Editing Proton–proton chain (section)

==The proton–proton chain==
The first step in all the branches is the fusion of two [[proton]]s into a [[deuteron]]. As the protons fuse, one of them undergoes [[beta plus decay]], converting into a [[neutron]] by emitting a [[positron]] and an [[electron neutrino]]<ref name=":0">{{Cite book|title=Nuclear Physics of Stars|last=Iliadis |first=Christian |date=2007|publisher=Wiley-VCH|isbn=9783527406029|location=Weinheim|oclc=85897502}}</ref> (though a small amount of deuterium nuclei is produced by the "pep" reaction, see below):

:{| border="0"
|- style="height:2em;"
|[[Proton|p]]&nbsp;||+&nbsp;||p||→&nbsp;||{{nuclide|link=yes|deuterium|2}}
| +||{{SubatomicParticle|link=yes|Positron}}
| +
|{{math|{{SubatomicParticle|link=yes|electron neutrino}}}}&nbsp;||+&nbsp;||{{val|0.42|u=MeV}}
|}

The [[positron]] will [[Annihilation|annihilate]] with an [[electron]] from the environment into two [[gamma rays]]. Including this [[annihilation]] and the energy of the neutrino, the net reaction

:{| border="0"
|- style="height:2em;"
|[[Proton|p]]&nbsp;||+&nbsp;||p|| + {{SubatomicParticle|link=yes|Electron}} →&nbsp;{{nuclide|link=yes|deuterium|2}}
| +
|{{math|{{SubatomicParticle|link=yes|electron neutrino}}}}&nbsp;||+&nbsp;||{{val|1.442|u=MeV}}
|}

has a [[Q value (nuclear science)|''Q'' value]] (released [[energy]]) of 1.442 [[Electronvolt|MeV]]:<ref name=":0" />
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 the [[Weak interaction|weak nuclear force]]. The average [[proton]] in the core of the [[Sun]] waits 9 billion years before it successfully fuses with another [[proton]]. It has not been possible to measure the [[Cross section (physics)|cross-section]] of this reaction experimentally because it is so low<ref>{{Cite book|title=The Physics of Stars|last=Phillips |first=Anthony C. |date=1999|publisher=John Wiley|isbn=0471987972|edition= 2nd|location=Chichester|oclc=40948449}}</ref> but it can be calculated from theory.<ref name= Adelberger/>

After it is formed, the deuteron produced in the first stage can fuse with another proton to produce the [[stable nuclide|stable]], light [[isotope]] of [[helium]], {{SimpleNuclide|link=yes|Helium|3}}:

<!-- Autogenerated using Phykiformulae 0.11 by [[User:SkyLined]]
D + H → He-3 + y (mass-energy of 5.493 MeV)
-->:{| border="0"
|- style="height:2em;"
|{{nuclide|link=no|deuterium|2}}&nbsp;||+&nbsp;||{{nuclide|link=no|hydrogen|1}}&nbsp;||→&nbsp;||{{nuclide|link=yes|helium|3}}&nbsp;||+&nbsp;||{{math|{{SubatomicParticle|link=no|Gamma}}}}&nbsp;||+&nbsp;||{{val|5.493|u=MeV}}
|}

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.<ref name= Adelberger/>

In the Sun, each helium-3 nucleus produced in these reactions exists for only about 400 years before it is converted into helium-4.<ref>This time and the two other times above come from: Byrne, J. ''Neutrons, Nuclei, and Matter'', Dover Publications, Mineola, NY, 2011, {{ISBN|0486482383}}, p 8.</ref> Once the helium-3 has been produced, there are four possible paths to generate {{SimpleNuclide|link=yes|Helium|4}}. In {{nowrap|p–p I}}, helium-4 is produced by fusing two helium-3 nuclei; the {{nowrap|p–p II}} and {{nowrap|p–p III}} branches fuse {{SimpleNuclide|Helium|3}} with pre-existing {{SimpleNuclide|Helium|4}} to form [[beryllium]]-7, which undergoes further reactions to produce two helium-4 nuclei.

About 99% of the energy output of the sun comes from the various {{nowrap|p–p}} chains, with the other 1% coming from the [[CNO cycle]]. According to one model of the sun, 83.3 percent of the {{SimpleNuclide|link=yes|Helium|4}} produced by the various {{nowrap|p–p}} branches is produced via branch I while {{nowrap|p–p II}} produces 16.68 percent and {{nowrap|p–p III}} 0.02 percent.<ref name= Adelberger>{{cite journal |last=Adelberger |first=Eric G. |display-authors=etal |date=12 April 2011 |title=Solar fusion cross sections. II. The pp chain and CNO cycles |journal=Reviews of Modern Physics |volume=83 |issue=1 |page=201 |doi=10.1103/RevModPhys.83.195|arxiv= 1004.2318 |bibcode= 2011RvMP...83..195A|s2cid=119117147 }} See Figure 2. The caption is not very clear but it has been confirmed that the percentages refer to how much of each reaction takes place, or equivalently how much helium-4 is produced by each branch.</ref> 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.<ref>{{cite journal |display-authors=etal|last1=Aldo Serenelli |title=New Solar Composition: The Problem With Solar Models Revisited |journal=The Astrophysical Journal Letters |date=Nov 2009 |volume=705 |issue=2 |pages=L123–L127 |doi=10.1088/0004-637X/705/2/L123 |arxiv=0909.2668 |bibcode=2009ApJ...705L.123S |s2cid=14323767 }} Calculated from model AGSS09 in Table 3.</ref> The difference is apparently due to slightly different assumptions about the composition and [[metallicity]] of the sun.

There is also the extremely rare {{nowrap|p–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:
:{{math|4 <sup>1</sup>H<sup>+</sup> + 2 e{{sup|-}} → <sup>4</sup>He<sup>2+</sup> + 2 ν<sub>e</sub>}}

releasing 26.73 MeV of energy, some of which is lost to the neutrinos.

===The {{nowrap|p–p I}} branch===<!-- Autogenerated using Phykiformulae 0.11 by [[User:SkyLined]]
He-3 + He-3 → He-4 + 2H + 12.86 MeV
-->
:{| border="0"
|- style="height:2em;"
|{{nuclide|link=no|helium|3}}&nbsp;||+&nbsp;||{{nuclide|link=no|helium|3}}&nbsp;||→&nbsp;||{{nuclide|link=yes|helium|4}}&nbsp;||+&nbsp;||2&nbsp;{{nuclide|link=no|hydrogen|1}}&nbsp;||+&nbsp;||{{val|12.859|u=MeV}}
|}

The complete chain releases a net energy of {{val|26.732|u=MeV}}<ref>{{cite book|last1=LeBlanc|first1=Francis|title=An Introduction to Stellar Astrophysics}}</ref> but 2.2 percent of this energy (0.59 MeV) is lost to the neutrinos that are produced.<ref>{{cite journal|last1=Burbidge|first1=E. |last2=Burbidge |first2=G. |last3=Fowler |first3=William |last4= Hoyle |first4=F. |title=Synthesis of the Elements in Stars|journal=Reviews of Modern Physics|date=1 October 1957|volume=29|issue=4|pages=547–650|doi=10.1103/RevModPhys.29.547|bibcode= 1957RvMP...29..547B|doi-access=free|url=https://authors.library.caltech.edu/records/bpgfm-dsq46/files/BURrmp57.pdf?download=1}}</ref>
The {{nowrap|p–p I}} branch is dominant at temperatures of 10 to {{val|18|ul=MK}}.<ref name=":1">{{Cite book |last=Iliadis |first=Christian |url=https://www.worldcat.org/oclc/908071061 |title=Nuclear physics of stars |date=2015 |isbn=978-3-527-33649-4 |edition=Second, revised and enlarged |location=Weinheim, Germany |oclc=908071061}}</ref>
Below {{val|10|u=MK}}, the {{nowrap|p–p}} chain proceeds at slow rate, resulting in a low production of {{SimpleNuclide|Helium|4}}.<ref>{{Cite journal |title=Solar fusion cross sections. II. Theppchain and CNO cycles |date=2010|doi=10.1103/RevModPhys.83.195 |arxiv=1004.2318 |last1=Adelberger |first1=E. G. |last2=García |first2=A. |last3=Robertson |first3=R. G. Hamish |last4=Snover |first4=K. A. |last5=Balantekin |first5=A. B. |last6=Heeger |first6=K. |last7=Ramsey-Musolf |first7=M. J. |last8=Bemmerer |first8=D. |last9=Junghans |first9=A. |last10=Bertulani |first10=C. A. |last11=Chen |first11=J.-W. |last12=Costantini |first12=H. |last13=Prati |first13=P. |last14=Couder |first14=M. |last15=Uberseder |first15=E. |last16=Wiescher |first16=M. |last17=Cyburt |first17=R. |last18=Davids |first18=B. |last19=Freedman |first19=S. J. |last20=Gai |first20=M. |last21=Gazit |first21=D. |last22=Gialanella |first22=L. |last23=Imbriani |first23=G. |last24=Greife |first24=U. |last25=Hass |first25=M. |last26=Haxton |first26=W. C. |last27=Itahashi |first27=T. |last28=Kubodera |first28=K. |last29=Langanke |first29=K. |last30=Leitner |first30=D. |journal=Reviews of Modern Physics |volume=83 |pages=195–245 |s2cid=119117147 |display-authors=1 }}</ref>

===The {{nowrap|p–p II}} branch===
[[File:Proton-Proton II chain reaction.svg|thumb|Proton–proton II chain]]
{{See also|Lithium burning}}
<!-- Autogenerated using Phykiformulae 0.11 by [[User:SkyLined]]
He-3 + He-4 → Be-7 + y
Be-7 + e- → Li-7 + ve (0.861 MeV / 0.383 MeV)
Li-7 + H → 2He
-->:{| border="0"
|- style="height:2em;"
|{{nuclide|link=no|helium|3}}&nbsp;||+&nbsp;||{{nuclide|link=no|helium|4}}&nbsp;||→&nbsp;||{{nuclide|link=yes|beryllium|7}}||+&nbsp;||{{math|{{SubatomicParticle|link=no|Gamma}}}}&nbsp;||+&nbsp;||{{val|1.59|u=MeV}}
|- style="height:2em;"
|{{nuclide|link=no|beryllium|7}}&nbsp;||+&nbsp;||{{SubatomicParticle|link=no|Electron}}&nbsp;||→&nbsp;||{{nuclide|link=yes|lithium|7}}||+&nbsp;||{{math|{{SubatomicParticle|link=no|Electron Neutrino}}}}&nbsp;||+&nbsp;||{{val|0.861|u=MeV}}&nbsp;||/&nbsp;||{{val|0.383|u=MeV}}
|- style="height:2em;"
|{{nuclide|link=no|lithium|7}}&nbsp;||+&nbsp;||{{nuclide|link=no|hydrogen|1}}&nbsp;||→&nbsp;||2 {{nuclide|link=no|helium|4}}&nbsp;||&nbsp;||&nbsp;||+&nbsp;||{{val|17.35|u=MeV}}
|}

The {{nowrap|p–p II}} branch is dominant at temperatures of 18 to {{val|25|u=MK}}.<ref name=":1" />

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 of {{SimpleNuclide|Beryllium|7}} to {{SimpleNuclide|Lithium|7}} carry an energy of {{val|0.861|u=MeV}}, while the remaining 10 percent carry {{val|0.383|u=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 from {{chem|7|Be}} to stable {{chem|7|Li}} is about 0.862&nbsp;MeV, almost all of which is lost to the neutrino if the decay goes directly to the stable lithium.

===The {{nowrap|p–p III}} branch===
[[File:Proton-Proton III chain reaction.svg|thumb|Proton–proton III chain]]
<!-- Autogenerated using Phykiformulae 0.11 by [[User:SkyLined]]
He-3 + He-4 → Be-7 + y
Be-7 + H → B-8 + y
B-8 _ _ → Be-8 + e+ + ve
Be-8 _ _ → 2He
-->:{| border="0"
|- style="height:2em;"
|{{nuclide|link=yes|helium|3}}&nbsp;||+&nbsp;||{{nuclide|link=yes|helium|4}}&nbsp;||→&nbsp;||{{nuclide|link=yes|beryllium|7}}&nbsp;||+&nbsp;||{{math|{{SubatomicParticle|link=yes|Gamma}}}}&nbsp;||&nbsp;||&nbsp;||+&nbsp;||{{val|1.59|u=MeV}}
|- style="height:2em;"
|{{nuclide|link=yes|beryllium|7}}&nbsp;||+&nbsp;||{{nuclide|link=yes|hydrogen|1}}&nbsp;||→&nbsp;||{{nuclide|link=yes|boron|8}}&nbsp;||+&nbsp;||{{math|{{SubatomicParticle|link=|Gamma}}}}
|- style="height:2em;"
|{{nuclide|link=yes|boron|8}}&nbsp;||&nbsp;||&nbsp;||→&nbsp;||{{nuclide|link=yes|beryllium|8}}&nbsp;||+&nbsp;||{{SubatomicParticle|link=yes|Positron}}&nbsp;||+&nbsp;||{{SubatomicParticle|link=yes|Electron Neutrino}}&nbsp;||
|- style="height:2em;"
|{{nuclide|link=yes|beryllium|8}}&nbsp;||&nbsp;||&nbsp;||→&nbsp;||2&nbsp;{{nuclide|link=yes|helium|4}}
|}

The last three stages of this chain, plus the positron annihilation, contribute a total of 18.209&nbsp;MeV, though much of this is lost to the neutrino.

The {{nowrap|p–p III}} chain is dominant if the temperature exceeds {{val|25|u=MK}}.<ref name=":1" />

The {{nowrap|p–p III}} chain is not a major source of energy in the Sun, but it was very important in the [[solar neutrino problem]] because it generates very high energy neutrinos (up to {{val|14.06|u=MeV}}).

===The {{nowrap|p–p IV}} (Hep) branch===
This reaction is predicted theoretically, but it has never been observed due to its rarity (about {{val|0.3|ul=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 to {{val|18.8|u=MeV}}{{citation needed|date=June 2020}}<!--Total energy is more, like 19.795-->).

<!-- Autogenerated using Phykiformulae 0.11 by [[User:SkyLined]]
He-3 + H → He + e+ + ve + 18.23 MeV
-->:{| border="0"
|- style="height:2em;"
|{{nuclide|link=yes|helium|3}}&nbsp;||+&nbsp;||{{nuclide|link=yes|hydrogen|1}}&nbsp;||→&nbsp;||{{nuclide|link=yes|helium|4}}&nbsp;||+&nbsp;||{{SubatomicParticle|link=yes|Positron}}&nbsp;||+&nbsp;||{{math|{{SubatomicParticle|link=yes|Electron Neutrino}}}}
|}

The mass–energy relationship gives {{val|19.795|u=MeV}} for the energy released by this reaction plus the ensuing annihilation, some of which is lost to the neutrino.

===Energy release===
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 is {{val|26.73|u=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 the {{nuclide|link=yes|helium|4}} from the {{nowrap|p–p I}} reaction) adds energy to the plasma in the Sun. This heating keeps the core of the Sun hot and prevents it from [[gravitational collapse|collapsing]] 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 the {{nowrap|p–p I}}, {{nowrap|p–p II}}, and {{nowrap|p–p III}} chains carry away 2.0%, 4.0%, and 28.3% of the energy in those reactions, respectively.<ref>Claus E. Rolfs and William S. Rodney, ''Cauldrons in the Cosmos'', The University of Chicago Press, 1988, p. 354.</ref>

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 as [[electromagnetic 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 the {{nowrap|p–p}} reactions, not the 1% coming from the CNO cycle.

{| class="wikitable"
|+ Luminosity production in the sun
|-
! Branch 
! Percent of<br><sup>4</sup>He produced 
! Percent loss<br>due to neutrino<br>production 
! Relative<br>amount of<br>energy lost
! Relative amount<br>of luminosity<br>produced
! Percentage<br>of total<br>luminosity
|-
| Branch I   || {{0}}83.3%      || {{0}}2%   || 1.67%   || 81.6%   || {{0}}83.6%
|-
| Branch II  || {{0}}16.68%     || {{0}}4%   || 0.67%   || 16.0%   || {{0}}16.4%
|-
| Branch III || {{0|00}}0.02% || 28.3%     || 0.0057% || {{0}}0.014% || {{0|00}}0.015%
|-
| total      || 100%       ||           || 2.34%   || 97.7%   || 100%
|}