# Where does the energy from a nuclear bomb come from?

I'll break this down to two related questions:

With a fission bomb, Uranium or Plutonium atoms are split by a high energy neutron, thus releasing energy (and more neutrons). Where does the energy come from? Most books I've ever come across simply state $E=mc^2$ and leave it at that. Is it that matter (say a proton or neutron) is actually converted into energy? That would obviously affect the elements produced by the fission since there would be less matter at the end. It would also indicate exactly how energy could be released per atom given the fixed weight of a subatomic particle. I remember hearing once that the energy released is actually the binding energy that previously held the nucleus together.

With a fusion bomb, two hydrogen isotopes are pushed together to form helium and release energy - same question: where does this energy come from? Is matter actually converted or are we talking about something else here?

Sorry is this is rather basic - I haven't done physics since high school.

Energy of a fission nuclear bomb comes from the gravitational energy of the stars.

Protons and neutrons can coalesce into different kinds of bound states. We call these states atomic nuclei. The ones with the same number of protons are called isotopes, the ones with different number are nuclei of atoms of different kinds.

There are many possible different stable states (that is, stable nuclei), with different number of nucleons and different binding energies. However there are also some general tendencies for the specific binding energy per one nucleon (proton or neutron) in the nuclei. States of simple nuclei (like hidrogen or helium) have the lowest specific nucleon binding energy amongst all elements, but the higher is the atomic number, the higher the specific energy gets. However, for the very heavy nuclei the specific binding energy starts to drop again.

Here is a graph that sums it up:

It means that when nucleons are in the medium-atomic number nuclei, they have the highest possible binding energy. When they sit in very light elements (hidrogen) or very heavy ones (uranium), they have weaker binding. Thus, one can say that for the low "every-day" temperatures, the very heavy elements (like the very light ones) are quasistable in a sense.

Fission bomb effectively "lets" the very heavy atomic nuclei (plutonium, or uranium) to resettle to the atoms with lower number of nucleons, that is, with higher bound energies. The released binding energy difference makes the notorious effect. In terms of the graph cited above, it corresponds to nucleons moving from the right end closer to the peak.

Yet this is not the only way to let nucleons switch to the higher binding energy state than the initial one. We can "resettle" very light elements (like hydrogen) and let nucleons move to the peak from the left. That would be fusion.

Heavy nucleons emerge in the stars. Here the gravitational energy is high enough to let the nucleons "unite" into whatever nuclei they like. Stars usually are formed from the very light elements and the nucleons inside, again, tend to get to the states with lower energies, and form more "medium-number" nuclei. The energy difference powers stars and we see the light emission, high temperatures and all other fun effects.

However, sometimes the temperatures in the stars are so high, that nucleons form the very heavy nuclei from the medium-number nuclei. even though there is no immediate "energy" benefit.

These heavy elements then disseminate everywhere with the death of the star. This stored star energy can then be released in the fission bomb.

• I suppose that this is technically correct, but it is very misleading. That first sentence is going to give people some very strange ideas. – Colin K Mar 2 '11 at 21:19
• Care to elaborate? What strange ideas? – Dmitry Borzov Mar 3 '11 at 1:23
• The first sentence does sound odd at first glance, but it is absolutely correct and sums up the situation across cosmological time-scales very neatly. Gravitational collapse leads to star formation and the formation of heavy elements. The energy liberated via their decay is thus precisely the gravitational potential energy of the atoms of the primordial gas before it underwent gravitational collapse. But, yes, it can trip you up on a first read. – user346 Mar 3 '11 at 4:00
• The first sentence is great, except to prejudiced people – HDE Mar 3 '11 at 15:17
• Atomic nuclei > Fe are simply very efficient rechargeable batteries – Martin Beckett Mar 18 '11 at 18:39

Yes, matter of mass $m$ is directly converted to energy $E=mc^2$ which literally means that the weight of the remnants of the atomic bomb is smaller than the weight of the atomic bomb at the beginning. Fission reduces the mass by 0.1 percent or so; for fusion, you may get closer to 1 percent of mass difference.

In principle, you may release 100 percent of the $E=mc^2$ energy stored in the mass $m$ - for example by annihilation of matter with antimatter; or by creating a black hole and waiting until it evaporates into pure Hawking radiation - which is pure energy.

And yes, the differences in the energies - and the corresponding masses because $E=mc^2$ is true universally (in the rest frame), there is really just one independent quantity - may be attributed to the (changing) interaction energy between the nucleons (or quarks). Alternatively, you may imagine that the extra energy was stored in extra gluons and quark-antiquark pairs inside the nuclei. Those two descriptions are actually equivalent although this fact is not self-evident.

• So, in the case of splitting a plutonium atom, what gets lost? Presumably electrons are not touched. So, do we lose protons or neutrons? That would determine the elements produced. I assume the resultant atoms would be in plasma form given the temperatures involved. – dave Mar 2 '11 at 19:35
• And in the case of fusion: we have tritium + deuterium fusing to make helium + a neutron + energy. Doesn't that leave us with still 3 neutrons and 2 protons at the end, ie: the same at the start? – dave Mar 2 '11 at 19:37
• @dave: you are confusing particles and mass. No particles need to be destroyed (although they can be) for the mass of the reaction products to be lower. The binding energy you mentioned before is the source for the released energy, and it also represents the mass in the fission fuel which is lost. E=mc^2 is true in a very literal sense. – Colin K Mar 2 '11 at 19:42
• The mass of a nucleus is not equal to the sum of the masses of the protons and neutrons that make it up. So during a radioactive decay, the mass can change even if no particles are created or destroyed. Strange but true. Moreover, the same thing is true in principle for chemical reactions: the mass of an H$_2$O molecule is less than the masses of two H's and an O separately. But in chemical reactions, unlike nuclear reactions, that mass difference is too small to measure. – Ted Bunn Mar 2 '11 at 20:15
• @Dave- The Making of the Atomic Bomb , a book by Richard Rhodes is an excellent historical and popular technical account of physics in the first half of the 20th century and culminates with the Manhattan Project and the fission bomb. The sequel, Dark Suns is about the fusion bomb, Teller, Ulam and the Soviets. – Gordon Mar 2 '11 at 20:55

I'm not sure what you're looking for here, but I'll try a different (and simpler) approach than the other answers:

In a "traditional" chemical bomb, the energy comes from the electromagnetic force: you're breaking bonds between atoms and making more stable (lower-energy) bonds.

In a fission bomb, the energy comes from the strong nuclear force: you're breaking bonds between nucleons, producing final products that are in a lower-energy state*.

In a fusion bomb, the energy is also coming from the strong nuclear force: you're making bonds between nucleons, producing final products that are in a lower-energy state.

*To go up one level of complexity (and accuracy), Janne808 is correct: in fission bombs, the electromagnetic force is a significant contributor. In fact, in the absence of coulomb repulsion between protons, I can't think of any fission reactions that would be exothermic.

• In the last paragraph, do you mean "fusion"? Otherwise, could you please elaborate? – T. Verron Feb 9 '18 at 9:09

Coulomb repulsion between like charges is the biggest contributor to the amount of energy released.

• This is a case where a downvote should really include a comment. The answer is not obviously flawed, its not a troll, the grammar is fine, etc. So why the down vote? Is it simply incorrect? Whats the deal? – Colin K Mar 2 '11 at 21:18
• I don't think this claim is defensible: the nuclear binding energy clearly exceeds the Coulomb repulsion or the original nucleus would never have existed in the first case. – dmckee --- ex-moderator kitten Mar 2 '11 at 22:40
• Clearly, but the attractive Yukawa potential falls off exponentially and the repulsive Coulomb potential dominates by far at short distances. Seems to make sense, although the short scale issues and many-body physics isn't 100% clear yet. – Janne808 Mar 2 '11 at 23:54