<p>Contrary to common myth, the addition of alpha particles to iron-peak nuclei is exothermic. However, in the nickel/iron core of a massive star, there are no free alpha particles, they must be broken off a nickel/iron nucleus and then fused with another nickel/iron nucleus to form anything heavier. It is this <em>two-stage</em> process that is endothermic (see <a href="https://astronomy.stackexchange.com/questions/36719/what-effects-besides-mass-defect-cause-the-alpha-ladder-beyond-iron-56-nickel">What effects besides &quot;mass defect&quot; cause the alpha ladder beyond iron-56/nickel-56 to be endothermic?</a>).</p>
<p>Now, given the possibility of rearrangement of the nucleons amongst the nuclei in the core then what should happen is that they equilibriate to the nucleus which has the highest <em>binding energy per nucleon</em> - which is indeed (just) 62Ni.</p>
<p>However, there must be a simple route to produce 62Ni in order for this to happen. The reason that most of the nuclei end up as 56Ni (and then decay to 56Fe) is that there is no easy route to produce 62Ni from 56Ni. It would require the addition of 1-2 alpha particles followed by decays or something similar. This doesn't happen because (i) there is a small net energy penalty for producing a nucleus even heavier than 56Ni (like 60Zn) but more importantly (ii) the increased Coulomb barrier to initiate nuclear fusion in nuclei with <span class="math-container">$&gt;25$</span> protons would require temperatures that are so high that the nuclei photodisintegrate before they can decay into something else.</p>

CC BY-SA 4.0
2023-01-17T11:09:51.650Z
2023-02-13T17:00:16.240Z
2023-02-13T17:00:16.240Z
ProfRob
<p>We know that iron is often regarded as 'nuclear ash' because of its inability to fuse with other atoms, as it has a high binding energy <em>per nucleon</em>. However I found that <a href="https://en.wikipedia.org/wiki/Nickel-62" rel="nofollow noreferrer">Nickel-62</a>, grabs the title of having highest binding energy per nucleon: <span class="math-container">$8.7945$</span> MeV. So my confusion arises is that can iron fuse any further? Why do we see iron as more abundant in becoming nuclear ash rather than nickel-62? Is it related to nuclear fusion reaction steps?</p>

CC BY-SA 4.0
2022-01-03T17:22:50.897Z
https://physics.stackexchange.com/questions/460457/why-is-iron-the-peak-of-the-binding-energy-curve this should help you
Kavin Ishwaran
CC BY-SA 4.0
2022-01-03T18:55:54.003Z
See https://astronomy.stackexchange.com/q/21284/16685 & the questions it links to.
PM 2Ring
<star><nucleosynthesis><core><fusion>
Why do stars usually stop fusion at iron, even though nickel-62 has the highest binding energy per nucleon?
Kshitij Kumar
# Why do stars usually stop fusion at iron, even though nickel-62 has the highest binding energy per nucleon?

(Asked by: username_0 on 2022-01-03)

We know that iron is often regarded as 'nuclear ash' because of its inability to fuse with other atoms, as it has a high binding energy *per nucleon*. However I found that [Nickel-62](https://en.wikipedia.org/wiki/Nickel-62), grabs the title of having highest binding energy per nucleon: $8.7945$ MeV. So my confusion arises is that can iron fuse any further? Why do we see iron as more abundant in becoming nuclear ash rather than nickel-62? Is it related to nuclear fusion reaction steps?

**Tags:** star, nucleosynthesis, core, fusion
> **Comments:**
> **username_1**: https://physics.stackexchange.com/questions/460457/why-is-iron-the-peak-of-the-binding-energy-curve this should help you  
> **username_2**: See https://astronomy.stackexchange.com/q/21284/16685 & the questions it links to.  

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## 1 answers

### Answer by username_3 on 2023-01-17. Score: 10 | Accepted answer

Contrary to common myth, the addition of alpha particles to iron-peak nuclei is exothermic. However, in the nickel/iron core of a massive star, there are no free alpha particles, they must be broken off a nickel/iron nucleus and then fused with another nickel/iron nucleus to form anything heavier. It is this *two-stage* process that is endothermic (see [What effects besides "mass defect" cause the alpha ladder beyond iron-56/nickel-56 to be endothermic?](https://astronomy.stackexchange.com/questions/36719/what-effects-besides-mass-defect-cause-the-alpha-ladder-beyond-iron-56-nickel)).

Now, given the possibility of rearrangement of the nucleons amongst the nuclei in the core then what should happen is that they equilibriate to the nucleus which has the highest *binding energy per nucleon* - which is indeed (just) 62Ni.

However, there must be a simple route to produce 62Ni in order for this to happen. The reason that most of the nuclei end up as 56Ni (and then decay to 56Fe) is that there is no easy route to produce 62Ni from 56Ni. It would require the addition of 1-2 alpha particles followed by decays or something similar. This doesn't happen because (i) there is a small net energy penalty for producing a nucleus even heavier than 56Ni (like 60Zn) but more importantly (ii) the increased Coulomb barrier to initiate nuclear fusion in nuclei with $>25$ protons would require temperatures that are so high that the nuclei photodisintegrate before they can decay into something else.


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