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Just like what the title says, I would like to ask how it is possible to breed sufficient amount of Pu239 via breeding process using fast neutrons and U238. Due to its inherently larger fission cross section when reacting with fast neutorns, wouldn't majority of U238 just split up into the likes of I153 and other various fission products rather than capturing the neutron and become Pu239 which is the goal of breeding process? This is also the case for other actinide, transuranium elements which is the reason fast reactors are inherently more "clean" and could potentially even burn what we have as nuclear wastes today. This is especially intriguing for me since I've learnt that only around 11% of the U238 reacting with fast neutron ends up splitting despite the bigger cross section.

From what I know the main fuel for FBRs is the MOX or pyroprocessed fuel which is akin to MOX in terms of composition. So I understand that the supply of the fast neutron would be sufficient thanks to Pu239 fission but can't get my head around the mechanism behind U238 reaction.

I suspect that it may be that my understanding of neutron cross section is faulty or the neutrons caused by fission are actually slower than 1MeV threshhold and therefore leading to a false conclusion.

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    $\begingroup$ You need to pay attention to the cross section vs energy of neutrons. $\endgroup$
    – Jon Custer
    Sep 19, 2021 at 14:04

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Consider a uranium fueled reactor, consisting of enriched U235 and U238. U235 is fissile in that it can fission with a zero energy neutron. U238 is fissionable with high energy neutrons and is also fertile in that through absorption of a neutron and subsequent beta decay a fissile isotope is produced, specifically Pu239. So in operation some of the U238 undergoes fission with high energy neutrons and some of the U238 transmutes to Pu239 through neutron capture. Although most of the fission occurs in the U235, fission of the U238 contributing to the overall criticality of the reactor is addressed by including a fast fission factor in the relationship for the multiplication factor.

Some of the U238 is converted to Pu239, and if the amount of Pu239 produced is greater than the initial amount of U235, the reactor is a breeder reactor: it produces more fissile material (Pu239) than is uses (U235).

Breeding occurs if the value of $\eta$ is sufficiently large where $\eta$ is the average number of neutrons emitted in fission per neutron absorbed in the fissile isotope. $\eta$ must be greater than 2 for breeding; one neutron must be absorbed by the fissile isotope and produce one neutron to sustain the chain reaction, and at the same time more than one neutron must be absorbed by the fertile isotope to breed more fissile material than used in the fission process. $\eta$ is greater in a fast reactor (higher energy neutron energy where fission occurs) than in a thermal reactor, so reactors that breed are fast breeder reactors.

The energy distribution of neutrons from fission is shown in the figure below.

enter image description here

The $ (n,\gamma)$ neutron capture and fission cross sections for U238 are shown in the following figure.

enter image description here

For the neutrons from fission with energies greater than about 1 MeV, the fission cross section of U238 is greater than the capture cross section as your question indicates.

However, the energy distribution of neutrons in the reactor is not the same as the energy distribution of the neutrons from fission due to slowing down of the fission neutrons due to interaction with materials in the reactor. The following figure shows the energy distribution of neutrons for a thermal reactor and for a fast reactor, specifically a liquid metal fast reactor (LMFBR).

enter image description here

For the thermal reactor, a moderator, such as water or graphite, is used to purposefully slow the fission neutrons down to low energy where the fission cross section of the fissile material- U235- is the greatest to allow criticality with the least amount of fissile material. For a fast reactor, no moderator is used, nonetheless the fission neutrons do lose energy due to interactions with the coolant (liquid sodium for the LMFBR) and other structural materials. The fission cross section is less for the fast reactor neutron spectrum than for the thermal neutron energy spectrum, so the fast reactor requires more fissile material than does the thermal reactor. However, the fast reactor is a breeder reactor due to the sufficiently high value of $\eta$ in the fast reactor neutron energy spectrum.

A substantial fraction of the neutrons in the fast reactor have energies less than 1 MeV and at these lower energies neutron capture is much more likely than fission. Considering the energy of the neutrons in the fast reactor, neutron capture is much more likely than fission.

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  • $\begingroup$ Thanks a lot! So those fast reactors are in fact more of a intermediate-fast neutron reactors if I could put it that way; or rather "fast" in general terms that the neutrons are faster than thermal neutrons. Now this makes total sense. $\endgroup$
    – MK.s
    Sep 21, 2021 at 13:57
  • $\begingroup$ Exactly. Glad I could help. $\endgroup$
    – John Darby
    Sep 21, 2021 at 14:21

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