Editing Nuclear fusion (section)

==Important reactions==
<!-- This section is linked from [[Fusion power]] -->
{{unreferenced section|date=August 2023}}

===Stellar reaction chains===
At the temperatures and densities in stellar cores, the rates of fusion reactions are notoriously slow. For example, at solar core temperature (''T'' ≈ 15&nbsp;MK) and density (160&nbsp;g/cm<sup>3</sup>), the energy release rate is only 276&nbsp;μW/cm<sup>3</sup>—about a quarter of the volumetric rate at which a resting human body generates heat.<ref>[http://fusedweb.pppl.gov/CPEP/Chart_Pages/5.Plasmas/SunLayers.html FusEdWeb | Fusion Education]. Fusedweb.pppl.gov (9 November 1998). Retrieved 17 August 2011. {{Webarchive|url=https://web.archive.org/web/20071024012758/http://fusedweb.pppl.gov/CPEP/Chart_pages/5.Plasmas/SunLayers.html |date=24 October 2007 }}</ref> Thus, reproduction of stellar core conditions in a lab for nuclear fusion power production is completely impractical. Because nuclear reaction rates depend on density as well as temperature, and most fusion schemes operate at relatively low densities, those methods are strongly dependent on higher temperatures. The fusion rate as a function of temperature (exp(−''E''/''kT'')), leads to the need to achieve temperatures in terrestrial reactors 10–100 times higher than in stellar interiors: ''T'' ≈ {{val|0.1|–|1.0|e=9|u=K}}.

===Criteria and candidates for terrestrial reactions===
{{main article|Fusion power#Fuels}}
In artificial fusion, the primary fuel is not constrained to be protons and higher temperatures can be used, so reactions with larger cross-sections are chosen. Another concern is the production of neutrons, which activate the reactor structure radiologically, but also have the advantages of allowing volumetric extraction of the fusion energy and [[tritium]] breeding. Reactions that release no neutrons are referred to as [[Aneutronic fusion|''aneutronic'']].

To be a useful energy source, a fusion reaction must satisfy several criteria. It must:

;Be [[exothermic]]: This limits the reactants to the low ''Z'' (number of protons) side of the [[Nuclear binding energy#Nuclear binding energy curve|curve of binding energy]]. It also makes helium {{SimpleNuclide|link=yes|Helium|4}} the most common product because of its extraordinarily tight binding, although {{SimpleNuclide|link=yes|Helium|3}} and {{SimpleNuclide|link=yes|Hydrogen|3}} also show up.
;Involve low atomic number (''Z'') nuclei: This is because the electrostatic repulsion that must be overcome before the nuclei are close enough to fuse ([[Coulomb barrier]]) is directly related to the number of protons it contains – its atomic number.
;Have two reactants: At anything less than stellar densities, three-body collisions are too improbable. In inertial confinement, both stellar densities and temperatures are exceeded to compensate for the shortcomings of the third parameter of the Lawson criterion, ICF's very short confinement time.
;Have two or more products: This allows simultaneous conservation of energy and momentum without relying on the electromagnetic force.
;Conserve both protons and neutrons: The cross sections for the weak interaction are too small.

Few reactions meet these criteria. The following are those with the largest cross sections:<ref>{{cite book
 |author1         = M. Kikuchi, K. Lackner
 |author2         = M. Q. Tran
 |name-list-style = amp
 |year            = 2012
 |title           = Fusion Physics
 |url             = http://www-pub.iaea.org/books/IAEABooks/8879/Fusion-Physics
 |page            = 22
 |publisher       = [[International Atomic Energy Agency]]
 |isbn            = 9789201304100
 |access-date     = 8 December 2015
 |archive-date    = 8 December 2015
 |archive-url     = https://web.archive.org/web/20151208122642/http://www-pub.iaea.org/books/IAEABooks/8879/Fusion-Physics
 |url-status      = live
}}</ref><ref>{{cite book
|author1=K. Miyamoto
|year=2005
|title=Plasma Physics and Controlled Nuclear Fusion
|publisher=[[Springer-Verlag]]
|isbn=3-540-24217-1
}}</ref>

<!-- Autogenerated using Phykiformulae 0.10 by [[User:SkyLined]]
(1)     D + T    → He   (3.52 MeV ) + n   (14.06 MeV )
(2i)    D + D    → T    (1.01 MeV) + p    (3.02 MeV ) _ _ _ _       _ _50%
(2ii)   _ _ _    → He-3 (0.82 MeV) + n    (3.27 MeV ) _ _ _ _       _ _50%
(3)     D + He-3 → He   (3.6 MeV ) + p    (14.7 MeV )
(4)     T + T    → He   _ _       _ + 2n   _ _        _ _ _ + 11.3MeV
(5)  He-3 + He-3 → He   _ _       _ + 2p   _ _        _ _ _ + 12.9MeV
(6i) He-3 + T    → He   _ _       _ + p    + n        _ _ _ + 12.1MeV _ _57%
(6ii)   _ _ _    → He   (4.8 MeV ) + D    (9.5 MeV  ) _ _ _ _       _ _43%
(7i)    D + Li-6 → 2He  + 22.3 MeV
(7ii)   _ _ _    → He-3 + He      _ + n    _ _        _ _ _ + 1.8 MeV
(7iii)  _ _ _    → Li-7 (0.625 MeV) + p (4.375 MeV)
(7iv)   _ _ _    → Be-7 (0.425 MeV )+ n (2.975 MeV)
(8)     p + Li-6 → He   (1.7 MeV ) + He-3 (2.3 MeV)
(9)  He-3 + Li-6 → 2He  + p       _ _ _    _ _        _ _ _ + 16.9 MeV
(10)    p + B-11 → 3He  _ _       _ _ _    _ _        _ _ _ + 8.7 MeV

-->:{| border="0"
|- style="height:2em;"
|(1)&nbsp;||{{nuclide|link=yes|deuterium}}&nbsp;||+&nbsp;||{{nuclide|link=yes|tritium}}&nbsp;||→&nbsp;||{{nuclide|link=yes|helium|4}}&nbsp;||(&nbsp;||style="text-align:right;"|{{val|3.52|u=MeV}}||)&nbsp;||+&nbsp;||[[Neutron|n<sup>0</sup>]]&nbsp;||(&nbsp;||style="text-align:right;"|{{val|14.06|u=MeV}}||)
|- style="height:2em;"
|(2i)&nbsp;||{{nuclide|link=yes|deuterium}}&nbsp;||+&nbsp;||{{nuclide|link=yes|deuterium}}&nbsp;||→&nbsp;||{{nuclide|link=yes|tritium}}&nbsp;||(&nbsp;||style="text-align:right;"|{{val|1.01|u=MeV}}||)&nbsp;||+&nbsp;||[[Proton|p<sup>+</sup>]]&nbsp;||(&nbsp;||style="text-align:right;"|{{val|3.02|u=MeV}}||)&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;|| 50%
|- style="height:2em;"
|(2ii)&nbsp;||&nbsp;||&nbsp;||&nbsp;||→&nbsp;||{{nuclide|link=yes|helium|3}}&nbsp;||(&nbsp;||style="text-align:right;"|{{val|0.82|u=MeV}}||)&nbsp;||+&nbsp;||[[Neutron|n<sup>0</sup>]]&nbsp;||(&nbsp;||style="text-align:right;"|{{val|2.45|u=MeV}}||)&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;|| 50%
|- style="height:2em;"
|(3)&nbsp;||{{nuclide|link=yes|deuterium}}&nbsp;||+&nbsp;||{{nuclide|link=yes|helium|3}}&nbsp;||→&nbsp;||{{nuclide|link=yes|helium|4}}&nbsp;||(&nbsp;||style="text-align:right;"|{{val|3.6|u=MeV}}||)&nbsp;||+&nbsp;||[[Proton|p<sup>+</sup>]]&nbsp;||(&nbsp;||style="text-align:right;"|{{val|14.7|u=MeV}}||)
|- style="height:2em;"
|(4)&nbsp;||{{nuclide|link=yes|tritium}}&nbsp;||+&nbsp;||{{nuclide|link=yes|tritium}}&nbsp;||→&nbsp;||{{nuclide|link=yes|helium|4}}&nbsp;||&nbsp;||&nbsp;||&nbsp;||+&nbsp;||2&nbsp;[[Neutron|n<sup>0</sup>]]&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||+&nbsp;||style="text-align:right;"|{{val|11.3|u=MeV}}
|- style="height:2em;"
|(5)&nbsp;||{{nuclide|link=yes|helium|3}}&nbsp;||+&nbsp;||{{nuclide|link=yes|helium|3}}&nbsp;||→&nbsp;||{{nuclide|link=yes|helium|4}}&nbsp;||&nbsp;||&nbsp;||&nbsp;||+&nbsp;||2&nbsp;[[Proton|p<sup>+</sup>]]&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||+&nbsp;||style="text-align:right;"|{{val|12.9|u=MeV}}
|- style="height:2em;"
|(6i)&nbsp;||{{nuclide|link=yes|helium|3}}&nbsp;||+&nbsp;||{{nuclide|link=yes|tritium}}&nbsp;||→&nbsp;||{{nuclide|link=yes|helium|4}}&nbsp;||&nbsp;||&nbsp;||&nbsp;||+&nbsp;||[[Proton|p<sup>+</sup>]]&nbsp;||+&nbsp;||[[Neutron|n<sup>0</sup>]]&nbsp;||&nbsp;||&nbsp;||&nbsp;||+&nbsp;||style="text-align:right;"|{{val|12.1|u=MeV}}||&nbsp;|| 57%
|- style="height:2em;"
|(6ii)&nbsp;||&nbsp;||&nbsp;||&nbsp;||→&nbsp;||{{nuclide|link=yes|helium|4}}&nbsp;||(&nbsp;||style="text-align:right;"|{{val|4.8|u=MeV}}||)&nbsp;||+&nbsp;||{{nuclide|link=yes|deuterium}}&nbsp;||(&nbsp;||style="text-align:right;"|{{val|9.5|u=MeV}}||)&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;|| 43%
|- style="height:2em;"
|(7i)&nbsp;||{{nuclide|link=yes|deuterium}}&nbsp;||+&nbsp;||{{nuclide|link=yes|lithium|6}}&nbsp;||→&nbsp;||2&nbsp;{{nuclide|link=yes|helium|4}}&nbsp;||+&nbsp;||style="text-align:right;"|{{val|22.4|u=MeV}}
|- style="height:2em;"
|(7ii)&nbsp;||&nbsp;||&nbsp;||&nbsp;||→&nbsp;||{{nuclide|link=yes|helium|3}}&nbsp;||+&nbsp;||{{nuclide|link=yes|helium|4}}&nbsp;||&nbsp;||+&nbsp;||[[Neutron|n<sup>0</sup>]]&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||+&nbsp;||style="text-align:right;"|{{val|2.56|u=MeV}}
|- style="height:2em;"
|(7iii)&nbsp;||&nbsp;||&nbsp;||&nbsp;||→&nbsp;||{{nuclide|link=yes|lithium|7}}&nbsp;||+&nbsp;||[[Proton|p<sup>+</sup>]]&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||+&nbsp;||style="text-align:right;"|{{val|5.0|u=MeV}}
|- style="height:2em;"
|(7iv)&nbsp;||&nbsp;||&nbsp;||&nbsp;||→&nbsp;||{{nuclide|link=yes|beryllium|7}}&nbsp;||+&nbsp;||[[Neutron|n<sup>0</sup>]]&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||+&nbsp;||style="text-align:right;"|{{val|3.4|u=MeV}}
|- style="height:2em;"
|(8)&nbsp;||[[Proton|p<sup>+</sup>]]&nbsp;||+&nbsp;||{{nuclide|link=yes|lithium|6}}&nbsp;||→&nbsp;||{{nuclide|link=yes|helium|4}}&nbsp;||(&nbsp;||style="text-align:right;"|{{val|1.7|u=MeV}}||)&nbsp;||+&nbsp;||{{nuclide|link=yes|helium|3}}&nbsp;||(&nbsp;||style="text-align:right;"|{{val|2.3|u=MeV}}||)
|- style="height:2em;"
|(9)&nbsp;||{{nuclide|link=yes|helium|3}}&nbsp;||+&nbsp;||{{nuclide|link=yes|lithium|6}}&nbsp;||→&nbsp;||2&nbsp;{{nuclide|link=yes|helium|4}}&nbsp;||+&nbsp;||[[Proton|p<sup>+</sup>]]&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||+&nbsp;||style="text-align:right;"|{{val|16.9|u=MeV}}
|- style="height:2em;"
|(10)&nbsp;||[[Proton|p<sup>+</sup>]]&nbsp;||+&nbsp;||{{nuclide|link=yes|boron|11}}&nbsp;||→&nbsp;||3&nbsp;{{nuclide|link=yes|helium|4}}&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||&nbsp;||+&nbsp;||style="text-align:right;"|{{val|8.7|u=MeV}}
|}
{{Nucleosynthesis}}

For reactions with two products, the energy is divided between them in inverse proportion to their masses, as shown. In most reactions with three products, the distribution of energy varies. For reactions that can result in more than one set of products, the branching ratios are given.

Some reaction candidates can be eliminated at once. The D–<sup>6</sup>Li reaction has no advantage compared to [[Proton|p<sup>+</sup>]]–{{nuclide|link=yes|Boron|11}} because it is roughly as difficult to burn but produces substantially more neutrons through {{nuclide|link=yes|Deuterium}}–{{nuclide|link=yes|Deuterium}} side reactions. There is also a [[Proton|p<sup>+</sup>]]–{{nuclide|link=yes|Lithium|7}} reaction, but the cross section is far too low, except possibly when ''T''<sub>''i''</sub> > 1 MeV, but at such high temperatures an endothermic, direct neutron-producing reaction also becomes very significant. Finally there is also a [[Proton|p<sup>+</sup>]]–{{nuclide|link=yes|Beryllium|9}} reaction, which is not only difficult to burn, but {{nuclide|link=yes|Beryllium|9}} can be easily induced to split into two alpha particles and a neutron.

In addition to the fusion reactions, the following reactions with neutrons are important in order to "breed" tritium in "dry" fusion bombs and some proposed fusion reactors:
<!-- Autogenerated using Phykiformulae 0.10 by [[User:SkyLined]]
{{Nuclide2|neutron}} + {{Nuclide2|lithium|6}} -> T + He + 4.784 MeV
n + {{Nuclide2|lithium|7}} -> T + He + n − 2.467 MeV

-->:{| border="0"
|- style="height:2em;"
|[[Neutron|n<sup>0</sup>]]&nbsp;||+&nbsp;||{{nuclide|link=yes|lithium|6}}&nbsp;||→&nbsp;||{{nuclide|link=yes|tritium}}&nbsp;||+&nbsp;||{{nuclide|link=yes|helium|4}} + 4.784 MeV
|- style="height:2em;"
|[[Neutron|n<sup>0</sup>]]&nbsp;||+&nbsp;||{{nuclide|link=yes|lithium|7}}&nbsp;||→&nbsp;||{{nuclide|link=yes|tritium}}&nbsp;||+&nbsp;||{{nuclide|link=yes|helium|4}} + [[Neutron|n<sup>0</sup>]] − 2.467 MeV
|}

The latter of the two equations was unknown when the U.S. conducted the [[Castle Bravo]] fusion bomb test in 1954. Being just the second fusion bomb ever tested (and the first to use lithium), the designers of the Castle Bravo "Shrimp" had understood the usefulness of <sup>6</sup>Li in tritium production, but had failed to recognize that <sup>7</sup>Li fission would greatly increase the yield of the bomb. While <sup>7</sup>Li has a small neutron cross-section for low neutron energies, it has a higher cross section above 5 MeV.<ref name=cross_section>[http://www.kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_7/4_7_4c.html Subsection 4.7.4c] {{Webarchive|url=https://web.archive.org/web/20180816195425/http://www.kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_7/4_7_4c.html |date=16 August 2018 }}. Kayelaby.npl.co.uk. Retrieved 19 December 2012.</ref> The 15 Mt yield was 150% greater than the predicted 6 Mt and caused unexpected exposure to fallout.

To evaluate the usefulness of these reactions, in addition to the reactants, the products, and the energy released, one needs to know something about the [[nuclear cross section]]. Any given fusion device has a maximum plasma pressure it can sustain, and an economical device would always operate near this maximum. Given this pressure, the largest fusion output is obtained when the temperature is chosen so that {{math|{{angbr|''σv''}}/''T''<sup>2</sup>}} is a maximum. This is also the temperature at which the value of the triple product {{mvar|nTτ}} required for [[Fusion energy gain factor#Ignition|ignition]] is a minimum, since that required value is inversely proportional to {{math|{{angbr|''σv''}}/''T''<sup>2</sup>}} (see [[Lawson criterion]]). (A plasma is "ignited" if the fusion reactions produce enough power to maintain the temperature without external heating.) This optimum temperature and the value of {{math|{{angbr|''σv''}}/''T''<sup>2</sup>}} at that temperature is given for a few of these reactions in the following table.

{| class="wikitable" style="margin:auto;"
|-
!fuel                                           !! ''T'' [keV] !! {{math|{{angbr|''σv''}}/''T''<sup>2</sup>}} [m<sup>3</sup>/s/keV<sup>2</sup>]
|-
|{{nuclide|deuterium}}–{{nuclide|tritium}}      || 13.6        || {{val|1.24|e=-24}}
|-
|{{nuclide|deuterium}}–{{nuclide|deuterium}}    || 15          || {{val|1.28|e=-26}}
|-
|{{nuclide|deuterium}}–{{nuclide|helium|3}}     || 58          || {{val|2.24|e=-26}}
|-
|p<sup>+</sup>–{{nuclide|lithium|6}}            || 66          || {{val|1.46|e=-27}}
|-
|p<sup>+</sup>–{{nuclide|boron|11}}             || 123         || {{val|3.01|e=-27}}
|}

Note that many of the reactions form chains. For instance, a reactor fueled with {{nuclide|tritium}} and {{nuclide|helium|3}} creates some {{nuclide|deuterium}}, which is then possible to use in the {{nuclide|deuterium}}–{{nuclide|helium|3}} reaction if the energies are "right". An elegant idea is to combine the reactions (8) and (9). The {{nuclide|helium|3}} from reaction (8) can react with {{nuclide|lithium|6}} in reaction (9) before completely thermalizing. This produces an energetic proton, which in turn undergoes reaction (8) before thermalizing. Detailed analysis shows that this idea would not work well,{{Citation needed|date=April 2010}} but it is a good example of a case where the usual assumption of a [[Maxwell–Boltzmann distribution|Maxwellian]] plasma is not appropriate.

=== Abundance of the nuclear fusion fuels ===
{{See also|Abundance of the chemical elements|Abundance of elements in Earth's crust|Abundances of the elements (data page)|Aneutronic fusion|CNO cycle|Cold fusion}}
{| class="wikitable"
|+
!Nuclear Fusion Fuel Isotope
!Half-Life
!Abundance
|-
|{{Nuclide|hydrogen|1|link=yes}}<ref>[[CNO cycle#]]</ref><ref name=":5">[[CNO cycle#CNO-I]]</ref><ref name=":6">[[CNO cycle#CNO-II]]</ref><ref name=":7">[[CNO cycle#CNO-III]]</ref><ref name=":8">[[CNO cycle#CNO-IV]]</ref><ref name=":9">[[CNO cycle#HCNO-I]]</ref><ref name=":10">[[CNO cycle#HCNO-II]]</ref><ref name=":11">[[CNO cycle#HCNO-III]]</ref>{{Excessive citations inline|date=December 2024}}
|Stable
|99.98%
|-
|{{Nuclide|deuterium|link=yes}}<ref name=":2">[[Nuclear fusion#Criteria and candidates for terrestrial reactions]]</ref><ref name=":3">[[Aneutronic fusion#Coulomb barrier]]</ref><ref name=":4">[[Aneutronic fusion#Candidate reactions]]</ref><ref name=":1">[[Cold fusion#Lack of expected reaction products]]</ref>
|Stable
|0.02%
|-
|{{Nuclide|tritium|link=yes}}<ref name=":2" /><ref name=":3" />
|12.32(2) y
|trace
|-
|{{Nuclide|Helium|3|link=yes}}<ref name=":2" /><ref name=":3" /><ref name=":4" />
|stable
|0.0002%
|-
|{{Nuclide|Helium|4|link=yes}}<ref name=":1" />
|stable
|99.9998%
|-
|{{Nuclide|lithium|6|link=yes}}<ref name=":2" /><ref name=":3" /><ref name=":4" />
|stable
|7.59%
|-
|{{Nuclide|lithium|7|link=yes}}<ref name=":2" /><ref name=":4" />
|stable
|92.41%
|-
|{{Nuclide|boron|11|link=yes}}<ref name=":2" /><ref name=":3" /><ref name=":4" /><ref>[[Aneutronic fusion#Residual radiation]]</ref>
|stable
|80%
|-
|{{Nuclide|carbon|12|link=yes}}<ref name=":5" /><ref name=":9" />
|stable
|98.9%
|-
|{{Nuclide|carbon|13|link=yes}}<ref name=":5" />
|stable
|1.1%
|-
|{{Nuclide|nitrogen|13|link=yes}}<ref name=":5" /><ref name=":9" />
|9.965(4) min
|syn
|-
|{{Nuclide|nitrogen|14|link=yes}}<ref name=":5" /><ref name=":6" /><ref name=":9" />
|stable
|99.6%
|-
|{{Nuclide|nitrogen|15|link=yes}}<ref name=":5" /><ref name=":6" /><ref name=":7" /><ref name=":9" /><ref name=":10" /><ref name=":4" />{{Excessive citations inline|date=December 2024}}
|stable
|0.4%
|-
|{{Nuclide|oxygen|14|link=yes}}<ref name=":9" />
|70.621(11) s
|syn
|-
|{{Nuclide|oxygen|15|link=yes}}<ref name=":5" /><ref name=":6" /><ref name=":9" /><ref name=":10" />
|122.266(43) s
|syn
|-
|{{Nuclide|oxygen|16|link=yes}}<ref name=":6" /><ref name=":7" /><ref name=":8" /><ref name=":10" /><ref name=":11" />{{Excessive citations inline|date=December 2024}}
|stable
|99.76%
|-
|{{Nuclide|oxygen|17|link=yes}}<ref name=":6" /><ref name=":7" /><ref name=":8" />
|stable
|0.04%
|-
|{{Nuclide|oxygen|18|link=yes}}<ref name=":7" /><ref name=":8" />
|stable
|0.20%
|-
|{{Nuclide|fluorine|17|link=yes}}<ref name=":6" /><ref name=":7" /><ref name=":8" /><ref name=":10" /><ref name=":11" />
|64.370(27) s
|syn
|-
|{{Nuclide|fluorine|18|link=yes}}<ref name=":7" /><ref name=":8" /><ref name=":10" /><ref name=":11" />
|109.734(8) min
|trace
|-
|{{Nuclide|fluorine|19|link=yes}}<ref name=":8" /><ref name=":11" />
|stable
|100%
|-
|{{Nuclide|Neon|18|link=yes}}<ref name=":10" /><ref name=":11" />
|1664.20(47) ms
|trace
|-
|{{Nuclide|Neon|19|link=yes}}<ref name=":11" />
|17.2569(19) s
|trace
|}

=== Neutronicity, confinement requirement, and power density ===
Any of the reactions above can in principle be the basis of [[fusion power]] production. In addition to the temperature and cross section discussed above, we must consider the total energy of the fusion products ''E''<sub>fus</sub>, the energy of the charged fusion products ''E''<sub>ch</sub>, and the atomic number ''Z'' of the non-hydrogenic reactant.

Specification of the {{nuclide|deuterium}}–{{nuclide|deuterium}} reaction entails some difficulties, though. To begin with, one must average over the two branches (2i) and (2ii). More difficult is to decide how to treat the {{nuclide|tritium}} and {{nuclide|helium|3}} products. {{nuclide|tritium}} burns so well in a deuterium plasma that it is almost impossible to extract from the plasma. The {{nuclide|deuterium}}–{{nuclide|helium|3}} reaction is optimized at a much higher temperature, so the burnup at the optimum {{nuclide|deuterium}}–{{nuclide|deuterium}} temperature may be low. Therefore, it seems reasonable to assume the {{nuclide|tritium}} but not the {{nuclide|helium|3}} gets burned up and adds its energy to the net reaction, which means the total reaction would be the sum of (2i), (2ii), and (1):

:5 {{nuclide|link=yes|deuterium}} → {{nuclide|link=yes|helium|4}} + 2 [[Neutron|n<sup>0</sup>]] + {{nuclide|link=yes|helium|3}} + [[Proton|p<sup>+</sup>]],  ''E''<sub>fus</sub> = 4.03 + 17.6 + 3.27 = 24.9&nbsp;MeV, ''E''<sub>ch</sub> = 4.03 + 3.5 + 0.82 = 8.35&nbsp;MeV.

For calculating the power of a reactor (in which the reaction rate is determined by the D–D step), we count the {{nuclide|deuterium}}–{{nuclide|deuterium}} fusion energy ''per D–D reaction'' as ''E''<sub>fus</sub> = (4.03&nbsp;MeV + 17.6&nbsp;MeV) × 50% + (3.27&nbsp;MeV) × 50% = 12.5&nbsp;MeV and the energy in charged particles as ''E''<sub>ch</sub> = (4.03&nbsp;MeV + 3.5&nbsp;MeV) × 50% + (0.82&nbsp;MeV) × 50% = 4.2&nbsp;MeV. (Note: if the tritium ion reacts with a deuteron while it still has a large kinetic energy, then the kinetic energy of the helium-4 produced may be quite different from 3.5 MeV,<ref>A momentum and energy balance shows that if the tritium has an energy of E<sub>T</sub> (and using relative masses of 1, 3, and 4 for the neutron, tritium, and helium) then the energy of the helium can be anything from [(12E<sub>T</sub>)<sup>1/2</sup>−(5×17.6MeV+2×E<sub>T</sub>)<sup>1/2</sup>]<sup>2</sup>/25 to [(12E<sub>T</sub>)<sup>1/2</sup>+(5×17.6MeV+2×E<sub>T</sub>)<sup>1/2</sup>]<sup>2</sup>/25. For E<sub>T</sub>=1.01&nbsp;MeV this gives a range from 1.44&nbsp;MeV to 6.73&nbsp;MeV.</ref> so this calculation of energy in charged particles is only an approximation of the average.) The amount of energy per deuteron consumed is 2/5 of this, or 5.0&nbsp;MeV (a [[specific energy]] of about 225&nbsp;million [[Megajoule|MJ]] per kilogram of deuterium).

Another unique aspect of the {{nuclide|deuterium}}–{{nuclide|deuterium}} reaction is that there is only one reactant, which must be taken into account when calculating the reaction rate.

With this choice, we tabulate parameters for four of the most important reactions

{| class="wikitable" style="margin:auto; text-align:right;"
|-
!fuel                                          !!''Z''!!''E''<sub>fus</sub> [MeV]!!''E''<sub>ch</sub> [MeV]!!neutronicity
|-
|{{nuclide|deuterium}}–{{nuclide|tritium}}     ||  1  || 17.6 || 3.5 || 0.80
|-
|{{nuclide|deuterium}}–{{nuclide|deuterium}}   ||  1  || 12.5 || 4.2 || 0.66
|-
|{{nuclide|deuterium}}–{{nuclide|helium|3}}    ||  2  || 18.3 ||18.3 || ≈0.05
|-
|p<sup>+</sup>–{{nuclide|boron|11}}            ||  5  || 8.7  || 8.7 || ≈0.001
|}

The last column is the [[aneutronic fusion|neutronicity]] of the reaction, the fraction of the fusion energy released as neutrons. This is an important indicator of the magnitude of the problems associated with neutrons like radiation damage, biological shielding, remote handling, and safety. For the first two reactions it is calculated as {{nowrap|(''E''<sub>fus</sub> − ''E''<sub>ch</sub>)/''E''<sub>fus</sub>}}. For the last two reactions, where this calculation would give zero, the values quoted are rough estimates based on side reactions that produce neutrons in a plasma in thermal equilibrium.

Of course, the reactants should also be mixed in the optimal proportions. This is the case when each reactant ion plus its associated electrons accounts for half the pressure. Assuming that the total pressure is fixed, this means that particle density of the non-hydrogenic ion is smaller than that of the hydrogenic ion by a factor {{nowrap|2/(''Z'' + 1)}}. Therefore, the rate for these reactions is reduced by the same factor, on top of any differences in the values of {{math|{{angbr|''σv''}}/''T''<sup>2</sup>}}. On the other hand, because the {{nuclide|deuterium}}–{{nuclide|deuterium}} reaction has only one reactant, its rate is twice as high as when the fuel is divided between two different hydrogenic species, thus creating a more efficient reaction.

Thus there is a "penalty" of {{nowrap|2/(''Z'' + 1)}} for non-hydrogenic fuels arising from the fact that they require more electrons, which take up pressure without participating in the fusion reaction. (It is usually a good assumption that the electron temperature will be nearly equal to the ion temperature. Some authors, however, discuss the possibility that the electrons could be maintained substantially colder than the ions. In such a case, known as a "hot ion mode", the "penalty" would not apply.) There is at the same time a "bonus" of a factor 2 for {{nuclide|deuterium}}–{{nuclide|deuterium}} because each ion can react with any of the other ions, not just a fraction of them.

We can now compare these reactions in the following table.

{| class="wikitable" style="margin:auto; text-align:right;"
|-
!fuel                                         !!{{math|{{angbr|''σv''}}/''T''<sup>2</sup>}}!!penalty/bonus !!inverse reactivity!!Lawson criterion!!power density [W/m<sup>3</sup>/kPa<sup>2</sup>]!!inverse ratio of power density
|-
|{{nuclide|deuterium}}–{{nuclide|tritium}}    || {{val|1.24|e=-24}} ||  1  ||    1 ||   1 || 34     ||    1
|-
|{{nuclide|deuterium}}–{{nuclide|deuterium}}  || {{val|1.28|e=-26}} ||  2  ||   48 ||  30 ||  0.5   ||   68
|-
|{{nuclide|deuterium}}–{{nuclide|helium|3}}   || {{val|2.24|e=-26}} || 2/3 ||   83 ||  16 ||  0.43  ||   80
|-
|p<sup>+</sup>–{{nuclide|lithium|6}}          || {{val|1.46|e=-27}} || 1/2 || 1700 ||     || 0.005  || 6800
|-
|p<sup>+</sup>–{{nuclide|boron|11}}           || {{val|3.01|e=-27}} || 1/3 || 1240 || 500 || 0.014  || 2500
|}

The maximum value of {{math|{{angbr|''σv''}}/''T''<sup>2</sup>}} is taken from a previous table. The "penalty/bonus" factor is that related to a non-hydrogenic reactant or a single-species reaction. The values in the column "inverse reactivity" are found by dividing {{val|1.24|e=-24}} by the product of the second and third columns. It indicates the factor by which the other reactions occur more slowly than the {{nuclide|link=yes|deuterium}}–{{nuclide|link=yes|tritium}} reaction under comparable conditions. The column "[[Lawson criterion]]" weights these results with ''E''<sub>ch</sub> and gives an indication of how much more difficult it is to achieve ignition with these reactions, relative to the difficulty for the {{nuclide|link=yes|deuterium}}–{{nuclide|link=yes|tritium}} reaction. The next-to-last column is labeled "power density" and weights the practical reactivity by ''E''<sub>fus</sub>. The final column indicates how much lower the fusion power density of the other reactions is compared to the {{nuclide|link=yes|deuterium}}–{{nuclide|link=yes|tritium}} reaction and can be considered a measure of the economic potential.

===Bremsstrahlung losses in quasineutral, isotropic plasmas===
{{unreferenced section|date=August 2023}}
The ions undergoing fusion in many systems will essentially never occur alone but will be mixed with [[electron]]s that in aggregate neutralize the ions' bulk [[electrical charge]] and form a [[Plasma (physics)|plasma]]. The electrons will generally have a temperature comparable to or greater than that of the ions, so they will collide with the ions and emit [[x-ray]] radiation of 10–30 keV energy, a process known as [[Bremsstrahlung]].

The huge size of the Sun and stars means that the x-rays produced in this process will not escape and will deposit their energy back into the plasma. They are said to be [[Opacity (optics)|opaque]] to x-rays. But any terrestrial fusion reactor will be [[Optical depth|optically thin]] for x-rays of this energy range. X-rays are difficult to reflect but they are effectively absorbed (and converted into heat) in less than mm thickness of stainless steel (which is part of a reactor's shield). This means the bremsstrahlung process is carrying energy out of the plasma, cooling it.
 
The ratio of fusion power produced to x-ray radiation lost to walls is an important figure of merit. This ratio is generally maximized at a much higher temperature than that which maximizes the power density (see the previous subsection). The following table shows estimates of the optimum temperature and the power ratio at that temperature for several reactions:

{| class="wikitable" style="margin:auto;"
|-
!fuel                                        !!''T''<sub>i</sub> [keV]!!''P''<sub>fusion</sub>/''P''<sub>Bremsstrahlung</sub>
|-
|{{nuclide|deuterium}}–{{nuclide|tritium}}   ||   50 || 140
|-
|{{nuclide|deuterium}}–{{nuclide|deuterium}} ||  500 ||   2.9
|-
|{{nuclide|deuterium}}–{{nuclide|helium|3}}  ||  100 ||   5.3
|-
|{{nuclide|helium|3}}–{{nuclide|helium|3}}   || 1000 ||   0.72
|-
|p<sup>+</sup>–{{nuclide|lithium|6}}         ||  800 ||   0.21
|-
|p<sup>+</sup>–{{nuclide|boron|11}}          ||  300 ||   0.57
|}

The actual ratios of fusion to Bremsstrahlung power will likely be significantly lower for several reasons. For one, the calculation assumes that the energy of the fusion products is transmitted completely to the fuel ions, which then lose energy to the electrons by collisions, which in turn lose energy by Bremsstrahlung. However, because the fusion products move much faster than the fuel ions, they will give up a significant fraction of their energy directly to the electrons. Secondly, the ions in the plasma are assumed to be purely fuel ions. In practice, there will be a significant proportion of impurity ions, which will then lower the ratio. In particular, the fusion products themselves ''must'' remain in the plasma until they have given up their energy, and ''will'' remain for some time after that in any proposed confinement scheme. Finally, all channels of energy loss other than Bremsstrahlung have been neglected. The last two factors are related. On theoretical and experimental grounds, particle and energy confinement seem to be closely related. In a confinement scheme that does a good job of retaining energy, fusion products will build up. If the fusion products are efficiently ejected, then energy confinement will be poor, too.

The temperatures maximizing the fusion power compared to the Bremsstrahlung are in every case higher than the temperature that maximizes the power density and minimizes the required value of the [[Lawson criterion|fusion triple product]]. This will not change the optimum operating point for {{nuclide|link=yes|deuterium}}–{{nuclide|link=yes|tritium}} very much because the Bremsstrahlung fraction is low, but it will push the other fuels into regimes where the power density relative to {{nuclide|link=yes|deuterium}}–{{nuclide|link=yes|tritium}} is even lower and the required confinement even more difficult to achieve. For {{nuclide|link=yes|deuterium}}–{{nuclide|link=yes|deuterium}} and {{nuclide|link=yes|deuterium}}–{{nuclide|link=yes|helium|3}}, Bremsstrahlung losses will be a serious, possibly prohibitive problem. For {{nuclide|link=yes|helium|3}}–{{nuclide|link=yes|helium|3}}, [[Proton|p<sup>+</sup>]]–{{nuclide|link=yes|lithium|6}} and [[Proton|p<sup>+</sup>]]–{{nuclide|link=yes|boron|11}} the Bremsstrahlung losses appear to make a fusion reactor using these fuels with a quasineutral, isotropic plasma impossible. Some ways out of this dilemma have been considered but rejected.<ref>{{cite journal|title=Fundamental Limitations on Plasma Fusion Systems not in Thermodynamic Equilibrium|journal=Dissertation Abstracts International|volume= 56-07 |issue=Section B|page= 3820|author=Rider, Todd Harrison|bibcode=1995PhDT........45R|year=1995}}</ref><ref>Rostoker, Norman; Binderbauer, Michl and Qerushi,
Artan. [https://web.archive.org/web/20051223133818/http://fusion.ps.uci.edu/artan/Posters/aps_poster_2.pdf Fundamental limitations on plasma fusion systems not in thermodynamic equilibrium]. fusion.ps.uci.edu</ref> This limitation does not apply to [[Non-neutral plasmas|non-neutral and anisotropic plasmas]]; however, these have their own challenges to contend with.