# Why does fusion above nickel 56 require energy?

I've always struggled with the concept of fission and fusion. I mean I can show a mass deficit with math easily enough but I have always struggled understanding the fundamental concept of why things are the way they are....maybe someone can help me find the error in my reasoning. Please no math or B/E charts in your explanations!

As we build a nucleus, you have to overcome electrostatic repulsion between protons, but after you get them close enough they bond because of the nuclear strong force (neutrons don't have this problem). Free floating nucleons are higher energy than bound nucleons just like a ball in the air is higher energy than a ball on the ground, so once bonding happens the nucleus sheds the excess energy IAW conservation of energy and the lower energy state is reflected as a mass deficit.

As we continue to grow the atom the nuclear force increases at first because more immediate nucleon neighbors, however eventually the nucleus gets too fat and the effects of the strong force for each additional nucleon diminished due to range while the repulsion is essentially unaffected. Eventually the largest atoms are made and are unstable because the strong force more or less perfectly balances out the electrostatic repulsion.

As the atoms vibrate they change shape slightly and parts fly off because they are so close to exploding (natural radiation).

So that explains why fusion releases energy and why there is a max atom size. What it doesn't explain is why after iron fission releases energy and fusion requires energy.

It seems to me like a smaller "unbound" particle should always release energy when bound and saying that after Nickel-56 fusion requires energy seems like a reversal of physics (obviously its not but I don't understand why).

At one point I thought I had it all figured out. I reasoned that fusion is an infinite potential well due to the strong force which is much stronger than the electrostatic forces. Fission was not the result of binding energy but actually just the result of the electrostatic repulsion releasing i.e. fission and fusion really have two different sources where one is nuclear strong force from binding and one is electrostatic tension being released due to deformation. Is that accurate?

• en.wikipedia.org/wiki/Semi-empirical_mass_formula is a good start. In particular, the surface term favours larger nuclei (reduces surface area) whilst the Coulomb term favours smaller ones. Commented Jun 11, 2019 at 18:43
• You asked basically the same question a few weeks ago (or somebody else did with very similar phrasing). Start with physics.stackexchange.com/questions/482264/… and go from there. B/E charts are pretty useful, so I'm not sure why you are averse to them. Commented Jun 11, 2019 at 18:45
• You mention multiple times that fission is due to the Coulomb repulsion making large nuclei energetically unfavourable over multiple smaller ones. Why do you then say that it doesn't explain 'why after ion fission releases energy'? Commented Jun 11, 2019 at 18:47
• Thanks Jacob..I guess I am thinking that fission and fusion have the same mechanism. Is it a true statement that one is due to the nuclear strong force and one is due to Coulombic force? ... I still don't see why fusing two or four nucleons releases energy but fusing two fission fragments together (fission in reverse) requires energy.....Is it two separate components at play? Like does the nuclear force still result in a release of energy during fusion of large particles but the net result due to electrostatic repulsion makes it net negative? Commented Jun 11, 2019 at 19:01
• Jon I couldn't find the one you were talking about, but if someone answered it already that works. Commented Jun 11, 2019 at 19:04

If you work through the masses and binding energies you will see that adding an alpha particle to $$^{56}$$Ni is in fact exothermic.

However, that alpha particle has to come from somewhere. What is endothermic is to break an alpha particle off one $$^{56}$$Ni nucleus and add it to another. In other words, once you have a stellar core of $$^{56}$$Ni then that is the configuration with (roughly) the largest binding energy per nucleon and to rearrange those nucleons in a different way will require energy from somewhere (e.g. gravitational collapse leading to higher temperatures and photodisintegration of the nuclei).

You might well ask, well why doesn't this fusion take place in places where there are losts of alpha particles already, like in a helium-burning shell for example? The answer is that the temperatures are not high enough to overcome the Coulomb repulsion and initiate the reaction.

Here is the binding energy curve.

This is an experimental curve.

What does this curve say for the increasing number of nucleons in the periodic table?

It says that up to nickel56 on average the nucleus is bound in such a way that adding two nuclei together will release energy, and this is called fusion. After Nickel , the higher the number of nucleons, the more the binding energy of the average nucleon, more energy is needed. Only if a higher A number nucleus breaks into fragments with lower A, there will be energy released.

This is "explained" by the two forces of the nucleon-nucleon interactions, the strong force and the electromagnetic force. Two protons repel each other but are strongly attracted by the strong nuclear force, a residual force from the strong interactions modeled with pion exchanges. When neutrons are added in the mix the atoms can become stable .

At the number of 50+ nucleons the binding in a nucleus still is due to the interplay of the nuclear force with the electromagnetic repulsion, but to get a bound nucleus, more energy has to be expended as the A number gets higher.

Thus a high A number nucleus will give energy if it breaks into fragments with smaller A, which have a lower binding energy.It is a many body problem that is modeled in various ways. The interplay of electromagnetic and nuclear forces is the same , the number of nucleons changes the possibility of getting energy out of fusion or fission.

• Adding an alpha to Nickel-56 is exothermic. Commented May 28, 2021 at 7:19