12
$\begingroup$

My understanding is that stellar fusion naturally stops at iron because it is energetically unfavourable to grow the nucleus further.

But iron is only the third most tightly-bound nucleus, nickel is number one, so shouldn't iron favourably fuse with helium?

$\endgroup$
7
$\begingroup$

The iron-peak elements are mostly the product of alpha capture reactions onto nuclei that begin with a similar number of neutrons and protons ($Z = N$).

The nuclear burning associated with carbon and oxygen (in type Ia supernovae) or silicon (in the cores of massive stars at the ends of their lives) is very fast or even explosive. The important reactions in determining the immediate final products are those that proceed on fast timescales.

In these cases, there is a competition between alpha capture and photodisintegration with the constraint that $Z \simeq N$, since weak, flavour-changing reactions are generally too slow to move the neutron/proton ratio far from unity before an equilibrium has been reached between alpha capture and photodisintegration.

Subject to these constraints, then Ni56 turns out to be the most stable nucleus. Further alpha captures to Zn60 (or beyond) are not favored, because these nuclei have lower binding energy and the higher temperatures that would be required to overcome the greater coulomb barrier results in photodisintegration to smaller nuclei.

Hence there is no easy route to Ni62.

The fact that Fe56 dominates the iron-peak element abundances that are seen in the atmospheres of stars and the interstellar medium is because the Ni56 in supernovae ejecta decays to Co56 and then Fe56 on half-lives of 6 and 77 days respectively. Inside the dense core of a massive star then electron capture by Ni56 can be more rapid, but still results in Fe56.

$\endgroup$
  • $\begingroup$ Regarding Fe-56 in the interstellar medium I understand the explanation. For Fe-56 to dominate in stars, the half-lives of 6 and 77 days respectively might be too long if the Si-burning phase only lasts a few days. I would like to find information about the (shorter?) half-lives of Ni-56 and Co-56 under the conditions of an Si-burning stellar core. This would change my (incorrect?) idea that weak decay rates are insensitive to external conditions. $\endgroup$ – gamma1954 May 7 '15 at 22:45
  • $\begingroup$ @gamma1954 The rates of weak interactions are certainly dependent on the external conditions. beta decay can be blocked by electron degeneracy for example. It could well be that the electron capture rates are very enhanced when the electron density is high. I'm not sure you are following my argument. The Fe we see in stars was produced in type Ia and II supernovae - not formed within those stars. We don't "see" what's in the core of a massive star at the end of its life and most of it ends up in a neutron star or black hole. $\endgroup$ – Rob Jeffries May 7 '15 at 23:09
  • $\begingroup$ Thank you for clarifying: the iron we see in stellar atmospheres was produced in supernovae of other stars. I knew it, but misinterpreted your answer. Dredge-up processes may reveal elements like technetium being formed within a star, but they do not have the time to reveal iron production. $\endgroup$ – gamma1954 May 8 '15 at 8:16
5
$\begingroup$

$^{56}Ni$ is produced in silicon-fusion stars. The fusion process doesn't "stop" at $Fe$. Several A=56 nuclides show up. See the Wiki-pedia article on :Silicon burning.

Also, Introductory Nuclear Physics by Krane, Chapter 19, Section 4.

$\endgroup$
  • $\begingroup$ Any idea why Fe is so often quoted as being the stop point? $\endgroup$ – spraff Mar 4 '15 at 19:53
  • $\begingroup$ After looking at some mass tables, I'm puzzled at your assertion regarding which nuclide is most tightly bound. What two nuclides do you think are more tightly bound than $^{56}Fe$? I'm having trouble finding them. EDIT: Never mind. I found them. Ni-62 and Fe-58. It has to do with whether an alpha reaction can get to those two nuclei within the star. I'll have to do some calculations. $\endgroup$ – Bill N Mar 4 '15 at 20:37
-3
$\begingroup$

From Wikipedia:

Because the nuclear force is stronger than the Coulomb force for atomic nuclei smaller than iron and nickel, building up these nuclei from lighter nuclei by fusion releases the extra energy from the net attraction of these particles. For larger nuclei, however, no energy is released, since the nuclear force is short-range and cannot continue to act across still larger atomic nuclei. Thus, energy is no longer released when such nuclei are made by fusion; instead, energy is absorbed in such processes.

$\endgroup$

protected by ACuriousMind Mar 27 '17 at 13:21

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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