So $^3$H turns into $^3$He through $\beta^-$ decay, right? The energy of this reaction is only 18.6 keV.

However, according to the constituent quarks model, the only thing different between $^3$H and $^3$He nuclei is that the first one has $d$ quark where the second one has $u$ quark (so the decay of this $d$ into $u$ is what creates an extra proton instead of the initial neutron).

Now, my question is thus: if the mass of $d$ quark is (by some modern estimates) around 5 MeV (plus-minus one - doesn't really matter), and the mass of $u$ quark is about 2.5 MeV (again, rounding a bit), how is the turning of one into other only releases 18.6 keV of energy? Where does the most part of the energy vanish instead? As far as I know, for strong forces neutrons and protons are pretty much the same - so what is going on?

  • $\begingroup$ You might peruse the TUNL evaluation of A=3 that can be found at tunl.duke.edu/nucldata/chain/03.shtml $\endgroup$ – Jon Custer Apr 17 at 20:31
  • $\begingroup$ Nuclear physics is going on. $\endgroup$ – my2cts Apr 17 at 20:45
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    $\begingroup$ On factor you're missing is the rest mass/energy of the electron, being roughly half an Mev. You can't look at one quark alone, The binding/interaction energy is really the difference, The difference between a tritium atom's mass and helium 3 is roughly 18.6 ev. $\endgroup$ – R. Rankin Apr 17 at 23:45
  • $\begingroup$ So why is this difference there in the first place? Wouldn't one expect these nuclei being basically the same? Or the proton in $^3$He somehow behaves differently than the neutron in $^3$H? $\endgroup$ – Eugene B. Apr 20 at 15:15

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