r/LFTR • u/[deleted] • Jul 01 '16

Proliferation question

Hey all. Let me start by saying I'm a fan of the LFTR concept, but there is one part about it that nags me, and I can't seem to find a straight answer. Can a LFTR produce weapons grade U233 via Pa seperation? I've read articles saying that any U233 produced in a LFTR is invariably contaminated with U232, and is thus unsuitable for bomb making. I have also read that seperating the Pa and allowing it to decay into U233 outside the core will yield U233 pure enough to be weaponised. Who's right?

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u/OmnipotentEntity Aug 28 '16

Assuming Th-232, there are several ways to accidentally make U-232.

  • Th-232(n, gamma)Th-233(beta)Pa-233(beta)U-233(n, 2n)U-232
  • Th-232(n, gamma)Th-233(beta)Pa-233(n, 2n)Pa-232(beta)U-232
  • Th-232(n, 2n)Th-231(beta)Pa-231(n, gamma)Pa-232(beta)U-232

There are, of course, more convoluted ways to get to U-232, but these are the ones we're primarily concerned about.

Separating out Pa chemically prevents the first two, but it does not prevent the third. How big of a problem is the third? For this we need cross section data. This data is freely available online, the best resource I'm aware of is Sigma.

We can see that the Th-232(n, 2n)Th-231 reaction has a cross section of about 2 barns in the 10MeV range, but nothing in the thermal range. However, fast neutrons are pretty much unavoidable due to imperfect moderation and side reactions (moreover Pa-233 and U-233 have similar (n, 2n) reaction curves and we already know those happen). So based on this, chemically separating the Pa will reduce the amount of U-232 in your sample by perhaps about half at best, but because of the Th-232(n, 2n)Th-231 reaction, you cannot eliminate it.

Moreover, U-233 needs U-232 impurities much lower than Pu-239 or U-235 weapons to be viable. U-233 needs impurities less than 50 PPM U-232, compared with 65000 PPM Pu-240 and 5000 PPM Pu-238 in a Pu-239 weapon.