Editing Nuclear fusion (section)

=== 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.