# How weight and volume-efficient can you make a high-yield nuclear weapon for setting off a volcano?

Let's say that we need a nuke that can fit into a very small space while also providing a very large boom.

Now, the US's B-41 nuclear bomb was quite mass-efficient indeed; it put out 22 petajoules, or 5.2 megatons, per metric ton of mass, and was, as Wikipedia so succinctly put it, "the most efficient known thermonuclear weapon in terms of yield to actual weight". But is it possible to make a nuke that's even more mass-efficient?

Moreover, how small can you make a high-yield nuke in terms of physical dimensions? I want to fit mine down a borehole and crack the cap over a supervolcano, so it's got to be long, thin, and powerful (jokes at my expense). But can you make a nuclear bomb long and thin enough to fit down something like the Kola Superdeep Borehole, or even one that's inside of a foot in diameter and still in the megaton range? The B-41 was more than four feet across, for instance, and while the W-54 used for the "Davy Crockett" and Special Atomic Demolition Munition was 11 inches in diameter, it also had a yield of somewhere between 0.01 and 1 kilotons, with a commonly accepted one of 20 tons of TNT. That's nowhere near enough to blow through, say, Yellowstone's magma plug.

Remember, this hypothetical skinny nuke is allowed to be long; just not wide.

Assume a Teller–Ulam device with a yield of 1 gigaton, whose mechanism of operation is detonating one fission bomb, followed by increasingly large fusion detonations that are set off by the previous fusion detonation. In other words, it's an arbitrarily large multi-stage device.

If I had to guess, the problem is the diameter of the fission primary, since Ivy Mike used a liquid deuterium nuclear fuel, and liquid deuterium, is, well, a liquid; despite needing cooling equipment, it can be fit into pretty much any shape; the fuel of the second stage is less of a problem. I'm not well-educated on this, though.

• Most of us are probably already on the U.S. Homeland Security watchlist for the many WMD questions posited on this Stack and a realistic answer to this question would likely get everyone else listed. But really, you want a realistic analysis of how small a fictional bomb can be? You might want to consider whether or not details like this are really relevant.
– JBH
Nov 5, 2021 at 6:04
• @JoinJBHonCodidact I'd like to brag about my ideas being based in some kind of reality. As Adrian Colomitchi pointed out, there's gonna be some handwaving involved, but I'd prefer to keep it to a minimum. Also, I like writing interesting questions and getting interesting answers. Nov 5, 2021 at 6:09
• help center: "Some subjective questions are allowed, but 'subjective' does not mean 'anything goes.' All subjective questions are expected to be constructive. What does that mean? Constructive subjective questions: ...are more than just mindless social fun." How will you judge a best answer? Without that, the question is opinion-based.
– JBH
Nov 5, 2021 at 6:13
• @JoinJBHonCodidact I'm asking how small a nuclear weapon of a one-gigaton yield can be made. I don't believe that that's opinion-based, since the answer can be based off of physical principles and evidence rather than some vague definition of what I think works best. Nov 5, 2021 at 6:58
• @JoinJBHonCodidact Well, there are physical principles relevant to this. For instance, if the fission primary is an implosion-type weapon, it's going to have a minimum possible diameter, since the fissile elements need to be arranged in a sphere of a certain radius. Moreover (let's say this thing uses liquid deuterium), there might be a minimum size of cooling device required to keep that deuterium liquid, meaning that the thing needs to be longer and thinner in order to hold the necessary amount of deuterium. There are definitely factors this can be based off of other than my opinion. Nov 5, 2021 at 7:10

There's gonna be some handwaving involved, but you may be able to use the tunnel geometry in your advantage.

The inertial pressure of the tunnel walls are going to be immense be it only because of:

1. the mass of the walls
2. the fact that the walls are made from tough materials (resisting loss of energy by deformation) and refractory (do not sink a good amount of that energy in chemical and structural modification)

Have a stack of:

1. the hydrogen bomb in front
2. followed by the fission bomb triggering the hydrogen bomb
3. a neutron damper after
4. a small, kt range, tactical nuke to project the fissile projectile through the neutron damper and super-compress the fission bomb that triggers the fusion one (the one at point 2.)
5. a massive, shock-wave shaping, plug of steel at the end, to reflect/focus the tac nuke energy on the fissile projectile of the trigger.

Notes

• the "neutron damper" at point 3 - you don't want the neutrons of the tac nuke to produce a premature "dud" fission reaction in the trigger of the hydrogen bomb
• you may start the stack with a shockwave shaping plug too, to have a better compression of the hydrogen that the stuff is going to fuse.

All of that, in a tight fit with walls of the tunnel. With careful timing of everything, I think a solution is possible.

# The extreme limit from basic physics is about the size of a soda bottle.

A "gigaton" is not actually based on the yield of TNT, but a convention that a gram of TNT has one Calorie (kcal). Notice TNT is actually a low-calorie food - most meals are 4 kcal per gram (carbohydrate, protein) or 9 kcal/g for fat. Most people eat it for the kick, I suppose. Anyway, you need a meal with 1 giga (10^9) x ton (10^6 gTNT/ton) x 1 kcal/gTNT = 10^15 kcal = 4.184E+18 J. That sounds like a heavy meal, but J = kg m^2/s^2, so let's divide this by the speed of light squared (3E+8 m/s) to get 46 grams of energy. A pity my paunch is not so efficient. We can even halve this energy by providing just 23 grams of antimatter and borrowing the 23 grams of matter from whatever it hits. I looked up 180 TJ/gAntimatter and that agrees. Except the director wants an authentic antique nuclear firearm for some reason, so now we have to go to the curve of binding energy. So for example Cf-251, a rather heavy isotope which is hard to do much better than with any data we've seen, has a weight of 251.079587 g/mol = 1.000317 g/nucleon, versus 53.9396090 g/mol = 0.99888 g/nucleon for Fe-54. That's a difference of 0.001437 g/nucleon, meaning that we need 32 kg of nucleons to get our 23 grams of energy. A real nuke, that doesn't deposit neat bricks of iron in its wake, or using U-235 (only 235.0439299 = 1.000187 g/nucleon), would be considerably heftier.

The 32 kg of nucleons will take up space depending on density - for Cf that's 15.1 g/mL, within 75% of the most dense materials known, so it will take up 2.2 liters, about the size of a soda bottle. In theory.