Sorry for being ignorant, but I'm in high school and our chemistry teacher barely went over beta decay. I decided to do some research and learned that in β+ decay, positrons are emitted from protons in order to turn it into a neutron. But positrons have mass, so where does that mass come from? Do protons "give it mass"? If so, why wouldn't protons lose mass, and how could they become neutrons if neutrons are more massive?

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    $\begingroup$ $\beta^-$ decay turns a neutron into a proton and an electron, not a proton into a neutron and an electron. $\endgroup$ – HDE 226868 Oct 12 '15 at 18:18
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    $\begingroup$ Sorry, fixed what I meant. $\endgroup$ – Sir Cumference Oct 12 '15 at 18:19
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    $\begingroup$ Ah. See Wikipedia: However, β+ decay cannot occur in an isolated proton because it requires energy due to the mass of the neutron being greater than the mass of the proton. β+ decay can only happen inside nuclei when the daughter nucleus has a greater binding energy (and therefore a lower total energy) than the mother nucleus. $\endgroup$ – HDE 226868 Oct 12 '15 at 18:21
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    $\begingroup$ The up quark emits a W boson and turns into a down quark; the W boson then decays, producing a positron and a neutrino. The energy comes from the difference in binding energy. See also physics.stackexchange.com/questions/92747/…. $\endgroup$ – HDE 226868 Oct 12 '15 at 18:24
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    $\begingroup$ You can consider them to be virtual particles. $\endgroup$ – HDE 226868 Oct 12 '15 at 18:32

In beta decay, the mass difference between the parent and daughter particles is converted to the kinetic energy of the daughter particles. For instance, in the decay of the free neutron, $$ \rm n \to p + e^- + \bar\nu_e, \tag{$\beta^-$ decay} $$ the difference between the mass on the left and the mass on the right is about $0.78\,\mathrm{MeV}/c^2$, and this is the energy liberated in the decay. (If you're a chemistry person, an eV is a useful energy unit; the $E=mc^2$ conversion is roughly $1000\,\mathrm{MeV}\approx 1\,\mathrm{amu}\times c^2$.) Equivalent processes like $$ \rm p + \bar\nu_e \to n + e^+ \tag{neutrino capture} $$ don't occur unless the kinetic energy on the left side is already large enough to account for the extra mass on the right side. Since the electron/positron mass is about $0.51\,\mathrm{MeV}/c^2$, neutrino capture on protons at rest is impossible for neutrinos with less than $1.80\rm\,MeV$ kinetic energy. This means, among other things, that neutrinos emitted from neutron decay at rest will never have enough energy to cause positron emission on protons at rest elsewhere.

You get $\beta^-$ decay from free neutrons because free neutrons are heavier than free protons. However it's not the case for all nuclei that the more positive isobars are less massive. For instance, the mass difference between postassium-40 and argon-40 is about $1.50\,\mathrm{ MeV}/c^2$, with potassium (19 protons) heavier than argon (18 protons), so the decay $$ \rm ^{40}_{19}K \to {}^{40}_{18}Ar^- + \beta^+ + \nu_e + 0.48\, MeV $$ is allowed (though rarer than some other branches) and merrily proceeding inside the bananas on your kitchen counter.


In $ β^+$ decay, what generally happens is, the weak interaction converts an atomic nucleus into a nucleus with atomic number decreased by one while emitting a positron ($e^+$) and an electron neutrino ($ν_e$). $$ _A^ZX → ^A_{Z-1}Y + e^+ + v_e $$

This may be considered as the decay of a proton inside the nucleus to a neutron $$ p → n + e^+ + v_e $$

$β^+$ decay generally occurs in proton-rich nuclei. However, $β^+$ decay cannot occur in an isolated proton because it requires energy, due to the mass of the neutron being greater than the mass of the proton.

For example, let’s take the nuclear reaction

$$ \rm ^{40}_{19}K \to {}^{40}_{18}Ar^- + \beta^+ + \nu_e + 0.48\, MeV $$

In the above reaction, a proton of $^{40}_{19}K$ is converted into a neutron, decreasing the atomic number by 1 and making it $^{40}_{18}Ar^-$and keeping the mass number same. This also emits a positron ($β^+ $or $e^+$ ) and an electron neutrino and 0.48MeV of energy. Now It may seem like the mass and energy on RHS is greater than on LHS, arising doubt on the conservation of mass/energy.

But, that's not how $β^+$ decay works. $β^+$decay can only happen inside nuclei when the daughter nucleus has greater binding energy (and therefore lower total energy) than the mother nucleus. The difference between these energies goes into the reaction of converting a proton into a neutron (as energy is equivalent to mass by the relation $E=mc^2$), a positron and a neutrino and into the kinetic energy of these particles.

So, it’s a fact that binding energy of $^{40}_{18}Ar^-$ (8.595261375 MeV) is greater than that of the binding energy of$ ^{40}_{19}K$ (8.5380806 MeV), and that means, total energy of the nucleus of $^{40}_{18}Ar^-$ is lesser than the total energy of the nucleus of $ ^{40}_{19}K$. This difference in energy makes up that little extra mass needed for the proton to become a neutron and to emit a positron(has mass) while neutrino can be considered massless. This difference in energy also contributes to the extra kinetic energy attained by the emitted particles after the reaction and also that 0.48MeV is part of this energy.

Remember that, mass need not always be conserved and so is energy ( if you think mass and energy are 2 different entities). In reactions, especially in nuclear reactions, mass sometimes gets converted into energy and vice versa. But the total mass + energy of the reaction has to be conserved.


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