Detailed analysis of lithium blanket reactions

See Neutron Reactions in a Lithium Blanket for an introduction to this topic!

Reaction X-section (barns)
Type Product Fast Inter. Thermal
n + 6Li n,t 4He + 3H 0.028 A 424.9 B 940.3 C
n,2n 4He + H + 2n 0.078 D ~ 0 ~ 0
capture 7Li ~ 0 0.017 m 0.039 p
scatter 6Li + n 0.906 1.691 0.735
n,p 6Li + p 0.007 k ~ 0 ~ 0
n + 7Li n,Xt 4He + 3H + n 0.112 E 0.151 F ~ 0
capture 2 • 4He ~ 0 0.020 n 0.040 q
n,2n 6Li + 2n 0.070 G ~ 0 ~ 0
n,d 6Li + 2H ~ 0 ~ 0 0.010 s
scatter 7Li + n 1.025 1.663 0.972

The most important entries in the table are in bold.

A, B, and C produce one tritium nucleus.

D, E, F and G produce either one tritium nucleus plus a neutron, or two neutrons. Each neutron can then go on to produce a tritium nucleus via the reactions in B or C, thus producing two tritium nuclei for one initial fast neutron. Because of the relatively very large cross-sections in B and C, this is by far the most likely outcome of D, E, F or G – as long as there’s a reasonable amount of 6Li. Natural lithium contains 7.5% 6Li, which is enough to make this route highly probable – as long as there’s nothing else around that would absorb neutrons.

You could in fact enrich the 7Li to increase the probability of E, F and G – just as uranium can be enriched in 235U, but much easier.

However, at best E could contribute an extra 9.3% [1] to the amount of tritium produced, G could contribute an extra 5.8% [2], and F could contribute an extra 1.9% [3]. This totals a possible extra 17% – or in fact a little less, because this percentage should be reduced by the percentage of 6Li present.

Against that reduction, D gives a small improvement – but it really is small, just 7.7% [4] times the percentage of 6Li.

Now let’s consider the losses. There are only minuscule losses in the lithium itself:

These really are tiny losses. Their cross-sections are very small by comparison with the others.

But we’ve only got 17% of the initial fast neutrons going into reactions that can produce an extra tritium. All the rest produce just one, or bounce around until they escape or get absorbed by something other than lithium.

How many escape? That depends on the thickness of the blanket. The mean free path of a fast neutron in lithium is 147mm, so if the blanket is 147mm thick then 1/e fast neutrons will go straight through it. That’s 36.8%. If the blanket is 294mm thick, 1/e2 fast neutrons will go straight through it. That’s 13.5%.

But that’s just the ones that go straight through. The first thing that happens to most of the initial fast neutrons that don’t go straight through is a scattering collision, and some of those are glancing collisions that don’t reduce the energy very much, and don’t deflect the neutron’s direction very much either. They now have another chance to escape – and they’ve got less of the blanket left to stop them. So the blanket needs to be that much thicker...

Then there’s all the stuff that isn’t lithium to consider. Lithium melts at 181 °C, so it has to be contained in something. Perhaps it’ll be stainless steel. That’s mostly iron, typically with some nickel and chromium. The mean free path of 14.64 MeV neutrons in iron is 45.8mm, and not very different from that in nickel or chromium. It’s unlikely that the container for the lithium will be 45.8mm thick, but it could well be a tenth of that, so 1/e0.1 of the initial neutrons will make it through without a collision. That’s 90.5% – so 9.5% will have a collision.

The next table shows what will happen in these collisions.

56Fe (91.754%) 54Fe (5.845%) 58Ni (68.077%) 60Ni (26.223) 52Cr (83.789%)
barns % barns % barns % barns % barns %
Total 2.574 2.498 2.664 2.763 2.419
Elastic 1.137 44 1.08 43 1.297 49 1.481 54 1.029 43
Inelastic 0.769 30 0.482 19 0.297 11 0.657 24 0.919 38
n,2n 0.439 17 0.001 0 0.022 1 0.347 13 0.26 11
n,np 0.071 3 0.491 20 0.56 21 0.078 3 0.089 4
n,p 0.114 4 0.361 14 0.364 14 0.123 4 0.088 4
n,alpha 0.041 2 0.083 3 0.092 3 0.054 2 0.033 1

Iron, nickel and chromium all have other isotopes, but these are the most abundant ones. By far the most important is 56Fe, since iron forms the largest percentage of stainless steel, and 56Fe forms 91.754% of iron. As you can see from the table, the cross-sections for the various reactions with 14.64MeV neutrons are not enormously different between the different nuclides.

Glancing elastic collisions don’t reduce the energy of the neutron much, nor change its direction much, so there might as well not have been any collision at all. Other elastic collisions, and all inelastic collisions, change the direction of movement of the neutron, meaning about 50% of them won’t enter the lithium at all; they also reduce the energy of the neutron, meaning that even if it does enter the lithium, it’s no longer a fast neutron, and very possibly not even an intermediate one, so it will only produce one tritium at most.

Although 17% of collisions with 56Fe double up the neutron, both these will have low energy, and on average about half of them will head back into the reactor, rather than on into the lithium.

This is the point to mention using a beryllium alloy instead of stainless steel. Beryllium is one of those neutron multipliers – and if it was embedded within the lithium the way it could double up a fast neutron to two slower ones might be thought to be useful. Until you realize it’s no better than what the lithium itself can do. And as the container for the lithium, it’s like that stainless steel: many of the neutrons will head back into the reactor instead of going on into the lithium, so all you’ve done is convert a potentially useful fast neutron into, on average, one, less useful, slower one.

The bottom three rows of the table are relatively unimportant, with small figures. The n,np row results in about half the incident neutrons continuing as low energy neutrons, and half of them heading back. The last two rows both result in the elimination of neutrons.

So overall, if the wall of the container is 4.5mm of stainless steel, there’ll be a reduction of about 4.5% in the neutron flux in the lithium. Other materials and thicknesses will give similar (or worse) results.

Finally, there’s the question of what proportion of the spherical angle around the reactor can be covered. There must be side walls to the lithium containers, as well as access for other services to the reactor. It’s hard to imagine how the coverage could approach anywhere near 100%.

[1] 0.112/(0.112 + 0.070 + 1.025)

[2] 0.070/(0.112 + 0.070 + 1.025)

[3] 0.25 x 1.025/(0.112 + 0.070 + 1.025) x (0.151/1.663) – the 0.25 is an estimate of the proportion of scatterings that will result in an intermediate neutron. Glancing collisions leave the neutron going fast, with a reduced distance to go before it passes right through the blanket; inelastic collisions leave it too slow for F, as do head-on collisions; some but not all other collisions will leave it going at an appropriate velocity for F.

[4] 0.078/(0.028 + 0.078 + 0.906 + 0.007)