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When calculating the $Q$-value, $Q = \Delta M \cdot c^2$, of this reaction:

$$ ^6Li \ (\alpha, p)\ ^9Be \quad \iff \quad \alpha + \ ^6Li \ \longrightarrow \ ^9Be + p $$

The $Q$-value can also be written in terms of binding energies $BE$.

Should I consider the binding energy of the proton $p$ $ $ ~$1u\cdot c^2$ or $0$?

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Note that binding energy is relative, like the gravitational potential energy in certain respects. That said, the binding energy of the proton could be considered non-zero if its sub-particles, i.e. quarks, are studied. However, since in the energy region of your mentioned reaction these sub-particles play no direct role (except for perhaps the underlying physics), it is more reasonable if you consider the binding energy of the nucleons, i.e. protons and neutrons, as zero. Note that if you consider the binding energy to be non-zero for both sides of the reaction, similar final results would be obtained.

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  • $\begingroup$ I don't think this is right, for the reasons given in my answer. $\endgroup$
    – user4552
    Jun 18, 2019 at 20:06
  • $\begingroup$ A proton will bind with a neutron to form a deuteron, with a binding energy of 2.224 MeV. Two protons will bind with two neutrons to form an alpha particle or helium-4 nucleus, with a binding energy of 28.295 MeV. This sort of thing just doesn't apply for a single proton. In the Wikipedia quark article you can read that the up quark mass is 2.3 ± 0.7 MeV, and the down quark mass is 4.8 ± 0.5 MeV. That’s really not very much compared to the proton mass of 938.272 MeV. For the binding energy scenario to work for 3 quarks, you'd expect the quark masses to be circa 330 MeV. $\endgroup$
    – John Duffield
    Jun 20, 2019 at 21:12
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No, a proton doesn't have a well-defined binding energy. That would be the energy required in order to separate it into three quarks, but free quarks don't exist.

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  • $\begingroup$ What about en.wikipedia.org/wiki/Proton_decay ? $\endgroup$
    – user126527
    Jun 18, 2019 at 8:34
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    $\begingroup$ @JollyJoker a proton isn't a bound state of a pion and a positron. It is a bound state of three quarks. $\endgroup$
    – John Dvorak
    Jun 18, 2019 at 9:04
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    $\begingroup$ Free quarks don't exist, but the quarks still have a well-defined rest mass. Why isn't that sufficient? $\endgroup$
    – John Dvorak
    Jun 18, 2019 at 9:06
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    $\begingroup$ @JohnDvorak: Actually, a quark doesn't really have a well-defined rest mass. For a particle like an electron that can exist as a free particle, the rest mass is the mass of the isolated particle, which includes the energy stored in its electric field. This energy is classically infinite, but quantum effects make it finite. For a quark, the rest mass would include the energy in the gluon field, but that's not the same field as for a bound quark. The so-called rest masses of quarks are really more like model parameters, and their values are context-dependent. $\endgroup$
    – user4552
    Jun 18, 2019 at 13:38
  • $\begingroup$ Could you have a universe where the only thing was one single quark? The explanations I have seen about color confinement talk about what happens if you try to pull two quarks apart, so maybe starting out with a free quark is a loophole. $\endgroup$
    – Display Name
    Jun 18, 2019 at 21:04
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Yes. Protons aren't elementary particles, so they do have binding energy. What that means exactly is a bit trickier, since the structure of the proton isn't something we have a full consensus on yet.

There's two main ways of seeing protons:

  • Protons are made out of three quarks. Each of these quarks only has a tiny fraction of the mass of the proton, and the difference is the binding energy. Note that the binding energy is positive, and huge - almost all of the mass around you comes from this binding energy.
  • Protons are made out of a sea of quarks and their corresponding anti-quarks, except for three unpaired quarks. The mass of the proton comes from all of these quarks and anti-quarks, but since all except three are paired, most of the behaviour of the proton comes from the three unpaired quarks. But even in this case, the proton does have binding energy - it just doesn't have to account for 99% of the mass of the proton.

From UMD:

What one has learn about the nucleon structure through high-energy scattering? First of all, one learn there are indeed 2 up valence quarks and 1 down quark, with electric charge 2/3 and -1/3 of the proton, respectively.

(the properties of a proton are largely determined by the three valence (unpaired) quarks)

Second, the number of quarks is infinite because the integration does not seem to converge. This is because there are infinite number of quark and antiquark pairs in the proton.

Note that both are essentially the same as far as QFT is concerned - the massive binding energy would mean the spontaneous creation of quark-antiquark pairs; quark confinement explains why we can't ever observe these in isolation (the binding energy of quarks increases with distance, and eventually gets large enough that instead of more separation, new quarks are created). They're also essentially the same for outside observation - the mass of the proton is the same regardless of whether it comes from the binding energy of quarks inside the proton or the sum of the masses of the constituent quarks.

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  • $\begingroup$ I had never heard of the first way of seeing protons you described (solely made of 3 quarks). If both are correct, how can one decide which to apply to, say the $BE$ for $p$ in the $Q$-value for the reaction I described above? $\endgroup$
    – João Bravo
    Jun 18, 2019 at 14:40
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    $\begingroup$ @JoãoBravo You're not looking for the binding energy of the proton there. The thing to keep in mind that binding energy is a property of systems - and in your case, you're studying the binding energy of nuclei, not its constituent protons and neutrons. The proton in your reaction also has gravitational potential energy, but you don't care about that, do you? It doesn't change from the left side of the equation to the right side, just like the strong binding energy doesn't. $\endgroup$
    – Luaan
    Jun 18, 2019 at 17:42
  • $\begingroup$ Thank you, that remark was enlightening. It boils down to the energy region of the reaction I am studying. $\endgroup$
    – João Bravo
    Jun 18, 2019 at 19:05
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Yes, protons have binding energy when they combine with neutrons to form a nucleus. In the fusion process where nucleons combine to form helium, they require less mass within the nucleus than they did as free particles. This surplus mass is emitted as binding energy, which is where the sun gets its power. When a nucleon leaves the nucleus,it follows that this binding energy must be re-supplied to it before it can become a free particle.

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    $\begingroup$ I understand that, but I was just asking from the point of view of the formation of a proton $\endgroup$
    – João Bravo
    Jun 17, 2019 at 22:43

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