Energy from fusion beyond iron peak

Question

In the diagram of nuclear binding energy per nucleon Eb/A (vertical axis) and mass number A (horizontal axis), Fe-56 has one of the highest values. Many authors state that nuclear fusion can only produce energy going from lighter nuclei towards the iron peak, but not beyond the iron peak. Going beyond the iron peak, we end up with a nucleus having less binding energy per nucleon. Why would that prohibit energy production by fusion?

As an example, suppose an Fe-56 nucleus captures an alpha particle, resulting in a Ni-60 nucleus. Fe-56 and He-4 have a rest mass 55.9349+4.0026=59.9375 u which is more than 59.9308 u for Ni-60 (using three atomic masses so we have as many electrons before and after; ignoring the electron binding energy). Because some mass is lost, this "fusion beyond the iron peak" would seem to be possible and to produce energy. Where do I go wrong?

I understand that this fusion requires a very high temperature in the core of a massive star at the end of Si-burning because of the high Coulomb barrier, and that photodesintegration would quickly destroy the Ni-60. But my question is: is energy production through fusion beyond the iron peak possible in principle?

Answer 1

Yes it is possible. Much of the Ni in the universe is in this form. However, I understand that the main production mechanism is alpha capture onto 56 Ni (which is at the end of the "alpha chain") to produce 60 Zn (which is slightly exothermic), followed by electron capture.

In fact, the calculation you really need to do has to factor in the origin of the alpha particle as well as the route to getting to your final nucleus.

The end point of the Si burning alpha chain is 56Ni. It is exothermic to add an additional alpha particle, as you have found. However, where does the alpha particle come from? The alpha particles are "reactive" and will fuse rapidly with any lighter elements present. They can of course be stripped from heavier nuclei, but this has an energy penalty. If you calculate how much energy is required to take an alpha particle from 56Ni and then add it to another 56Ni nucleus to produce 60Zn, you will find that is a net endothermic process. In addition to this, in order to penetrate the higher Coulomb barrier with each additional proton, the temperature has to be higher. This in turn results in higher energy photons in the gas, which become capable of photodisintegrating heavier nuclei, which is certaintly endothermic.

For these reasons, the energy density of the whole gas is minimised when most of the nuclei produced are 56Ni, which then undergoes weak decays and electron captures to 56Fe.

Answer 2

59.9375 - 59.9308 = 0.0067

I don't have a ref handy, but I'm pretty sure the Coulomb barrier is higher than that, which means the reactions would be losing energy and cooling the core.

Answer 3

Fusion can continue beyond iron but energy must be added. When one looks at supernovae light curves, one can see a typical type 2 collapse brightness decay matches the decay of ni56 co56. All elements beyond iron are created and blasted out into space during supernovae.