3
$\begingroup$

I want to produce fissile material.

I want to produce a lot of fissile material.

Fortunately, I read the Wikipedia page on Ted Taylor. It states that:

Apart from bombs, Taylor also explored concepts of producing large amounts of nuclear fuel in an expedited manner. His plans, known as MICE (Megaton Ice Contained Explosions), essentially sought to plant a thermonuclear weapon deep in the ice and detonate it, resulting in a giant underground pool of radioactive materials that could then be retrieved. While his idea had merit, Taylor ultimately received little support for this concept and the project never came to fruition.

This idea got me thinking. Nuclear weapons produce some level of neutron radiation, so, presumably, said neutron radiation can make valuable fissile material/nuclear fuel out of the following materials:

However, nuclear weapons are extremely powerful [citation needed], and megaton-level explosions will damage the above materials if they're too close to said explosions, rendering them non-harvestable, so, instead of a megaton-level explosive, I'm going to use a ton-level explosive: the American W54 tactical nuclear warhead, which, in the specific configuration it's in in the Special Atomic Demolition Munition, has a yield of 10 tons of TNT. The W54 was used in the Davy Crockett nuclear recoilless rifle system due to its ability to neutron-activate the hulls of tanks - specifically, Soviet ones passing through the Fulda Gap. As such, I believe that it has a relatively low blast yield relative to its ability to emit neutron radiation.

I am going to place it into an arbitrarily large ice block and then detonate it. Said ice block is laced with solid chunks of both lithium-6 and uranium-238 that are a certain distance from the warhead. I shall call it N.I.C.E. - Nuclear Ice Contained Explosions, as a sort of miniature version of M.I.C.E.

The question: how far away from the detonation should these materials be placed to avoid being destroyed by the detonation? NUKEMAP says at least 10 meters, since the nuclear fireball of a 10-ton-of-TNT-level nuclear detonation is that size, and will vaporize anything within itself; however, thermal and pressure effects outside of that are harder to calculate, and I'd rather have the lithium-6 and uranium-238 as close to the bomb as possible, so as to maximize the neutron flux applied to them.

Note that there will be a solid layer of ice between the bomb and these materials, whose thickness will be equivalent to the distance between them and the bomb - so as to ensure that they aren't melted by the bomb's non-fireball heat effects.

I don't particularly care if this is practical. I just want it to happen, story-wise. Perhaps the people doing it are on an orbital habitat with no fissile materials and one tiny (relatively speaking), W54-sized nuclear bomb as their only neutron source.

$\endgroup$
2
  • $\begingroup$ The ice or water between the bomb and the materials being irradiated may also help thermalize the neutrons. The different elements have different cross sections. Fast (high energy) neutrons are captured as easily. $\endgroup$
    – UVphoton
    Jun 19, 2022 at 12:18
  • 2
    $\begingroup$ Would it matter if the materials were vaporized? Would they condense back into a solid afterwards if they are contained, although it probably would be easier if already in chunks for removal. I am assuming there would need to be some post processing of the materials after the explosion has cooled. $\endgroup$
    – UVphoton
    Jun 19, 2022 at 12:22

1 Answer 1

3
$\begingroup$

At roughly 2.89 neutrons per Pu fission, ignoring those that go back into the chain reaction as well as those that leak out of the system, your 'gain' is capped at 2.89 times as much Pu as was expended (fissioned) in the explosion. Which, particularly for low yield weapons, is pitifully small (they are very inefficient).

Lets run some numbers:

20T yield is 83GJ, which is 5E29 eV. At about 180MeV output per fission event, that means 2.9E21 fissions. At 2.89 neutrons per, that is 8.38E21 total neutrons released. If every single one hits a U-238 and ends up as Pu-239, you have produced 0.014 moles of Pu-239, or a little over 3 grams.

Sure, you only used up 0.004 moles in the explosion, but you now have to separate all the Pu (from the warhead and what you produced) from your affected volume, reassemble the warhead (with 0.01 more moles of Pu), and try again. Remember that your warhead starts with many moles (and multiple kilograms) of Pu-239.

Bottom line - infeasible.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .