Tritium decay energy too small for quark model

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?

• You might peruse the TUNL evaluation of A=3 that can be found at tunl.duke.edu/nucldata/chain/03.shtml – Jon Custer Apr 17 at 20:31
• Nuclear physics is going on. – my2cts Apr 17 at 20:45
• 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. – R. Rankin Apr 17 at 23:45
• 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? – Eugene B. Apr 20 at 15:15