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Nickel-58 and nickel-62 are four neutrons apart, not even just two, so this fact is especially confusing to me...

I have read about how nickel is synthesized in stars from silicon, so this is perhaps part of the reason...

Two Si-28 nuclides become Ni-56, perhaps... But why are 'only' two more neutrons added? Why not four or six?

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  • $\begingroup$ Generally more neutrons are required per proton for stability as the nucleus gets heavier. So, building up from lighter nuclei means there are fewer neutrons available then might be ideal. And good enough is, well, good enough. $\endgroup$
    – Jon Custer
    Commented Nov 16, 2022 at 15:01
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    $\begingroup$ Yes, possible duplicate of physics.stackexchange.com/q/182525 and physics.stackexchange.com/q/168237. $\endgroup$
    – David Bailey
    Commented Nov 16, 2022 at 16:58
  • $\begingroup$ Also see astronomy.stackexchange.com/q/36719/16685 for more on the alpha ladder. $\endgroup$
    – PM 2Ring
    Commented Nov 16, 2022 at 18:43
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    $\begingroup$ @DavidBailey Those questions are certainly related. But as Jon mentioned, there's also the issue of the neutron : proton ratio. There's usually not an excess of neutrons, except during supernovae, or neutron star collisions. $\endgroup$
    – PM 2Ring
    Commented Nov 16, 2022 at 18:47

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TLDR: The formation of more neutron-rich iron-peak nuclei (in significant quantities) is disfavoured because the rapid or explosive nucleosynthesis conditions do not allow sufficient time for weak processes to act to decrease the equilibrium $Z/A$ ratio.

Most nickel is in fact produced by type Ia supernovae - exploding white dwarfs, with some contributed in core collapse supernovae.

The critical factor determining which reactions are favoured and what end products are produced is the ratio of $Z/A$ in the reacting material. In white dwarfs (and also in the cores of massive stars), this ratio is close to 0.5. In white dwarfs and the cores of massive stars, this is because they are mostly made of carbon, oxygen and silicon (in massive stars) to begin with and then the iron-peak elements are built up by rapidly adding alpha particles (the "alpha chain"), which leaves the $Z/A \simeq 0.5$ ratio unchanged. Because the explosive burning that takes place is fast, there is insufficient time for weak nuclear processes (electron capture, beta decay) to alter the $Z/A$ ratio very much.

In terms of Ni, mostly it is 56-Ni that is produced, which is of course unstable and beta-decays to iron (via cobalt). 58-Ni can be produced from beta-plus decay of 60-Zn (the end of the alpha chain - see here) or the addition of a proton to 57-Co in type Ia explosions (Blondin et al. 2022). 62-Ni would require the decay from even heavier nuclei produced by alpha capture, but these are not produced in quantity because of competing photodisintegration in massive stellar cores. Instead, what 62-Ni is produced comes from synthesis in material with a lower $Z/A$, which can only be found towards the centre of an exploding white dwarf (due to electron capture in the higher density material). It can also be subsequently formed by s-process neutron captures onto iron, but this is slow and inefficient.

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  • $\begingroup$ That makes much more sense =) $\endgroup$
    – Emilio Pisanty
    Commented Nov 17, 2022 at 10:33

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