Nuclear Fission

At 2 a.m. on the morning of 28 October 1969, a control-rod operator at Tarapur Atomic Power Station near Mumbai lifted a bundle of cadmium rods a few centimetres out of a tank of boiling water the size of a cricket pitch. Inside that tank sat 132 tonnes of uranium dioxide fuel, stacked in Zircaloy tubes. The moment the rods cleared the fuel, something changed in the water. Neutrons that had been getting absorbed by the cadmium were now free to wander into uranium nuclei. Each absorption had a small probability of splitting a ^{235}\text{U} nucleus; each split released two-ish more neutrons; each of those had a small probability of splitting another ^{235}\text{U}. Within seconds, the number of fissions per second climbed by a factor of a thousand, then a million, then held steady at 4\times 10^{19} per second. The water started to boil. Steam shot up through the turbines. 160 megawatts of electrical power flowed out onto the Maharashtra grid.

That is what nuclear fission looks like, staged. A process that Enrico Fermi first demonstrated in a Chicago squash court in 1942, scaled up and turned into a power plant. To understand why the operator's action worked — and why a control rod has to move in centimetres rather than metres, and why the moderator has to be heavy water rather than ordinary water, and why uranium-235 is special but uranium-238 is not — you need three things: the binding energy curve, the cross-section of a nucleus for absorbing a neutron, and a single dimensionless number called k. This article builds all three.

The picture — a nucleus shaped like a waterdrop

Hold the uranium-235 nucleus in your head as a liquid drop. 92 protons, 143 neutrons, 236 when you add the captured neutron — all packed into a sphere about 7.4 \times 10^{-15} m across. Inside that sphere, two forces compete.

The nuclear force is attractive, short-ranged (it fades to nothing beyond about one nucleon diameter), and acts equally between protons and neutrons. It is what makes the drop want to stay spherical — surface tension, in the liquid analogy. Every nucleon on the surface of the drop has fewer neighbours than one in the interior, so surface nucleons cost energy, and the drop minimises energy by minimising surface area.

The electrostatic (Coulomb) force is repulsive, long-ranged, and acts only between protons. Every pair of the 92 protons pushes on every other pair, across the full diameter of the nucleus. Unlike the nuclear force, Coulomb repulsion does not fade with nucleon count — it grows as Z^2. For a nucleus as large as uranium, the Coulomb energy is very nearly equal to the surface-tension energy. The drop is on the verge of being unstable.

Now feed it one more neutron. The new nucleus is uranium-236 in an excited state — it has 6.5 MeV of extra energy, delivered as the neutron settled into its bound orbital. That energy has to go somewhere. In most nuclei, it would come off as a gamma ray and the nucleus would settle back to its ground state. But in ^{236}\text{U}, 6.5 MeV is almost exactly enough to deform the drop past the point of no return — past the fission barrier, which for uranium sits at about 6 MeV. The drop elongates into a peanut shape, then a dumbbell, then two separate drops joined by a thinning neck. The neck snaps. Coulomb repulsion, no longer held back by surface tension across the narrow waist, flings the two fragments apart at roughly 7% of the speed of light.

The liquid-drop picture was proposed by Bohr and Wheeler in 1939 — months after Lise Meitner and Otto Frisch interpreted Hahn and Strassmann's chemistry experiments as "splitting the uranium atom". It gets the broad physics right: fission is energetically favourable, but requires a small activation push to cross the barrier. Modern nuclear theory adds shell-model corrections (the fragments prefer certain magic-number configurations, making the mass distribution asymmetric), but for the energy balance, the liquid drop is all you need.

Where the 200 MeV comes from — read it off the binding curve

Every fission event of ^{235}\text{U} is different — the fragments come out with a distribution of masses, not fixed identities. But a representative reaction is:

Conservation of nucleon number: 235 + 1 = 141 + 92 + 3 \cdot 1 = 236. Conservation of charge: 92 + 0 = 56 + 36 + 0 = 92. Both sides balance.

Now compute the energy released using the binding-energy-per-nucleon curve. The relevant numbers, read off the standard curve:

Why: B is the energy that would be needed to rip the nucleus apart into free nucleons. A higher B/A means the nucleus is more tightly bound, i.e., further down the energy hill. The free neutrons have B = 0 — a free neutron is not bound to anything.

Step 1. Compute the total binding energy before fission.

Why: only ^{235}\text{U} is bound. The incoming neutron is free. Its 0.025 eV of thermal kinetic energy is negligible compared to the MeV scale of nuclear binding.

Step 2. Compute the total binding energy after fission.

Why: barium and krypton are both bound. The three emitted neutrons are free, so they contribute 0.

Step 3. The energy released is the increase in binding energy.

Why: if the fragments are more tightly bound than the parent, the system has fallen to a lower energy state. The energy difference has to come out somewhere — as kinetic energy of the fragments, neutrons, and gamma photons. The exact number varies from 180 to 210 MeV depending on which fragments are produced, so "200 MeV" is the textbook average.

Result: Each fission event liberates about 200 MeV.

Where does that 200 MeV end up? Roughly:

Everything except the antineutrinos stays inside the reactor fuel and moderator, gets converted to heat, and is what you eventually collect as steam-turbine power. The antineutrinos fly out through the reactor walls, through the Earth, through the Milky Way, and are lost. Kamiokande and SNOLAB (and, in India, the proposed India-based Neutrino Observatory at Theni) exist to catch a tiny fraction of them.

One fission = 200 MeV = 3.2 \times 10^{-11} J. One kilogram of ^{235}\text{U} contains N_A/0.235 \approx 2.56 \times 10^{24} atoms. If every one of them fissions, that is 8.2 \times 10^{13} J — the energy released by burning about 2,700 tonnes of coal. This is why the fuel loadings of reactors are so small compared to coal plants. Tarapur's 132-tonne fuel charge, replaced every few years, does the work of a coal mountain.

Why ^{235}\text{U} and not ^{238}\text{U} — the slow-neutron puzzle

Natural uranium is 99.27% ^{238}\text{U} and only 0.72% ^{235}\text{U}. Only the rare isotope fissions easily. To understand why, compare the fission barriers.

For ^{235}\text{U}, the barrier is \sim 5.8 MeV. Absorbing a thermal neutron delivers \sim 6.5 MeV of excitation. Excitation exceeds barrier by 0.7 MeV — fission is prompt and certain.

For ^{238}\text{U}, the barrier is \sim 7.0 MeV. Absorbing a thermal neutron delivers only \sim 4.8 MeV of excitation. Excitation is 2.2 MeV short — the drop sits in its excited state and relaxes by gamma emission, not fission.

Why the asymmetry: ^{235}\text{U} has an odd number of neutrons (143). When it absorbs one, the new neutron pairs up with the lone unpaired neutron, releasing the nucleon pairing energy of \sim 1.2 MeV — an extra energy bonus that is exactly what pushes the excitation over the barrier. ^{238}\text{U} already has paired neutrons (146), so the incoming neutron finds no partner and no pairing bonus. Nuclear pairing — the same physics that stabilises even-even nuclei like ^{4}\text{He} and ^{16}\text{O} — is ultimately why ^{235}\text{U} is the fuel of the nuclear age.

^{238}\text{U} can still fission if you hit it hard — with a "fast" neutron of a few MeV of kinetic energy. That kinetic energy, added to the excitation, does clear the barrier. This is why plutonium-fuelled fast breeder reactors (India's prototype FBR at Kalpakkam, Tamil Nadu) can burn ^{238}\text{U} indirectly: fast neutrons convert ^{238}\text{U} into fissile ^{239}\text{Pu}, which then fissions readily.

The bottom line: thermal (slow) neutrons fission ^{235}\text{U} and ^{239}\text{Pu}. Fast neutrons fission ^{238}\text{U} and ^{232}\text{Th} — but the cross-section is a hundred times smaller, so you need a dense neutron flux to make it work.

Chain reaction — the factor k

One fission creates on average \nu = 2.43 prompt neutrons (for thermal ^{235}\text{U} fission). Call the current generation n fissions per second. Most neutrons do not cause another fission — some escape the fuel, some get absorbed by ^{238}\text{U} without fissioning, some get swallowed by control rods or structural materials. Let k be the average number of next-generation fissions produced by each current fission.

where P_{\text{fission}} is the probability that a given prompt neutron survives long enough to cause another fission. Then the number of fissions per second grows as:

After g generations, starting from n_0:

Why: each generation multiplies the population by k. This is the same algebra as compound interest, or bacterial growth, or any other process where the rate is proportional to the amount.

Since neutron generations in a thermal reactor take about \tau \approx 0.1 millisecond on average — time to thermalise in the moderator, diffuse to a fuel nucleus, and get absorbed — the time evolution is:

Three regimes:

• Subcritical (k < 1, \ln k < 0): n decays exponentially. The reaction dies.
• Critical (k = 1): n stays constant. Steady power.
• Supercritical (k > 1): n grows exponentially with time constant \tau / \ln k.

A reactor is not trying to grow its power — it is trying to hold it steady. The operator's job, when starting up Tarapur that October morning, was to inch k from 0.99 to 1.000001 — just barely supercritical, so the power rose gently over seconds to the target level, then back to exactly 1 to hold it there. If the control rods slip and k reaches 1.01, the power doubles every \tau / \ln(1.01) \approx 10 ms. If k reaches 1.1, it doubles every 1 ms. No control system can catch that. The reactor would melt in seconds.

This is the margin that nuclear engineers live in. Why it has been practical at all is the subject of the next section.

Delayed neutrons — the grace that makes reactors controllable

Not all fission neutrons are prompt. A small fraction, called \beta \approx 0.0065 for ^{235}\text{U}, are delayed — they are emitted not at the moment of scission but seconds later, from the beta decay of certain neutron-rich fragment nuclei (notably ^{87}\text{Br} → ^{87}\text{Kr}^* → ^{86}\text{Kr} + n, with a 56-second half-life).

This 0.65\% sounds small, but it completely transforms the control problem. The effective generation time is no longer 0.1 ms (prompt neutrons only) but roughly

Why: the weighted average is dominated by the delayed contribution. Even though only 0.65% of neutrons are delayed, their 10-second mean delay makes them set the pace. Provided k stays within 1 \pm \beta, a reactor is delayed-critical — the effective time constant for power changes is seconds, not milliseconds, and mechanical control rods can comfortably catch it.

If k > 1 + \beta, the reactor becomes prompt critical — it achieves criticality on prompt neutrons alone, the delayed contribution becomes irrelevant, and the time constant crashes back to 0.1 ms. This is the runaway regime. Every reactor safety system is designed around one rule: never cross the prompt-critical boundary. The Chernobyl accident in 1986 was, in essence, a sudden transition to prompt critical. The Indian PHWR design has multiple independent shutdown systems — a fast-dropping shutoff rod bank and an emergency liquid-poison injection (gadolinium nitrate into the moderator) — specifically to keep this from happening.

Critical mass — geometry enters the story

A lump of pure ^{235}\text{U} at room temperature is not automatically supercritical. Some of the neutrons produced by a fission will escape the lump before they get absorbed, because neutrons travel a mean-free-path of a few centimetres through uranium before being absorbed. If the lump is small, most of the path is near the surface, so most neutrons escape. If the lump is large, most of the path is deep inside, so most neutrons get absorbed in fuel.

Balance these. Call \Sigma_f the macroscopic fission cross-section (fissions per neutron per cm travelled) and \Sigma_a the total absorption cross-section (fission + capture). The neutron-production-to-absorption ratio in an infinite lump is

For pure ^{235}\text{U}, k_\infty \approx 2.1 — comfortably above 1. In a finite lump of radius R, some neutrons leak out. A leakage factor L(R) reduces the multiplication:

where B_g \sim \pi/R is a geometric buckling that measures how curved the neutron-density distribution is (tighter for smaller R) and B_m is a material buckling that depends on k_\infty and the diffusion properties of the fuel. Setting k_{\text{eff}} = 1 and solving for R gives the critical radius.

For a bare sphere of pure ^{235}\text{U} metal, the critical radius is about 8.5 cm — a grapefruit. The corresponding mass is about 52 kg. Surround the sphere with a neutron reflector (beryllium, tungsten, or natural uranium), and neutrons that would have escaped get bounced back. Critical mass drops to about 15 kg — a softball.

Why this matters: critical mass is the hard engineering number. For ^{239}\text{Pu}, the bare critical mass is only 10 kg — which is why plutonium, not uranium, became the fuel of choice for compact bomb designs. For power reactors, which operate with dilute fuel (natural uranium is only 0.72% ^{235}\text{U}), the "critical mass" concept generalises to a critical core volume containing tonnes of fuel. Tarapur's core is a cylinder 4 m across and 3 m tall, holding 132 tonnes of \text{UO}_2.

Anatomy of a reactor — moderator, control rods, coolant

A power reactor has to solve three problems at once:

1. Slow the neutrons down. Prompt fission neutrons come out at \sim 2 MeV — too fast to be efficiently absorbed by ^{235}\text{U} (the thermal capture cross-section is 600 barns; the fast cross-section is 1 barn). You need to thermalise them to 0.025 eV, a factor of 10^8 reduction in kinetic energy. This is the job of the moderator.
2. Hold k at exactly 1. If k drifts high, the power grows; if low, it dies. You need a knob. This is the job of the control rods.
3. Carry the heat out. Each fission deposits 180 MeV of heat in the fuel. A 500-MW reactor is releasing 500 million joules per second of heat in a few cubic metres of fuel. Unless you extract it continuously, the fuel melts in seconds. This is the job of the coolant.

Moderator. The physics is elastic scattering. A neutron collides with a light nucleus and transfers kinetic energy; after about 20 collisions with hydrogen, or 120 with carbon, the neutron has lost enough energy to be thermal. But hydrogen — abundant and cheap in ordinary water (H₂O) — has a problem: it also absorbs neutrons, forming deuterium. In natural uranium (0.72% ^{235}\text{U}), this absorption is enough to push k_\infty below 1. You cannot build a natural-uranium reactor with ordinary water.

Two solutions: either enrich the uranium to 3–5% ^{235}\text{U} (the Tarapur approach, following the American BWR design imported in 1969) and use ordinary water; or keep natural uranium but use heavy water (D₂O), where the deuterium nucleus absorbs 600 times less efficiently than hydrogen. The price of heavy water is the cost of separating it from ordinary water — about 1 part in 6400 naturally, taking industrial distillation columns to concentrate.

India went heavy-water. The reason was geopolitics: after the 1974 Pokhran test, the Nuclear Suppliers Group embargoed enriched uranium to India. Enrichment technology became proliferation-controlled. But natural uranium and heavy water — the latter producible domestically — were not. The Bhabha Atomic Research Centre and the Department of Atomic Energy built a series of PHWRs at Kalpakkam, Narora, Kakrapar, Kaiga, and Rawatbhata, all running on natural-uranium fuel that India could mine at Jaduguda (Jharkhand) and enrich to D₂O at Vadodara (Gujarat). Kudankulam, built later with Russian cooperation, returned to enriched uranium. These two lineages — heavy-water and light-water — are the spine of the Indian nuclear programme.

Control rods. Cadmium or boron carbide, both of which have enormous neutron-absorption cross-sections (2450 barns for cadmium, 3840 barns for ^{10}\text{B} at thermal energies). Inserting a rod deeper into the core increases \Sigma_a, decreases k, and shuts the reaction down. Withdrawing opens k up. For fine control, a few rods move continuously; for shutdown, a separate bank drops fully in within seconds. The Indian PHWRs also have a secondary shutdown system: emergency injection of gadolinium nitrate poison directly into the moderator tank, guaranteeing shutdown even if every rod fails to drop.

Coolant. In a PHWR, heavy water at 100 atm pressure flows through the fuel channels at ~300 °C, never boiling. It carries heat to a steam generator where ordinary-water steam is raised. In a BWR (Tarapur), the coolant is ordinary water that is allowed to boil inside the reactor core itself — simpler, but the steam contains trace tritium. In India's fast breeder at Kalpakkam, the coolant is liquid sodium — needed because water would moderate the fast neutrons the breeder relies on.

Worked examples

Common confusions

• "Fission is the opposite of fusion." They are more like complementary tools. Both release energy by moving nuclei toward the peak of the binding-energy curve — fission comes down from uranium, fusion comes up from hydrogen. Both are governed by the same B/A curve. Read about the fusion side at Nuclear Fusion.

• "Slow neutrons are 'weaker' than fast neutrons, so they shouldn't fission easier." Counter-intuitive but true. "Slower" here means lower kinetic energy, which means longer de Broglie wavelength, which means more time spent in the neighbourhood of the nucleus per encounter — and therefore a much higher probability of being absorbed. The thermal absorption cross-section of ^{235}\text{U} is 580 barns (1 barn = 10^{-28}\text{ m}^2); the fast cross-section is about 1 barn. Slow neutrons fission ^{235}\text{U} 600 times more effectively per encounter than fast ones do.

• "A nuclear reactor can explode like a bomb." It cannot. A bomb requires prompt-critical geometry with k well above 1 + \beta, achieved by driving subcritical pieces together faster than they can blow themselves apart. A power reactor is engineered — by fuel dilution, delayed-neutron physics, and negative temperature feedback — to never reach prompt critical. A reactor accident can release radioactive fission products (Three Mile Island, Fukushima), cause a steam explosion (Chernobyl), or melt the core (all three), but it cannot detonate like a bomb.

• "The fission fragments are just barium and krypton." No — they are a distribution. Fission of ^{235}\text{U} produces dozens of fragment pairs; the mass distribution is asymmetric, peaking near A = 95 (zirconium, molybdenum) and A = 137 (caesium, barium). The asymmetry is a consequence of nuclear shell structure. Symmetric fission (two A = 118 fragments) is rare. The fragments are all neutron-rich and radioactive — they beta-decay back toward stability, which is where most of the long-term radiation from spent fuel comes from (^{137}\text{Cs}, ^{90}\text{Sr}, ^{131}\text{I}).

• "Neutrons from one fission fall on the next uranium nucleus right away." No — they wander. A prompt 2-MeV neutron has to travel through the moderator, bounce around 20–30 times losing energy, random-walk through the lattice, and eventually either get absorbed in fuel, get captured by something else (including control rods, ^{238}\text{U}, or coolant), or leak out. The whole process takes about 0.1 ms. The randomness is why k is defined as an average — a single neutron either causes a next fission or does not, but the average over many neutrons gives a well-defined multiplication factor.

• "The chain reaction is like dominos." The imagery is useful but misleading. Dominos propagate in a fixed pattern with a definite geometry. Fission neutrons propagate stochastically; whether a chain reaction sustains depends on bulk averages (k) and geometry (critical mass), not on any one "next" domino. A better analogy is a forest fire: whether it spreads depends on the density and moisture of the trees, not on the fate of a single leaf.

Where this leads next

• Nuclear Fusion — the other side of the binding-energy curve, where light nuclei combine to release energy. The sun's power source and the dream of tokamak-based electricity.
• Mass Defect and Binding Energy — the curve that underpins everything in this article, derived from Einstein's E=mc^2 and measured to parts per million in mass spectrometers.
• Radioactivity — Types of Decay — alpha, beta and gamma decay, the modes by which fission fragments relax to stability and the source of spent-fuel radiotoxicity.
• Radioactive Decay Law and Half-Life — the exponential mathematics that governs fission-product inventory, waste storage time, and the ratio of spent-fuel cooling ponds to dry-cask storage.
• Elastic Collisions — the neutron-moderator slowdown physics we invoked is a direct application of the elastic-collision energy-transfer formula.