kinematically impossible. They may be possible with some photon capture/emission, but I doubt it.

share|cite|improve this answer
Worth mentioning that all these other processes are kinematically forbidden. :) – Michael Brown Apr 3 '13 at 14:13
@MichaelBrown: Oh, right. Damn. Might be possible with some photon capture/emission, though (doubt it) – Manishearth Apr 3 '13 at 14:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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