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I caught a pretty well done 2 hour documentary on atomic bomb history yesterday on the local PBS station. In it, they go over the paths taken for design of the first bombs, including the Thin Man Pu239 based gun design. After Segrè found the spontaneous fission rate from the reactor-produced Pu was much too high (from Pu240 contamination), that design was abandoned.

It is stated by one of the speakers in the documentary (might have been Rhodes) that such a design is "impossible" (not "was impossible"), implying that it's not feasible at all.

I did a quick back-o-the-envelope calculation as follows (I'm a mathematics guy, not a physicist, hence the question...):

Taking the ~60Kg U235 of the Little Boy, and 3*10^-4 neutrons/gram-second, ~18 neutrons/sec.

Guessing at (since it won't be as efficient as implosion) 2.5X the 6Kg mass of the Fat Man Pu239 to be used in the hypothetical gun and 0.022 neutrons/gram-second, ~330 neutrons/sec.

Taking the insertion time as 1 millisecond (which I assume could be improved with current technology), I get ~0.018 and ~0.33 neutrons/ms average for U235 mass and Pu239 mass respectively.

Assuming Poisson distribution of arrivals, I end up with ~0.98 and ~0.72 probabilities of no stray neutrons during assembly for U235 mass and Pu239 mass respectively.

While that shows a much higher possibility of a fizzle in that latter, it seems far from "impossible".

My question: Given a sufficient mass of "pure" Pu239 (say cyclotron produced, or whatever the current state-of-the-art might be to produce it), and current state-of-the-art technologies to accelerate a projectile, is such a design really impossible (cost of Pu production, efficiency, practicality aside). As in, could an actual high-yield device (as opposed to a fizzle-yield) be produced?

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  • $\begingroup$ I thought the problem with a gun-type plutonium device was that as the pieces get close enough, fission starts to take place in the two pieces. By the time they meet and form one mass, there are already too many fission by-products to support a self-sustaining chain reaction. And if the smaller piece is fired rapidly enough to avoid the production of those fission by-products, it acts like a bullet and blasts apart the larger piece. $\endgroup$
    – BillDOe
    Jul 30, 2015 at 0:01
  • $\begingroup$ @BillOer: Same problem (pre-detonation) exists for U235 (e.g. Little Boy), hence the need to have short insertion time. Worse for PU available at the time due to Pu240 contamination, but question is re: "pure" Pu239. $\endgroup$
    – rasher
    Jul 30, 2015 at 0:19

3 Answers 3

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This is from The Nuclear Weapons Archive:

2.1.4.1.2 Gun Assembly

Assembling a critical mass by firing one piece of fissionable material at another is an obvious idea and was the first approach developed for designing atomic bombs. But it is probably not obvious how you take two subcritical masses and obtain the equivalent of three critical masses by bringing them together.

This can be made clear by conducting a thought-experiment. Imagine a spherical pit made up of about three critical masses of fissionable material. Now remove a core (like an apple core) from the pit with a mass slightly less than critical. Since the center of the pit is now hollow, its effective density has been reduced to 2/3 of the original density. Since we now have two critical masses remaining in the pit, and the reduction in density leads to a further reduction of (2/3)^2 = 4/9, the pit now contains only 2*(4/9) = 8/9 of a critical mass.

The two sub-critical pieces can be brought together by firing the cylindrical core down a gun barrel into the center of the hollowed-out pit. The insertion time is large - over 1 millisecond. This is the design used in Little Boy, the bomb dropped on Hiroshima (except that a slightly less efficient squat cylinder was used, rather than a spherical pit).

The primary advantage of gun assembly is simplicity. It is as close to a fool-proof design as ordinance technology allows.

The drawbacks are:

a. the lack of compression, which requires large amounts of fissionable material, and leads to low efficiency;

b. only uranium-235 (and possibly U-233) can be used due to the slow insertion speed;

c. the weight and length of the gun barrel makes the weapon heavy and fairly long.

Their website is here.

And this is from "Nuclear Weapons Design" at this website:

If plutonium -- even weapon-grade -- were used in a gun-assembly design, neutrons released from spontaneous fission of its even-numbered isotopes would likely trigger the nuclear chain reaction too soon, resulting in a "fizzle" of dramatically reduced yield.

Me again: so it seems like it might be possible to get a plutonium-based gun assembly device to actually explode rather than fizzle, but the timing would be so critical that if would fail more often that it would explode.

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  • $\begingroup$ if the hole was not cylindrical, but conical instead, you could arrange that the pieces went into contact nearly at the same time, rendering the slow insertion speed irrelevant $\endgroup$
    – lurscher
    Jul 30, 2015 at 2:19
  • $\begingroup$ Appreciate the response, but again, question is re: pure Pu239 (which current weapons-grade is not), combined with possibly better (faster) projectile assembly. +1 for the effort. $\endgroup$
    – rasher
    Jul 30, 2015 at 3:43
  • $\begingroup$ @rasher, pure Pu239 would make insertion time even more critical. It takes about 260 nanoseconds for the nuclear chain reaction to complete, so (I'm guessing) just +- 10 nanoseconds would result in an explosion or a fizzle. $\endgroup$
    – BillDOe
    Jul 30, 2015 at 5:10
  • $\begingroup$ @lurscher, actually I thought the original gun-assembly did use a conical piece, but this guy seems to know what he's talking about. I'm thinking that a conical piece makes the difference between blasting the larger piece into a lot of small pieces, and too slow an insertion speed causing a fizzle that much more critical, (i.e. too fast = blasted core; to slow = fizzle). A cylinder adds some small amount of fudge factor. $\endgroup$
    – BillDOe
    Jul 30, 2015 at 5:17
  • $\begingroup$ @BillOer you appear to have it backward - lower spontaoeous fission rate relaxes insertion time requirements. $\endgroup$
    – rasher
    Jul 30, 2015 at 5:37
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The "bullet" in Thin Man was a conical shell, which would go *over" the target, reducing contact time incredibly. It was molded on an old cone clutch design from an automobile, with two conical rings of plutonium meeting at high speed.

The problem was contamination. In effect, they would have to cast the core pieces, install, and drop within 28 days, else risk spontaneous fissile events. This lack of standby capability makes the Thin Man nearly worthless for anything but a surprise, spur of the moment attack. Like doing a raid on Moscow from forward bomber bases in Poland using Lancaster bombers, as Churchill proposed.

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  • $\begingroup$ Interesting info re: Churchill. +1 $\endgroup$
    – rasher
    Aug 14, 2015 at 9:55
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A typical implosion assembly of a Pu pit is time limited by the maximum detonation velocity of the high explosive used. As a ball park figure, you are not going to get a shock a compression rate of more than 10 km/s because of that limit. So in theory a gun assembly might work if you could accelerate the projectile to twice that velocity (the implosion is coming in at max 10km/s from all sides). Maybe you could halve this with a "double ended" gun, with each projectile moving towards centre of mass at that speed. Basically, it is probably possible but too much of an engineering nightmare. The nearest you might get is two separated plates of Pu each coated with an explosive charge, but that is a planar implosion assembly and not what anyone would recognize as a "gun".

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    $\begingroup$ Agreed. I've been fiddling with more detailed calculations based on Reed's work in "Physics of the Manhattan Project": my initial back-o-envelope were not too off, and given high purity Pu239 and SOTA "gun" technology like a light-gas gun, it is no longer "impossible", but utterly impractical, costly, and certainly unnecessary with the superiority of implosion. I assumed this, but wanted to see if "impossible" held water. Thanks for the reply. +1 $\endgroup$
    – rasher
    Aug 14, 2015 at 9:53

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