Fission, in layman's (or "initiate's") terms, is easy enough to understand; a large atom with a lot of protons and neutrons requires a large amount of force provided by the strong interaction to overcome electromagnetism and keep the nucleus together. The necessary energies were imparted to the atom during its formation, which we can replicate to some degree within a nuclear reactor; a combination of heat and fast-moving free protons/neutrons recaptured by the atoms of the fuel turns uranium into plutonium, even as the actual atom-splitting produces much lighter krypton and barium isotopes (or a host of other possibilities, some more likely than others, as with many other types of reactions). That atom-splitting, by the way, reduces the size of the resulting nuclei, and therefore reduces the amount of strong interaction force required to contain them; the leftover energy is released as a high-frequency gamma photon.

Fusion, however, is an odder beast. A fusion reaction requires a large amount of starting energy; enough to strip the electrons off of the deuterium/tritium nuclei and form plasma. That heat is also enough to accelerate the particles fast enough that when they collide, the initial electromagnetic repulsion is overcome and the strong interaction binds them together.

Now, that's a lot of energy required; the Sun's inner core where most of the fusion occurs is estimated to be about 15 million Kelvin. The reaction, however, doesn't seem to release any energy, based at least on this simplistic explanation. Now, obviously that's wrong; pretty much all the energy we have available to us right now is, however indirectly, a result of the Sun bathing us in the energy from nuclear fusion. The rest of it, such as from nuclear fission, is also star-based, via creation of superheavy elements in stellar nucleogenesis.

But where is the source of this energy on the subatomic level? Surely more strong force is required to contain a helium atom than a hydrogen one, because the helium now actually has two protons and must actually overcome the Coulomb force (as opposed to hydrogen, which only has one proton and so the Coulomb force between the proton and its zero-charge neutrons is lesser). So, what imbalance does fusion induce within the nucleus that is then resolved with the release of a photon?

  • $\begingroup$ You are right to observe that "more strong force is required to contain a helium atom" - and this is exactly the work done by this strong attractive force to bring the nucleons so close together that gives us the net surplus of energy in fusion! $\endgroup$
    – Slaviks
    Commented Feb 12, 2013 at 7:50

4 Answers 4


The basic idea

  1. Fuse two relatively small nuclei together; say they have masses $m_1$ and $m_2$.

  2. Get out a larger particle that has mass that's less than the sum of the original masses $M<m_1+m_2$.

  3. The missing mass $\Delta m = M-(m_1+m_2)$ is released as energy via $\Delta E=\Delta mc^2$.

Why does the fused particle have less mass than the sum of the original two? See ChrisWhite's post above.

Disclaimer: this is a vastly simplified explanation; for more details see here.


There are other interactions to consider besides the Coulomb interaction. A very nice model of the nucleus is the liquid drop model, in which one models it as a constant-density liquid with various interparticle interactions. The result is known as the semi-empirical mass formula, which I summarize here.

Let $Z$ be the number of protons, $N$ the number of neutrons, and $A = N + Z$ the atomic number. The "volume" of the drop scales as $A$, the "surface area" scales as $A^{2/3}$, and the "radius" and typical interparticle separation scale as $A^{1/3}$. Coulomb repulsion therefore scales as $Z^2/A^{1/3}$, but there are other energy considerations. In general, we expect the binding energy to be something like $$ E_\text{B} = a_\text{V} A - a_\text{S} A^{2/3} - a_\text{C} \frac{Z^2}{A^{1/3}} - a_\text{A} \frac{(N-Z)^2}{A} + \chi a_\text{P} \frac{1}{A^{1/2}}, $$ where the five parameters $a_\text{X}$ are (hopefully) constant across all nuclei so we can fit this formula to empirical data. The energy has contributions from volume (A), surface area (S), the Coulomb interaction (C), proton-neutron asymmetry (A). The last term tries to take into account spin pairing, and in my notation $$ \chi = \begin{cases} 1, & \text{$Z,N$ even} \\ 0, & \text{$A$ odd} \\ -1, & \text{$Z,N$ odd.} \end{cases} $$ The linked article contains more detail and further references if you are interested.

While this formula does have its limitations (after all, it neglects all sorts of messy quantum-ness), it is still rather illustrative and it works remarkably well. For simplicity, let's neglect the pairing term. Also, let's take $Z = N = A/2$, which is pretty good for the most stable light elements, which is what you are most interested in. One particular fit cited in that article gives $a_\text{V} = 15.8~\mathrm{MeV}$, $a_\text{S} = 18.3~\mathrm{MeV}$, and $a_\text{C} = 0.714~\mathrm{MeV}$. With these numbers we can plot the energy as a function of $A$. However, you have to be careful to plot the energy per nucleon, which I'll denote $\epsilon = E_\text{B}/A$. This is useful because the number of nucleons is conserved in all fission/fusion reactions. The result is below.

binding energy per nucleon

As you can see, fusing light elements makes for tighter bound products. This is because of those other terms. You can think of it as intermediate-mass elements striking a balance between long-range Coulomb repulsion and strong force nearest-neighbor attraction. Nucleons like to have neighbors on all sides but not to have too many neighbors off in the distance.

You can also look at the nuclear mass per nucleon, $m = (Z/A) m_\text{p} + (N/A) m_\text{n} - \epsilon$, plotted below.

enter image description here

If we allowed the proton-neutron ratio to vary, this would give a slightly different answer to which nucleus is most stable. Indeed, direct measurements tell us Ni-62 has the greatest binding energy per nucleon, while Fe-56 has the lowest mass per nucleon.

  • $\begingroup$ I would add that this is a successful fit to the experimental binding energy curve per nucleon upload.wikimedia.org/wikipedia/commons/thumb/5/53/…common_isotopes.svg/671px-Binding_energy_curve-_common_isotopes.svg.png $\endgroup$
    – anna v
    Commented Feb 12, 2013 at 7:42
  • $\begingroup$ Your answer explains pretty well the binding energies in the nucleus (your first graph), but I still don't see where the fusion-released energy comes from. In fact, from the graph it seems to me, that until 56-62 you need to actually add energy so that the binding energy grows (which is obviously wrong). Could you please correct my misunderstanding? Otherwise great answer and I really like your explanation and simplification of the equation. $\endgroup$
    – Vojtěch
    Commented Oct 5, 2015 at 21:58
  • 1
    $\begingroup$ @Vojtěch There's always an ambiguity in the sign of binding energy, and not just in nuclear physics. Here more positive means more tightly bound, which means you have to extract energy from the system to get there. Others use the opposite convention. $\endgroup$
    – user10851
    Commented Oct 5, 2015 at 22:41
  • $\begingroup$ It is still a bit hard to get the point. Is this understanding correct: binding energy E of nuclid A and created nuclid B: E(A) + E(A) = E(B) + E_released? $\endgroup$
    – Vojtěch
    Commented Oct 9, 2015 at 8:19

The nuclei of atoms are quantum systems, and as such, the nucleons have certain energy levels associated with them inside the nucleus. This is best understood with the nuclear shell model. If the repulsion due to the electromagnetic force can be overcome, the nucleons of two colliding atoms will attempt to configure themselves into the most stable low energy state that is possible.

The nuclear shell model is not perfect, this is one of the reasons that caused nuclear physicists to look for better models of nucleons themselves. Experimental evidence shows that the quark model is the most accurate model of the nucleons to date.

Based on the quark model, it has been determined that nearly 99% of the mass of nucleons like the proton is due to the kinetic energy of quarks and the associated gluon fields.

So when atoms fuse and energy is given up, its principle source is kinetic energy of quarks and energy from the gluon field for the associated atoms.

  • 1
    $\begingroup$ Doh - I was just bringing my mind around to Josh's strong force conceptual explanation: what's the deal, Hal? Thanks. $\endgroup$
    – Fattie
    Commented Nov 6, 2014 at 16:25

When fusion occurs in the sun, its due to quantum tunneling causing hydrogen atoms to bind, forming helium. Energy is released because two hydrogen atoms have more energy then one helium atom, and when they bind the excess is released into space. The energy itself comes from some of the mass deteriorating into photons.


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