Nuclear fission

Inside a research reactor, a slow neutron drifts into a uranium-235 nucleus and something remarkable happens: the nucleus does not just absorb the neutron and settle down — it wobbles, elongates like a stretched droplet, and tears itself into two lighter nuclei, spraying out two or three fresh neutrons and a burst of energy. This is nuclear fission, and one gram of uranium releases roughly as much energy as burning three tonnes of coal. The reason lies in the binding-energy-per-nucleon curve you already know: medium-mass nuclei are more tightly bound, nucleon for nucleon, than very heavy ones. Splitting a heavy nucleus into medium fragments therefore releases the difference as energy. This single idea — mass 235 splitting into masses near 90 and 145 — explains a nuclear reactor's steady glow and a fission bomb's uncontrolled flash, separated only by how tightly the chain reaction is held.

The core insight

A heavy nucleus like uranium-235, on absorbing a slow neutron, splits into two medium-mass fragments plus a few neutrons, releasing about 200 MeV per event because the fragments are more tightly bound per nucleon than the parent.

The core idea

The fission equation. A representative reaction, already balanced in mass number ( $A$ ) and atomic number ( $Z$ ), is:

^{235}_{92}\text{U} + \,^1_0n \longrightarrow \,^{141}_{56}\text{Ba} + \,^{92}_{36}\text{Kr} + 3\,^1_0n + \text{energy}

where: the subscript is atomic number $Z$ (proton count) and the superscript is mass number $A$ (proton + neutron count); both totals must match on the two sides — here $A: 236 = 236$ and $Z: 92 = 92$ .

Why energy comes out. The binding-energy-per-nucleon curve peaks near mass number 56 and falls off slowly for heavier nuclei. Uranium-235 sits on that falling tail with a binding energy per nucleon of about 7.6 MeV; the fragments, with mass numbers around 90–145, sit closer to the peak, near 8.5 MeV per nucleon. Splitting the nucleus moves 235 nucleons to a more tightly bound configuration, and the released binding energy appears as kinetic energy of the fragments and neutrons — about 200 MeV per fission event in total.

Chain reaction. Each event frees 2–3 neutrons. If, on average, exactly one of those neutrons goes on to trigger another fission, the reaction sustains itself at a steady rate — the reactor is critical. If more than one does, on average, the fission rate grows exponentially — supercritical, the regime of an uncontrolled chain reaction (a fission weapon). If fewer than one does, the reaction dies out — subcritical.

Reactor parts (qualitative). A power reactor holds the chain reaction critical using four components: the fuel (enriched U-235, which fissions), the moderator (water or graphite, which slows fast neutrons to the thermal speeds at which U-235 fissions most readily), control rods (boron or cadmium, which absorb excess neutrons to hold the reaction at exactly critical), and the coolant (a fluid that carries the fission heat out to a turbine).

Figure

A uranium 235 nucleus after absorbing a slow neutron splits into barium 141 and krypton 92 plus three released neutrons with an energy label

Derivation AEnergy released in fission from binding energy per nucleon

The goal is to turn a binding-energy-per-nucleon difference between a fissioning nucleus and its fragments into a total energy release for a bulk sample of fuel, using only the mass number, the two binding-energy figures, and Avogadro's number — no calculus, just bookkeeping of energy per nucleon multiplied by nucleon count and then by the number of nuclei.

Step 1. Find the binding energy per nucleon gained. If the parent nucleus (mass number $A$ ) has binding energy per nucleon $\varepsilon_i$ and the fragments have average binding energy per nucleon $\varepsilon_f$ , each of the $A$ nucleons becomes more tightly bound by $(\varepsilon_f - \varepsilon_i)$ .

Step 2. Multiply by the nucleon count to get energy per fission event. Since all $A$ nucleons gain this extra binding energy,

\Delta E_{\text{event}} = A(\varepsilon_f - \varepsilon_i).

where: $A$ is the mass number of the parent (235 for uranium-235), $\varepsilon_i$ and $\varepsilon_f$ are the binding energies per nucleon (in MeV) of the parent and the fragments respectively, and $\Delta E_{\text{event}}$ is the energy released per fission event in MeV.

Step 3. Convert to joules and scale to a bulk sample. One MeV equals $1.6\times10^{-13}$ J, so $\Delta E_{\text{event}}$ in joules follows directly. For $N$ fissioning nuclei (for example one mole, $N = N_A = 6.02\times10^{23}$ ), the total energy released is

E_{\text{total}} = N \times \Delta E_{\text{event}}.

This is the same bookkeeping used to estimate reactor fuel energy budgets: fix $A$ , $\varepsilon_i$ , $\varepsilon_f$ from the binding-energy curve, and scale by however many nuclei actually fission. Verified: this reproduces the Pass-1 worked calculation, $\Delta E_{\text{event}} = 235(8.5-7.6) = 211.5\ \text{MeV}$ per event and $E_{\text{total}} \approx 2.04\times10^{13}\ \text{J}$ for one mole.

Worked example

Estimate the fission power output of a reactor consuming uranium-235 at a rate of 1 gram per day, given binding energy per nucleon of U-235 is 7.6 MeV and of the fragments is 8.5 MeV. (Molar mass of U-235 is 235 g/mol, $N_A = 6.02\times10^{23}$ .)

Step 1. Find the number of nuclei fissioning per day. One gram of U-235 is $1/235$ mol, so the number of nuclei is $N = (1/235)\times6.02\times10^{23} = 2.56\times10^{21}$ nuclei.

Step 2. Find the energy released per fission event, reusing the Pass-1 result: $\Delta E_{\text{event}} = 235\times(8.5-7.6) = 211.5\ \text{MeV} = 211.5\times1.6\times10^{-13}\ \text{J} = 3.384\times10^{-11}\ \text{J}$ .

Step 3. Find the total energy released per day: $E = N\times\Delta E_{\text{event}} = 2.56\times10^{21}\times3.384\times10^{-11} = 8.66\times10^{10}\ \text{J}$ .

Step 4. Convert to power: one day is $86{,}400$ s, so $P = E/t = 8.66\times10^{10}/86{,}400 \approx 1.0\times10^{6}\ \text{W}$ .

Burning just 1 gram of U-235 per day sustains about 1 megawatt of thermal power — a scale that shows why nuclear fuel is used sparingly compared with tonnes of coal for a similar output.

Common traps

Writing a fission equation without checking that both mass number and atomic number balance on the two sides — always verify both conservation laws, not just one.
Thinking a chain reaction needs every released neutron to cause further fission — only an average of one neutron per event is needed for criticality; the rest are absorbed, escape, or are captured without fission.
Swapping the moderator's role (slows neutrons to increase fission probability) with the control rods' role (absorbs neutrons to reduce reaction rate) — their effects are opposite.
Assuming "critical" means dangerous or about to explode — in reactor language it means the chain reaction is exactly self-sustaining at a steady rate, which is the normal operating state of a power reactor; "supercritical" is the runaway case.

Ask the tutor: Want to practice balancing more fission equations or work through a different reactor-energy calculation?

Why does splitting a heavy nucleus release energy instead of absorbing it?

It comes down to the binding-energy-per-nucleon curve. That curve peaks around mass number 56 and slopes gently downward for heavier nuclei like uranium-235. When U-235 splits into fragments with mass numbers near 90 and 145, those fragments sit closer to the peak — each of their nucleons is bound more tightly than it was in the parent. The system releases the extra binding energy as kinetic energy of the fragments and neutrons, which is why fission is exothermic even though it looks like "breaking" a nucleus apart.

Why do reactors need a moderator if fast neutrons are already produced by fission?

Fast neutrons are far less likely to cause further fission in U-235 than slow (thermal) neutrons — the fission probability rises sharply once neutrons are slowed down. Without a moderator, most fast neutrons would escape or be absorbed without triggering another fission, and the chain reaction would die out. Materials like water or graphite are good moderators because their light nuclei can absorb kinetic energy from neutrons in repeated collisions without themselves absorbing the neutron.

Is a critical reactor about to go out of control?

No — this is a common misreading of the word. "Critical" simply describes the steady, self-sustaining state where the fission rate stays constant: exactly one neutron per event, on average, causes the next fission. This is the normal, safe operating condition every power reactor is designed to sit in. The dangerous state is "supercritical," where more than one neutron per event causes further fission and the rate grows rapidly; control rods are pushed in specifically to prevent a reactor from drifting into that regime.

Nuclear fission

The core idea

Worked example

Common traps

Before this

Where this leads