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It is difficult for me to understand how beta decay releases energy. The neutron is more massive than a proton, electron and antineutrino combined. Would this not imply that energy is being converted to mass rather than being released?

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  • $\begingroup$ For a free neutron the decay $n \to p + e + \bar{\nu}$ is energetically possible. I think you mean to ask about the reverse decay $p \to n + \bar{e} + \nu$? $\endgroup$ – jacob1729 Oct 20 '19 at 17:41
  • $\begingroup$ Yes. I understand this can happen especially inside atomic nuclei. Excuse my ignorance I'm no physicist by profession. $\endgroup$ – Derek Seabrooke Oct 20 '19 at 17:43
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The free decay:

$$ p^+ \to n + \bar{e}^+ + \nu_e$$

cannot occur in isolation for the reason the OP has noted: in the proton's rest frame the system has energy $m_p$ whilst the products have at least $m_n+m_e$ of energy (plus whatever the kinetic energy is) - since this exceeds $m_p$ the reaction violates energy-momentum conservation. It is kinematically disallowed.

However the same reason does not necessarily forbid the similar decay:

$$ ^A_Z X \to ^A_{Z-1}Y + \bar{e}^+ + \nu_e$$

since the mass of a nuclide is not given by the sum of the neutron and proton masses but instead also has a mass deficit caused by binding energy. It is thus possible for the masses in this equation to work out.

In particular, the energy of $X$ will be larger than that of $Y$ due to Coulomb interactions amongst the protons. Decreasing the number of protons will reduce this repulsion and lower the overall energy. Of course, there are lots of other contributions to the mass which is why nuclei tend to end up with roughly equal numbers of protons and neutrons, with slightly more of the latter. A first step towards understanding this is provided by the liquid drop model of the nucleus and in particular by the semi-empirical mass formula.

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  • $\begingroup$ So you're implying nucleons are actually less massive inside a nucleus and than they are free? Wouldn't that be similar to saying that a person has less mass on the Earth than they do in space? Also this would seem to imply that within the nucleus protons have more mass than neutrons which is the opposite of normal. $\endgroup$ – Derek Seabrooke Oct 20 '19 at 19:29
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    $\begingroup$ No, the mass of a nucleus is a measure of its energy content. That energy comes from a number of sources: rest mass energy, kinetic energy, as well as potential contributions from the electrostatic and strong forces. The net effect is that the energy of a bound state is less than the energies of the separated constituents, because if it were otherwise the state wouldn't be bound. $\endgroup$ – jacob1729 Oct 20 '19 at 19:32
  • $\begingroup$ @DerekSeabrooke You have to realize that it is not Galilean mechanics that can describe quantum mechanical systems, but special relativity. The formalism of four vectors helps in understanding this. An aggregate of quantum entities (in this case protons and neutrons) has the added four vectors of the constituents, and the length of that four vector is the $mass^2$ hyperphysics.phy-astr.gsu.edu/hbase/Relativ/…. $\endgroup$ – anna v Nov 11 '19 at 5:37

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