# What matter in the original atom bomb is converted to energy?

When an atom bomb goes off some matter is converted to energy according to $E = m c^2$.

I'd like to know exactly what matter in the original atom bomb is converted to energy. Is it protons, neutrons, electrons? Is it a collection of atoms? What goes?

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The answers telling you it's binding energy are correct. It might be worth mentioning that the same is true for ordinary chemical reactions; there's no qualitative difference between chemical and nuclear reactions, it's just that nuclear binding energies are much larger than chemical binding energies. –  Matt Reece Dec 7 '12 at 12:54
@MattReece True, though it should be emphasized that chemical reactions are purely due to electromagnetism (and mostly electron exchange), while nuclear reactions involve the (electro-)weak interaction (it's called weak because of the low interaction distance, at short distances it's very dominant however) –  Tobias Kienzler Dec 10 '12 at 7:56

Take a nucleus of U-235 and determine its mass. Induce it to fission by firing a neutron at it. When it does so, collect all the pieces (except the extra neutron) and determine their total mass. You will find that all the pieces weigh just a hair less than original nucleus. The difference is the "binding energy", also previously known as the "packing fraction", and is the mass that was converted to energy.

Iron has the highest binding energy. Atoms heavier than iron will release energy when split. Atoms lighter than iron will release energy when fused together. Fuse two deuterium nuclei into a helium nucleus and you will find that the helium weighs just a little less than two deuteriums (deuteria?). It takes a lot of energy to get the two deuteriums close enough to fuse, but if you manage to do it (with a fission bomb as the trigger of a hydrogen bomb) you get quite a bit of energy out.

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I'd +1 your answer if you fixed that statement about iron having the lowest binding energy - the convention is to call the energy required to split the binding "binding energy", and that of iron is highest: en.wikipedia.org/wiki/… –  Tobias Kienzler Dec 7 '12 at 9:24
Curious. I'm used to the old packing fraction curve which is inverted from the binding energy and shows iron at the bottom. I've changed it. –  Jim Garrison Dec 7 '12 at 9:32
Sorry for the nitpicking - most sign conventions are a matter of history anyway... xkcd.com/567 –  Tobias Kienzler Dec 7 '12 at 9:38
actually, it should be 'binding energy per nucleon' –  Vineet Menon Dec 7 '12 at 14:56
One deuteron - two deuterons. en.wiktionary.org/wiki/deuteron –  Christoffer Hammarström Dec 7 '12 at 15:35

One does not simply vanish mass, you'd have to provide the respective antiparticles to obtain Annihilation, converting the entire mass into energy. What does happen in nuclear reactions is that the elements obtained afterwards have a total mass $m_\text{new}$ that is lower* than the one $m_\text{old}$ before. This is due to the different binding energies involved, causing the so-called mass-defect. This missing mass $\Delta m=m_\text{new}-m_\text{old}$ is what is converted into energy $E=\Delta m c^2$, which is partially set free as kinetic energy of the new matter (including $\alpha$ (He nuclei) and $\beta^\pm$ (Positrons, electrons) particles), and partly into the creation of $\gamma$ particles (Photons, i.e. radiation).

* Not always, it can be higher as well, the basic rule is that the binding energy per nucleus particle increases towards iron from both sides, i.e. in order to convert iron into hydrogen or oxygen, you'd need to invest energy. This is shown in the following image, with the notable exceptions of the rather important elements Oxygen, Carbon and Helium:

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I'm a bit hesitant about your first sentence. Obviously you can't vanish mass when that would violate conservation laws, e.g. you can't make an electron vanish because its charge must be conserved. But particle-antiparticle pairs are just one way to balance conserved properties. –  MSalters Dec 7 '12 at 10:30
@MSalters True, but in most cases it boils down to immediate steps involving the anti-particles. E.g. a neutron can annihiliate anti-hydrogen, but it decays into hydrogen first and then its proton and electron annihilate the anti-proton and positron, while a (anti-)neutrino remains. Electric charge is not the only conserved quantity though, there's also e.g. spin (which is why that neutrino is involved) –  Tobias Kienzler Dec 7 '12 at 10:58
@TobiasKienzler "a neutron can annihiliate anti-hydrogen, but it decays into hydrogen first and then its proton and electron annihilate the anti-proton and positron" I think you can have direct proton-antineutrom "annihilation" via strong interactions. –  mmc Dec 10 '12 at 16:03

If you take a proton or something and make it go really, really fast – give it lots of kinetic energy then relativity says it gets heavier. In the same way if you’ve got lots of potential energy if you put it into a really big heavy atom like uranium then it will also get heavier. When you split up that big atom you get less potential energy and that means that the resultants (all of the protons and all of the neutrons inside) are lighter than they were before because they have less potential energy.

So there are actually the same number of constituents, but the sum total weight is decreased.

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A relativistic particle does not become heavier, it's only the relation between momentum and velocity that shows non-linear effects which can be compensated by claiming "hey if I replace the $m$ in $p=mv$ by $\gamma m$ the relation is correct, so $\gamma m$ must be a mass". What relativity does say is that at high velocities you cannot double velocity by doubling momentum anymore. –  Tobias Kienzler Dec 7 '12 at 9:52
The amount of constituents changes, e.g. a neutron can decay into a proton and an electron, and $\gamma$ particles (high energy photons) can be created as well. The part about the potential energy (usually explained as nuclear binding energy) is correct though, as long as you stay above iron –  Tobias Kienzler Dec 7 '12 at 9:54

The rest mass of Nuclear Binding Energy is converted into external energy of destruction in fission-based nuclear bombs.

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Correct, but without any further elaboration or at least a Wikipedia link e.g. to en.wikipedia.org/wiki/Nuclear_binding_energy this is not a very helpful answer... –  Tobias Kienzler Dec 7 '12 at 9:58