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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, 2016 at 15:48
  • $\begingroup$ it may be electron or positron $\endgroup$
    – user46925
    Jan 25, 2016 at 15:49
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    $\begingroup$ No, its an electron (note the superscript -_. $\endgroup$
    – Lewis Miller
    Jan 25, 2016 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, 2016 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, 2016 at 16:32

2 Answers 2

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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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  • $\begingroup$ For #1, isnt lepton number conserved by the anti-neutrino for the newly created electron? Why include virtual neutrinos and imply that a neutrino can change into an electron? If a neutrino is required to start beta decay then does that mean half life of radioactive material is proportional to the local density of neutrinos? $\endgroup$
    – Jason
    Jan 28, 2022 at 1:32
  • $\begingroup$ @Jason no it does not react with a real neutrino that is passing by. The amount of energy is available to pull it from the vacuum. $\endgroup$
    – JDługosz
    Jan 31, 2022 at 15:17
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Leptons are not present inside the nucleus or in the nucleons(protons and neutrons). Instead what happens is the W- boson is created and this turns the down quark into an up quark. The W- boson then gets converted into an electron and an electron antineutrino.

The leptons are created from W- boson and are fundamental.

The mass of the ejected particles comes from the energy of turning the neutron into a proton.

The creation of leptons are similar except opposite if the W+ is used and there are no changes if Z boson is used.

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