Fission and Fusion: The Numbers
A single gram of uranium-235, fissioned completely, releases about
8\times 10^{10}\ \text{J} — the same energy as burning roughly
2.5 tonnes of coal. A thimbleful of nuclear fuel out-punches a truckload of the
chemical kind by a factor of a few million. And over your head, every second, the Sun fuses some
6\times 10^{11}\ \text{kg} of hydrogen into helium and pours the leftover
binding energy out as sunlight. Fission and fusion look like opposites — one splits heavy nuclei
apart, the other welds light ones together — yet both hand back enormous energy.
You already met the reason qualitatively in
the binding-energy-per-nucleon curve. This
page is about the quantities: exactly how much energy comes out, why, and what it
takes to get it. We will do the energy accounting three ways, split a real uranium nucleus on paper,
fuse deuterium and tritium, count neutrons through a chain reaction to see where "critical mass"
comes from, and finish at the temperatures and confinement a star — or a tokamak — needs to make
fusion go. One idea runs through all of it: energy is released by any move toward the peak of
the binding-energy curve.
The master key: climbing toward the peak
The binding energy B of a nucleus is how much energy you would have to
pour in to pull it completely apart into free protons and neutrons. Equivalently — this is
the whole trick — a bound nucleus weighs less than its separate parts by the
mass defect \Delta m = B/c^2. What matters for energy
release is not B itself but the binding energy per nucleon
B/A, and how it changes across the curve.
That curve rises steeply from hydrogen, peaks near B/A \approx 8.8\ \text{MeV}
around iron and nickel (A \approx 56–62), then
sags gently down to uranium at B/A \approx 7.6\ \text{MeV}. A nucleus near
the peak is the most tightly bound — the most stable. So there are two ways to move
uphill toward that peak and release the difference:
- Fusion — join light nuclei (left of the peak) into a heavier, more
tightly bound one;
- Fission — split a heavy nucleus (right of the peak) into two
medium-mass, more tightly bound fragments.
Both end up with products whose nucleons are more tightly bound than the reactants'. The
energy released is simply the improvement in total binding energy, and by Einstein's relation that is
the same number as the mass that "went missing":
-
Binding-energy bookkeeping. The energy released is the gain in total binding
energy of the system,
Q = B(\text{products}) - B(\text{reactants}).
-
Mass–energy bookkeeping. Identically, it equals the mass defect times
c^2,
Q = \Delta m\,c^2 = \big[\textstyle\sum m_{\text{reactants}} - \sum m_{\text{products}}\big]\,c^2,
and in practical units Q\,[\text{MeV}] = \Delta m\,[\text{u}] \times 931.494.
-
The toward-the-peak rule. Q > 0 (energy is
released) whenever the products sit higher on the B/A curve
than the reactants — i.e. closer to the iron peak. Light nuclei fuse, heavy nuclei fission; both
climb toward iron.
The unit conversion is worth memorising: 1\ \text{u} of mass defect is
worth 931.494\ \text{MeV} of energy. Every number on this page comes out of
that one line.
Two worked examples
Example 1 — splitting uranium-235. A slow neutron strikes a
{}^{235}\text{U} nucleus; the compound nucleus wobbles and snaps in two. One
typical channel is
{}^{235}\text{U} + n \;\longrightarrow\; {}^{141}\text{Ba} + {}^{92}\text{Kr} + 3n.
Add up the masses (in atomic mass units \text{u}):
- reactants: 235.043930 + 1.008665 = 236.052595\ \text{u};
- products: 140.914411 + 91.926156 + 3(1.008665) = 235.866562\ \text{u}.
The mass defect and its energy are
\Delta m = 236.052595 - 235.866562 = 0.186033\ \text{u},
Q = 0.186033 \times 931.494 \approx 173\ \text{MeV}.
That 173\ \text{MeV} is the prompt energy — mostly the kinetic
energy of the two fragments flying apart. The fragments are neutron-rich and radioactive, and as they
beta-decay toward stability they release the rest, so the total climbs to the famous
\approx 200\ \text{MeV} per fission. Spread over
236 nucleons that is only
200/236 \approx 0.85\ \text{MeV} per nucleon — exactly the little downhill
step from uranium's B/A to the fragments' higher one.
Example 2 — deuterium–tritium fusion. The reaction chosen for the first fusion
reactors, because it ignites at the lowest temperature:
{}^{2}\text{H} + {}^{3}\text{H} \;\longrightarrow\; {}^{4}\text{He} + n.
Masses again:
- reactants: 2.014102 + 3.016049 = 5.030151\ \text{u};
- products: 4.002602 + 1.008665 = 5.011267\ \text{u}.
\Delta m = 0.018884\ \text{u}, \qquad Q = 0.018884 \times 931.494 \approx 17.6\ \text{MeV}.
Only 17.6\ \text{MeV} per event — one-eleventh of a fission. But look
per nucleon: 17.6/5 \approx 3.5\ \text{MeV} per nucleon, more than
four times fission's 0.85. Because energy per kilogram of
fuel scales with energy per nucleon, D–T fusion delivers far more punch per kilogram than fission
does. (The 14.1\ \text{MeV} neutron carries most of that energy away and is
what a reactor's blanket must catch.)
Per kilogram, and versus chemistry. One kilogram of
{}^{235}\text{U} contains about
2.56\times10^{24} nuclei, so fissioning it all yields
(2.56\times10^{24})(200\ \text{MeV})(1.602\times10^{-13}\ \tfrac{\text{J}}{\text{MeV}}) \approx 8.2\times10^{13}\ \text{J/kg}.
The same sum for D–T fusion gives about 3.4\times10^{14}\ \text{J/kg} —
four times more. Compare burning coal at roughly
3\times10^{7}\ \text{J/kg}: nuclear fuel beats chemical fuel by a factor of
a few million. That is the whole reason we bother with the enormous difficulty of
fission and fusion — the energy density is in a different universe. (This is also why the hook is true:
one gram of uranium \approx 8\times10^{10}\ \text{J} \approx 2.5 tonnes of
coal.)
Chain reactions and critical mass
Fission is special because each event spits out spare neutrons (three, in our example) — and neutrons
trigger more fissions. Whether that snowballs, holds steady, or fizzles is captured by a single number,
the neutron multiplication factor k: the average number of
neutrons from one fission that go on to cause the next fission. After
g generations the neutron population is
N(g) = N_0\,k^{g}.
- k < 1 — subcritical. Each generation is smaller;
the reaction dies out.
- k = 1 — critical. Exactly self-sustaining; a steady
power output. This is where a reactor is held.
- k > 1 — supercritical. Each generation is bigger;
the population grows exponentially — a rising reactor, or a bomb.
Sweep the slider below through k=1 and watch the curve flip from a fading
decay to a runaway climb. The knife-edge at k=1 is dramatic: a fraction of
a percent either side is the difference between "off" and "exponential".
Why a minimum mass exists. Neutrons are only useful if they hit another nucleus
before escaping through the surface. Neutron production happens throughout the volume, so it
scales as r^3; neutron leakage happens at the surface, so it scales
as r^2. The ratio of loss to gain goes as
r^2/r^3 = 1/r — it shrinks as the lump grows. Below a
critical radius too many neutrons leak and k < 1; above
it, production wins and k \ge 1. The corresponding mass is the
critical mass.
Two more pieces make a controllable reactor. A moderator (water, graphite) slows the
fast fission neutrons to low "thermal" speeds, where {}^{235}\text{U} is
vastly more likely to grab them — this is what lets natural or lightly enriched uranium reach
k=1 at all. And crucially, a small fraction (\sim0.65\%
for {}^{235}\text{U}) of fission neutrons are delayed,
emitted seconds later by decaying fragments rather than promptly. Reactors are run so that they are
subcritical on prompt neutrons alone and reach k=1 only with the
delayed ones. That stretches the effective generation time from microseconds to seconds — slow enough
for control rods (and human operators) to keep k pinned at
1. Without delayed neutrons, a reactor would be uncontrollable.
Making fusion go: stars and laboratories
Fusion gives more energy per nucleon, so why is it so much harder than fission? Because before two
nuclei can fuse, their positive charges must be pushed together against a fierce electrostatic
repulsion — the Coulomb barrier. For two protons at the range of the strong force
(\sim10^{-15}\ \text{m}) that barrier is of order
\sim1\ \text{MeV}. To slam nuclei together that hard by sheer thermal
motion needs temperatures around 10^{7}–10^{8}\ \text{K},
where matter is a fully ionised plasma.
Remarkably, the Sun's core is "only" about 1.5\times10^{7}\ \text{K} — the
typical particle energy (\sim1\ \text{keV}) is a thousand times below
the barrier top. Fusion still happens because of two quantum rescues: nuclei in the high-energy
tail of the thermal distribution, and quantum tunnelling straight through
the barrier. The two effects pull in opposite directions with energy, and their product peaks at the
Gamow peak — a narrow window of energies where almost all stellar fusion actually
occurs.
In stars, hydrogen becomes helium two ways. In the Sun and cooler stars the
proton–proton chain dominates; in hotter, heavier stars the CNO cycle
uses carbon, nitrogen and oxygen as catalysts. Both have the same bottom line:
4\,{}^{1}\text{H} \;\longrightarrow\; {}^{4}\text{He} + 2e^{+} + 2\nu_e + \text{energy}, \qquad Q \approx 26.7\ \text{MeV}.
That 26.7\ \text{MeV} is again just a mass defect: four hydrogen atoms
outweigh one helium-4 by about 0.7\% of their mass, and that
0.7\% is what lights the sky.
On Earth, we cannot use gravity to confine a plasma the way a star does, so a reactor
must both heat the fuel and hold it together long enough and densely enough to get
net energy out. The condition is the Lawson criterion, usually stated as a
triple product of density n, confinement time
\tau and temperature T:
n\,\tau\,T \;\gtrsim\; 3\times10^{21}\ \text{keV}\cdot\text{s}\cdot\text{m}^{-3} \quad(\text{for D–T ignition}).
Two engineering routes chase that number. Magnetic confinement (the doughnut-shaped
tokamak) traps a thin, hot plasma in magnetic fields for seconds at a time — low
n, large \tau. Inertial confinement
crushes a tiny fuel pellet with lasers to enormous density for a few nanoseconds — huge
n, tiny \tau. Both are aiming at the same triple
product from opposite ends.
Because the road to iron gets harder the closer you get, and eventually stops paying. Fusion in stars
stalls at iron: fusing iron into anything heavier would go down the
B/A curve, so it absorbs energy rather than releasing it. A massive
star that builds an iron core has no fuel left to hold itself up — the core collapses, and the rebound
is a supernova. Elements heavier than iron aren't made by ordinary fusion at all;
they're forged in the neutron-rich fury of supernovae and neutron-star mergers, by rapid neutron
capture. So iron is a peak, not a destination: the universe is slowly drifting toward iron,
but the timescales are astronomical and most matter is still hydrogen. The gold in a ring and the
uranium in a reactor are both "uphill" of iron — leftovers from cosmic events violent enough to climb
the wrong way.
No — and treating it as a universal constant is a classic error. The "critical mass of
{}^{235}\text{U} is about 52\ \text{kg}" figure
you see quoted is for one specific case: a bare sphere of pure metal. Change the situation and the
number moves a lot:
- Shape. A sphere minimises surface area for its volume, so it leaks least and has
the smallest critical mass. Flatten it into a pancake and leakage soars — the same mass can be
perfectly subcritical. (This is a genuine safety principle.)
- Density. Compress the metal and the neutrons find nuclei sooner; critical mass
falls roughly as 1/\rho^{2}. Implosion weapons exploit exactly this.
- Enrichment. Natural uranium is 0.7\%\ {}^{235}\text{U}
and has no critical mass as bare metal at all — it simply can't sustain a fast chain
reaction.
- A reflector. Surround the lump with a neutron-reflecting "tamper" and you bounce
escaping neutrons back in, cutting the critical mass substantially.
"Critical mass" is a property of a whole configuration — geometry, density, purity, surroundings — not
a fixed label on an element.
It's a harmless slogan but a misleading picture. Nothing gets annihilated in a fission or fusion
reaction — you have exactly the same protons and neutrons before and after. What changes is how tightly
they're bound. A more tightly bound system sits in a deeper potential well, and a deeper well
means less total energy — and, since E = mc^2, less
mass. The "missing" mass is just the binding energy that was carried off by the fast fragments,
neutrons and photons. So the honest statement isn't "mass turned into energy"; it's that the products
are more tightly bound, and the released binding energy shows up as kinetic energy and radiation —
while the whole system's mass drops by exactly Q/c^2. Mass and energy were
never two different things being swapped; they're two labels for the same conserved quantity.