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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?

James K
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Jeff Cohen
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  • This is rather a physic question that can be answered by quantic mechanics considerations... In short Iron is a very stable state for matter. – J. Chomel Jun 14 '17 at 14:54
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    Also see https://physics.stackexchange.com/questions/168237/why-does-fusion-stop-at-iron-when-nickel-is-most-tightly-bound – ProfRob Jun 14 '17 at 21:21
  • See https://astronomy.stackexchange.com/questions/36719/what-effects-besides-mass-defect-cause-the-alpha-ladder-beyond-iron-56-nickel – ProfRob Sep 12 '20 at 09:06

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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.

David Hammen
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  • Alpha capture would continue (at a very small energy cost), but requires temperatures that produce photons that can photodisintegrate the nuclei. – ProfRob Jun 14 '17 at 21:07
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    Actually, the addition of an alpha particle to 56Ni is exothermic. – ProfRob Sep 12 '20 at 09:05
  • @ProfRob Indeed! I calculate 2.69 MeV for that reaction, which is substantially smaller than the energy released in the other alpha ladder reactions, but it's certainly exothermic. :) (I used atomic masses not nuclide masses in my calculation, I assume that's ok). A slightly confusing note in the silicon-burning article mentions it's exothermic. Actually, both of those articles could do with some clean-up & a clearer explanation of the significance of the photodisintegration energy. – PM 2Ring May 28 '21 at 07:24
  • FWIW, here's the details of my calculation – PM 2Ring May 28 '21 at 07:26
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    @PM2Ring see https://astronomy.stackexchange.com/questions/36719/what-effects-besides-mass-defect-cause-the-alpha-ladder-beyond-iron-56-nickel – ProfRob May 28 '21 at 07:28
  • @ProfRob Thanks for reminding me about that question. I should've checked my bookmarks... – PM 2Ring May 28 '21 at 07:47