Theproton–proton chain, also commonly referred to as thep–p chain, is one of two known sets ofnuclear fusion reactions by whichstars converthydrogen tohelium. It dominates in stars with masses less than or equal to that of theSun,[2] whereas theCNO cycle, the other known reaction, is suggested by theoretical models to dominate in stars with masses greater than about 1.3 solar masses.[3]
In general, proton–proton fusion can occur only if thekinetic energy (temperature) of the protons is high enough to overcome their mutualelectrostatic repulsion.[4]
In the Sun,deuteron-producing events are rare.Diprotons are the much more common result of proton–proton reactions within the star, and diprotons almost immediately decay back into two protons. Since the conversion of hydrogen to helium is slow, the complete conversion of the hydrogen initially in thecore of the Sun is calculated to take more than ten billion years.[5]
Although sometimes called the "proton–proton chain reaction", it is not achain reaction in the normal sense. In most nuclear reactions, a chain reaction designates a reaction that produces a product, such as neutrons given off duringfission, that quickly induces another such reaction. The proton–proton chain is, like adecay chain, a series of reactions. The product of one reaction is the starting material of the next reaction. There are two main chains leading from hydrogen to helium in the Sun. One chain has five reactions, the other chain has six.
The theory that proton–proton reactions are the basic principle by which the Sun and other stars burn was advocated byArthur Eddington in the 1920s. At the time, the temperature of the Sun was considered to be too low to overcome theCoulomb barrier. After the development ofquantum mechanics, it was discovered thattunneling of thewavefunctions of the protons through the repulsive barrier allows for fusion at a lower temperature than theclassical prediction.
In 1939,Hans Bethe attempted to calculate the rates of various reactions in stars. Starting with two protons combining to give adeuterium nucleus and apositron he found what we now call Branch II of the proton–proton chain. But he did not consider the reaction of two3
He nuclei (Branch I) which we now know to be important.[6] This was part of the body of work instellar nucleosynthesis for which Bethe won theNobel Prize in Physics in 1967.
The first step in all the branches is the fusion of twoprotons into adeuteron. As the protons fuse, one of them undergoesbeta plus decay, converting into aneutron by emitting apositron and anelectron neutrino[7] (though a small amount of deuterium nuclei is produced by the "pep" reaction, see below):
Thepositron willannihilate with anelectron from the environment into twogamma rays. Including thisannihilation and the energy of the neutrino, the net reaction
has aQ value (releasedenergy) of 1.442MeV:[7]The relative amounts of energy going to the neutrino and to the other products is variable.
This is the rate-limiting reaction and is extremely slow due to it being initiated by theweak nuclear force. The averageproton in the core of theSun waits 9 billion years before it successfully fuses with anotherproton. It has not been possible to measure thecross-section of this reaction experimentally because it is so low[8] but it can be calculated from theory.[1]
After it is formed, the deuteron produced in the first stage can fuse with another proton to produce thestable, lightisotope ofhelium,3
He:
This process, mediated by the strong nuclear force rather than the weak force, is extremely fast by comparison to the first step. It is estimated that, under the conditions in the Sun's core, each newly created deuterium nucleus exists for only about one second before it is converted into helium-3.[1]
In the Sun, each helium-3 nucleus produced in these reactions exists for only about 400 years before it is converted into helium-4.[9] Once the helium-3 has been produced, there are four possible paths to generate4
He. Inp–p I, helium-4 is produced by fusing two helium-3 nuclei intoberyllium-6, which immediatelyemits two protons to become helium-4. Thep–p II andp–p III branches fuse3
He with pre-existing4
He to formberyllium-7, which undergoes further reactions to produce two helium-4 nuclei.
About 99% of the energy output of the sun comes from the variousp–p chains, with the other 1% coming from theCNO cycle. According to one model of the sun, 83.3 percent of the4
He produced by the variousp–p branches is produced via branch I whilep–p II produces 16.68 percent andp–p III 0.02 percent.[1] Since half the neutrinos produced in branches II and III are produced in the first step (synthesis of a deuteron), only about 8.35 percent of neutrinos come from the later steps (see below), and about 91.65 percent are from deuteron synthesis. However, another solar model from around the same time gives only 7.14 percent of neutrinos from the later steps and 92.86 percent from the synthesis of deuterium nuclei.[10] The difference is apparently due to slightly different assumptions about the composition andmetallicity of the sun.
There is also the extremely rarep–p IV branch. Other even rarer reactions may occur. The rate of these reactions is very low due to very small cross-sections, or because the number of reacting particles is so low that any reactions that might happen are statistically insignificant.
The overall reaction is:
releasing 26.73 MeV of energy, some of which is lost to the neutrinos.
The fusion of two3
2He nuclei produces a6
4Be nucleus, which promptly ejects two protons. The complete chain releases a net energy of26.732 MeV[11] but 2.2 percent of this energy (0.59 MeV) is lost to the neutrinos that are produced.[12]Thep–p I branch is dominant at temperatures of 10 to18 MK.[13]Below10 MK, thep–p chain proceeds at slow rate, resulting in a low production of4
He.[14]
| 3 2He | + | 4 2He | → | 7 4Be | + | γ | + | 1.59 MeV | ||
| 7 4Be | + | e− | → | 7 3Li | + | ν e | + | 0.861 MeV | / | 0.383 MeV |
| 7 3Li | + | 1 1H | → | 24 2He | + | 17.35 MeV |
Thep–p II branch is dominant at temperatures of 18 to25 MK.[13]
Note that the energies in the second reaction above are the energies of the neutrinos that are produced by the reaction. 90 percent of the neutrinos produced in the reaction of7
Be to7
Li carry an energy of0.861 MeV, while the remaining 10 percent carry0.383 MeV. The difference is whether the lithium-7 produced is in the ground state or an excited (metastable) state, respectively. The total energy released going from7
Be to stable7
Li is about 0.862 MeV, almost all of which is lost to the neutrino if the decay goes directly to the stable lithium.
The last three stages of this chain, plus the positron annihilation, contribute a total of 18.209 MeV, though much of this is lost to the neutrino.
Thep–p III chain is dominant if the temperature exceeds25 MK.[13]
Thep–p III chain is not a major source of energy in the Sun, but it was very important in thesolar neutrino problem because it generates very high energy neutrinos (up to14.06 MeV).
This reaction is predicted theoretically, but it has never been observed due to its rarity (about0.3 ppm in the Sun). In this reaction, helium-3 captures a proton directly to give helium-4, with an even higher possible neutrino energy (up to18.8 MeV[citation needed]).
The mass–energy relationship gives19.795 MeV for the energy released by this reaction plus the ensuing annihilation, some of which is lost to the neutrino.
Comparing the mass of the final helium-4 nucleus with the masses of the four protons reveals that 0.7 percent of the mass of the original protons has been lost. This mass has been converted into energy, in the form of kinetic energy of produced particles, gamma rays, and neutrinos released during each of the individual reactions. The total energy yield of one whole chain is26.73 MeV.
Energy released as gamma rays will interact with electrons and protons and heat the interior of the Sun. Also kinetic energy of fusion products (e.g. of the two protons and the4
2He from thep–p I reaction) adds energy to the plasma in the Sun. This heating keeps the core of the Sun hot and prevents it fromcollapsing under its own weight as it would if the sun were to cool down.
Neutrinos do not interact significantly with matter and therefore do not heat the interior and thereby help support the Sun against gravitational collapse. Their energy is lost: the neutrinos in thep–p I,p–p II, andp–p III chains carry away 2.0%, 4.0%, and 28.3% of the energy in those reactions, respectively.[15]
The following table calculates the amount of energy lost to neutrinos and the amount of "solar luminosity" coming from the three branches. "Luminosity" here means the amount of energy given off by the Sun aselectromagnetic radiation rather than as neutrinos. The starting figures used are the ones mentioned higher in this article. The table concerns only the 99% of the power and neutrinos that come from thep–p reactions, not the 1% coming from the CNO cycle.
| Branch | Percent of 4He produced | Percent loss due to neutrino production | Relative amount of energy lost | Relative amount of luminosity produced | Percentage of total luminosity |
|---|---|---|---|---|---|
| Branch I | 83.3% | 2.2% | 1.67% | 81.6% | 83.6% |
| Branch II | 16.68% | 4% | 0.67% | 16.0% | 16.4% |
| Branch III | 0.02% | 28.3% | 0.0057% | 0.014% | 0.015% |
| total | 100% | 2.34% | 97.7% | 100% |
Adeuteron can also be produced by the rare pep (proton–electron–proton) reaction (electron capture):
In the Sun, the frequency ratio of the pep reaction versus thep–p reaction is 1:400. However, theneutrinos released by the pep reaction are far more energetic: while neutrinos produced in the first step of thep–p reaction range in energy up to0.42 MeV, the pep reaction produces sharp-energy-line neutrinos of1.44 MeV. Detection of solar neutrinos from this reaction were reported by theBorexino collaboration in 2012.[16]
Both the pep andp–p reactions can be seen as two differentFeynman representations of the same basic interaction, where the electron passes to the right side of the reaction as a positron. This is represented in the figure of proton–proton and electron-capture reactions in a star, available at the NDM'06 web site.[17]
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