## The proton-proton 1 chain

In a previous article, I have written about the binding energy. If we look again at the diagram of binding energies per nucleus, we see that the first big peak is at the $^{4}\textrm{He}$ isotope, so that is where we expect the fusion reaction to go, because, as I also wrote in that article, the system wants to reach a lower energy state, i.e., a state with higher binding energy between protons and neutrons in the nucleus.

When using classical physics in order to estimate the temperature of the Sun, it was discovered that the Sun’s temperature was considered too low to overcome the Coulomb barrier between protons. After the development of quantum mechanics, it was discovered that quantum tunneling of the wavefunctions of the protons through the repulsive Coulomb barrier allows for fusion at a lower temperature than the classical prediction. However, even with quantum tunneling, it was unclear how proton-proton fusion might proceed since creating $^{4}\textrm{He}$ starting from 2 hydrogen nuclei (or 2 protons), is, as we will see in the next paragraph not easy. It all starts with this reaction:

$\large _{1}^{1}\textrm{H} + _{1}^{1}\textrm{H} \rightarrow _{2}^{2}\textrm{He} + \gamma$

The result of the reaction above is a diproton (with Z=2 protons and N=0 neutrons in the nuclei). This nuclei is unstable and immediately dissociates back into a pair of protons.

As we have seen in the article on binding energies and the weak force, neutrons are required to have a stable nuclei, in order to counteract the effect of the electromagnetic repulsion between protons. And in order to reduce the disruptive energy of the electric force, the weak interaction allows the number of neutrons to exceed that of protons. Therefore, without any neutrons, $_{2}^{2}\textrm{He}$ almost immediately disintegrates into two protons, producing no net energy release, and that also explains why $_{2}^{2}\textrm{He}$ is not in the diagram of binding energies but only $^{3}\textrm{He}$ (Z=2 and N=1) and $^{4}\textrm{He}$ (Z=2 and N=2) are because they have N > 0.

Thus for a reaction to generate energy, one must find a way to bypass $_{2}^{2}\textrm{He}$ and jump to a stable state. One possible solution to this problem was discovered by Hans Bethe in 1939: during the very brief period that $_{2}^{2}\textrm{He}$ lives, a weak nuclear reaction (also called a beta-plus decay) can occur that converts one of the protons into a neutron plus a positron (the antiparticle equivalent to an electron) plus a neutrino, a nearly massless particle with a spectrum of energies between 0.0 and 0.42 MeV.

$\large _{2}^{2}\textrm{He} \rightarrow _{1}^{2}\textrm{D} + {e}^{+} + \nu_{e} (\small\textrm{0.0 .. 0.42MeV})$

That is because the only nucleus that can exist with 2 nucleons is the deuteron. So when 2 protons combine, one of the proton has to convert into a neutron through beta+ decay simultaneously that the fusion reaction happens. 2 processes occurring simultaneously one driven by strong forces between proton and proton in order to get into one nucleus in the 10^-20 seconds timescale. The other process, a proton converting into neutron is driven by weak interaction, with a timescale that is about similar to the lifetime of a free neutron 14 minutes or 840 seconds. But here we are demanding that this beta decay happens in the same timescale as 10^-20 seconds. Therefore, the probability that this process happens in this timescale is very very low. So that first step of the proton-proton chain is very very slow because of this demand that the beta decay occurs together with the fusing of the two protons. This is known as the deuteron bottleneck. The probability that two protons turn into a deuteron is very very very small. Let’s say you can wait 10^10 years to see 1 proton out of 10^18 fuse with another proton! But because the total number of protons in the sun is so large, 6×10^11kg of hydrogen, approx 3.7×10^38 protons fuse each second inside the core of the sun.

(1) $\large _{1}^{1}\textrm{H} + _{1}^{1}\textrm{H} \rightarrow _{1}^{2}\textrm{D} + {e}^{+} + \nu_{e} (\small\textrm{0.0 .. 0.42MeV})$

The positron and neutrino, which do not feel the strong nuclear force, immediately escape from the nucleus, leaving behind a proton plus a neutron. The proton plus neutron do constitute a stable nucleus: deuteron (Z=1 and N=1) ($_{1}^{2}\textrm{D}$ OR $^{2}\textrm{H}$). The net reaction is exothermic, and the excess energy mostly goes into the recoil of the deuteron and positron. This excess energy is then turned into heat when the nuclei collide with other particle. Additionally, the positron will rapidly be attracted to any nearby electron. Since these are anti particles to each other they annihilate each other producing a power gamma ray (a form of electromagnetic energy far more violent than x-rays).

As soon as the first deuteron is created, it combines with another proton to form 3-Helium, strong interaction only no weak interaction this time.

(2)$\large _{1}^{2}\textrm{D} + _{1}^{1}\textrm{H} \rightarrow _{2}^{3}\textrm{He} + \gamma (\small\textrm{5.49MeV})$

This 3-He cannot combine with one more proton because it’s 4-Li and it will not exist with 3 p and 1 n. Combine with a D is almost unlikely because as soon as D is produced it combines with one p. Combine with n is also not likely because there aren’t any free n around. D has to WAIT for another 3He to be produced in order to fuse with it and produce 4He.

(3)$\large _{1}^{2}\textrm{D} + _{1}^{2}\textrm{D} \rightarrow _{2}^{4}\textrm{He} + _{1}^{1}\textrm{H} + _{1}^{1}\textrm{H} + 12.86MeV$

85% of the time it will follow this path.

The other 15% of the time, we have deuteron combining with existing 4-Helium: D + 4He -> 7Be + gamma because in the core there are enough 4He already produced during the first 4.5 billion years to take this path in the core.

So, to summarize:

2 times reaction (1), 2 times reaction (2) and one time reaction (3) is the p-p chain which happens 85% of the time.

That means 4 protons into one 4-Helium 2 positrons and 2 neutrinos with a total energy output of 26.7 MeV: 4p -> 4He + 2positrons+ 2neutrinos (26.7MeV)

The other paths also called ppII and ppIII in the remaining 15% also produce 4-Helium but I’ll leave them for the moment. Here is the 3 complete pp chains, of which I mainly talked about ppI in this post:

The only direct experimental evidence of this process are the neutrinos produced, why?

Because the energy we are getting in the form of “sunlight” and that is observed with spectroscopy is not coming from the fusion in the core! Fusion takes place inside the core and from there the energy goes through the radiative zone and the convective zone and then to photosphere.

The photosphere transparent to electromagnetic waves. The photons from the photonsphere they cross the photosphere and come to the earth but the passing from the core to the phostosphere may take some million years. So what we are receiving is only remotely connected to the fusion inside the core.

In the radiative zone, photons are emitted, absorbed than reemitted, and so on with different energies and frequencies. so what reach the photoshpere is very different from what was produced inside the core. So the IR and UV light we receive is not only very different than what is issued from the core it is also very old. So how do I know that fusion is still going on in the core because if it stops today, it will take one million year before we notice… These neutrinos very weakly interact with matter, million and million of neutrinos keep going through our bodies every seconds and we do not notice anything. However, scientists have built huge neutrinos detectors and instead of getting answers, they were puzzled by the experimental results. It took until 2001 to actually get an answer on the missing solar neutrinos.