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View Full Version : How much of the fissile material in an atom bomb is converted to energy?

Sapo
10-31-2008, 09:53 AM
Inspired by two other ongoing threads, and not wanting to hijack either.

I understand that in a nuclear bomb, there is matter being converted into energy, but they mentioned 53Kg of Uranium for critical mass, which sounds like a lot to turn to energy via E=mc2. How much of it is actually turned to energy and how much of it is left as matter?

(feel free to fight any obvious ignorance you spot on this question)

And in case you find this interesting and haven't found the two other threads:

beowulff
10-31-2008, 10:05 AM
This is what is known as "efficiency." It appears that for a pure fission bomb, the upper limit is around 25%, meaning that 25% of the available atoms actually undergo fission: http://nuclearweaponarchive.org/Nwfaq/Nfaq2.html (scroll down to 2.1.4.3).

There are techniques to increase bomb yield without needing to increase efficiency - they are discussed in the link.

Pleonast
10-31-2008, 10:06 AM
It's a fairly straight-forward calculation.

Say we have a 1 Megaton nuclear explosion. That's 4.184×1015 J according to the Wikipedia (http://en.wikipedia.org/wiki/Megaton). Then using Einstein's equation, m = E / c2 = 4.184×1015 J / (299792458 m/s)2 = 0.04655 kg = 46.55 g.

That's per Megaton; scale linearly for other values.

Edit: punched in the wrong exponent. :p

Walker in Eternity
10-31-2008, 10:12 AM

mks57
10-31-2008, 10:30 AM
Using Pleonast's numbers, converting one gram of mass to energy is the rough equivalent of a Hiroshima-class nuclear weapon. Even in efficient nuclear weapons, the amount of mass converted to energy is very small.

Sapo
10-31-2008, 11:01 AM
Wait, from the wiki link on Little Boy, it says 600 mg (!) were converted for a 14 kiloton explosion from a 64Kg total mass. Which doesn't fit anyone's numbers so far. What am I missing?

beowulff
10-31-2008, 11:09 AM
Wait, from the wiki link on Little Boy, it says 600 mg (!) were converted for a 14 kiloton explosion from a 64Kg total mass. Which doesn't fit anyone's numbers so far. What am I missing?

600mg of Atoms may have been split, but no known process converts matter to energy with 100% efficiency. So, even a fission bomb that has 100% efficiency (splits 100% of it's fuel) will only convert a small fraction of the mass of those atoms to energy (something like 2%, as I recall). A matter-antimatter reaction would have a much higher mass to energy conversion factor.

mks57
10-31-2008, 11:20 AM
Wait, from the wiki link on Little Boy, it says 600 mg (!) were converted for a 14 kiloton explosion from a 64Kg total mass. Which doesn't fit anyone's numbers so far. What am I missing?

That's consistent with Pleonast's numbers, sources on the web, and my post that a gram of mass was roughly equivalent to a Hiroshima-class (20 kt) weapon.

Polycarp
10-31-2008, 11:22 AM
Two things:

1) Only a small fraction of the mass of the warhead is converted to energy. In fission, some quantity A of uranium is converted to several other quantities B+C+D+E of breakdown fission products, the actual weights of B, C, D, and E approaching but not quite equalling A and the (rather miniscule in mass) difference being what is converted to energy -- a LOT of energy, due to the E=mc^2 equivalence.

2) The real-world results of a theoretical thought experiment are never ideal. In theory, the oxidizing of carbon produces CO2 and a bunch of heat; in practice, it produces CO2, heat, and some leftover carbon as soot Likewise there is an efficiency rate that is not 100% in an atomic weapon -- some proportion of the uranium will not fission, but just be scattered to hellangone by the explosion resulting from the greater proportion that does.

The yield of an atomic weapon can be calculated after the fact, and estimated beforehand, by taking both these into account: the first gives you the ideal theoretical yield in ergs, which can be expressed in joules or in the equivalent of kilo- or megatons of TNT. The second gives you the proportion of the theoretical yield that will actually result from real-world conditions.

Pleonast
10-31-2008, 11:22 AM
A generic mass conversion reaction goes like this:
[matter with some potential energy] -> [matter with less potential energy] + [energy]

The mass of the initial matter is equal to the mass of the end matter plus the mass-equivalent of the energy.

Note that this holds for all reactions that produce energy--even chemical reactions. The change in the matter's mass is small enough that we can ignore it for most purposes.

Sapo
10-31-2008, 11:30 AM
ok. Then we have several steps each with its own efficiency:

Some Uranium remains as Uranium scattered to the winds
Some Uranium is turned into whatever it is that Uranium turns when if splits.
When Uranium splits, the leftovers don't add up to the original mass, and that is the amount that is turned to energy.

From the previous posts, about 25% of Uranium splits (75% blows away). Of that 25%, about 1 gram per kiloton is turned to energy while the rest turns into whatever it is.

Right?

mks57
10-31-2008, 11:55 AM
About 50 mg of matter per kiloton of yield is converted to energy.

Sapo
10-31-2008, 12:34 PM
About 50 mg of matter per kiloton of yield is converted to energy.

Roughly, what is Uranium decaying into in an atom bomb (Hiroshima)? I remember seeing the graphic in a book a long time ago, but wiki is not helping me.

mlees
10-31-2008, 12:46 PM

Roughly, what is Uranium decaying into in an atom bomb (Hiroshima)? I remember seeing the graphic in a book a long time ago, but wiki is not helping me.

http://en.wikipedia.org/wiki/Nuclear_fission

Not all fissionable isotopes can sustain a chain reaction. For example, 238U, the most abundant form of uranium, is fissionable but not fissile: it undergoes induced fission when impacted by an energetic neutron with over 1 MeV of kinetic energy. But too few of the neutrons produced by 238U fission are energetic enough to induce further fissions in 238U, so no chain reaction is possible with this isotope. Instead, bombarding 238U with slow neutrons causes it to absorb them (becoming 239U) and decay by beta emission to 239Np which then decays again by the same process to 239Pu; that process is used to manufacture 239Pu in breeder reactors, but does not contribute to a neutron chain reaction.

Fissionable, non-fissile isotopes can be used as fission energy source even without a chain reaction. Bombarding 238U with fast neutrons induces fissions, releasing energy as long as the external neutron source is present. That effect is used to augment the energy released by modern thermonuclear weapons, by jacketing the weapon with 238U to react with neutrons released by nuclear fusion at the center of the device.

I am not a physicist, but I hope that helps...

Sapo
10-31-2008, 12:49 PM
http://en.wikipedia.org/wiki/Nuclear_fission

I am not a physicist, but I hope that helps...

So Np and Pu? I was expecting something more radical (but mostly because I don't know what I am talking about).

mlees
10-31-2008, 12:57 PM
Actually, according to the U235 link, it appears to decay into Thorium-231 (which I thought was only a WoW fantasy metal), then into Protactinium-231. Those are more exotic sounding metals. :)

Sapo
10-31-2008, 01:22 PM
Neptunium sounds pretty bad ass as it is. What I meant was have the atom split into more even halves way high on the Periodic Table.

Precambrianmollusc
10-31-2008, 01:24 PM
So Np and Pu? I was expecting something more radical (but mostly because I don't know what I am talking about).

I think the confusion may arise from the term 'decay'
Decay is normally used to describe radioactive decay where by an unstable isotope tries to loose energy by giving off alpha beta gamma radiation and all permutations of.
Hence if you ask what Uranium is decaying into, you will get the appropriate uranium isotope decay chain, which is a sequence of small steps. Also what was described in the link provided was what happens when a uranium atom absorbs a slow moving (thermal) neutron and gets a bit more energy, so it can then in turn decay into Np and Pu.

If you ask what the fission products of uranium are the answer will be a bit more dramatic. The fission products are the two, almost equal mass atoms formed by the spitting of the initiating nucleus. If you add up the mass of the fission products and the mass of the neutrons , alpha beta particles and neutrons, add in the kinetic energy of those plus the mas equivalence of the energy released in gamma radiation and heat etc you should get back to the original uranium mass (plus or minus some picky points).
The exact split depends on the energy of the incident neutrons and the parent atoms, but generally speaking the fission products will lie in two groups , one with atomic mass in the 90-100 range and the other group in the 110-140 range. For any given big fission event you will have quite a mix of product elements. These will all be quite energetic and have a lot of excess neutrons themselves so are quite likely to decay into more stable isotopes (more being a relative term) so the exact answer to what is left after an fission event is more of a mix of isotopes rather than any two specific isotopes.

beowulff
10-31-2008, 01:27 PM

Roughly, what is Uranium decaying into in an atom bomb (Hiroshima)? I remember seeing the graphic in a book a long time ago, but wiki is not helping me.

U235 splits into all kinds of elements, including Beryllium, Xenon and Zirconium: http://en.wikipedia.org/wiki/Fission_product

Precambrianmollusc
10-31-2008, 01:30 PM
spooky delayed double post - deleted

Sapo
10-31-2008, 01:30 PM
The exact split depends on the energy of the incident neutrons and the parent atoms, but generally speaking the fission products will lie in two groups , one with atomic mass in the 90-100 range and the other group in the 110-140 range

Thanks, that's more in line with what I remembered.

What is the appropriate term if decay isn't it? Just split?
As in "in an atom bomb Uranium x into this and that"

beowulff
10-31-2008, 01:39 PM
Thanks, that's more in line with what I remembered.

What is the appropriate term if decay isn't it? Just split?
As in "in an atom bomb Uranium x into this and that"

Fissions.

Precambrianmollusc
10-31-2008, 01:47 PM
Thanks, that's more in line with what I remembered.

What is the appropriate term if decay isn't it? Just split?
As in "in an atom bomb Uranium x into this and that"

Well I used to be a physicist, but a terribly bad one, so don't trust me too much, but decay is generally accepted to be a small change in atomic mass through one of the common forms of decay (alpha, beta (in all its forms) and gamma)
When it is undergoing fission , which is to say splitting into two substantial parts that are not covered by the above decay mechanisms, I beleive it would be correct, if a little clumsy,to say the uranium atom fissioned into x and y, but most of the time people go with 'split'.

Chronos
10-31-2008, 01:49 PM
600mg of Atoms may have been split, but no known process converts matter to energy with 100% efficiency.Addendum: There are at least two known processes which can convert matter to energy with 100% efficiency, and one more which can achieve 50% efficiency. All three of them, of course, are far beyond our level of technology, and only the third one has ever been observed in nature.

mlees
10-31-2008, 01:51 PM
Well, my apologies for spreading mis-info with post #14. :)

For some reason, I failed to consider that the article(s) was discussing natural decay vice chain reaction fission.

Sapo
10-31-2008, 02:01 PM
Addendum: There are at least two known processes which can convert matter to energy with 100% efficiency, and one more which can achieve 50% efficiency. All three of them, of course, are far beyond our level of technology, and only the third one has ever been observed in nature.
Isn't there a rule against teasing? Tell us what they are.

Ludovic
10-31-2008, 02:05 PM
Isn't there a rule against teasing? Tell us what they are.I also would like to know what the other way of converting 100% to energy is. One of them is fairly obvious.

As to the 50% one, I'm guessing it's either gravity waves of black holes encircling each other or fusion/x-rays from matter falling into a black hole.

Chronos
10-31-2008, 04:47 PM
Sorry, I guess I should follow up on that. The first way is to feed the matter into a black hole, and wait for it to come back out as Hawking radiation. Hawking radiation is mostly massless particles (what most folks would consider "energy"), and has no preference for matter vs. antimatter (no matter what was fed in), so anything that wasn't already massless, you could annihilate to produce energy. Unfortunately, Hawking radiation depends on the size of the black hole, and for any black hole known to exist in the Universe, it's pathetically weak. You'd need a very small hole to make this practical.

The second method is proton decay: Most current theories predict that the proton will eventually, after a very long time, decay, producing a positron and a bunch of other stuff. Again, anything that isn't already massless could be annihilated with other reaction products to produce stuff that is. The problem here is that proton decay is phenomenally slow, and would ordinarily take many times the lifespan of the Universe. On the other hand, the same theories which predict it also predict that it should occur much, much faster (fast enough to be practical) in the presence of a magnetic monopole. So the problem is just one of finding or making a monopole.

The third process is to drop things into a black hole, and extract the gravitational potential energy as you do so. Some inevitably gets into the black hole, but you can in principle get an output of up to half of the mass you started dropping in. You can use any size black hole whatsoever for this one. This is basically the process which powers quasars, though there it's uncontrolled and a lot less efficient.

Der Trihs
11-01-2008, 03:09 AM
What about antimatter ? That's a 100% conversion process, and it's one we can perform even with our technology on a tiny scale.