|12C + 12C||→||20Ne + α||4.621||44%|
|→||23Na + p||2.242||56%|
|23Na + p||→||20Ne + α||2.379|
|23Na + p||→||24Mg + γ||11.691|
|16O + α||→||20Ne + γ|
|20Ne + α||→||24Mg + γ|
|End result:||20Ne, 23Na,||24Mg|
1.43×1042 ηQ ρ
ergs g−1 s−1,
where ηQ is an appropriate fractional energy release per reaction.
12C + 16O reactions occur, but the rate is rather slower than both C + C and at higher temperatures O + O. Besides, the carbon is rapidly eliminated by the C + C reaction sequences.
|16O + 16O||→||28Si + α||9.593||21%|
|→||31P + p||7.678||61%|
|→||31S + n||1.500||18%|
|31S||→||31P + e+ + ν|
|31P + p||→||28Si + α|
|28Si + α||→||32S + γ, etc.|
|End result:||28Si, 30Si,||34S|
ε16O = 1.3×1052 ηQ ρ X16O2 T9−2/3 e−[139.53/T91/3 +corrections] ergs g−1 s−1.
1. 20Ne (γ,α) 16O, followed by 20Ne (α,γ) 24Mg (α,γ) 28Si
2. 28Si (γ,α) 24Mg (α,p) 27Al (α,p) 30Si
The former happens at temperatures intermediate between carbon burning and oxygen burning, and converts all Ne to O and Si.
Homework: Show that converting two 20Ne into one 16O + one 24Mg is exothermic, and three 20Ne into two 16O + one 28Si is even more exothermic.
Below is figure 2.26 from H&K:
The nuclear statistical equilibrium (often refered to by the acronym NSE in the literature) of the iron group depends on the time available to achieve equilibrium. The group elements start out predominantly as α-nuclei; i.e. equal (and even) numbers of protons and neutrons. These nuclei rapidly shuttle α's back and forth via photodisintegration reactions. The NSE may be calculated with the usual nuclear Saha equations, and favors 56Ni. But, 56Ni is proton-rich and beta decays with a half-life of 6.077 days to 56Co, which in turn decays with a half-life of 77.27 days to 56Fe. If weeks are available for these and other beta decays to happen, the equilibrium eventually favors 56Fe. The table above suggests that weeks are not available. In fact 56Co often appears in the spectra of supernovae, indicating its very recent origin from 56Ni.