Nickel hydrogen fusion

Does the fusion of nickel and hydrogen produce energy, or does it consume energy?

The question arises since that fusion in stars stops at iron.  Heavier elements are formed in a different way.  So, in that case, why is it an advantage to fuse iron?  A little thought may give some insight into a possible answer for that.  In a star, the possibility of fusion with hydrogen can no longer exist.  That's because all of the hydrogen was  fused long before this point.  Therefore, the most likely form of  fusion is with iron and another heavier element than hydrogen.  Fusion stops when there is nothing left but iron.  Since fusion of iron with iron isn't exothermic, fusion can't continue.  The star dies.

Fusion of hydrogen with itself is exothermic, of course.  But what about with iron?  Is that exothermic, or endothermic?  This may be a question for someone more knowledgeable than myself, but I attempted to find out  anyway.  How?  By calculating mass defects for the proposed isotopes with a fusion of hydrogen.  Take a proton, which is a hydrogen nucleus, and calculate what the mass would be if it were not lost.  The difference in mass between the actual new isotope after the fusion and what the two element's mass was before the fusion gives the mass defect.  Presumably, if mass was lost in the fusion, it had to go somewhere, right?  It so happens that it does.  Therefore, it must be exothermic, since mass was lost in its formation.

The energy gained from the mass lost is calculated from the famous equation E=mc squared.

The amount of energy does not detract from the subsequent energy gained from the decay of the unstable copper isotopes back into nickel.  Because this results in a further loss of mass, which implies another exothermic reaction.  Therefore, there are two energy productions: first from the fusion itself, and then from the subsequent beta decay.

I did this calculation with nickel 58 and hydrogen fusion.  I did away with my work, so I can't show it here.  If you wish, you can calculate it yourself.  It is a straightforward calculation.  Although I could be incorrect, I presume the fusions of other heavier isotopes of nickel proceed in the same fashion: as exothermic dual reactions, as with this example.