What are the stages of thermonuclear fusion that occurs in the Sun's core?

Answer 1

There are two theories for the thermonuclear cycle. In both, the overall reaction is 4 hydrogen nuclei combining to give 1 helium nucleus, 2 positrons and appx. 26.7 MeV of energy.

Proton - Proton Cycle:
This cycle can be explained in following steps:

Step I: When two protons fuse, they form deuteron nucleus (1D2 or 1H1), a positively charged electron called as positron (1e0) and some energy Q1 is released.

1H1 + 1H1 ---Fusion--> 1D1 + 1e0 + Q1



Step II: The deuteron then fuses with another proton to give 2He3 (i.e. a light isotope of Helium) and γ-quantum radiations.


1D2 + 1H1 ---Fusion--> 2He3 + γ



Step III: For the reaction in Step III to occur, the first two step reactions must occur twice to get two light isotopes of Helium. These two light isotopes fuse to produce stable 2He4.

2He3 + 2He3 ---Fusion--> 2He4 + 21H1 + Q2





The overall reaction is as follows:
41H1 ---Fusion--> 2He4 + 21e0 + 2γ + (2Q1 + Q2)
Total energy released = (2Q1+Q2) = 26.7 MeV.

Carbon - Nitrogen Cycle:
Step I: A proton of Hydrogen first interacts with C12 nucleus to produce N13 isotope and Q1 amount of energy.

6C12 + 1H1 ---Fusion--> 7N13 + Q1



Step II: Nitrogen N13 decays due to its radioactive nature and emits a positron 1e0.

7N13 ---Radioactive decay--> 6C13 + 1e0



Step III: Stable C13 nucleus reacts with another proton, releasing N14 and Q2 amount of energy

6C13 + 1H1 ---Fusion--> 7N14 + Q2


Step IV: Stable N14 now fuses with third proton to form O15 with Q3 amount of energy.

7N14 + 1H1 ---Fusion--> 8O15 + Q3



Step V: The O15 nucleus decays to give positron.

8O15 ---Radioactive decay--> 7N15 + 1e0



Step VI: Finally N15 nucleus interacts with forth proton to produce He Nucleus.

7N15 + 1H1 ---Fusion--> 2He4 + 6C12





The overall reaction is as follows:
41H1 ---Fusion--> 2He4 + 21e0 + (Q1 + Q2 + Q3)
Total energy released = (Q1+Q2 Q3) = 26.7 MeV.