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:
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: Introduction to Nuclear Weapon Physics and Design (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.
It’s a fairly straight-forward calculation.
Say we have a 1 Megaton nuclear explosion. That’s 4.184×10[sup]15[/sup] J according to the Wikipedia. Then using Einstein’s equation, m = E / c[sup]2[/sup] = 4.184×10[sup]15[/sup] J / (299792458 m/s)[sup]2[/sup] = 0.04655 kg = 46.55 g.
That’s per Megaton; scale linearly for other values.
Edit: punched in the wrong exponent. 
Sorry bad post.
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.
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.
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.
Two things:
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.
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.
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.
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?
About 50 mg of matter per kiloton of yield is converted to energy.
my bad, yes. I had seen that on wiki, already.
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.
So Np and Pu? I was expecting something more radical (but mostly because I don’t know what I am talking about).
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. 
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.
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.
U235 splits into all kinds of elements, including Beryllium, Xenon and Zirconium: Nuclear fission product - Wikipedia
spooky delayed double post - deleted