This diagram goes with Nuclear Fusion?.
It shows all the relevant nuclear reactions involved in creating tritium from lithium, using neutrons from the deuterium-tritium fusion reaction. The reaction cross sections for all these reactions (using which you can work out the proportion of the reactants that will undergo that particular reaction) are shown in the table (data obtained from Los Alamos National Laboratory t2.lanl.gov/nis/data/endf/endfvii.1-n.html and Korean Atomic Energy Research Institute (KAERI) atom.kaeri.re.kr:8080/ton).
| Reaction | X-section (barns) | ||||
|---|---|---|---|---|---|
| Type | Product | Fast | Inter. | Thermal | |
| n + 6Li | n,t | 4He + 3H | 0.028 | 424.9 | 940.3 |
| n,2n | 4He + H + 2n [1] | 0.078 | ~ 0 | ~ 0 | |
| capture | 7Li | ~ 0 | 0.017 | 0.039 | |
| scatter | 6Li + n | 0.906 | 1.691 | 0.735 | |
| n,p | 6Li + p [2] | 0.007 | ~ 0 | ~ 0 | |
| n + 7Li | n,Xt | 4He + 3H + n | 0.112 | 0.151 | ~ 0 |
| capture | 2 • 4He [3] | ~ 0 | 0.020 | 0.040 | |
| n,2n | 6Li + 2n | 0.070 | ~ 0 | ~ 0 | |
| n,d | 6Li + 2H | ~ 0 | ~ 0 | 0.010 | |
| scatter | 7Li + n | 1.025 | 1.663 | 0.972 | |
[1] Actually 5Li + 2n → 4He + H + 2n
[2] Actually 6He + p → 6Li + p
[3] Actually 8Li (→ 8Be) → 2 • 4He
The “Intermediate” column is a gross simplification – this isn’t the place for too much detail! – the actual cross-section varies with energy and this is merely a sort of average or typical value. There are several possible algorithms for calculating the average, and none of them are “correct.” Where I’ve found a value in a reasonably reliable looking source I’ve used it, otherwise I’ve used my own algorithm to calculate a value based on the detailed data from Los Alamos. These values should be treated with considerable scepticism whether it’s my value or anyone else’s.
For accurate work you have to consider the different cross-sections of all the alternative reactions at every energy from fast right down to thermal, and calculate the neutron flux at each energy – and that will depend on not only what nuclei are present, but also on the geometry of the blanket. I’ve calculated some constraints on what is possible, but it would be excessive to present the calculations here, and undoubtedly tighter constraints could be calculated anyway.
The “Fast” and “Thermal” figures are all taken directly from KAERI and checked against Los Alamos (differences are minor), and are much more robust because the energy of a neutron fresh from a D-T reaction is 14.64MeV with very little variation, and there’s very little variation in cross-sections at thermal energies (~0.025eV).
There are tricks that could be played with neutron multipliers such as lead or beryllium, but they don’t really alter the picture much. The best they can achieve is similar in effect to the n,2n and n,Xt reactions shown in the table. They would provide an alternative to those reactions, not an addition, since in either case the incident neutron has to be fast (or fast intermediate), and the resulting neutrons are slower. For more information on this, see Neutron Multipliers. (Of course if you put fissile nuclei in your blanket things become very different, but then you’re producing radioactive fission products in large quantities again.)
The end result of all the calculations is that IF you could ensure that none of the neutrons from the fusion reaction escaped or reacted with anything other than Lithium nuclei (or, possibly, lead or beryllium nuclei), you could probably just manage to produce a few percent more tritium nuclei than the number of neutrons you started with. (See details.) Unfortunately, that IF is a killer: a 14 MeV neutron from the fusion reaction has a mean free path in lithium of 147 mm[4], so with any credible thickness of lithium blanket, many of your neutrons will go straight through. A lot more will bounce around a few times, losing energy without reacting[5], before finally escaping or being lost to some side reaction.
They’ve got to get through the wall of the vessel containing the lithium first, without reacting with it or losing energy by bouncing off nuclei in it.
‘X’ in the diagram is those non-lithium nuclei in the vessel wall and structure – and whatever is beyond the lithium blanket, such as maybe superconducting magnets or powerful lasers.
Ultimately, it’s simply not credible that you can get anywhere near one tritium for each neutron – which leaves the deficit having to be made up using neutrons from fission reactors or particle accelerators, neither of which produce the huge numbers required very efficiently at all, and both of which produce large quantities of radioactive waste. Particle accelerators don’t normally produce “large quantities” of radioactive waste – but the kind of unbelievably heavy-duty particle accelerators you’d need to produce the necessary quantities of neutrons would, in the process of producing those neutrons. For more information see Particle Accelerators for Tritium Production.)
I originally drew this diagram, and did the relevant calculations, when I was a student studying Nuclear Engineering in the late 1960s. It was the subject of one of my blazing rows with my professor. Now redrawn using Acorn !Draw on a Raspberry Pi. The data in the table is up-to-date. The up-to-date data is only slightly different from the data I used originally and doesn’t affect the conclusions (it marginally tightens the constraints). The diagram was not affected.
[4] This is for natural lithium. Changes in the isotope ratio would make some difference, but very little as the neutron cross-sections are very similar at fusion energy. See also Neutron Mean Free Path Calculations.
[5] The details of this are interesting: a billiard ball analogy is actually quite apposite for “elastic” collisions. There can be “glancing” collisions, or “head on” collisions, or anything in between. Glancing collisions leave the incident neutron with most of its energy – but not deflected much, so if it’s already x% of the way through the blanket, it’s only got (100-x)% of the blanket to go through before it escapes. Other collisions deflect the neutron more, so it may be deflected onto a new course that doesn’t escape the blanket so soon – but its energy will be much reduced, and it will no longer have enough energy for any of the reactions that can result in the production of two tritium nuclei.
And then there’s “inelastic” scattering – which reduces the energy regardless of the angle of incidence.