$beta^+$ decay question

Question

I read that all baryons apart from the proton itself decay into protons (why though?) and that mesons do not decay into protons due to having less mass than protons.

Thus it makes sense for the $beta^-$ decay to occur as the neutron decays to the proton, but how does the $beta^+$ decay take place if the proton does not decay at all?

Bill N · Answer

Under certain conditions, such as having a high charge/mass ratio (not enough neutron constituents) the nucleus will cast off some charge, but doesn't have enough energy to eject a whole proton.  Effectively, in the field of a large mass, a positron and neutrino are ejected, the atomic number (Z) of the daughter nucleus is reduced by 1 compared to the parent, and the mass number (A) is unchanged. The daughter nucleus becomes more tightly bound (it has less mass energy than before).  This reaction does not appear to take place without the presence of other mass.

The proton by itself is mass restricted from decaying into a neutron plus positron ($Q = -1.804$ MeV) or even electron capture ($Q = -782$ keV). But the proton-proton $to e^+ + nu$ has a $Q= +420$ keV, so there is enough mass-energy present in the center of mass for the deuteron and positron to form.

The neutrino appears because in addition to conserving energy, momentum, charge, and baryon number, something called lepton number is conserved in the reaction.

$beta^+$ decay question

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