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Uranium-$235$ has a half-life of about $700$ million years and a critical mass of about $52\,$kg. That means if you take a large number of $U$-$235$ atoms, say a $26 \,$kg amount, and put each atom in a separate sealed container, then after $700$ million years about half of the atoms will have decayed.

Now suppose instead we have just a $26 \,$kg lump of the stuff. In this case the atoms are allowed interact with each other. And by definition of fissile material they do very strongly: Whenever one atom decays it shoots off particles with hit the other atoms and make them more likely to decay.

If we had a critical mass this effect would compound itself and we'd get a nuclear explosion: There would be a sudden explosive decay that would continue until we get below critical mass.

Since we only have half a critical mass the decay rate will not be so fast or sudden. But we'd still expect the interaction make half the atoms decay in less than $700$ million years.

How does one compute this half life (edit: time for half the atoms to decay)? Feel free to make whatever assumptions about the lump of material as desired. For example an idealized shape/density or some external force pushing the atoms together to prevent the lump blowing itself apart.

Daron
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    The concept of "critical mass" is somewhat mis-named or, at least, the name suggests a much simpler concept than reality. Saying that the chain reaction continues "until we get below critical mass" suggests that the chain reaction continues until so many atoms have fissioned that not enough remain to keep the reaction going. In reality, the chain reaction stops because the force of the explosion has spread the remaining U-235 atoms over such a large volume that neutrons no longer cause new fissions. Only a small fraction of the atoms actually fission in a nuclear explosion. – David Richerby Feb 03 '19 at 18:30
  • My answer to a previous question addresses another issue which has is usually discussed in similar language. – dmckee --- ex-moderator kitten Feb 03 '19 at 20:32
  • Critical mass doesn't mean explosion by the way. I mean it does not suffice. Butt it is not important to the current question. – Alchimista Feb 06 '19 at 09:33
  • Critical mass is just the point at which a self-sutaining reaction takes place. Each neutron released by fission will, on average, release one more neutron. I am also curious if anyone's done the math on what happens when each neutron has the maximum opportunity to cause fission. I expect it would easily keep things molten, and a continual plume of hydrogen, helium, iodine, cesium, and strontium would flow out of the continually shrinking radioactive core. Calculating the rate of decay would be a huge challenge. – Robert Rapplean Dec 23 '23 at 06:38

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As the other answers point out, half-life is a property of the isotope rather than the chunk. It can be defined as the time when there is 50% chance of any given nuclei having decayed spontaneously: there is no reference to any outside effects. The spontaneous decay of nuclei can be affected by their surroundings and obviously if you irradiate them a lot of things can happen.

The term "critical mass" is also a problem. If you have 52 kg of uranium in the form of a long wire it will not spontaneously explode. The neutrons released from spontaneous fission will mostly shoot away into the air. If you roll it up, things might get hotter. Alex Wellerstein has a very good post about why the term "critical mass" confuses things. A better term would be "critical condition": a condition where the spontaneous emission of neutrons will cause a chain reaction. This depends not just on mass, but the shape, presence of neutron reflectors, and other factors.

So, how do you actually calculate things? Basically you need to calculate the rate of emission of neutrons, and how these neutrons then diffuse in the chunk and cause more emission. This gets complicated quickly: the rate of stimulated emission depends on a cross-section that depends on neutron energy (this cross section can be measured but is not a simple function), and produce neutrons with random energies according to some probability distribution (another thing to measure). These neutrons then diffuse through the material as they bounce off other atoms, and one has to solve the diffusion equation for them (a fairly standard boundary value problem). Things of course get messier since they do produce more neutrons, so the actual calculation will be a diffusion equation with multiplication.

This is still a calculation for a steady-state situation rather than a critical situation where things grow over time; that adds extra mathematical wrinkles that I suspect are not readily available in the open literature. In any case, the total calculation is basically a model of the rate of neutron generation and absorption given the conditions, and we learn the critical condition by seeing where it shoots off to infinity. In practice the calculations are done numerically on a computer rather than analytically because of all the above complications but there are likely approximations one could run with pen and paper to get a rough sense of what is going on (after all, the Manhattan project did it successfully, although they sometimes used analogue computers like FERMIAC).

Anders Sandberg
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    As the other answers point out, half-life is a property of the isotope rather than the chunk. A half-life can be a property of a chunk of matter. Just not this chunk of matter, because its decay isn't a tailing exponential. –  Feb 03 '19 at 17:00
  • @BenCrowell can you clarify how half-life can be a property of a chunk of matter rather than just property of the actual element? After all half-lives are ultimately related to tunnelling probabilities, and that's not a function of how much stuff there is in the chunk. – ZeroTheHero Feb 03 '19 at 17:36
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    Is it at least possible to venture a – quite appropriately – Fermi computation for the length of time after which half the atoms in the OP's 26 kg of uranium aren't uranium anymore? – DaG Feb 03 '19 at 21:23
  • @DaG easy: one half-life, by definition. According to the OP that would be 700 million years. – ZeroTheHero Feb 03 '19 at 22:58
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    @ZeroTheHero: Even taking into account the more frequent interactions as suggested in the question? That is, can it be shown that the time magnitude is the same? – DaG Feb 04 '19 at 06:34
  • @DaG The interactions have nothing to do with this. Radioactive decay is (crudely) quantum tunnelling from the ground state. The fission in a nuclear explosion is from neutron capture and breakup of a highly excited state, which also produces neutrons to fee a chain reaction. – ZeroTheHero Feb 04 '19 at 13:50
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    @ZeroTheHero: Thanks, but please don't assume everybody else is utterly ignorant about this. What the OP asks, and I with them, is different: given that in a nuclear device (which, of course, has a very precise structure and initiator) the uranium is in large part destroyed in a fraction of a second by fission, and in a nuclear reactor in a few years (by fission and other phenomena), would any of these processes reduce the timing of disappearance of uranium in a, say, spherical, compact, 26-kg mass? I was hoping in a clarification by AndersSandberg, I must admit. – DaG Feb 04 '19 at 16:37
  • @DaG You are asking for a really different question as 1) this still doesn't change half-life, and 2) the amount left would depend on how the reactor is operated. The amount left would be a combination of decay due to half-life and reaction product. – ZeroTheHero Feb 04 '19 at 16:49
  • @ZeroTheHero Exactly! (Indeed, I never used the word “half-life”: I happen to know its definition.) Apparently, at last I have managed to explain myself. As to “how the reactor is operated”, I refer to OP's “Feel free to make whatever assumptions etc.” Anyway, the matter is even simpler than that: Anders mentions a calculation that is nigh impossible to perform analytically, so I was just asking him about a possible Fermi-like approach to it. – DaG Feb 04 '19 at 16:57
  • That is a fair point about the meaning of critical mass being misunderstood. How does your claim the calculation is fiendishly difficult, interact with Roger's claim the half-life of the chunk is the same as the half life of the constituent atoms? – Daron Feb 13 '19 at 13:45
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A fissionable material undergoing a chain reaction doesn't have a half-life. A half-life is a parameter that describes a tailing exponential curve. That's not what we have in the case of a chain reaction.

  • This answer is missing the point of the question. Terminology aside, what I'm clearly asking for is the length of time it takes until there is only 26kg of Uranium atoms in the chunk, the rest of the atoms are no longer uranium. – Daron Feb 05 '19 at 16:39
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    @Daron you really need to edit your question as the community clearly doesn’t understand it in the way you mean it. Ben’s answer is exactly on point. – ZeroTheHero Feb 05 '19 at 21:51
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Your definition of "fissile" only applies after they have acheived critical mass. And critical mass is a slippery number; if 52kg is called out as the critical mass (at STP), that probably presumes spherical arrangement; a cube would be subcritical.

And even if it was critical, that would be delayed critical - that is, some of the neutrons that keep it critical actually occur from fission products which have already been split, decaying further by neutron release, and subject to their own half-lives. This level of criticality will not yield an appreciable explosion.

That will result in an excursion, an exponential increase in power at human observable time scales (seconds), until the increase in power causes its own Rapid Unscheduled Disassembly.

Which is what happened at Chornobyl, except, the only reason Chornobyl was so unbelievably toxic is that this reactor had been running for quite some time and had old fuel with lots of fission and decay products like Cs137 in it, and that's what went kablooey all over Europe. If the Chornobyl excursion had happened with all brand new virgin reactor fuel that had never been subject to fission except for this excursion, the only radioactive fission products would've been from those few seconds of excursion -- not enough for anyone in Pripyat to notice, let alone Sweden. It wouldn't have even been detected by the west, and would've been a radiological nothingburger.

And even if there was prompt criticality - enough past critical mass that the fission could be sustained exclusively with U235 splits alone -- still, this would only result in a RUD.

Actually holding a critical mass together long enough to make Sodom & Gomorrah amounts of power is actually a hard problem. You're not going to do it by accident. And even then, it isn't a question of "long enough" because the time scale we're talking about is in shakes.

So the answer to your question is "the half-life doesn't change", because it is impossible for the criticality to remain assembled long enough to matter.

Harper - Reinstate Monica
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The half-life does not depend on the amount of material, but only on the material itself. The half-life of 10kg of U235 is the same as the half-life of 1kg of U235.

The half-life is usually measured experimentally by watching the decay of the initial lump of material. In a fission reaction what matters are cross-sections. Although there are theoretical models for computing these cross-sections (which are basically probabilities of interaction and depend on energy) you can bet that the ones critical to nuclear explosions have also been measured as accurately as possible.

ZeroTheHero
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  • But how long would it take until half the uranium atoms in the sample have decayed? The interaction between atoms would make that time less than the half life of U235. But by how much? – Daron Feb 05 '19 at 16:41
  • No... the interactions between the nuclei does not change the half-life. If you stick your sample in a reactor, then depending on the reaction rate, flux etc you will reduce the number of Uranium in the original sample, but that's not through radioactive decay. – ZeroTheHero Feb 05 '19 at 18:32
  • Yeah, but how long until half the uranium atoms have decayed? – Daron Feb 07 '19 at 10:15
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This is a sensible question in principle. If one decaying nucleus were to cause decays to its neighbours then nuclear decay in a large chunk of fissile material would be more rapid than one would predict from isolated nuclei, and the concept of 'half life' would become more complicated. In practice this doesn't happen because ${}^{235}{\rm U}$ decays by $\alpha$ emission, and you need neutrons to induce chain-reaction type fissions. So one doesn't have to take the effect into account, thank goodness.

RogerJBarlow
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