In layman's terms, why does the cold C-N-O process end in Carbon?

Question

I read up a little bit on fusion in stars, layman's articles only and the P-P chain makes hydrogen or Alpha particles. (error removed on triple-alpha), then the C-N-O process adds hydrogen to the stars carbon (I'm limiting this to the cold C-N-O process for simplicity), this adds protons to the C-12s until it gets to N-15 then it fuses with a proton and from what I've read this becomes C-12 and He-4, not O-16.

This seems odd to me, as O-16 is slightly lower energy and fusion likes to reach the lower energy state, but I'm guessing there's something about the super-stability of the He-4 and/or the velocity or energy of the incoming proton that splits the O-16. But that's just a guess.

Is that correct? Does the final stage of the cold C-N-O process yield primarily Carbon + Helium or does it produce some Oxygen as well?

This related question indicates that this might not be well understood.

Answer 1

Both 15N($p,\alpha$)12C and 15N($p,\gamma$)16O can and do take place. However, the former is a strong force interaction, whilst the latter is mediated by the electromagnetic force and is known as a "radiative capture" reaction.

Electromagnetic nuclear reactions are much slower than strong force nuclear reactions. So it is just a question of reaction rates and the first, quicker, reaction dominates. According to the wikipedia page on the CNO cycle (but with no citation), the production of 16O rather than 12C + $\alpha$ occurs just 0.04% of the time at solar core densities and temperatures. Simpson (2006) calculates a ratio of astrophysical S-factors (that determine the reaction rates) of 1:1360.

I guess the way you can think about it is: adding the proton to the nitrogen produces an intermediate excited state that can either alpha decay into a carbon nucleus or radiatively decay into an oxygen nucleus. Since the former happens much more quickly, then that's mostly what happens.

Answer 2

This question has several misunderstandings.

The CNO cycles (there is more than one CNO cycle) in a main sequence star have nothing to do with the triple alpha process. A star that is using the triple alpha process is fusing helium to form carbon. This is a star that is no longer on the main sequence.

Main sequence stars that use the CNO cycles to fuse hydrogen into helium do so using primordial carbon, nitrogen, and oxygen, where "primordial" means elements that were present when the star first formed. The very first stars had no carbon, nitrogen, or oxygen; they could only use the p-p chain while on the main sequence. Later generation stars had non-zero metallicity; "metals" to astronomers are any elements other than hydrogen and helium. In the Sun, and in most main sequence stars, carbon, nitrogen, and oxygen are very common "metals".

The CNO cycles are cycles. As such, they do not have a starting point or ending point. On reaching steady state, the abundance of the various nuclides involved in the CNO cycles will likely differ from the primordial abundance, but that's due to likelihoods of different transitions and due to bottlenecks.

Answer 3

The CNO process does not end in carbon. It's part of the hydron burning processes. Let's look at the different processes inside main sequence stars:

Hydrogen burning describes the fusion process from the lightest possible element (hydrogen) into the 2nd lightest element (helium). Or if you look at it at the level of protons and neutrons it converts 4 separate protons into a one nucleus which consists of 2 protons and 2 neutrons. The difference in binding energy is released in this process.

Now, if you go into the details even further, there are different ways this process can happen as it is not a single reaction but involves several intermediate steps which may be different and occur with a different likelyhood but which all result in essence in the same.

proton-proton-chain (pp-chain):

Let's look at the processes which occur there:

1. happens twice: $p + p \rightarrow {}^2H + \nu_e + e^+ $ This is the limiting reaction as it requires a weak interaction (it converts a proton into a neutron). It takes on average for a single proton 9 billion years inside the Sun. We are lucky there are so many protons inside the Sun.
2. happens twice: ${}^2H + p \rightarrow {}^3He + \gamma$ This reaction is comparatively quick as it is based on the strong nuclear force (no conversion of protons into neutrons, just nucleus transformation) and estimated to happen within one second after the generation of the ${}^2H$.
3. ${}^3He + {}^3He \rightarrow {}^4He + p + p$ In (1) you gain additional radiation energy of 1.442MeVif you consider the annihilation of the positron with a free electron which will occur immediately after. This the so-called ppI branch. If the collision of the ${}^3He$ was with one of the existing (and here produced) ${}^4He$ nuclei we would get the (in the Sun less likely) ppII and ppIII chains.

So overall we have the equation of $4{}^1_1H + 2e^- \rightarrow {}^4_2He + 2\nu_e + 7\gamma + 26.7MeV$

Now the CNO cycle (or also called Bethe-Weizsäcker-cycle in honour of the two people who suggested its existence independently and simultaneously) which effectively yields the same and also is a series of reactions. I will only illustrate the main cycle here:

1. ${}^{12}_6C + {}^1_1H \rightarrow {}^{13}_7N + \gamma$ (strong force)
2. ${}^{13}_7N \rightarrow {}^13_6C + e^+ +\nu_e$ (weak force, but decay, ~10 minutes)
3. ${}^{13}_6C + {}^1_1H \rightarrow {}^{14}_7N + \gamma$ (strong force)
4. ${}^{14}_7N + {}^1_1H \rightarrow {}^{15}_8O + \gamma$ (strong force)
5. ${}^{15}_8O \rightarrow {}^{15}_7N + e^+ + \nu_e$ (*weak force, but decay, ~2 minutes)
6. ${}^{15}_7N + {}^1_1H \rightarrow {}^{12}_6C + {}^4_2He$ (strong force)

Now, the weak force must not be overcome here anywhere in fusion but only plays a role on larg(er) instable nuclei which decay into nuclei, thus is not the time-limiting factor. Yet, it requires a proton to "collide" with a much heavier nucleus which already combines 6 (C) or 7 (N) protons along with some neutrons. Much more energy is needed to overcome this coulomb barrier. Thus this CNO cylce is strongly temperature (and pressure dependent):

And this is the reason that in the sun (and other dwarf stars) the pp process is the dominant one. But this process does not yield any other material than ${}^4_2He$, especially it does not yield $C$, $N$ or $O$ which are only used as catalyst; they are neither effectively produced nor consumed; their presence is a mere requirement for this process to happen and thus this process cannot have happened in the first stars (which only consisted of hydrogen and helium) in their main sequence regardless of mass and pressure.

Note also that the CNO cycle does only rarely create a (stable) ${}^{16}_8O$, but there ist most often only ever the non-stable ${}^{15}_8O$ which decays within 2 minutes. In the last step, theoretically (and with a small probability) a ${}^{16}_8O$ could be produced, but this is energetically less favourable.

However, even then the ${}^{12}C$ is not permanently destroyed, because except in about one out of $5\cdot 10^7$ cases, ${}^{16}O$ will again return to ${}^{12}C$ (cf. $8)

(quoted from Bethe's original publication)

Carbon is produced only via the triple alpha process. It happens when the amount of helium in a star's core becomes so little that the likelyhood of a hydrogen fusion sinks and density and temperature rose enough to start the triple alpha process which fuses helium into carbon. This requires MUCH higher pressure and densities than hydrogen fusion as you need to bring together three particles of twice the charge than previously within $10^{-16}s$ as ${}^8_4Be$ decays quickly and its creation is endothermic. The star leaves the main sequence phase at this point. The other alpha processes to create higher-order elements can start when enough carbon has been produced by the triple alpha process.