Energy in nuclear fusion - mass change vs potential energy explanation Trying to understand how the energy related to nuclear reactions. In particular why particular fusion of small nuclei release energy. So far, there seems to be two explanations which I have encountered that I understand (I'm in High School currently, so have stayed away from the large number of super scary, equation-heavy quantum mechanics ones).
One involves potential energy (and potential energy wells), and the other involves changes in mass/mass energy. I think I understand both of these separately:
During fusion of small nuclei, energy is required to overcome the coulombic repulsion between the nuclei, and get over the potential 'hump'. Once this is done, the strong interaction takes over as they fall into the well, and energy is released as potential energy is lost (due to positive work done by the strong interaction).

The other is the binding energy per nucleon explanation which essentially comes down to the fact that small nuclei have lower mass nucleons than slightly larger ones (when they're smaller than iron). So, when small nuclei smash together and form the larger one, mass is lost, this mass being converted into the kinetic energy of the products (by E=mc^2) - so that mass-energy is conserved.
My question is essentially whether these explanations can be combined into one (so is there some hidden link between them), or is one is more correct than the other, or have I just gotten something wrong (if there's a better explanation which is reasonably intuitive). I thought perhaps you could think of the increase in mass as you remove a nucleon from its nucleus as the manifestation for the potential energy associated with the strong interaction? but not sure if that is correct at all.
 A: The potential energy explanation is correct. The mass explanation, as you have stated it, is not quite correct.

The other is the binding energy per nucleon explanation which essentially comes down to the fact that small nuclei have lower mass nucleons than slightly larger ones

Here, it is not the nucleons that have a lower mass, it is the nucleus. The child nucleus has a lower mass than the sum of the parent nuclei masses. But this mass deficit cannot be attributed to individual nucleons, it is the whole nucleus.

So, when small nuclei smash together and form the larger one, mass is lost, this mass being converted into the kinetic energy of the products (by E=mc^2) - so that mass-energy is conserved.

The whole idea of “converting” mass into energy is a poor explanation of the actual physics. If $E=mc^2$ and you convert some $m$ into $E$ then you would get an increase in $E$ and a decrease in $m$ so then the left hand and right hand sides of $E= mc^2$ would no longer be equal.
There is no “conversion” involved. Energy has mass and mass has energy. The full formula is $m^2 c^2=E^2/c^2-p^2$. But if $p=0$ you get the famous expression. So the correct way to understand that is not about “converting” energy or mass. Instead, energy at rest has mass and mass is energy at rest.
So, what actually happens is that a system of particles with energy and mass is converted into a system of different particles with the same system mass and energy. Then part of that system leaves and is no longer counted as part of the system. And the remaining pieces of the system have a lower mass snd energy (and maybe momentum).
For example, a deuterium and a proton fuse to make a helium 3 and a photon. The system of the helium 3 and the photon have the same energy and mass as the system of the deuterium and proton. But the photon soon leaves and is no longer counted. So the remaining helium nucleus has less mass than the whole system did. No mass was converted, the particles converted and some left the system.
