Fusion in the Sun step by step

The overall fusion reaction 4 ¹H + 2 e --> ⁴He + 2 neutrinos + 6 photons happens in several steps. (This is not different from methane burning, which also happens in steps.)

Step 1

¹H + ¹H --> ²H + antielectron + neutrino

This step is nearly impossible.

The protons repel each other because they both have + 1 charge.
They have to get within 10^-15 m of each other for the strong interactions to hold them together.
One of the protons has to change into a neutron (emitting an anti-electron and neutrino).
- This requires a weak interaction.
- But the chance of this weak interaction happening just when the protons are together is almost zero.

This near impossiblity has two consequences:

The gas must be very hot, so that the protons hit each other with high speed.
- This lets them get near to each other, even though they repel each other.
Even so, the reaction is very rare. That is why the Sun is still burning after 4.6 billion years!

Step 2

electron + antielectron --> photon + photon

The antielectron soon hits an electron. They annihilate to make two photons. (These high energy photons will be absorbed in the gas, eventually giving lots of low energy photons.)

Step 3

²H + ¹H --> ³He + photon

The ²H hits a proton and sticks to it, making a ³He and emitting a photon in the process. (This high energy photon will be absorbed in the gas, eventually giving lots of low energy photons.)

Step 4

³He + ³He --> ⁴He + ¹H+ ¹H

Since these reactions make ³He nuclei, there are some of these bouncing around in the gas. When two of them hit each other, the 2 neutrons and 4 protons of rearrange themselves into one ⁴He nucleus and two free protons.

Note that steps 1,2 and 3 each happen twice for each time step 4 happens.

The net result is

¹H + ¹H --> ²H + antielectron + neutrino
¹H + ¹H --> ²H + antielectron + neutrino
electron + antielectron --> photon + photon
electron + antielectron --> photon + photon
²H + ¹H --> ³He + photon
²H + ¹H --> ³He + photon
³He + ³He --> ⁴He + ¹H+ ¹H

That is

6 ¹H + 2 e --> ⁴He + 2 ¹H + 2 neutrinos + 6 photons

4 ¹H + 2 e --> ⁴He + 2 neutrinos + 6 photons

The amount of energy released in each separate reaction has been measured. The net energy release is 26 MeV.

There are some other reactions also. They are less significant because they don't produce much energy. [For this reason, please don't try to memorize them.] But they have a special significance because they produce neutrinos, which we will study.

³He + ⁴He --> ⁷Be + photon
- ⁷Be = (4 p + 3 n)
⁷Be + electron --> ⁷Li + neutrino
- ⁷Li = (3 p + 4 n)
⁷Li + ¹H --> 2 ⁴He
alternatively
- ⁷Be + ¹H --> ⁸B + photon
  - ⁸B = (5 p + 3 n)
- ⁸B --> 2 ⁴He + antielectron + neutrino

There is another energy producing cycle that is important in hot stars (temperatures bigger than 16 x 10⁶K). It is called the CNO cycle. In it, carbon is a catalyst. It is needed to make the reaction work, but is not used up. [Again, I would suggest that you not memorize the cycle, but do understand the concept of a catalyst, which is important in practical chemical reactions.]

start with ¹²C
- ¹²C = (6 p + 6 n)
¹²C + ¹H --> ¹³N + photon
- ¹³N = (7 p + 6 n)
¹³N --> ¹³C + antielectron + neutrino
- ¹³C = (6 p + 7 n)
¹³C + ¹H --> ¹⁴N + photon
- ¹⁴N = (7 p + 7 n)
¹⁴N + ¹H --> ¹⁵O + photon
- ¹⁵O = (8 p + 7 n)
¹⁵O --> ¹⁵N + antielectron + neutrino
- ¹⁵N = (7 p + 8 n)
¹⁵N + ¹H --> ¹²C + ⁴He

The net result is that you get the ¹²C back and you have used up 4 protons to make one ⁴He, along with two antielectrons and three photons. The antielectrons will annihilate with electrons to make photons.

So hydrogen is burned to make helium with carbon as a catalyst.

Davison E. Soper, Institute of Theoretical Science, University of Oregon, Eugene OR 97403 USA soper@bovine.uoregon.edu