Nuclear Fission vs Fusion: The Binding Energy Curve Explained

The Sun is about 4.26 million tonnes lighter than it was a second ago. Not burned, not blown off into space — gone, traded away for the sunlight currently warming the back of your hand, at the exchange rate $E = \Delta mc^2$. It has been settling that bill every second for four and a half billion years, and it will keep settling it for another five.

We run the trick in reverse down here and get numbers in the same league. Split the nuclei in one kilogram of uranium-235 and you release roughly what you would get from burning 3,400 tonnes of coal: an ingot you could hold in one hand against a coal train too long to see the end of. Fusion glues small nuclei together. Fission rips big ones apart. Those really are opposite operations, and both of them pay out — which is the thing that makes no sense until you have seen the one graph that explains it. So here are three things to pull on until it does: a curve you can drag a nucleus along to watch which way it wants to move, a beam balance fine enough to catch the missing mass in the act, and a reactor core whose control rods are yours to get wrong.

What Is Binding Energy Per Nucleon?

Weigh a helium-4 nucleus and it comes out light — lighter than the two protons and two neutrons that went into it. Doing that sum honestly means working with whole atoms, so the electrons cancel on both sides: two hydrogen atoms plus two neutrons come to 4.032980 u, while the finished helium-4 atom weighs 4.002602 u. The missing 0.030378 u is not a measurement error and it did not leak away. It left as energy at the moment the nucleus formed, and by $E = \Delta mc^2$ that sliver of mass is worth 28.297 MeV.

That 28.297 MeV is the binding energy: the energy you would have to pour back in to pull the nucleus apart into four loose nucleons again. Divide it among the four and you get the binding energy per nucleon, 7.074 MeV, and that per-nucleon figure is the number that actually decides what reactions are worth doing. It is a measure of how deep each nucleon is sitting in its own hole. Plot it against mass number $A$ for every nuclide we know and you get a single curve that quietly determines which nuclear reactions release energy and which ones cost it — the bookkeeping one level in from the electron energy levels of the Bohr model, where the same logic decides which photons an atom can emit.

Why Do Both Fission and Fusion Release Energy?

Here is where intuition puts its foot wrong. Splitting apart and squeezing together are opposites, so surely one of them has to cost what the other pays?

They would, if the curve were a straight line. It isn't. It's a hill — a steep climb through the lightest elements, a long flat summit around iron and nickel at roughly 8.79 MeV per nucleon, then a slow slide down to uranium and beyond. A nucleon in a tightly bound nucleus is deeper in its hole, which is to say lower in energy, than a nucleon in a loosely bound one. Any rearrangement that moves nucleons toward the summit therefore leaves them lower down than they started, and the difference has to go somewhere. It leaves as kinetic energy.

Uranium sits well to the right of the summit, so breaking it into two mid-weight fragments walks its nucleons uphill on the graph and downhill in energy. Deuterium sits far to the left, so joining two of them does the same thing from the other side. Same destination, opposite directions of travel, and neither one is the reverse of the other in any useful sense — they are both just falling toward iron.

The curve below is the real one, drawn from tabulated binding energies for 39 nuclides with straight-line interpolation between them. The amber marker is a nucleus you can pick up and move, either by dragging it along the curve, tapping anywhere on the chart, or using the A slider. Two arrows grow out of wherever you put it, one for each move that nucleus could make: split into two equal halves, or fuse with an identical twin. Green means the move releases energy. Red means it would cost you.

What Is the Iron Trap, and Why Does Fusion in Stars Stop There?

Drag slowly through the middle of the chart and you find a band where both arrows stay red at once: from about $A = 47$ to $A = 92$, splitting costs energy and fusing costs energy. Every nucleus in that band is already sitting at the bottom of the hole. There is nowhere left to fall.

That band is why a massive star dies. A star is a slow-motion argument between gravity pulling in and fusion pushing out, and fusion only wins while it has something profitable left to burn. Hydrogen becomes helium, helium becomes carbon and oxygen, and in the heaviest stars the chain grinds on through neon, oxygen and silicon in shells like an onion — each stage hotter, faster and shorter than the last. Then the core turns to iron and nickel, and the argument ends. Fusing iron does not release energy; it absorbs it. Within about a second the core, no longer held up by anything, collapses, and the star comes apart as a supernova. Every iron atom in your blood was manufactured at the exact point in the curve where the manufacturing had to stop.

Iron-56 gets the credit, though nickel-62 is fractionally more tightly bound — 8.795 MeV per nucleon against iron's 8.790. Iron wins the fame because it is enormously more abundant, being where the silicon-burning chain actually lands. Look closely at the top of the curve in the simulation and you'll see both marked: a green star on iron-56 and a faint tick on nickel-62, a whisker higher.

How Is Binding Energy Calculated? The Mass Defect Formula

The recipe is the one we just used on helium, written down properly. For a nucleus with $Z$ protons and $N$ neutrons, the mass defect is what the separate ingredients weigh minus what the finished nucleus weighs:

$\Delta m = \left[ Z\, m_p + N\, m_n \right] - M_{\text{nucleus}}$

In practice nobody uses bare nuclear masses, because mass tables list whole atoms. Swap in $Z$ hydrogen atoms for the $Z$ protons and the $Z$ electrons you have just smuggled in on the left cancel against the $Z$ already in the atom on the right:

$\Delta m = \left[ Z\, m(^1\text{H}) + N\, m_n \right] - M_{\text{atom}}$

Then convert:

$BE = \Delta m\, c^2$

Working in SI units here is a waste of a Sunday. Atomic mass units come with their own conversion built in — $1\ \text{u} \cdot c^2 = 931.494$ MeV — so multiply the mass defect in u by 931.494 and read the answer straight off in MeV. Divide by $A$ and you have the number that goes on the curve.

Where Does the Missing Mass Go? Weighing E = mc²

Most explanations wave the mass defect through in a sentence, and that is where the misconception breeds: mass is conserved, so the products must weigh the same, so the energy must be coming from somewhere else — chemical bonds, or the neutron, or "the strong force". None of that is right. The products genuinely weigh less. Weigh both sides of a nuclear reaction accurately enough and the scale does not balance, and the shortfall, multiplied by $c^2$, is the energy released. Exactly, not approximately.

Nobody notices because $c^2$ is such a brutal exchange rate that the mass involved is pitiful. Even a reaction violent enough to run a city throws away less than half of one percent of its own mass. So the interactive below is a beam balance with a magnification dial, and the dial is the whole point: it doesn't change the reaction, only how hard the instrument is squinting.

That last step is the one worth sitting with. Burning methane converts mass into energy by precisely the same mechanism a reactor does. Your gas hob is running $E = \Delta mc^2$; it is simply running it about thirty million times more grudgingly, because chemistry only rearranges the electrons on the outside of atoms while nuclear reactions rearrange the nucleons at the centre, where the binding energies are millions of times larger. There is no separate "chemical energy" and "nuclear energy" — only one conversion, at two wildly different exchange rates.

And the exchange rate itself is not a nuclear-physics rule. $E = mc^2$ falls straight out of special relativity, from the same short piece of algebra that produces the time dilation of fast-moving clocks. Einstein wrote it down in 1905 with no nuclear reaction in sight; it took nearly three decades before Cockcroft and Walton, in 1932, checked it directly against a nuclear reaction.

What Happens in Nuclear Fission?

Fission is a heavy nucleus falling apart into two mid-weight pieces. Left alone, uranium-235 does this spontaneously about as often as never — its half-life against spontaneous fission runs into the quintillions of years. But feed it a slow neutron and it becomes uranium-236 in an excited state, wobbling like a water droplet that has just been flicked, and the wobble is enough. Stretch a nucleus far enough out of round and the electrostatic repulsion between the two ends of it beats the short-range nuclear force holding them together, and the droplet necks and snaps:

$n + {}^{235}\text{U} \rightarrow {}^{141}\text{Ba} + {}^{92}\text{Kr} + 3n$

Weigh both sides and the products come up 0.186033 u short, which is 173.3 MeV of prompt energy — almost all of it kinetic, carried by two fragments flying apart at a few percent of the speed of light. That is the number the beam balance above is weighing, and it is worth being careful about, because the figure quoted for U-235 fission in most places is about 200 MeV. Both are right. The 173.3 MeV is what comes out immediately; barium-141 and krypton-92 are themselves unstable and go on beta- and gamma-decaying toward stability over seconds, hours and days afterwards, delivering another 20 to 30 MeV. Prompt energy heats a reactor now. Decay energy is why a shut-down reactor still needs cooling for years, and why the fuel that comes out is dangerous long after the chain reaction stops. If the decay side is what you're after, the half-life calculator and radioactive decay simulator covers how fast that clock runs.

Why Doesn't Lead Split on Its Own?

Because energetically downhill and actually likely are different questions. The curve says splitting lead-208 would release energy. Lead never does it. What stands in the way is the fission barrier: to get from one round nucleus to two separate fragments, the thing has to pass through elongated shapes that are less tightly bound than either the start or the end. It is a ball in a valley on the far side of a low ridge — the destination is lower, and it still isn't going anywhere without a shove.

Uranium-235's barrier is unusually low, and the energy an absorbed neutron brings with it happens to be just enough to clear it. That single accident of nuclear structure is why uranium-235 fuels reactors and lead makes good radiation shielding, despite both sitting on the same downhill slope of the same curve.

How Does a Fission Chain Reaction Work?

Each fission throws out two or three free neutrons — on average 2.43 for thermal fission of U-235. If exactly one of them, on average, goes on to split another nucleus, the reaction sustains itself at a constant rate forever. If slightly more than one does, every generation is bigger than the last and the whole thing runs away. If slightly fewer, it dies.

That average is called $k$, the multiplication factor, and everything about nuclear engineering lives in the third decimal place of it. $k < 1$ is subcritical, a chain going out. $k = 1$ is critical, which despite how the word sounds in English is the ordinary, boring, desirable state of every power reactor on Earth. $k > 1$ is supercritical. Most of the 2.43 neutrons never fission anything — they escape through the surface of the core, or get swallowed by something that absorbs neutrons without splitting. Control rods are that second thing, made deliberately: boron or cadmium on a motor, driven into the core to eat the surplus.

The core below is a cartoon, but the arithmetic it runs is real. Fuel nuclei sit on a jittered grid, four control rods hang from the top edge on one slider, and you fire the first neutron yourself. $k$ is not asserted anywhere in the code — it is read live off the fates of the last 60 neutrons, exactly as a real instrument would infer it.

What Are Delayed Neutrons, and Why Do They Make Reactors Controllable?

You could not hold $k$ steady, and the simulation was never going to let you. Every neutron in it is a prompt neutron, released within microseconds of the fission that made it. A core running on prompt neutrons alone has a generation time measured in millionths of a second, so a 1% error in $k$ compounds itself thousands of times over before a human has finished noticing. No operator, and no motor, could steer that.

Real reactors are not steering that. About 0.65% of the neutrons from U-235 fission do not come from the fission itself — they come out of certain fission fragments that beta-decay first, arriving seconds to a minute late. That fraction sounds negligible and is anything but. Hold a reactor just below prompt criticality, so the chain can only sustain itself with the stragglers included, and the stragglers set the pace. Those precursors take about 12.5 seconds on average to hand over their neutron, and even at 0.65% of the total that drags the effective generation time out from millionths of a second to roughly a tenth of a second — enough that the reactor's power now takes tens of seconds to go anywhere. Suddenly the thing responds on a timescale a person can watch, a rod can travel, and a control system can damp. Every power reactor in the world is deliberately operated inside that narrow window, relying on the slowest 0.65% of its own neutrons to be steerable at all.

Why Can't a Nuclear Reactor Explode Like a Nuclear Bomb?

Three separate reasons, and any one of them is sufficient.

Enrichment is the first. Natural uranium is about 0.7% U-235; power reactor fuel is enriched to a few percent; a weapon needs over 90%. Below roughly 20% the fuel physically cannot assemble a fast enough chain reaction no matter how you arrange it, which is why the enrichment percentage, rather than the quantity of uranium, is what non-proliferation treaties actually count.

Geometry is the second. A bomb has to slam its fuel into a supercritical configuration and hold it there for the microseconds a prompt-neutron chain needs to consume a meaningful fraction of it. That takes precision explosives and careful design. A reactor core is the opposite shape by intention — spread out, full of neutron-absorbing structure, and built so that heating the fuel and the coolant reduces reactivity, which makes an overheating core a self-correcting one.

Burnout is the third, and you already watched it happen. Pull the rods out and the population explodes, then dies with most of the fuel unburned, because every fission permanently retires a nucleus and a runaway chain eats the fuel it needs to keep running. Reactor accidents are real and serious — Chernobyl blew a building apart with a steam explosion and burning graphite, and it dispersed a great deal of radioactive material — but the failure mode is heat, pressure and fire, not a nuclear detonation.

What Happens in Nuclear Fusion?

Fusion is the same energy accounting from the light end of the curve. The easiest reaction to run on Earth joins the two heavy isotopes of hydrogen:

$^2\text{H} + {}^3\text{H} \rightarrow {}^4\text{He} + n$

Deuterium and tritium, in and out, with 17.6 MeV released — per event that is a tenth of what a fission gives you, but per kilogram of fuel it is about four times more, because these nuclei are so much lighter that a kilogram contains vastly more of them.

The problem is getting them to touch. Both nuclei are positively charged, so they repel each other harder the closer they get, and the nuclear force that will eventually bind them has a range of about a femtometre — it does nothing at all until the nuclei are practically in contact. That electrostatic wall is the Coulomb barrier, and climbing it needs kinetic energy corresponding to temperatures of tens of millions of kelvin: hot enough that atoms are stripped bare and matter is a plasma of nuclei and loose electrons.

Why Is Fusion Harder to Achieve Than Fission?

Stars manage the barrier by cheating twice. The Sun's core runs at roughly fifteen million kelvin, which is not actually hot enough to clear the barrier head-on — but quantum mechanics lets a proton occasionally tunnel through a wall it cannot climb, and the core contains such an absurd number of protons that even a fantastically improbable event happens constantly. The Sun is in no hurry, either. Its first step, two protons fusing into deuterium, is so slow that an average proton waits billions of years for its turn, and that sluggishness is exactly why the Sun has lasted long enough for anything to evolve to notice it.

Fission needs none of this. A room-temperature neutron carries no charge, feels no Coulomb barrier, and can wander straight into a U-235 nucleus — which is why fission reactors have been generating commercial electricity since the 1950s while fusion power plants still do not exist. On Earth the two serious approaches are magnetic and inertial confinement. ITER, under construction in southern France, is a tokamak that holds a D-T plasma in a magnetic bottle for long stretches and aims to release ten times the power injected to heat it. The National Ignition Facility takes the opposite route, crushing a peppercorn-sized fuel capsule with 192 lasers in a few billionths of a second, and in December 2022 it became the first experiment anywhere to get more energy out of the fuel than the lasers delivered to it.

Nuclear Fission vs Fusion: What's the Difference?

* The 173 MeV figure is the prompt energy for the specific reaction weighed on the balance above, calculated from its mass defect. The ≈200 MeV usually quoted adds the 20–30 MeV that the unstable fragments release later as they beta- and gamma-decay toward stability. The per-kilogram row uses the 200 MeV total, since a reactor eventually collects all of it.

Radioactive decay, for comparison, is a much smaller transaction: a typical alpha or beta decay gives up a few MeV, because a decaying nucleus is only adjusting itself slightly rather than splitting in half or merging with a neighbour. Same curve, same accounting, roughly a fortieth of the energy per event.

Worked Examples for Physics Exams