Neutron Multipliers
Beryllium and lead each have the potentially useful property that a very high energy neutron, such as a 14.64 MeV neutron fresh from a fusion reaction, can knock a second neutron out of their nucleus, thus converting one high energy neutron into two medium energy neutrons, each of which could then go on to react with a lithium nucleus to produce tritium.
The cross sections for these reactions can be obtained from Los Alamos National Laboratory t2.lanl.gov/nis/data/endf/endfvii.1-n.html or the Korean Atomic Energy Research Institute (KAERI) atom.kaeri.re.kr:8080/ton).
The mean free path of a 14.64 MeV neutron is 5.66cm in lead, and 5.34cm in beryllium. (See Neutron Mean Free Path Calculations.)
However, in the case of lead the probability of the interaction being n,2n (neutron multiplication) is about 46%. Most of the other interactions are elastic collisions (very like billiard balls). In the case of beryllium, it’s only 22%, and again most of the other interactions are elastic collisions.
A beryllium nucleus is only nine times more massive than the neutron, so an elastic collision significantly reduces the neutron’s energy, and even if it’s still in the blanket, after just a very few such collisions it will no longer have enough energy for neutron multiplication. Inelastic collisions – a smaller but significant proportion of collisions – reduce the neutron energy very much more.
Lead nuclei are much more massive than neutrons, and the neutrons lose little energy in the elastic collisions. However, in inelastic collisions they do lose much of their energy.
In both cases the effective mean path length to an n,2n reaction is longer than the calculated mean free path. You can calculate a theoretical mean path length based on the cross-section for the n,2n reaction (12.3cm for lead, 30.1cm for beryllium) but particularly in the case of lead it’s not really quite as bad as that, because the elastic collisions scatter the neutrons, and some of them may therefore end up travelling in the plane of the blanket, or obliquely, rather than straight through it. Some of them will end up heading in the opposite direction, back towards the reactor instead of forward to the lithium blanket. This gets truly complicated, and dependent not only on the percentages of neutrons being scattered at different angles (the data for these calculations are available on the Los Alamos site) but also on the precise geometry of the blanket.
There are other possible reactions, apart from elastic and inelastic scattering and neutron multiplication, but especially in the case of lead, the cross-sections are very much smaller, so very few neutrons will be lost to them or contributed by them. Neutron capture and n,3n reactions are shown, but there are others. None are significant.
| σ/barn for 14.64 MeV neutrons | σ/barn, thermal neutrons | ||||||
|---|---|---|---|---|---|---|---|
| Material | Percentage | Elastic | Inelastic | n,2n | Capture | n,3n | Capture |
| 9Be | 100 | 0.959 | 0.023 | 0.269 | 0.32μ | ~0 | 7.60m |
| 208Pb | 52.4 | 2.410 | 0.480 | 2.467 | 1.10m | 2.35μ | 501μ |
| 207Pb | 22.1 | 2.376 | 0.325 | 2.651 | 3.73μ | 2.14μ | 0.712 |
| 206Pb | 24.1 | 2.396 | 0.411 | 2.539 | 18.4μ | 1.20μ | 0.031 |
| 204Pb | 1.4 | 2.754 | 0.488 | 2.095 | 8.15μ | 0.52μ | 0.661 |
I’ve included the last column for interest. 207Pb, which is 22.1% of natural lead, will absorb some of the neutrons once they reach thermal energies, taking them away from tritium production. It’s not a major effect, but not completely trivial either. This turns it into 208Pb, which is 52.4% of natural lead anyway. 204Pb is only 1.4% of natural lead, but if it absorbs a neutron it’s converted into 205Pb, which is radioactive with a half-life of 15.3 million years. 205Pb is also produced when 206Pb undergoes an n,2n reaction.