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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?

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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 endothermic), followed by electron capture.

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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.

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    $\begingroup$ Right. As long as you have a 'free' source of high energy alpha particles, yes you can 'produce' energy. Until you start wondering how to get high energy alpha particles... $\endgroup$ – Jon Custer Sep 16 at 21:33
  • $\begingroup$ The value of the rest mass is given for the assembled nucleus. Therefore it includes any energy required to get to that point, such as Coulomb repulsion. It doesn't need to be considered a second time. $\endgroup$ – BowlOfRed Sep 16 at 21:33
  • $\begingroup$ @Maury Markowitz My hypothetical fusion Fe56+alpha to Ni60 is meant to proceed through quantum mechanical tunneling, like most fusion reactions in stars (except neutron capture). This means that the Coulomb barrier doesn't enter the energy equation, if I'm correct. $\endgroup$ – gamma1954 Sep 16 at 21:49
  • $\begingroup$ @JonCuster photodisintegration ensures there are plenty of alpha particles around, however the same photodisintegration process limits the build up of more massive nuclei by alpha capture! $\endgroup$ – Rob Jeffries Sep 17 at 15:16
  • $\begingroup$ @RobJeffries - fine, if you have a free source of gammas you can produce energy! In either case, you are having to drive the fusion process with energy from somewhere else. $\endgroup$ – Jon Custer Sep 17 at 17:13
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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.

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