I only have high school physics knowledge, but here is my understanding:

Fusion: 2 atoms come together to form a new atom. This process releases the energy keeping them apart, and is very energetic. Like the sun!

Fission: Something fast (like an electron) smashes into an atom breaking it apart. Somehow this also releases energy. Less energy than fusion, and it's like a nuclear reactor.

Now my understanding is that the lowest energy state is when everything is tightly stuck together (as per fusion), and it costs energy to break them apart..

So.. why do both fusion and fission release energy?

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    $\begingroup$ By the way: Fission does not only occur when smashing something fast into the nucleus. It can also happen when a single slow neutron merges with the nucleus. $\endgroup$ Commented Jan 30, 2019 at 10:28
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    $\begingroup$ I would actually really like to see an answer in terms of entropy. How does the entropy of the system change when fission or fusion occur and why does this depend on the size of the nucleus? $\endgroup$
    – HRSE
    Commented Jan 30, 2019 at 11:21
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    $\begingroup$ I've deleted some comments that were answering the question. Please keep in mind that comments are to be used for suggesting improvements and requesting clarification on their parent post (i.e. the question), not for answering. $\endgroup$
    – David Z
    Commented Jan 30, 2019 at 21:12

11 Answers 11


In general, both fusion and fission may either require or release energy.

Purely classical model

Nucleons are bound together with the strong (and some weak) nuclear force. The nuclear binding is very short range; this means that we can think of nucleons as "sticking" together due to this force. Additionally the protons repel due to their electric charge.

As geometry means that a nucleon has only a limited number of other nucleons it can "stick" to, the attractive force per nucleon is more or less fixed.

The repulsive electric field is long range. That means that as the nucleus grows, the repulsion grows, so that eventually that repulsion exceeds the attractive effect and one cannot grow the nucleus further. Hence a limited number of possible elements.

Effectively this means the attractive force per nucleon increases rapidly for a small number of nucleons, then tops out and starts to fall.

Equivalently, the binding energy per nucleon behaves similarly.

As @cuckoo noted, iron and nickel have the most tightly bound nuclei; iron-56 having lowest mass per nucleon and nickel-62 having most binding energy.

This image (from Wikipedia) illustrates the curve in the typically presented manner:

enter image description here

However, I prefer to think of binding energy as negative and therefore better visualize iron as being the lowest energy state:

enter image description here For lighter elements:

  • Fission requires energy
  • Fusion releases energy

For heavier elements, the opposite is true.

The reason we mainly observe the release energy cases is because:

  • It is easier to do
  • It is more "useful"
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    $\begingroup$ Apparently iron 56 is not quite the most tightly bound nucleus, with iron 58 and nickel 62 being a bit more tightly bound, but iron 56 is somehow easier to "reach" using the fusion/decay mechanisms common in stars. $\endgroup$
    – Ian
    Commented Jan 31, 2019 at 5:12
  • $\begingroup$ So this cold fusion process I've been hearing about in science fiction for years is already possible, it's just not an energy source? Interesting. $\endgroup$
    – Aubreal
    Commented Jan 31, 2019 at 17:29
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    $\begingroup$ @AlexandreAubrey: No, "cold fusion", if it existed, would have the same net effect as ordinary energy-releasing fusion of light nuclei. What makes it "cold" is that supposedly you could somehow get it to happen without going through an intermediate stage that costs a lot of energy to get to -- namely where two nuclei are close enough together that you've spent a lot of energy overcoming their electrostatic repulsion but not yet so close together that the strong force has begun attracting them. $\endgroup$ Commented Jan 31, 2019 at 21:11
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    $\begingroup$ ... This energy cost is not lost; it will be paid back either when fusion happens, or if fusion fails to happen, then when the nuclei are forced apart again at high speed by electrostatic forces. But getting that much energy concentrated in a pair of nuclei in the first place is the major technological problem in fusion power, and the only known way to achieve it in quantity is to heat up a plasma to insane temperatures. Cold fusion would -- in some way never quite explained -- mean a way to avoid this barrier. $\endgroup$ Commented Jan 31, 2019 at 21:13
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    $\begingroup$ It should be pointed out however, it still consumes more energy that it releases... Muons decay fairly rapidly so they cannot be stored, and they take a lot of energy to produce. $\endgroup$
    – Aaron
    Commented Feb 1, 2019 at 21:30

Fission releases energy, because a heavy nucleus (like Uranium-235) is like a cocked mouse trap: it took energy to squeeze all those protons and neutrons hard enough together to make them barely stick (by the nuclear force) against the natural tendency for all those protons to fly violently apart because of their electrostatic repulsion. When struck by an incoming neutron, it is like a mouse touching the trigger pedal of the trap: BANG goes the nucleus.

In the case of fusion, the mechanism is different: the nuclear force between protons and between neutrons is very powerfully attractive but only kicks in when the particles are so close to each other that they are "touching". That attraction is not quite enough to stick two protons together against their electrostatic repulsion but if you add two neutrons to the recipe, you get enough mutually attractive nuclear force to overcome electrostatics and the particles then violently suck themselves together with a very powerful BANG.

Other fusion reactions in which the (2 protons plus two neutrons) get pressed onto a heavier nucleus (like carbon, nitrogen, oxygen, fluorine, ...) release progressively less energy, because it takes more and more work input to overcome the repulsion effect as the nucleus accumulates more protons. By the time you get to iron, further fusion reactions actually consume energy instead of releasing it, because the electrostatic repulsion effect gets bigger and bigger- and you are in the province of fission instead.

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    $\begingroup$ I would add that in both cases, the resultant products are in sum lighter than its reactants - and that is the whole trick about it. $E=mc^2$ $\endgroup$
    – Stian
    Commented Jan 30, 2019 at 7:49
  • $\begingroup$ yes yes........ $\endgroup$ Commented Jan 30, 2019 at 8:38
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    $\begingroup$ @StianYttervik E=mc2 + hf Just because when there's mass hf can be ignored; doesn't mean it's not in the equation! $\endgroup$
    – UKMonkey
    Commented Jan 30, 2019 at 9:52
  • $\begingroup$ @UKMonkey Fair enough! But the issue I was trying to shed some light on, which deserves it, is that in fission and fusion, the universe loses a tiny little bit of mass. This mass loss is what enables us to extract useful work. However, if you try to fuse anything above iron, or fise.. fissionate... dang it split something below - you usually don't get that mass loss. (seeing as the iron core is so stable) $\endgroup$
    – Stian
    Commented Jan 30, 2019 at 9:57
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    $\begingroup$ Could you generalize this as "light elements generate energy when they undergo fusion and heavy elements generate energy when they undergo fission"? $\endgroup$
    – Philipp
    Commented Jan 30, 2019 at 14:44

Your assumption about the lowest energy state when everything is tightly stuck together is incorrect.

It only goes this way until you get iron nuclei - and this is why iron is the heaviest element created by fusion.

Creating nuclei heavier than iron consumes energy rather than releasing it. And this is why these elements are only created in supernova explosions and other highly energetic events where there is abundant energy input.

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    $\begingroup$ The nuclear binding energy diagram shows this fact very nicely $\endgroup$ Commented Jan 30, 2019 at 3:00
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    $\begingroup$ @WorldSEnder The diagram also shows that fusion of elements lighter than iron releases significantly more energy than fission of elements heavier than iron. $\endgroup$
    – Vaelus
    Commented Jan 30, 2019 at 8:54
  • $\begingroup$ @Vaelus at least counting per nucleon. On a nucleus basis, fission triumphs again. $\endgroup$
    – Chieron
    Commented Jan 30, 2019 at 12:16
  • $\begingroup$ @Chieron - No. Fusion far exceeds fission in power output. The only reason we use fission plants is that fusion is much, much harder to do. $\endgroup$ Commented Jan 30, 2019 at 17:48
  • $\begingroup$ @PaulSinclair look at that graph. Splitting one uranium nucleus releases more energy than fusing two lighter nuclei of any kind. Fusion is more effective, because it is mass/output per nucleon which really counts. $\endgroup$
    – Chieron
    Commented Jan 31, 2019 at 10:13

I wanted to add another answer to show an important plot - binding energy per nucleon versus the atomic number (number of nucleons [protons + neutrons]).

Binding energy per nucleon versus nucleon number

The binding energy is the amount of energy required to break apart a nucleus. If, after a change, the amount of binding energy decreases, we must have supplied energy to break apart a nucleus. If, on the other hand, it increases, it must have released energy.

We can see from the plot that there are two ways to increase the binding energy per nucleon: first, start from the right, beyond iron, and break apart nuclei, moving to the left and up the slope. This is fission. Second, starting from the left, fusing nuclei together, climbing up the slope towards the right. This is fusion. You can see that the rewards are particularly big if you go from hydrogen to helium.

So, I guess the question now is: why isn't the plot monotonic? Why isn't it either always increasing or always decreasing? I think the other answers have already shed light on that.

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    $\begingroup$ Just showing that graph makes this the best answer. +1 $\endgroup$
    – bogl
    Commented Feb 1, 2019 at 17:27

In a small nucleus there is a relatively large fraction of nucleons at the surface, which lowers the total binding energy. The fusion of 2 very small nuclei to one medium-sized nucleus releases energy, mainly because in the resulting bigger nucleus there are fewer nucleons at the surface than before. This is analogous to the surface tension effect by which two water drops can fuse and release some energy due to the lowered total surface area.

In a big nucleus there is much Coulomb repulsion due to the many protons. The fission of a very big nucleus into 2 medium-sized nuclei releases energy, mainly because the total Coulomb repulsion within the 2 resulting nuclei is smaller than before.

Therefore, medium-sized nuclei (~ 55 nucleons) have the biggest binding energy per nucleon.

The Bethe-Weizsäcker formula for the binding energy of a nucleus gives a more quantitative explanation for this.


I think it's worth specifically addressing the word 'both' in the question. If you say 'both release energy', you're implying some contradiction, but there actually isn't any such issue. You're comparing apples and oranges.

For a substance which has an endothermic fusion reaction, the fission of that substance will probably be a net exothermic process. This is applicable for elements heavier than Iron.

For elements with exothermic fusion reactions, the opposite is true. So eventually, for a given element, only one of the two processes (fission and fusion) will be a net exothermic process. You could also look at this Physics SE post: Are all nuclear fusion reactions exothermic and fission reactions endothermic?


Because attraction of the strong nuclear force has a short range, while electrostatic repulsion has a long range.

As a consequence, electrostatic repulsion will grow faster with number of nucleons than nuclear attraction (protons throughout the nucleus will repel each other, while only neighboring nucleons will attract). This causes less binding energy per nucleon as their number increases and at some point this reaches maximum and starts to decrease.

Iron just happens to be at the maximum of the binding energy per nucleon. Therefore, for heavier elements than iron, fission releases energy, while for lighter elements than iron, fusion releases energy.

This can't be completely circumvented by adding more neutrons, mainly because weak nuclear force makes them unstable, but it works to some degree, which is why heavier nuclei usually have higher neutron to proton ratio. But at some point it is no longer possible to add more nucleons without making the whole thing unstable, which is why very heavy elements are radioactive.


Here is a qualitative view.

The net energy is positive only for fusion of elements lighter than iron, i.e. energy is released by fusion. For elements heavier than iron, fusion consumes energy, i.e. the net energy is negative.

We can use fission today to release energy due to the fact, that some process in the past (e.g. in a supernova) put energy into the heavy nuclei.


There's energy involved in keeping atoms apart, but also energy involved in holding an atom together. When you smash an atom, this bonding energy is released.

In fact, one of the four fundamental forces is responsible for binding atoms together - the strong nuclear force.


Fusion works by banging together the same two elements and sticking them together to form a new heavier element. When you add the masses of the two original elements it is greater than the new element.

It is this mass difference that becomes energy. You can calculate the amount of energy from Einstein famous equation E=mc(squared). Here m is (2 x Mass of original element) - (Mass of new element) and c is the speed of light.

Example: Two Hydrogen atoms together form one Helium atom.

When the mass of the two original elements becomes heavier the difference between their masses and the new element gets smaller.

Fission works by splitting one element into two new lighter elements. When you add the masses of the two new elements it is less than the original element.

It is this mass difference that becomes energy. You can calculate the amount of energy from Einstein famous equation E=mc(squared). Here m is (Mass of original element) - (Mass of new elements) and c is the speed of light.

I wanted to mention that is technically much more complicated than what I say here. The short answer is still the same: Mass is converted into Energy.

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    $\begingroup$ "Fission works by splitting one element into two new lighter elements. When you add the masses of the two new elements it is greater than the original element. " No, you got this backwards. The sum of the new element masses must be less or you won't have an energy release! $\endgroup$
    – Bill N
    Commented Jan 30, 2019 at 21:51

Wanted to provide a quick answer, but apparently now is frowned upon to give quick answers in comments, so here it is:

Roughly speaking, nuclear fission is endothermic for nuclei where nuclear fusion would be exothermic, and viceversa. For nuclei smaller than Iron, fission is typically endothermic, while fusion is exothermic. For nuclei heavier than Iron, the situation reverses.


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