Why is hydrogen-boron called a fusion reaction?

Question

As we know, we usually call fusion such a reaction in which two light nuclei make a heavier one and release energy. For fission, a heavy nucleus breaks into light ones. My question is, in proton-boron reaction, the product is He, which is not heavier than boron. It is more like fission rather than fusion. Why do we call it fusion? Is it just because it is on the left part of the binding energy diagram lighter than iron?

Answer 1

So the basic question is where the power is coming from.

The total binding energy of boron-11 is 6.928 MeV/nucleon while the proton obviously has a binding energy of 0 MeV/nucleon (it isn't bound to anything) while the resulting heliums have a binding energy of 7.074 MeV/nucleon.

Generally the nucleons will want to go from low to high binding energy; these amounts are secretly negative much like the gravitational potential energy is negative, so in their entropic “goal” to spread as much energy as possible across the universe by minimizing their energy state, the nucleons “want” to increase their binding energy.

Now you are right that part of this reaction qualifies as a sort of fission and we can actually put numbers to that, those eleven nucleons gain 0.146 MeV each for a total release of 1.6 MeV. But the lone proton achieves fusion into helium and increases its binding energy by 7.1 MeV. So of the 8.7 MeV total generated, something like 80% of it comes from the fusion part.

I actually think that the reason for calling this fusion rather than fission is probably more simple than that. Helium is actually extremely weirdly stable given its atomic mass. Pretty much all the other isotopes fit on a nice curve where we plot the binding energy per nucleon against the total mass of the nucleus. This increases approximately linearly from zero to the binding energy of neon-20, before leveling out with a maximum at iron-56 and a plateau until strontium-86, and then a much milder line of decreasing slope for isotopes of even higher mass.

The default convention is to basically ignore helium's weird stability and look at the rest of the curve, in which case things heavier than iron-56 undergo “fission” while things lighter than it undergo “fusion.” I imagine other folks don't even look at the fact that the boron is breaking apart into the helium, much less calculate what percentage of the energy comes from that, when classifying it as fusion energy. What matter is is not what it is turning into, what matters for classification purposes is just how big it is, “it is way way smaller than iron so it must be fusion energy, not fission energy.” For all we care, maybe those nucleons could be losing energy and the proton is getting enough energy to compensate. But probably a third term is actually required for these anomalous “fissions” where lithium beryllium and boron turn into heliums.

Answer 2

You are right that a nuclear fusion reaction occurs when two or more atomic nuclei combine to form a heavier one. However, a fusion reaction does not need to release energy. It's also worth highlighting that the fusion reactants are atomic nuclei, so when a neutron is added to a nucleus like Uranium we do not talk of nuclear fusion but rather neutron absorption. Nuclear fission occurs when a nucleus splits into two or more atomic nuclei, with the difference that it can also do so spontaneously during radioactive decay.

So when Hydrogen and Boron-11 nuclei combine it's called a fusion reaction, even if it only momentarily creates an excited Carbon-12 nucleus. This excited nucleus undergoes radioactive decay via fission into three alpha particles as you mention (compare the Hoyle state that has a very short half-life of $2.4\times10^{-16}~\mathrm{s}$). Ultimately, both fusion and radioactive decay via fission occur. Since the first step that has to be artificially induced is fusion, it's sufficient to call it Hydrogen-Boron fusion and the fission part is taken for granted even if it is very important from an energy calculation standpoint.