<p>There are plenty of misleading statements in Wikipedia and elsewhere on the internet about nucleosynthesis (I am busy searching to see if I have said something similar in the past!)</p>
<p>The reason that the alpha chain does not proceed significantly beyond <span class="math-container">$^{56}$</span>Ni is that in order to overcome the Coulomb barrier the temperatures need to be so high that the iron-peak nuclei are disintegrated by photons at this temperatures.</p>
<p>I suppose the sense in which the endothermic statement is true is when considering a core made of nickel. In order to produce alpha particles you need to disintegrate some Ni nuclei. This process is highly endothermic and can't be balanced by subsequent fusion.</p>
<p>e.g (And this is a bit simplistic) Photodisintegration of a Ni nucleus into 14 alpha particles requires 88.62 MeV. Then 14 fusion reactions with Ni nuclei, producing Zinc, would give back just 37.9 MeV. In contrast disintegrating <span class="math-container">$^{52}$</span>Fe into 13 alpha particles needs 80.5 MeV, but 13 fusion reactions of <span class="math-container">$^{52}$</span>Fe to Ni yields <span class="math-container">$8.1\times 13= 105.3$</span> MeV.</p>

CC BY-SA 4.0
2020-06-28T20:15:03.967Z
2020-09-12T08:52:55.447Z
2020-09-12T08:52:55.447Z
ProfRob
<p>Many sources state that fusion beyond iron-56/nickel-56 (and certainly beyond nickel-62) is impossible due to them being among the most tightly bound nuclei. For example, in the Wikipedia article on the iron peak (<a href="https://en.wikipedia.org/wiki/Iron_peak" rel="noreferrer">https://en.wikipedia.org/wiki/Iron_peak</a>), it is said that:</p>
<blockquote>
<p>For elements lighter than iron on the periodic table, nuclear fusion releases energy. For iron, and for all of the heavier elements, nuclear fusion consumes energy.</p>
</blockquote>
<p>However, when you actually compute the mass defect, the alpha ladder would be exothermic up to Tin.</p>
<p><span class="math-container">$$ Q=[m(Ni_{28}^{56})+m(He_{2}^{4})-m(Zn_{30}^{60})]c^2 $$</span>
<span class="math-container">$$ Q=[55.942132022u+4.00260325415u-59.941827035u]m_uc^2 $$</span>
<span class="math-container">$$ Q \approx 2.709 MeV $$</span>
<span class="math-container">$$$$</span>
<span class="math-container">$$ Ni_{28}^{56} + He_{2}^{4} \rightarrow Zn_{30}^{60} (+2.709 MeV)$$</span>
<span class="math-container">$$ Zn_{30}^{60} + He_{2}^{4} \rightarrow Ge_{32}^{64} (+2.587 MeV)$$</span>
<span class="math-container">$$ Ge_{32}^{66} + He_{2}^{4} \rightarrow Se_{34}^{68} (+2.290 MeV)$$</span>
<span class="math-container">$$ Se_{34}^{68} + He_{2}^{4} \rightarrow Kr_{36}^{72} (+2.151 MeV)$$</span>
<span class="math-container">$$ Kr_{36}^{72} + He_{2}^{4} \rightarrow Sr_{38}^{76} (+2.728 MeV)$$</span>
<span class="math-container">$$ Sr_{38}^{76} + He_{2}^{4} \rightarrow Zr_{40}^{80} (+3.698 MeV)$$</span>
<span class="math-container">$$ Zr_{40}^{80} + He_{2}^{4} \rightarrow Mo_{42}^{84} (+2.714 MeV)$$</span>
<span class="math-container">$$ Mo_{42}^{84} + He_{2}^{4} \rightarrow Ru_{44}^{88} (+2.267 MeV)$$</span>
<span class="math-container">$$ Ru_{44}^{88} + He_{2}^{4} \rightarrow Pd_{46}^{92} (+2.276 MeV)$$</span>
<span class="math-container">$$ Pd_{46}^{92} + He_{2}^{4} \rightarrow Cd_{48}^{96} (+3.030 MeV)$$</span>
<span class="math-container">$$ Cd_{48}^{96} + He_{2}^{4} \rightarrow Sn_{50}^{100} (+3.101 MeV)$$</span></p>
<p>I ended my calculation here because I wasn't able to find the masses of other isotopes that would, theoretically, follow the chain. I understand that these are highly unstable and their fusion would need an immense amount of energy to overcome the Coulomb barrier. However, my point is that, according to the calculations above, once the barrier is overcome, the fusion would actually <em>release</em> energy, not consume it. So, is the notion of fusion beyond the iron peak elements being endothermic false or am I missing something?</p>

CC BY-SA 4.0
2020-06-28T18:12:01.337Z
Related: https://astronomy.stackexchange.com/a/21288/16685
PM 2Ring
CC BY-SA 4.0
2020-06-28T18:23:01.980Z
Interesting! I don't know the complete answer to your question, but 1) those alphas mainly come from photodisintegration, which *is* endothermic. 2) those fusion products are progressively more & more deficit in neutrons and have short half-lives.
PM 2Ring
<nucleosynthesis>
What effects besides "mass defect" cause the alpha ladder beyond iron-56/nickel-56 to be endothermic?
HeatherB
# What effects besides "mass defect" cause the alpha ladder beyond iron-56/nickel-56 to be endothermic?

(Asked by: username_0 on 2020-06-28)

Many sources state that fusion beyond iron-56/nickel-56 (and certainly beyond nickel-62) is impossible due to them being among the most tightly bound nuclei. For example, in the Wikipedia article on the iron peak (<https://en.wikipedia.org/wiki/Iron_peak>), it is said that:

> For elements lighter than iron on the periodic table, nuclear fusion releases energy. For iron, and for all of the heavier elements, nuclear fusion consumes energy.

However, when you actually compute the mass defect, the alpha ladder would be exothermic up to Tin.

$$ Q=[m(Ni\_{28}^{56})+m(He\_{2}^{4})-m(Zn\_{30}^{60})]c^2 $$
$$ Q=[55.942132022u+4.00260325415u-59.941827035u]m\_uc^2 $$
$$ Q \approx 2.709 MeV $$
$$$$
$$ Ni\_{28}^{56} + He\_{2}^{4} \rightarrow Zn\_{30}^{60} (+2.709 MeV)$$
$$ Zn\_{30}^{60} + He\_{2}^{4} \rightarrow Ge\_{32}^{64} (+2.587 MeV)$$
$$ Ge\_{32}^{66} + He\_{2}^{4} \rightarrow Se\_{34}^{68} (+2.290 MeV)$$
$$ Se\_{34}^{68} + He\_{2}^{4} \rightarrow Kr\_{36}^{72} (+2.151 MeV)$$
$$ Kr\_{36}^{72} + He\_{2}^{4} \rightarrow Sr\_{38}^{76} (+2.728 MeV)$$
$$ Sr\_{38}^{76} + He\_{2}^{4} \rightarrow Zr\_{40}^{80} (+3.698 MeV)$$
$$ Zr\_{40}^{80} + He\_{2}^{4} \rightarrow Mo\_{42}^{84} (+2.714 MeV)$$
$$ Mo\_{42}^{84} + He\_{2}^{4} \rightarrow Ru\_{44}^{88} (+2.267 MeV)$$
$$ Ru\_{44}^{88} + He\_{2}^{4} \rightarrow Pd\_{46}^{92} (+2.276 MeV)$$
$$ Pd\_{46}^{92} + He\_{2}^{4} \rightarrow Cd\_{48}^{96} (+3.030 MeV)$$
$$ Cd\_{48}^{96} + He\_{2}^{4} \rightarrow Sn\_{50}^{100} (+3.101 MeV)$$

I ended my calculation here because I wasn't able to find the masses of other isotopes that would, theoretically, follow the chain. I understand that these are highly unstable and their fusion would need an immense amount of energy to overcome the Coulomb barrier. However, my point is that, according to the calculations above, once the barrier is overcome, the fusion would actually *release* energy, not consume it. So, is the notion of fusion beyond the iron peak elements being endothermic false or am I missing something?

**Tags:** nucleosynthesis

> **Comments:**
> **username_1**: Related: https://astronomy.stackexchange.com/a/21288/16685  
> **username_1**: Interesting! I don't know the complete answer to your question, but 1) those alphas mainly come from photodisintegration, which *is* endothermic. 2) those fusion products are progressively more & more deficit in neutrons and have short half-lives.  

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## 1 answers

### Answer by username_2 on 2020-06-28. Score: 16

There are plenty of misleading statements in Wikipedia and elsewhere on the internet about nucleosynthesis (I am busy searching to see if I have said something similar in the past!)

The reason that the alpha chain does not proceed significantly beyond $^{56}$Ni is that in order to overcome the Coulomb barrier the temperatures need to be so high that the iron-peak nuclei are disintegrated by photons at this temperatures.

I suppose the sense in which the endothermic statement is true is when considering a core made of nickel. In order to produce alpha particles you need to disintegrate some Ni nuclei. This process is highly endothermic and can't be balanced by subsequent fusion.

e.g (And this is a bit simplistic) Photodisintegration of a Ni nucleus into 14 alpha particles requires 88.62 MeV. Then 14 fusion reactions with Ni nuclei, producing Zinc, would give back just 37.9 MeV. In contrast disintegrating $^{52}$Fe into 13 alpha particles needs 80.5 MeV, but 13 fusion reactions of $^{52}$Fe to Ni yields $8.1\times 13= 105.3$ MeV.


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