<blockquote>
  <p>Why does fusing iron in a stellar core use more energy than it releases?</p>
</blockquote>

<p>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.</p>

<p>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.)</p>

<p>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.</p>

CC BY-SA 3.0
2017-06-14T21:07:22.667Z
Alpha capture would continue (at a very small energy cost), but requires temperatures that produce photons that can photodisintegrate the nuclei.
ProfRob
CC BY-SA 4.0
2020-09-12T09:05:07.633Z
Actually, the addition of an alpha particle to 56Ni is exothermic.
ProfRob
CC BY-SA 4.0
2021-05-28T07:24:54.793Z
@ProfRob Indeed! I calculate 2.69 MeV for that reaction, which is substantially smaller than the energy released in the other [alpha ladder reactions](https://en.wikipedia.org/wiki/Alpha_process), 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](https://en.wikipedia.org/wiki/Silicon-burning_process) 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
CC BY-SA 4.0
2021-05-28T07:26:34.763Z
FWIW, here's the [details of my calculation](https://sagecell.sagemath.org/?z=eJwtzkFuwyAQBdA9pxgpmzgFzGBIIVJXlZftBaosrJrEjlRsg2tLOX3HSdl8ad58NDv4aHIOGS5p-IFunsd8Kst1XWP7LW9NaOR1kLex_Px9L1ns4Q2sld5o1M46xcG-6lp4do-beBJ0hJXncPRmky6QGKmUPqpKW4MVUS0QWduQ-AqloZbSRsNOu1o4NtE89l_qDC_QhS0F3CMlCyk9Df_tkWR4ZmPq47yfONBSwS5Dggx9hL1ADnQoFicG9Bb6YKJSPtDeY_IsLhyWQ9sUfzxsRjA=&lang=sage)
PM 2Ring
CC BY-SA 4.0
2021-05-28T07:28:10.117Z
@PM2Ring see https://astronomy.stackexchange.com/questions/36719/what-effects-besides-mass-defect-cause-the-alpha-ladder-beyond-iron-56-nickel
ProfRob
CC BY-SA 4.0
2021-05-28T07:47:04.923Z
@ProfRob Thanks for reminding me about that question. I should've checked my bookmarks...
PM 2Ring
CC BY-SA 3.0
2017-06-14T20:53:58.917Z
2017-06-14T22:41:40.250Z
2017-06-14T22:41:40.250Z
David Hammen
<p>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.</p>

<p>Why does fusing iron in a stellar core use more energy than it releases?</p>

CC BY-SA 3.0
2017-06-14T14:54:42.803Z
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
CC BY-SA 3.0
2017-06-14T21:21:19.540Z
Also see https://physics.stackexchange.com/questions/168237/why-does-fusion-stop-at-iron-when-nickel-is-most-tightly-bound
ProfRob
CC BY-SA 4.0
2020-09-12T09:06:55.903Z
See https://astronomy.stackexchange.com/questions/36719/what-effects-besides-mass-defect-cause-the-alpha-ladder-beyond-iron-56-nickel
ProfRob
<supernova>
Why does fusing iron in a stellar core use more energy than it releases
Jeff Cohen
# Why does fusing iron in a stellar core use more energy than it releases

(Asked by: username_0 on 2017-06-14)

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?

**Tags:** supernova

> **Comments:**
> **username_1**: This is rather a physic question that can be answered by quantic mechanics considerations... In short Iron is a very stable state for matter.  
> **username_2**: Also see https://physics.stackexchange.com/questions/168237/why-does-fusion-stop-at-iron-when-nickel-is-most-tightly-bound  
> **username_2**: See https://astronomy.stackexchange.com/questions/36719/what-effects-besides-mass-defect-cause-the-alpha-ladder-beyond-iron-56-nickel  

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## 1 answers

### Answer by username_3 on 2017-06-14. Score: 3

> 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.

> **Comments:**
> **username_2**: Alpha capture would continue (at a very small energy cost), but requires temperatures that produce photons that can photodisintegrate the nuclei.  
> **username_2**: Actually, the addition of an alpha particle to 56Ni is exothermic.  
> **username_4**: @username_2 Indeed! I calculate 2.69 MeV for that reaction, which is substantially smaller than the energy released in the other [alpha ladder reactions](https://en.wikipedia.org/wiki/Alpha_process), 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](https://en.wikipedia.org/wiki/Silicon-burning_process) 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.  
> **username_4**: FWIW, here's the [details of my calculation](https://sagecell.sagemath.org/?z=eJwtzkFuwyAQBdA9pxgpmzgFzGBIIVJXlZftBaosrJrEjlRsg2tLOX3HSdl8ad58NDv4aHIOGS5p-IFunsd8Kst1XWP7LW9NaOR1kLex_Px9L1ns4Q2sld5o1M46xcG-6lp4do-beBJ0hJXncPRmky6QGKmUPqpKW4MVUS0QWduQ-AqloZbSRsNOu1o4NtE89l_qDC_QhS0F3CMlCyk9Df_tkWR4ZmPq47yfONBSwS5Dggx9hL1ADnQoFicG9Bb6YKJSPtDeY_IsLhyWQ9sUfzxsRjA=&lang=sage)  
> **username_2**: @PM2Ring see https://astronomy.stackexchange.com/questions/36719/what-effects-besides-mass-defect-cause-the-alpha-ladder-beyond-iron-56-nickel  
> **username_4**: @username_2 Thanks for reminding me about that question. I should've checked my bookmarks...  

---

