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In beta minus decay, beta-minus particle and anti-neutrino are ejected, leaving behind daughter nucleus. $\beta^-$ and anti-neutrino both are leptons.

  1. Were the leptons already present in the nucleus in some form?
  2. Weak interactions are responsible for various processes here (and transformation of bosons).But, anyhow, if the above leptons are created, then can we call leptons elementary & indivisible?
  3. Are above leptons mass equivalent of some released energy?
  4. Is the transformation of quarks (neutron to proton conversion) and bosons ($W^+,W^-,Z$) the only cause of creation of above leptons? ...SIMILAR process for beta-minus and plus decay. Only neutron-proton conversion opposite, there's positron instead of electron, neutrino instead of anti-neutrino. My question about creation of leptons remains the same.
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  • $\begingroup$ What's a $\beta^-$ particle? $\endgroup$ – pfnuesel Jan 25 '16 at 15:48
  • $\begingroup$ it may be electron or positron $\endgroup$ – user46925 Jan 25 '16 at 15:49
  • $\begingroup$ No, its an electron (note the superscript -_. $\endgroup$ – Lewis Miller Jan 25 '16 at 16:10
  • $\begingroup$ SIMILAR process for beta-minus and plus decay. Only neutron-proton conversion opposite, there's positron instead of electron, neutrino instead of anti-neutrino. My question about creation of leptons remains the same. $\endgroup$ – HEU Jan 25 '16 at 16:21
  • $\begingroup$ all "decay"s ( bad word ) are possible if merely the reactions respect the conservations and the exclusions rules $\endgroup$ – user46925 Jan 25 '16 at 16:32
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  1. No.

    A pair of neutrinos is pulled from the vacuum. One of them interacts with one of the quarks via the weak force, and they both change identity: the quark to another kind, thus changing the neucleon; the neutrino to an electron, which escapes. (The negative charge unit also moved from the quark to the lepton.) The electron escapes as the beta radiation, along with the anti-neutrino that goes unnoticed.

  2. They are created as opposed pairs. They are elementary. The W doesn't leave the diagram if you draw a Feynman diagram. If you elaborate more you can show that this can be represented as a temporary pair, but we usually don't.

  3. The mass of the created particles indeed is counted. It comes from the potential energy in the binding energy of the neucleus: that's why it decays! Changing a neutron to a proton releases energy as it's bound tighter, and that more than pays for the particle mass and kenetic energy.

  4. Yes, the weak interaction is "the same" between pairs of quarks or leptons of various kinds. That's a key symmetry and organizational principle in understanding the standard model particles. In all cases using W, one particle changes to its partner with the different charge. With Z it's just like electrtic force in that it doesn't change the types or move the charge around.

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