2

In the book "Horizons: exploring the universe-Cengage learning (2018)", p200, it states that:

Helium fusion produces carbon, and some of the carbon nuclei absorb helium nuclei to form oxygen. A few of the oxygen nuclei can absorb helium nuclei and form neon and then magnesium. Some of these reactions release neutrons, which, having no charge, are more easily absorbed by nuclei to gradually build even heavier nuclei. These reactions are not important as energy producers, but they are slow-cooker processes that form small traces of heavier elements right up to bismuth with atomic weight 209, nearly four times heavier than iron. Many of the atoms in your body were produced this way.

Base on my question previously, I assume we are talking about the alpha process here. And when I check the Wikipedia of the Alpha Process, it says that the alpha process ends at $^{56}_{28}\mathrm{Ni}$ due to photodisintegration, but as the suggests, the process should produce $^{209}_{83}\mathrm{Bi}$. The question is, why do Bismuth, an atom with atomic number $83$ are produced here? Shouldn't atom like $^{}_{82}\mathrm{Pb}$ with even atomic number being produced here?

Jack the Ranger
  • 1,296
  • 6
  • 24
  • 1
    See https://en.wikipedia.org/wiki/S-process and https://astronomy.stackexchange.com/search?q=s-process – PM 2Ring May 27 '21 at 04:17
  • That quote is a bit misleading. "These reactions" refers to the absorption of neutrons (the s-process). Those neutrons are released by some alpha process reactions. – PM 2Ring May 27 '21 at 04:22
  • @PM2Ring So "These reactions" means both the alpha process and the s-process, is that correct? – Jack the Ranger May 27 '21 at 06:14
  • 1
    No, it just means the s-process reactions. The alpha process reactions are all exothermic (they produce energy), a few of them also produce free neutrons. But the heavier alpha process reactions (from neon burning onwards) operate at such high temperatures that photodisintegration can occur, and that process is endothermic (it absorbs energy). Photodisintegration knocks a proton or alpha out of a nucleus. It's especially important for the last stages of the alpha ladder (the burning of silicon & higher nuclei) because it's the main source of the alpha particles used in those reactions. – PM 2Ring May 27 '21 at 07:03

2 Answers2

5

Bismuth is produced via the s-process.

It is the slow absorption of neutrons that can produce heavier elements beyond the peak of the binding energy per nucleon curve at iron. The alpha process isn't capable of producing much of anything beyond nickel.

ProfRob
  • 151,483
  • 9
  • 359
  • 566
2

I'll (attempt to) add to @ProfRob's answer:

The alpha process isn't capable of producing much of anything beyond nickel.

The alpha particles (helium nuclei) and the target nuclei are both positively charged and face a huge barrier due to electrostatic (Coulomb) repulsion.

They successfully tunnel through this Coulomb barrier from time to time only because it is energetically favorable. This is generally called nuclear fusion.

Once you get well above iron1 there is no longer an energy advantage. The nuclei are in a lower energy state separate from each other. You'd need to add energy to the system to make them stick. So the coulomb barrier tunneling will be unsuccessful unless you have high energy helium nuclei. We do that on Earth with particle accelerators, generally to tens of MeV energy to get it to happen.

That's why alpha decay happens for the much heavier nuclei. Many of them will spontaneously "spit out" alpha particles; getting rid of them. They are in no mood to accept some low energy alpha particle knocking at the door. We use this spitting out of energetic alpha particles for heaters in spacecraft like Mars rovers Curiosity and Perseverance and deep space probes like the Voyagers and even on the moon.

They are called radioisotope thermoelectric generators and are a good concrete example of nuclei much heavier than iron having no interest whatsoever in accepting more alpha particles.

1Thanks to @ProfRob and @PM 2Ring for their helpful comments. For people who call anything bigger than helium "metal" I was surprised at the sudden attention to detail ;-)

uhoh
  • 31,151
  • 9
  • 89
  • 293
  • 1
    You have to be careful with your fusion energy argument. Sticking alpha particles to nuclei is still an exothermic process beyond iron. The key is where those alpha particles come from. – ProfRob May 28 '21 at 07:07
  • 1
    See https://astronomy.stackexchange.com/questions/36719/what-effects-besides-mass-defect-cause-the-alpha-ladder-beyond-iron-56-nickel – ProfRob May 28 '21 at 07:33
  • 1
    I calculate that $$^{56}{28}\mathrm{Ni}+,^{4}{2}\mathrm{He}\longrightarrow,^{60}_{30}\mathrm{Zn}$$ releases ~2.69 MeV, slightly less than HeatherB's value in the post ProfRob just linked. – PM 2Ring May 28 '21 at 07:44
  • @ProfRob yes I've adjusted the language a bit and added a (humorous) footnote. Thanks for pointing this out! – uhoh May 28 '21 at 07:55
  • @PM2Ring ditto. – uhoh May 28 '21 at 07:55
  • 2
    https://xkcd.com/2340/ ;) – PM 2Ring May 28 '21 at 08:09
  • 2
    I think the serious point is that it is only when you try and rearrange the nucleons that the iron-peak emerges. Just sticking alpha particles to nuclei tends to be exothermic. But not if those alpha particles come from breaking up other nuclei. – ProfRob May 28 '21 at 09:00
  • @ProfRob these are discrete helium nuclei, no? On a nuclear timescale they've long since "forgotten" where they come from right? They don't "know" where they came from, do they? These are not four nucleon transfer reactions between two heavier nuclei, are they? So the scope of my answer is only on the capture (or "fusion") of 4He with heavy nuclei. Whether they happen to have enough energy to overcome the coulomb barrier or not if so, how they got it is outside the scope of what I've tried to explain here, but certainly someone doing a full simulation must worry about it. – uhoh May 28 '21 at 12:56
  • @ProfRob So certainly my supplementary answer is an incomplete view, but it's all I can muster and as far as I can tell not wrong as written. – uhoh May 28 '21 at 12:57
  • On a nuclear timescale they've long since "forgotten" where they come from right? Sure, but that's not the point. The photodisintegration that produced those alphas is endothermic. So while the alpha reaction itself is exothermic, when you include the cost of producing the alpha, the total process is endothermic. See the last paragraph of ProfRob's linked answer for some actual (approximate) calculations. – PM 2Ring Jun 05 '21 at 08:45
  • @PM2Ring Don't worry, I appreciate now that ProfRob's point is external to what I've written about (which is where I was going there). I got it. I'm just sticking with "the scope of my answer (which) is only on the capture (or "fusion") of 4He with heavy nuclei." and went on to say "...how they got it is outside the scope of what I've tried to explain here, but certainly someone doing a full simulation must worry about it." – uhoh Jun 05 '21 at 10:47
  • @PM2Ring Their answer is 3 sentences and contains no link, sources or citations. If they or you or anyone else would like to write a more complete answer to the OP's question, then it belongs in an answer post, not buried in comments under mine. Otherwise this is just "I would have written something different". This question though needs all the attention it can get! Density and dimensionality of zeros in inverse square force fields of randomly distributed sources in (at least) 1, 2 and 3 dimensions? – uhoh Jun 05 '21 at 10:48
  • as does this one: How was the high rate of degradation of Hubble's main mirror initially predicted? Why did it turn out to be lower? Were cosmic rays really a concern? – uhoh Jun 05 '21 at 10:56