5
$\begingroup$

I'm currently studying the alpha process, specifically the triple alpha process.

I understand why helium-4 is stable, and why beryllium-8 'prefers' to split into two helium-4 atoms. I also understand why beryllium-8 and helium-4 form carbon-12. However, I don't understand why carbon-12 is stable. Why doesn't carbon-12 split into three helium-4 atoms?

$\endgroup$
  • $\begingroup$ Because it is not energetically favourable? Isn't that, aside from conservation laws, always the reason a particular process does or does not happen? Can you be more specific about what you want to know? $\endgroup$ – ACuriousMind Jul 20 '16 at 15:26
  • 2
    $\begingroup$ @ACuriousMind I think he essentially asks, why is it energetically favorable. $\endgroup$ – user259412 Jul 20 '16 at 16:03
  • 3
    $\begingroup$ It may be of interest to point out the following wiki reference en.wikipedia.org/wiki/Fred_Hoyle, Hoyle noticed that for the triple-alpha process, which generates carbon from helium, would require the carbon nucleus to have a very specific resonance energy and spin for it to work. The large amount of carbon in the universe, demonstrated to Hoyle that this nuclear reaction must work. Based on this Hoyle predicted the values of the energy, the nuclear spin and the parity of the compound state in the carbon nucleus formed by three alpha particles, which was later borne out by experiment. $\endgroup$ – jim Jul 20 '16 at 18:17
6
$\begingroup$

The possible ways in which Carbon-12 can decay into three He-4 atoms are:

  1. C-12$\rightarrow$ Be-8 + He-4

    followed by

    Be-8 $\rightarrow$ He-4 + He-4

This is not likely to happen as the first process is not energetically favourable. You can deduce that it is not favourable because you mentioned that Be combine with He to form C, hence suggesting that the reaction is energetically favourable. Therefore the reverse process cannot be favourable.

  1. In theory, another way of going from a C-12 to 3 He-4 nuclei is by a double nuclear fission. However, nuclear fission of light nuclei (like C-12) has almost zero probability of spontaneously occurring.

This is why C-12 nuclei are stable.

$\endgroup$
  • $\begingroup$ This answer simply re-states a fact stipulated in the question in more technical language and doesn't explain anything. The question could be stated "Why is C-12 energetically favorable relative 2 alpha particles while Be-8 is not energetically favorable relative 2 alpha particles?". Sure, it's an observed fact. But answers to "why" question (when they are available at all) come from lower level theories. What is it about the binding of nuclei that makes C-12 strongly enough bound to not fission and Be-8 weakly enough bound that it does fission? $\endgroup$ – dmckee Jul 21 '16 at 2:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.