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CNO cycle of stellar nucleosynthesis involves several steps in which different isotopes of Carbon, Nitrogen, and Oxygen are transmuted into each other when hit by a free protons - this is a very simplified description.

\begin{align} {}^{12}C + {}^{1}H &\to {}^{13}N + \gamma\\ {}^{13}N &\to {}^{13}C + e\\ {}^{13}C + {}^{1}H &\to {}^{14}N + \gamma\\ {}^{14}N + {}^{1}H &\to {}^{15}O + \gamma\\ {}^{15}O &\to {}^{15}N + e\\ {}^{15}N + {}^{1}H &\to {}^{12}C + {}^{4}He\\ \end{align}

My question is about the last step that produces an $\alpha$ particle. It looks like $^{16}O$ would be an intermediate product, which should be stable, similarly to two steps earlier which produce $^{15}O$ except $^{16}O$ is more stable so it wouldn't emit anything in the next step and break the cycle. Why and how does $^{16}O$ emit $\alpha$?

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As pointed in the comment sometimes, especially at higher temperatures $^{16}O$ does remain stable and we get CNO-II cycle:

\begin{align} {}^{14}N + {}^{1}H &\to {}^{15}O + \gamma\\ {}^{15}O &\to {}^{15}N + e\\ {}^{15}N + {}^{1}H &\to {}^{16}O + \gamma\\ {}^{16}O + {}^{1}H &\to {}^{17}F + \gamma\\ {}^{17}F &\to {}^{17}O + e\\ {}^{17}O + {}^{1}H &\to {}^{14}N + {}^{4}He\\ \end{align}

There are further modifications to this cycle at higher and higher temperatures, but we still call it CNO unfortunately, we could have really hot FOFO cycle instead of boring CNO-IV.

Anyway. Why does $^{16}O$ ever emit $\alpha$, and why is it more stable at higher temperatures? If anything I would expect protons with higher kinetic energy to create more unstable nuclei.

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    $\begingroup$ Wikipedia says that a minor branch, CNO-II, does produce O-16 and a photon at that step, instead of the usual C-12 + He-4. $\endgroup$ – PM 2Ring Jun 18 at 17:11
  • $\begingroup$ yes, good point. but it gives me more questions than answers. I will update my question. thanks $\endgroup$ – Milo Bem Jun 19 at 8:45
  • $\begingroup$ Oh, I totally agree! It seems backwards that the O-16 cycle is more stable at higher temperatures. Hopefully, one of the astrophysicists will give us a good answer. $\endgroup$ – PM 2Ring Jun 19 at 9:33

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