Suppose we create an Fe-56 nucleus and an Ni-62 nucleus, each from individual protons and neutrons. In the case of Ni-62, more mass per nucleon is converted to binding energy. Thus we could argue the Ni-62 nucleus to be more strongly bound than the Fe-56 nucleus, if I'm correct so far.

  1. Why is Fe-56 mentioned in many astrophysics texts as the most strongly bound of all nuclei?

Fe-56 is commonly mentioned as the dominant end product of fusion reactions in the core of massive stars. If I'm correct, fusion reactions beyond Si-28 are accompanied by partial disintegrations, resulting in a cocktail of fragments, not exclusively multiples of He-4 (nuclear statistical equilibrium).
2. Why is much more Fe-56 than Ni-62 produced in the core of a massive star, although Ni-62 is more tightly bound than Fe-56? What determines the share of each nuclide in the resulting iron group?

  • $\begingroup$ More on Ni-62. $\endgroup$
    – Qmechanic
    May 6, 2015 at 21:06
  • 2
    $\begingroup$ Possible duplicate in physics.stackexchange.com/q/168237 $\endgroup$
    – ProfRob
    May 6, 2015 at 21:30
  • 1
    $\begingroup$ Possible duplicate of physics.stackexchange.com/q/15943 $\endgroup$
    – user12262
    May 6, 2015 at 22:00
  • $\begingroup$ @Rob Jeffries Thank you, but it's not quite a duplicate. Both wikipedia sources seem to describe fusion beyond Si-28 only as He-4 addition. However, from what I understand of the process leading to a nuclear statistical equilibrium, (partial) disintegration produces quite a diversity of fragments. So not only He-4 is available for fusion and Ni-62 could be created. Still, Fe-56 dominates. Why? $\endgroup$
    – gamma1954
    May 6, 2015 at 22:13
  • $\begingroup$ @user12262 Thank you, but it doesn't seem to be a duplicate. The wikipedia source doesn't refer to the specific situation in the core of a massive star at the end of the fusion process. $\endgroup$
    – gamma1954
    May 6, 2015 at 22:17

2 Answers 2


The final stage of nucleosynthesis at the core of a massive star involves the production of iron-peak elements, mostly determined by competition between alpha capture and photodisintegration. The starting material is mostly Si28 and weak processes are unable to significantly alter the n/p ratio from unity on short enough timescales. Thus the expected outcome of these quasi-equilibrium reactions should be nuclei with $Z \simeq N$. Subject to that constraint, then the most stable nucleus resulting from alpha captures onto Si is Ni56.

To produce heavier nuclei (e.g. Zn60) by alpha capture requires higher temperatures (because of the higher Coulomb barrier) and at these higher temperatures photodisintegration drives the equilibrium back towards smaller nuclei.

So where does all the confusion arise? Most of the iron-peak material ejected in a supernova is formed slightly further out from the core in explosive Si burning. The major product is Ni56, as above, and this then undergoes weak decays to Co56 and then Fe56 with half lives of 6 days and 77 days respectively. Thus the most common iron-peak product that ends up in the interstellar medium is Fe56 (also from alpha capture in type Ia supernovae).

  • 1
    $\begingroup$ Thank you for indicating the distinction between the material that is and isn't ejected. I understand that most of the Fe-56 in the ISM is produced in the decay of Ni-56 > Co-56 > Fe-56 after the shell burning products of a massive star (core collapse supernova) or the fusion products of a type Ia supernova have been ejected. $\endgroup$
    – gamma1954
    May 7, 2015 at 20:17
  • $\begingroup$ On the other hand, how can Fe-56 become the dominant nuclide in the inner core of a massive star before it collapses? As you wrote, mostly Ni-56 is produced (Z/N=1) after the Si-disintegration and subsequent alpha captures. But how can Fe-56 become dominant in the inner core of a massive star before it collapses? If I'm correct, the rearrangement leading to the iron group after Si-disintegration in the inner core takes a relatively "long" time and proceeds at a relatively "low" temperature (perhaps $10^4-10^5$ s and 2 GK), compared to the explosive Si-burning in the shell. $\endgroup$
    – gamma1954
    May 7, 2015 at 20:19
  • $\begingroup$ According to Clayton, beta decay under core conditions can be much faster than in the laboratory, e.g. electron capture lifetime in the core 1 min for Ni-56. This could allow enough time for the Ni-56 > Co-56 > Fe-56 decay (reducing Z/N) before the inner core collapses. Most often this Fe-56 is "buried" and transformed in the neutron star or black hole and never contributes to the ISM. Could this be correct? Iliadis (Nuclear Physics of Stars, section 5.5.4 and 5.5.5) and Clayton (Principles of Stellar Evolution and Nucleosynthesis, section 7.2) give useful information. $\endgroup$
    – gamma1954
    May 7, 2015 at 20:20
  • $\begingroup$ @gamma1954 Yes, I think the overall timescale of Si burning (a few days) is long enough for significant electron capture to Fe 56 in the core. Possibly that invalidates what I say about flavour changes not being fast enough, but the point about photodisintegration driving the equilibrium towards smaller nuclei (and also that Zn 60 has lower binding energy) seems key in preventing significant Ni 62 formation. $\endgroup$
    – ProfRob
    May 7, 2015 at 21:40
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    $\begingroup$ @gamma1954 Further thoughts on this. Electron capture is more rapid in the core, but it is still slower than alpha capture and photodisintegration Therefore beta processes can only act on the equilibrium products available to them. Since it is difficult to get to Zn60 by alpha capture, let alone Ge64, then electron capture cannot produce much Ni62. $\endgroup$
    – ProfRob
    May 13, 2015 at 9:30

A very nice question about a common misconception in books on astrophysics (I've made the same mistake in a comment here). According to M.P. Fewell, the origin of this misconception lies in the theory of stellar nucleosynthesis and the abundance of the elements. While other nuclei have higher binding energy per nucleon, $^{56}\mathrm{Fe}$ is more abundant

because the competition between photodisintegration and charged-particle capture starts to favor photodisintegration at iron.

Once the chain of reactions reaches $\mathrm{Fe}$, there is no reason to examine heavier or more stable nuclei, because the conditions are such that they are barely produced.


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