# How many times do solar protons repeatedly fuse and fission before they form deuteron

In the proton-proton chain reaction in the Sun, the first step is $$p + p \rightarrow \; ^2_2{\rm He} .$$ After this, the most likely thing to happen next is that the reverse reaction occurs: the $$^2_2{\rm He}$$ splits up and we are back where we started. Sometimes, very much more rarely, there is a weak beta+ decay process which converts a proton to a neutron and then we get a deuteron and further reactions can occur. I know how to find the rate for the latter process, but I would like to find the rate for the first process. My question is: in the solar core, how many times does the fuse--fission sequence ($$p+p \rightarrow {\rm He} \rightarrow p+p$$) occur, on average, for any given proton, before we get the weak decay and a deuteron? Even an order-of-magnitude estimate from rough cross section formula would be helpful. (I have tried to find it with no luck yet).

Edit added 19 Feb 2018: My research so far has yielded numbers between $$10^{15}$$ and $$10^{28}$$. So anyone who can pin it down better than that range (13 orders of magnitude) is making progress!

• Lots and lots. :) I've never seen a raw number, or probability, just the estimated half-life of 1 billion years for a solar core proton to be converted to He-4. Wikipedia says that the deuteron converts to He-3 in about 4 seconds, and He-3 to He-4 can proceed in various ways, taking about 400 years. – PM 2Ring Feb 12 at 11:54

There is a nice analysis by Gillian Knapp here. She argues that it's natural to define the cross-section for collision as the cross-section for an S-wave collision, and this comes out to be four orders of magnitude bigger than the geometrical cross-section. Using this along with the number density of protons in the core, she gets a collision rate of $$\sim 10^{12}\ \text{s}^{-1}$$. This is the rate of assaults on the Coulomb barrier, not the rate of close collisions in which we actually penetrate the barrier. The observed fusion rate, based on how much of the sun's hydrogen remains unburned, is $$5\times10^{-18}\ \text{s}^{-1}$$. The WKB barrier penetration probability for S waves is $$\sim e^{-16}\sim10^{-7}$$. So we have
$$\frac{\text{rate of barrier-penetrating collisions}}{\text{rate of fusion}} = \frac{(\text{rate of S-wave collisions})(\text{WKB probability})}{\text{rate of fusion}}\sim 10^{23}.$$