Fission, Fusion and E = mc²

Beginner

Step 1. At the centre of every atom sits a tiny nucleus, a tight cluster of protons and neutrons. The number of protons tells you the element: 92 means uranium, 1 means hydrogen. Chemical reactions only shuffle the electrons outside; changing a nucleus takes a nuclear reaction.

Step 2. Fission means splitting. In the animation a blue neutron flies into a gold uranium-235 nucleus. The nucleus swallows it, wobbles and breaks into two smaller nuclei, with a flash of energy and two or three new neutrons. The two pieces shoot off in opposite directions.

Step 3. Each new neutron can split another uranium nucleus: a chain reaction. Follow the dashed paths as they branch, 1, 2, 4… In a power station, control rods soak up the spare neutrons so that each split causes just one more.

Step 4. Fusion means joining. On the right, two heavy-hydrogen nuclei rush at each other. Both are positive, so they repel and slow down; only a collision at enormous speed, at millions of degrees, lets them merge into helium and a fast neutron. The Sun shines by fusion too: it joins ordinary hydrogen into helium.

Step 5. Energy is never made from nothing; it only changes form. So where does it come from? Weigh everything very precisely: after either reaction the pieces are slightly lighter. The missing mass has turned into energy, $E=m{c}^2$. Because the speed of light $c$ is huge, a speck of mass gives a lot: one kilogram of uranium-235, fully split, gives as much heat as about 2,700 tonnes of coal.

Check yourself: do the helium nucleus and the neutron weigh more or less than the two heavy-hydrogen nuclei they came from? Answer: less. The difference has left as energy.

Intermediate

What we measure: nuclei, masses and energies

In a fission no proton or neutron is lost, yet the pieces come out slightly lighter than what went in. A nucleus is described by two whole numbers: the proton number $Z$, which fixes the element, and the neutron number $N$. Their sum $A=Z+N$, the mass number, counts the nucleons. Uranium-235, ${}^{235}\mathrm{U}$, has $Z=92$ protons and $N=143$ neutrons. Isotopes share $Z$ but differ in $N$: deuterium, ${}^2\mathrm{H}$, and tritium, ${}^3\mathrm{H}$, are hydrogen with one and two neutrons.

Nuclear masses are given in atomic mass units, $1\mathrm{u}=1.6605\times 10^{-27}\mathrm{kg}$, one twelfth of a carbon-12 atom. Nuclear energies are given in mega-electronvolts, $1\mathrm{MeV}=1.602\times 10^{-13}\mathrm{J}$, a million electronvolts, against the few electronvolts of a chemical bond. Tabulated masses are those of neutral atoms; in these reactions the electrons cancel.

Mass defect: where the energy comes from

Where does a reaction's energy come from? Special relativity answers: a body at rest holds the rest energy
$E=m{c}^2$
so a system that gives energy away becomes lighter by that energy divided by $c^2$. Since $c^2=8.99\times 10^{16}J/kg$, a tiny loss of mass carries a huge energy.

The mass defect $\Delta m$ is the total mass of the reactants minus that of the products, and the energy released, the $Q$ value, is
$Q=\Delta m{c}^2,1\mathrm{u}\cdot c^2=931.5\mathrm{MeV}$
A negative $Q$ means that the reaction needs energy to happen.

Nothing is destroyed: nucleons and charge are conserved ($1+235=141+92+3$ and $92=56+36$ in the fission), but the nucleons end up more tightly held. The energy needed to pull a nucleus apart into free protons and neutrons is its binding energy; per nucleon it climbs from 1.1 MeV in deuterium to about 8.8 MeV near iron-56, then sinks to 7.6 MeV in uranium. Fission and fusion both move nucleons towards that peak, and the extra binding leaves as energy.

The working equations

Besides $Q=\Delta m{c}^2$, four relations do most of the work: the two reactions of the animation, the growth of a chain reaction, and the kinetic energy of a body of mass $m$ and momentum $p$:

• $n+{}^{235}\mathrm{U}\to {}^{141}\mathrm{Ba}+{}^{92}\mathrm{Kr}+3n+173\mathrm{MeV}$
• ${}^2\mathrm{H}+{}^3\mathrm{H}\to {}^4\mathrm{He}+n+17.6\mathrm{MeV}$
• $N_g=N_0{k}^g$
• $K=\frac{p^2}{2m}$

To find a reaction's energy, balance it (the same mass numbers and charges on both sides), subtract the summed masses and multiply by 931.5 MeV per u. To find who carries it, use momentum: reactants almost at rest give products with equal and opposite momenta, and since $K=p^2/(2m)$ the lighter product gets more.

Worked example: the reactions in the animation

Start with the fission. With the atomic masses 235.0439 u for uranium-235, 140.9144 u for barium-141, 91.9262 u for krypton-92 and 1.0087 u for the neutron:
$m_{\mathrm{before}}=235.0439+1.0087=236.0526\mathrm{u}$
$m_{\mathrm{after}}=140.9144+91.9262+3\cdot 1.0087=235.8667\mathrm{u}$
$\Delta m=236.0526-235.8667\approx 0.186\mathrm{u},Q\approx 0.186\cdot 931.5\approx 173\mathrm{MeV}$

These are the $\Delta m$ and $Q$ labels at the first split. The fissions counter then climbs to 3 and 7 (passing 5 briefly) and the energy readout to 520 and 1213 MeV: one $Q$ per fission. The 173 MeV is forty million times the 4 eV that burning a carbon atom gives.

About 160 MeV becomes kinetic energy of the two fragments. The other 13 MeV goes to the three neutrons (about 6 MeV) and prompt gamma rays (about 7 MeV). The uranium was practically at rest, so the fragments leave with equal and opposite momenta and the lighter krypton takes the larger share: $160\cdot 141/233\approx 97\mathrm{MeV}$ against 63 MeV for barium. Their speeds, $0.048c$ and $0.031c$, differ by the factor $141/92\approx 1.5$ that the two green $v$ arrows show. The fuel soon stops the fragments, and their energy becomes heat.

The fusion works the same way, with a mass defect ten times smaller, so more decimals are needed: 2.014102 u for deuterium, 3.016049 u for tritium, 4.002603 u for helium-4 and 1.008665 u for the neutron.
$\Delta m=5.030151-5.011268=0.018883\mathrm{u},Q=0.018883\cdot 931.5\approx 17.6\mathrm{MeV}$

Again the momenta are equal and opposite. The neutron, four times lighter than the helium nucleus, moves four times faster and takes four fifths of the energy, $\frac{4}{5}\cdot 17.6\approx 14.1\mathrm{MeV}$, against 3.5 MeV for the helium: the two energy labels beside the products.

Worked example: a year in a power station

A large reactor produces 3000 MW of heat, which its turbines turn into about 1000 MW of electricity. How much uranium-235 does it split in a year, and how much mass disappears? With $3.156\times 10^7\mathrm{s}$ in a year,
$E=Pt=3.0\times 10^9\mathrm{W}\cdot 3.156\times 10^7\mathrm{s}=9.47\times 10^{16}\mathrm{J}$
Counting about 200 MeV, or $3.20\times 10^{-11}\mathrm{J}$, per fission (the later decays of the fragments included), the number of fissions is
$N_{\mathrm{fis}}=\frac{9.47\times 10^{16}\mathrm{J}}{3.20\times 10^{-11}\mathrm{J}}=2.96\times 10^{27}$

That is $2.96\times 10^{27}/(6.02\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1})=4.91\times 10^3\mathrm{mol}$ of uranium-235, or $4.91\times 10^3\mathrm{mol}\cdot 0.235kg/mol\approx 1150\mathrm{kg}$. The mass that disappears is far smaller, $\Delta m=E/c^2\approx 1.05\mathrm{kg}$, about a thousandth of the uranium split.

A coal station giving the same heat, at 30 MJ per kilogram, would burn over three million tonnes, yet lose the same 1.05 kg of mass.

Reading the arrows and the counters

Every green arrow is a velocity. With almost no total momentum to start with, two products move in opposite directions with speeds in the inverse ratio of their masses: 141 to 92 for krypton and barium, 4 to 1 for the neutron and the helium nucleus. On the fusion side the incoming arrows are in the ratio 3 : 2, because the lighter deuterium nucleus must move faster to match the momentum of the tritium nucleus. Both shrink as the repulsion between the positive nuclei slows them.

The dashed lines are neutron paths. Each early fission frees three neutrons, two of which split new nuclei, so each fission causes $k=2$ new ones. Generation $g$ then holds $N_g=N_0{k}^g=2^g$ fissions, starting from $N_0=1$: 1, 2 and 4. A bomb is a chain with $k>1$ in dense metal, its generations about ten nanoseconds apart. A power reactor is held at $k=1$: control rods of boron or cadmium absorb spare neutrons, and a moderator such as water slows the rest, since slow neutrons split uranium-235 much more readily.

Common mistakes

• Nucleons are not destroyed. The mass number balances, $1+235=141+92+3$; the mass drops because the fragments are more tightly bound, by about 0.8 MeV per nucleon.
• $E=m{c}^2$ is not special to nuclei. Burning coal also loses mass, about $3\times 10^{-10}$ of it; fission loses about a thousandth and D–T fusion four thousandths.
• Fission and fusion do not cancel. Both move nucleons towards the binding peak near iron; splitting iron, or fusing beyond it, would cost energy.
• Count every particle. Leaving the three outgoing neutrons out of the fission equation gives $\Delta m=3.21\mathrm{u}$ and almost 3000 MeV, seventeen times too much.
• Watch the units. Multiply a mass in u by 931.5 MeV, or convert to kilograms first: $0.0189\cdot 1.66\times 10^{-27}\mathrm{kg}\cdot 8.99\times 10^{16}{\mathrm{m}}^2/{\mathrm{s}}^2=2.82\times 10^{-12}\mathrm{J}$, which is again 17.6 MeV.

Where you meet it

About a tenth of the world's electricity comes from fission reactors. The Sun fuses hydrogen into helium and turns about 4.3 million tonnes of mass into light every second; fusion experiments aim to do the same on Earth with D–T fuel. Fission fragments are radioactive, which links back to Radioactivity and Half-Life, and the rest energy $m{c}^2$ comes from Special Relativity. Ahead, Schrödinger's Equation and Tunnelling explains how nuclei slip through the Coulomb barrier, and Particles and Forces: the Standard Model shows what holds protons and neutrons together.

Advanced

Mass as stored energy

Legend

$E$	Energy; the axis points towards higher energy.	$\mathrm{J}$
$Z{m}_p{c}^2$	Rest energy of the $Z$ protons of the reaction, far apart and at rest.	$\mathrm{J}$
$N{m}_n{c}^2$	Rest energy of the $N$ neutrons of the reaction, far apart and at rest.	$\mathrm{J}$
$E_i$	Rest energy of the reactants: the nucleus ${}^{235}\mathrm{U}$ and a neutron.	$\mathrm{J}$
$E_f$	Rest energy of the products: ${}^{141}\mathrm{Ba}$, ${}^{92}\mathrm{Kr}$ and three neutrons.	$\mathrm{J}$
$B_i$	Binding energy of the reactants, $\sum B(\text{before})$: how far their level lies below the loose nucleons.	$\mathrm{J}$
$B_f$	Binding energy of the products, $\sum B(\text{after})$: how far their level lies below the loose nucleons.	$\mathrm{J}$
$Q$	Energy released by the reaction: $Q=E_i-E_f=B_f-B_i$.	$\mathrm{J}$
${}^{235}U$	Uranium-235 nucleus, with 92 protons and 143 neutrons.
${}^{141}Ba$	Barium-141 nucleus, a fission fragment.
${}^{92}Kr$	Krypton-92 nucleus, the other fragment.
$n$	Neutron (blue); $3n$ is three free neutrons.

Fission and fusion both release energy by rearranging nucleons, and how much they release can be read off from masses alone. Figure 1 shows why. It is the energy ledger of the fission in the animation, $n+{}^{235}\mathrm{U}\to {}^{141}\mathrm{Ba}+{}^{92}\mathrm{Kr}+3n$, drawn to scale.

Start at the top line. Take all $Z=92$ protons and $N=144$ neutrons that take part in the reaction (the 143 of the uranium plus the incoming one) and imagine them far apart and at rest. Their energy is then just their rest energy, $Z{m}_p{c}^2+N{m}_n{c}^2$, where $m_p$ and $m_n$ are the proton and neutron masses.

Now let the nucleons come together into nuclei. The strong nuclear force pulls them in, so the energy drops. The size of that drop is the binding energy $B$: the energy you would have to supply to pull the nucleons apart again. On the left, the reactants, the ${}^{235}\mathrm{U}$ nucleus and the neutron $n$, sit at the level $E_i$, a depth $B_i$ below the top line. On the right, the products, ${}^{141}\mathrm{Ba}$, ${}^{92}\mathrm{Kr}$ and three free neutrons, sit at the level $E_f$, a depth $B_f$ below it. The short step between the two lower levels is the energy handed to the surroundings:
$Q=E_i-E_f=B_f-B_i$
Notice how small it is: about a tenth of either depth. The nucleons are bound almost as tightly before the split as after it; fission only tightens them a little.

Why should this energy show up as a change of mass? Picture a sealed box with perfectly insulating walls, holding a uranium-235 nucleus and a slow neutron, standing on a scale. Let the fission happen inside. The fragments fly apart and their kinetic energy becomes heat in the walls, but nothing leaves the box, so its total energy is unchanged, and so is its mass. Now open a window and let the heat out: the box becomes lighter by exactly $Q/c^2$. The missing mass, the mass defect, has not been destroyed; it is the mass carried away by the released energy.

The general rule is that the mass of any composite body is its total internal energy divided by $c^2$: the rest energies of its parts plus their kinetic and potential energies. Inside a nucleus the strong force makes the potential energy large and negative, so a nucleus weighs less than its free nucleons. That shortfall, times $c^2$, is the binding energy, the depth of a level below the top line of Figure 1:
$B=(Z{m}_p+N{m}_n-M)c^2$
Here $M$ is the mass of the nucleus. Tables list the masses of neutral atoms, electrons included; with those, use the mass of a hydrogen atom $m_{\mathrm{H}}$ in place of $m_p$, so that the electron masses cancel.

Now the energy balance of a whole reaction. Rearrange the definition as $M{c}^2=Z{m}_p{c}^2+N{m}_n{c}^2-B$ for every nucleus, and add these up before and after the reaction. Fission, and the fusion of deuterium with tritium (D–T fusion, ${}^2\mathrm{H}+{}^3\mathrm{H}\to {}^4\mathrm{He}+n$), keep the total number of protons and the total number of neutrons separately. So the nucleon rest energies are the same on both sides and cancel; a free neutron simply counts as a nucleus with $B=0$. Only the binding energies remain:
$Q=\sum B(\text{after})-\sum B(\text{before})$
In the symbols of Figure 1, $B_i=\sum B(\text{before})$ and $B_f=\sum B(\text{after})$, so this is the same $Q=B_f-B_i$. A reaction releases energy when its products are more tightly bound than its reactants, and a table of masses is enough to say how much, whatever the details of the forces or of the way the nucleus breaks.

Reading the binding-energy curve

Legend

$A$	Mass number: the number of nucleons in the nucleus.
$B/A$	Binding energy per nucleon: the total binding energy $B$ divided by $A$, the average binding of one nucleon. The dashed line marks $8\mathrm{MeV}$.	$\mathrm{J}$
${}^{2}H$	Deuterium, with $B/A=1.112\mathrm{MeV}$.
${}^{3}H$	Tritium, with $B/A=2.827\mathrm{MeV}$.
${}^{4}He$	Helium-4, with $B/A=7.074\mathrm{MeV}$: the product of D–T fusion.
${}^{56}Fe$	Iron-56, with $B/A=8.790\mathrm{MeV}$, near the top of the curve.
${}^{92}Kr$	Krypton-92, with $B/A=8.513\mathrm{MeV}$: a fission fragment.
${}^{141}Ba$	Barium-141, with $B/A=8.326\mathrm{MeV}$: the other fragment.
${}^{235}U$	Uranium-235, with $B/A=7.591\mathrm{MeV}$: the fissioning nucleus.

The binding-energy curve tells you how tightly bound every nucleus is, and therefore which reactions can release energy at all. Figure 2 plots the binding energy per nucleon, $B/A$, against the mass number $A$, the number of nucleons, on a logarithmic axis so that light and heavy nuclei both fit. The dots are measured nuclei. The purple curve is the liquid-drop formula of the next section, drawn from $A=10$ (the expert graph of the animation shows the same plot, with the curve in gold).

Read it from left to right. It climbs steeply through the lightest nuclei: 1.112 MeV per nucleon for deuterium, ${}^2\mathrm{H}$, 2.827 MeV for tritium, ${}^3\mathrm{H}$, and 7.074 MeV for helium-4. It peaks at 8.790 MeV near iron-56, ${}^{56}\mathrm{Fe}$, then sinks slowly to 7.591 MeV for uranium-235. Any reaction that moves nucleons up the curve releases energy. On the steep left side that means joining light nuclei: fusion. On the gentle right side it means splitting heavy ones: fission.

The arrows in Figure 2 are the two reactions of the animation, whose markers climb the curve in the same way. The fusion arrows run from ${}^2\mathrm{H}$ and ${}^3\mathrm{H}$ up to ${}^4\mathrm{He}$. The fission arrows run from ${}^{235}\mathrm{U}$ to its two fragments, krypton-92 at 8.513 MeV and barium-141 at 8.326 MeV, both higher than uranium. Since $B/A$ times $A$ is the binding energy $B$ of a nucleus, the rule $Q=\sum B(\text{after})-\sum B(\text{before})$ gives both energies without a single mass, free neutrons counting zero:
$Q_{\mathrm{fus}}=4\cdot 7.074-2\cdot 1.112-3\cdot 2.827=17.59\mathrm{MeV}$
$Q_{\mathrm{fis}}=141\cdot 8.326+92\cdot 8.513-235\cdot 7.591=173.3\mathrm{MeV}$

Per nucleon the two processes are very different. D–T fusion releases about 3.5 MeV for each of its five nucleons, in one steep climb. Fission gains much less per nucleon. Of the 235 nucleons of uranium, 233 stay bound in the fragments, each about 0.81 MeV more tightly than before (the fragments' average $B/A$ of 8.400 MeV, minus 7.591 MeV). The other two leave as free neutrons and give up their 7.591 MeV of binding each. The balance, $233\cdot 0.809-2\cdot 7.591\approx 173\mathrm{MeV}$, is the same $Q$: fission's large total comes from the large number of nucleons, not from a large gain for each.

Iron sets the limit for both. Fusing nuclei beyond the peak, or splitting nuclei below it, moves nucleons down the slope, to less tight binding, so the reaction costs energy instead of releasing it.

Why the curve peaks: the liquid drop

The shape of the curve comes from a contest between two forces: the nuclear force, which binds each nucleon to its neighbours, and the electric repulsion between protons. A simple model that treats the nucleus as a drop of charged liquid captures both and gives the curve of Figure 2.

Start with the nuclear force. It is strong but short-ranged, acting only over about $10^{-15}\mathrm{m}$, and it saturates: each nucleon binds only its nearest neighbours, however many others there are. Two things follow. First, nucleons pack the same way in every nucleus, so all nuclei have about the same density; the volume grows in proportion to $A$, and the radius is $R=r_0{A}^{1/3}$ with $r_0\approx 1.2\times 10^{-15}\mathrm{m}$.

Second, the binding behaves like that of a liquid. Every inside nucleon contributes the same amount, so the binding grows with the volume, as $a_{V}A$. Nucleons at the surface have fewer neighbours, so a term proportional to the surface area, $a_S{A}^{2/3}$, is subtracted. In light nuclei a large share of the nucleons sit at the surface: that is why the curve rises so steeply on the left.

Next, the repulsion. A sphere of radius $R$ with charge $Ze$ spread evenly through it stores the electric energy $\frac{3}{5}{Z}^2{e}^2/(4\pi {\epsilon }_{0}R)$. Putting in $R=r_0{A}^{1/3}$ gives a term $a_C{Z}^2/A^{1/3}$, with the coefficient
$a_C=\frac{3}{5}\frac{e^2}{4\pi {\epsilon }_0{r}_0}\approx 0.72\mathrm{MeV}$

Last, the Pauli exclusion principle: no two neutrons, and no two protons, can occupy the same state, so each kind fills its own ladder of energy levels. A nucleus with more neutrons than protons must put its surplus neutrons on higher rungs, which costs energy. This asymmetry term vanishes when $N=Z$. Together the four terms give
$B=a_{V}A-a_S{A}^{2/3}-\frac{a_C{Z}^2}{A^{1/3}}-\frac{a_A(N-Z{)}^2}{A}$
with $a_V=15.75$, $a_S=17.8$, $a_C=0.711$ and $a_A=23.7\mathrm{MeV}$, fitted to measured masses; these are the values behind the curve of Figure 2. The fitted $a_C$ agrees well with the estimate above.

Now the peak. On its own, the volume term gives every nucleon the same binding and would make $B/A$ flat. The surface term pulls $B/A$ down most in light nuclei. The Coulomb term pulls it down most in heavy ones: with $Z$ roughly proportional to $A$, it grows about as $A^{5/3}$, faster than the volume term. In uranium-235 it costs 975 MeV out of a total binding of 1784 MeV. Between these two losses lies the maximum near iron. Two smaller nuclei hold the same 92 protons with less electric energy in total, because each proton is now repelled only by the protons of its own fragment; the surface energy rises a little, but the Coulomb saving is larger, and fission releases the difference.

The same contest explains why heavy nuclei carry extra neutrons. The Coulomb term favours fewer protons, the asymmetry term favours $N=Z$, and as nuclei grow the balance shifts towards neutrons: uranium-235 has 143 neutrons for 92 protons. The fragments inherit this surplus. Barium-141 has 85 neutrons, while the heaviest stable barium, barium-138, has only 82; with too many neutrons for their size, the fragments are radioactive.

The fission barrier

Legend

$\epsilon$	Elongation: how far the drop is stretched from a sphere into an elongated shape of the same volume.
$E$	Energy of the nucleus at that shape, measured from that of the sphere.	$\mathrm{J}$
$E_B$	Height of the fission barrier above the sphere: about 6 MeV for uranium.	$\mathrm{J}$
${}^{236}{U}^{\ast }$	Excitation energy of ${}^{236}{\mathrm{U}}^{\ast }$, formed by ${}^{235}\mathrm{U}$ and a slow neutron: 6.5 MeV, above the barrier.	$\mathrm{J}$
${}^{239}{U}^{\ast }$	Excitation energy of ${}^{239}{\mathrm{U}}^{\ast }$, formed by ${}^{238}\mathrm{U}$ and a slow neutron: 4.8 MeV, below the barrier.	$\mathrm{J}$

The curve says uranium would release energy by splitting, so why does a uranium nucleus not split by itself? Figure 3 answers this. It plots the energy $E$ of the nucleus, measured from that of the sphere, against its elongation $\epsilon$: a small number that says how far the drop has been stretched from a sphere into an elongated shape of the same volume. The shapes under the axis show the sphere, the stretched drop and the drop about to break.

Two terms of the liquid drop change as it stretches. The surface grows, so the surface energy $E_S$ rises, by $\frac{2}{5}{\epsilon }^2{E}_S$. The protons move farther apart on average, so the Coulomb energy $E_C$ falls, by $\frac{1}{5}{\epsilon }^2{E}_C$. The net change, $(\frac{2}{5}{E}_S-\frac{1}{5}{E}_C){\epsilon }^2$, is positive as long as $E_C<2E_S$: then the curve starts upward from the sphere, and a small stretch is pulled back. From the liquid drop, $E_C/E_S=(a_C/a_S)Z^2/A$, so the condition reads $Z^2/A<2a_S/a_C\approx 50$.

Uranium, with $Z^2/A\approx 36$, passes this test: it sits in the hollow on the left of Figure 3. To split, it must first be stretched over the hump, the fission barrier, whose height $E_B$ is about 6 MeV. Beyond the top the Coulomb energy wins, and the energy falls steadily as the nucleus forms a neck and breaks in two: the arrow running down on the right.

Where does the energy to cross the barrier come from? A captured neutron brings its binding energy into the nucleus as excitation energy, and the two blue arrows of Figure 3 show how much. Uranium-235 and a slow neutron form ${}^{236}{\mathrm{U}}^{\ast }$ (the star marks an excited nucleus) with 6.5 MeV of excitation, more than the barrier: its arrow clears the hump, and the nucleus splits. This is the brief quiver of the nucleus in the animation. Uranium-238 and a slow neutron form ${}^{239}{\mathrm{U}}^{\ast }$ with only 4.8 MeV: its arrow runs into the wall of the hump and stops.

The difference is pairing. Nucleons bind more strongly in pairs, and the neutron added to uranium-235 completes a pair, while the one added to uranium-238 is left on its own. To split uranium-238, the incoming neutron must bring the missing energy as kinetic energy, about 1 MeV.

That is why reactors run on the rare isotope, uranium-235, only 0.7 % of natural uranium: it splits with neutrons of any energy. Slow neutrons are captured by it far more readily than fast ones, so reactors slow their neutrons down.

Chain reactions and delayed neutrons

Each fission sets free two or three neutrons, three in the animation's reaction, and these can split further nuclei. The multiplication factor $k$ is the average number of neutrons from one fission that go on to cause another fission. After $g$ generations, $N_0$ neutrons have become $N_g=N_0{k}^g$. With $k<1$ the chain dies out, with $k=1$ it holds steady, and with $k>1$ it grows. A reactor runs at $k=1$.

How fast the chain grows depends on the time between generations. If every neutron were emitted at the moment of fission (a prompt neutron), a generation would last about $10^{-4}\mathrm{s}$, which makes $10^4$ generations per second. Then even $k=1.001$ would multiply the power by $1.001^{10^4}\approx 2\times 10^4$ in one second, far too fast for any mechanical control.

Control rests on delayed neutrons. About 0.65 % of the fission neutrons are not emitted at the split but seconds later, by fragments as they decay. Waiting for them stretches the average generation to about 0.1 s, so the same $k=1.001$ raises the power by only about 1 % per second, slow enough to move control rods. This works only while the chain needs the delayed neutrons. Above $k\approx 1.0065$ the prompt neutrons alone keep the chain growing and the fast multiplication returns, so a reactor must always stay below that value.

Fusion: the Coulomb barrier and tunnelling

Legend

$r$	Distance between the centres of the two nuclei.	$\mathrm{m}$
$U$	Potential energy of the pair: electric repulsion outside $r_n$, nuclear attraction inside.	$\mathrm{J}$
$U_C$	Height of the Coulomb hill at contact: about 0.4 MeV.	$\mathrm{J}$
$E$	Collision energy of the pair: 10 keV in the example of the text.	$\mathrm{J}$
$r_n$	Contact distance, where the nuclear force grips: about 3.5 fm.	$\mathrm{m}$
$r_t$	Turning point, where $U=E$: about 144 fm for 10 keV.	$\mathrm{m}$
$P$	Probability that a pair tunnels through the barrier from $r_t$ to $r_n$.

Fusion faces the opposite problem. The nuclei to be joined are both positive, so they repel each other, and the nuclear force grips only once they touch. Figure 4 shows the obstacle. It plots the potential energy $U$ of a deuterium–tritium pair against the distance $r$ between their centres. Far apart, $U$ is their electric repulsion, $e^2/(4\pi {\epsilon }_{0}r)$: the red curve, rising towards the left. At the contact distance $r_n$, about 3.5 fm, the nuclear force takes over and the pair drops into the deep nuclear well at the bottom left. The top of the red hill, at $r_n$, is the Coulomb barrier $U_C$. With $e^2/(4\pi {\epsilon }_0)=1.44\mathrm{MeV}\mathrm{fm}$,
$U_C=\frac{e^2}{4\pi {\epsilon }_0{r}_n}=\frac{1.44\mathrm{MeV}\mathrm{fm}}{3.5\mathrm{fm}}\approx 0.4\mathrm{MeV}$

Now follow a pair that comes in from the right with collision energy $E$, the purple arrow. Classically it climbs the hill until all its kinetic energy has become potential energy, where $U=E$; that is the turning point $r_t$, where it stops and turns back. In a plasma at $1.2\times 10^8\mathrm{K}$ the thermal energy is $k_{\mathrm{B}}T=10\mathrm{keV}$. A head-on pair with 10 keV turns back at $r_t=1.44\mathrm{MeV}\mathrm{fm}/0.010\mathrm{MeV}=144\mathrm{fm}$, about 40 times farther out than $r_n$. Classically it never fuses.

Quantum mechanics changes this. A nucleus also behaves as a wave, and a wave can leak through a region that a particle cannot classically enter. This is tunnelling: the dashed arrow of Figure 4 from $r_t$ to $r_n$, crossed with probability $P$. In the animation it is the step labelled tunnelling, where the slowed nuclei jump the last stretch. For D–T at collision energy $E$,
$P\approx e^{-\sqrt{E_G/E}}$
where the Gamow energy $E_G\approx 1.2\mathrm{MeV}$ is a constant fixed by the charges and masses of the pair. $P$ is tiny but rises steeply with energy: $2\times 10^{-5}$ at 10 keV, $2\times 10^{-3}$ at 30 keV.

Two effects therefore compete. In a plasma, fast nuclei are rare, and their number falls off quickly with energy; slow nuclei are common but hardly ever tunnel. The product of the two peaks well above the thermal energy, so in a plasma at 10 keV most fusions come from the few pairs with energies near 30 keV.

A fusion plasma gives out more energy than it loses only when it is dense enough, hot enough and held together long enough. The condition is on the product of the density $n$, the temperature $T$ and the confinement time $\tau$: $nT\tau$ must reach about $3\times 10^{21}\mathrm{keV}\cdot s/{\mathrm{m}}^3$, for instance $10^{20}$ ions per cubic metre at 10 keV, held for 3 s.

Worked problem: what drives the fragments apart

Problem: in the animation's fission the two fragments carry away about 160 MeV of kinetic energy, most of the 173 MeV; the neutrons and gamma rays take the rest. Which force supplies it, how far apart are the fragments when they separate, and how do they share the energy?

The force must act over distances much larger than the range of the nuclear force, so it is the electric repulsion between the 56 protons of barium and the 36 of krypton. Model the fragments as two charges that start at rest with their centres a distance $d$ apart. As they fly apart, their electric potential energy turns entirely into kinetic energy:
$K_{\mathrm{tot}}=\frac{Z_1{Z}_2{e}^2}{4\pi {\epsilon }_{0}d}=\frac{56\cdot 36\cdot 1.44\mathrm{MeV}\mathrm{fm}}{d}=\frac{2903\mathrm{MeV}\mathrm{fm}}{d}$

Step 1, the separation. If the fragments were touching spheres, of radii $r_0{A}^{1/3}=6.2$ and $5.4\mathrm{fm}$, then $d=11.7\mathrm{fm}$, and the formula would give about 250 MeV, more than the whole $Q$. The actual 160 MeV needs a larger starting distance, $d=2903/160=18.1\mathrm{fm}$. So when they separate, the fragments are not spheres but elongated shapes joined by a thin neck, their centres about 6 fm farther apart than touching spheres. Their deformation energy is not lost: it heats the fragments, which cool by boiling off neutrons.

Step 2, the sharing. The fissioning nucleus is almost at rest, so the fragments fly off with equal and opposite momenta. With $K=p^2/(2m)$ and the same $p$, kinetic energy is inversely proportional to mass, so the lighter fragment takes the larger share: $K_{\mathrm{Kr}}=160\cdot 141/233=97\mathrm{MeV}$ and $K_{\mathrm{Ba}}=63\mathrm{MeV}$. Both have the same momentum, $p=\sqrt{2mK}\approx 4070\mathrm{MeV}/c$.

Step 3, how quickly. At separation $r$ the kinetic energy equals the potential energy lost so far, $K(r)=K_{\mathrm{tot}}(1-d/r)$. The fragments therefore have 90 % of their energy by $r=10d\approx 180\mathrm{fm}$, still deep inside the electron cloud of the atom, which is about $10^5\mathrm{fm}$ across. The energy of fission is delivered by electric repulsion, almost entirely within a tiny region around the nucleus.

Limits of validity

The liquid drop is a fit, not a theory. It matches heavy nuclei to about 0.5 % (1786 MeV for uranium-235, against 1784 MeV measured), but it ignores shell structure: like electrons in atoms, nucleons fill shells, and nuclei with 2, 8, 20, 28, 50, 82 or 126 protons or neutrons, the magic numbers, are extra stable. So the drop gives ${}^4\mathrm{He}$ only 16 MeV of binding instead of the measured 28.3 MeV, predicts 162 MeV instead of 173 MeV for the animation's fission, and would split uranium into two equal halves, whereas shell effects make it split unequally, mostly into fragments near $A=95$ and $A=140$.

Two parts of the energy calculation are approximations too. Sharing energy by $K=p^2/(2m)$ is non-relativistic. In D–T fusion the neutron takes four fifths of the 17.59 MeV, that is 14.07 MeV, and moves at $0.17c$; the exact relativistic two-body calculation with the true masses gives 14.03 MeV, 0.3 % less.

The second approximation is in writing $Q$ as a difference of binding energies, which needs $Z$ and $N$ to be conserved separately. Where the weak force turns protons into neutrons, as in the Sun's net reaction $4{}^1\mathrm{H}\to {}^4\mathrm{He}+2e^{+}+2\nu$, the mass difference between a neutron and a hydrogen atom, 0.78 MeV, enters for each converted proton: the energy released is $28.3-2\cdot 0.78=26.7\mathrm{MeV}$, 0.7 % of the mass. What stays exact at these energies is the conservation of energy, momentum, charge and nucleon number.

Where this leads

Schrödinger's Equation and Tunnelling turns the factor $e^{-\sqrt{E_G/E}}$ into a calculation. Particles and Forces: the Standard Model shows where the nuclear force comes from, as a leftover of the strong interaction that binds quarks with gluons, and describes the weak force that lets the Sun turn protons into neutrons. Looking back, Special Relativity supplies $E=m{c}^2$, Radioactivity and Half-Life describes the decaying fragments that provide the delayed neutrons, Coulomb's Law gives the barrier, and Momentum and Collisions shares the energy among the products.