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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$\begingroup$ This is rather a physic question that can be answered by quantic mechanics considerations... In short Iron is a very stable state for matter. $\endgroup$– J. ChomelJun 14, 2017 at 14:54
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1$\begingroup$ Also see physics.stackexchange.com/questions/168237/… $\endgroup$– ProfRobJun 14, 2017 at 21:21
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$\begingroup$ See astronomy.stackexchange.com/questions/36719/… $\endgroup$– ProfRobSep 12, 2020 at 9:06
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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.
answered Jun 14, 2017 at 20:53
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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. $\endgroup$– ProfRobJun 14, 2017 at 21:07
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1$\begingroup$ Actually, the addition of an alpha particle to 56Ni is exothermic. $\endgroup$– ProfRobSep 12, 2020 at 9:05
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$\begingroup$ @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. $\endgroup$– PM 2RingMay 28, 2021 at 7:24
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$\begingroup$ FWIW, here's the details of my calculation $\endgroup$– PM 2RingMay 28, 2021 at 7:26
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1$\begingroup$ @PM2Ring see astronomy.stackexchange.com/questions/36719/… $\endgroup$– ProfRobMay 28, 2021 at 7:28