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This question already has an answer here:

I know larger stars can fuse heavier and heavier elements up to iron where it stops because fusing iron requires more energy than it releases, causing a collapse and supernova.

Why does fusing iron in a stellar core use more energy than it releases?

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marked as duplicate by Rob Jeffries, Hohmannfan, James K, zephyr, Sir Cumference Jun 15 '17 at 14:09

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An atomic nucleus contains protons and neutrons. There is a balance of forces: the (residual) strong nuclear force attracts protons and neutrons together. The electro-magnetic force pushes them apart.

When nuclei fuse, energy is required to push them together, as you have to push against the electromagnetic repulsion. But when the nuclei come close enough, the nuclei are attracted by the strong force, and energy is released as the nuclei fall into each other. For small nuclei, the energy released in fusion is greater than the energy needed to overcome electromagnetic repulsion. For large nuclei (which have a larger positive charge) the reverse is true, more energy is required to push the nuclei together than is released.

Iron is at the balance point, at which the energy required to overcome electomagnetic repulsion just exceeds what is given out on fusion. So iron is the smallest nuclei to absorb energy when it engages in fusion. And so when substantial amounts of iron form in the core of a star, the core will no longer be able to heat up to maintain the balance with gravity, the core collapses and the star explodes. See Why is iron responsible for causing a supernova?

Iron is therefore one of the most stable nuclei (not the most stable: that is Nickle-62, but for various reasons this is not likely for form inside stars) Iron acts like the "ash" of nuclear fusion.

A more detailed answer would require understanding the quantum mechanics of how binding energy in nuclei works, which can explain why it is Iron that has close to maximum binding energy.

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Why does fusing iron in a stellar core use more energy than it releases?

It doesn't, at least not in the alpha ladder. The alpha ladder starts with the carbon-12 produced by the triple alpha process. A carbon-12 and an alpha particle (helium-4) combine to form oxygen-16, which in turn combines with an alpha particle to form neon-20, and so on, up to titanium-44, chromium-48, then iron-52, then nickel-56, and then zinc-60.

It's the production of zinc-60 that kills stars rather than the production of nickel-56. The reactions up to and including the production of nickel-56 are exothermic (i.e., they release heat). The production of zinc-60 is endothermic (it consumes heat, in the form of a gamma particle). The energy-producing reactions that kept the star from collapsing on itself end. Moreover, the temperatures needed to produce zinc-60 are so very high that photons can photodisintegrate zinc-60, recreating the nickel-56 nuclei and alpha particles that created those zinc-60 nuclei. The alpha ladder pretty much stops with nickel-56. (Other processes create elements beyond nickel.)

The nickel-56 ejected from supernovae is rather short-lived, decaying with a half life of 6 days into cobalt-56. This too is radioactive, decaying with a half life of 77 days into iron-56. The unique signatures of these two decays, and the slightly delayed transition from nickel-56 decay to cobalt-56 decay, are one of the key signs that a supernova has occurred.

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  • $\begingroup$ Alpha capture would continue (at a very small energy cost), but requires temperatures that produce photons that can photodisintegrate the nuclei. $\endgroup$ – Rob Jeffries Jun 14 '17 at 21:07

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