Nuclear Fission & Chain Reactions

Fission is not history; it runs about a tenth of the world's electricity right now. Every commercial power reactor — pressurised-water, boiling-water, heavy-water, and the small modular reactors now being built — is a chain reaction pinned at $k=1$, boiling water with $200\text{-MeV}$ events to spin a turbine. Submarines and carriers carry compact reactors that run for years without refuelling. The fission fragments are the feedstock of nuclear medicine: molybdenum-99, milked from fission products, decays to the technetium-99m used in tens of millions of scans a year. Fission also breeds new fuel — uranium-238 absorbing a neutron becomes fissile plutonium-239 — the basis of breeder reactors and, less happily, of proliferation. Its hard problems are equally real: the long-lived radioactive waste of the neutron-rich fragments, meltdown risk, and the weapons legacy that began with the very calculation Meitner did in the snow. The cleaner sibling at the other end of the binding-energy curve — fusion, which powers the Sun and leaves no long-lived waste — is the next article.

Picture the uranium-235 nucleus as the binding-energy curve of the previous article invites you to: a droplet of nuclear fluid, $92$ protons and $143$ neutrons crammed into a ball a few femtometres across, held together by the strong force but strained by the electrical repulsion of all those protons. It is one of the heaviest droplets nature makes, and it sits well down the far slope of the binding-energy curve — at about $7.6\ \text{MeV}$ per nucleon, roughly $1.2\ \text{MeV}$ short of the iron-56 peak. Every nucleon in it would "prefer" to live in a middle-weight nucleus nearer that peak, where binding is tightest. Left alone, uranium-235 barely notices; its own fission is fantastically rare. But feed it one extra neutron and the droplet becomes uranium-236 in a violently excited state. The captured neutron brings in about $6.5\ \text{MeV}$ of binding energy, and that is enough to set the whole drop oscillating. It stretches into a dumbbell; the two ends, each positively charged, shove each other apart electrically faster than the short-range strong force can pull them back; the neck pinches off; and the drop snaps into two fragments that fly apart at a fair fraction of light speed, trailing two or three loose neutrons and a burst of gamma rays.

The two fragments are never a fixed pair. A given uranium-235 nucleus might split into barium and krypton, into caesium and rubidium, into a hundred other combinations — the split is lopsided, typically about $95$ and $140$ in mass number, and always into two pieces that together carry the $92$ protons and (minus the escaping neutrons) the $235+1$ nucleons. And the spare neutrons are no accident: uranium-235 sits at a much higher neutron-to-proton ratio ($\sim\!1.55$) than mid-mass fragments need for stability ($\sim\!1.3$), so the split leaves them several neutrons too many — most shed at once as the $\sim\!2.5$ free neutrons, the rest by later $\beta^-$ decay. What is always the same is the bottom line: middle-weight fragments, a few spare neutrons, and about $200\ \text{MeV}$ of energy released, mostly as the kinetic energy of the recoiling fragments — which is to say, as heat.

The energy of fission does not come from the incoming neutron — that neutron can be almost stationary, carrying a fraction of an electronvolt. It comes from the rearrangement of the nucleons into a more tightly bound configuration, exactly as the binding-energy curve predicts. Reading that curve from right to left, uranium sits at about $7.6\ \text{MeV}$ per nucleon and the mid-mass fragments near $8.4$–$8.5\ \text{MeV}$ per nucleon. Averaged over the split, each of the $\sim\!236$ nucleons ends up about $0.85\ \text{MeV}$ more deeply bound, and released binding energy is energy handed to the outside world:

where: $A \approx 236$ is the nucleon count of the fissioning system, $\Delta(\text{BE}/A) \approx 0.85\ \text{MeV}$ is the average gain in binding energy per nucleon on going from uranium up toward the iron peak, and $Q$ is the energy released. The physical reading is that fission is downhill on the binding-energy curve just as fusion is downhill at the light end — both processes move nucleons toward the iron peak, and both therefore release energy. The same $Q$ can be read as a mass defect: $200\ \text{MeV}$ divided by $931.5\ \text{MeV}$ per atomic mass unit is only about $0.21\ \text{u}$, so less than one part in a thousand of the uranium's mass is converted — a whisper of mass, an avalanche of energy.

The single most consequential fact about fission is buried in the reaction above: it is triggered by one neutron and produces several. On average uranium-235 gives about $2.5$ neutrons per fission. If even one of those, on average, survives to be absorbed by another uranium-235 nucleus and split it, the process feeds itself — a chain reaction, whose possibility was confirmed in 1939 when Joliot's Paris group and, independently, Szilard and Fermi showed each fission emits spare neutrons. Whether the chain grows, holds, or dies is captured by a single number, the multiplication factor $k$: the average number of neutrons from one fission that go on to cause a further fission. Starting from $N_0$ fissions in the first "generation," the number in the $g$-th generation follows a geometric law:

where: $N_0$ is the number of fissions in the starting generation, $k$ is the multiplication factor, and $g$ counts generations. The physical reading is stark: $k$ appears as a power, so the chain is exquisitely sensitive to it. If $k<1$ the powers of $k$ shrink to nothing and the chain dies (subcritical). If $k=1$ every generation exactly reproduces the last and the reaction ticks over steadily (critical) — this is a power reactor. If $k>1$ the powers of $k$ explode and the fission rate runs away (supercritical).

Why can you hold a lump of uranium in your hand without it exploding? Because in a small piece most neutrons escape through the surface before they meet another nucleus. A neutron born inside the metal has four possible fates: it can cause a new fission, be absorbed without fissioning (by uranium-238 or by impurities), leak out through the surface, or — in a reactor — be captured after being slowed. The chain sustains only if, on balance, at least one neutron per fission finds a new nucleus, i.e. $k \geq 1$. Whether it does is a competition between volume, where fissions happen (scaling as the cube of the size), and surface, where neutrons leak (scaling as the square). Make the lump bigger and the volume-to-surface ratio grows, so proportionally fewer neutrons escape and $k$ rises. There is therefore a critical mass — the smallest mass that reaches $k=1$. For a bare sphere of pure uranium-235 it is about $52\ \text{kg}$ (a ball roughly $17\ \text{cm}$ across); wrap it in a neutron-reflecting "tamper" and the figure drops to around $15\ \text{kg}$; for plutonium-239 it is around $10\ \text{kg}$ bare. These are quoted figures, set by neutron-transport geometry beyond this article to derive; the point you must own is why a critical mass exists at all — the surface-to-volume leakage argument. Natural uranium will not by itself sustain a fast chain: it is $99.3\%$ uranium-238, which mostly absorbs fast neutrons without fissioning, and only $0.7\%$ fissile uranium-235. Two escape routes exist. Enrich the uranium-235 fraction — to a few percent for reactor fuel, above $90\%$ for a weapon — or slow the neutrons down so dramatically that they fission the sparse uranium-235 far more readily than uranium-238 absorbs them. That second route is what a reactor does.

A power reactor and a bomb run the same physics with opposite intentions. The bomb wants $k$ as far above $1$ as possible for as long as it can hold together; the reactor wants $k$ pinned at exactly $1$, releasing $200\ \text{MeV}$ at a time, steadily, as heat to boil water and drive a turbine. Three components achieve that pinning. The moderator — ordinary water, heavy water, or graphite — surrounds the fuel and slows the fast fission neutrons (born at about $2\ \text{MeV}$) by elastic collisions down to thermal speeds, energies around $0.025\ \text{eV}$, matching the jostling of room-temperature atoms. Slow neutrons are absorbed by uranium-235 far more readily, so moderation is what lets low-enriched fuel sustain a chain at all. The control rods — cadmium or boron, both greedy neutron absorbers — lower into the core to soak up surplus neutrons and pull $k$ down, or withdraw to let it rise; they are the throttle and brake. And a reactor leans on a gift of physics: a small fraction of the fission neutrons are delayed, emitted seconds later by the decaying fragments. Those delayed neutrons stretch the effective generation time from microseconds to seconds — the only reason a mechanical rod can keep up. A bomb, on prompt neutrons alone, gets no such grace.

Board level — energy from one gram of uranium-235 (CBSE and ICSE pattern)

Each fission of uranium-235 releases about $200\ \text{MeV}$. Taking $N_A = 6.022\times10^{23}\ \text{mol}^{-1}$ and $1\ \text{MeV} = 1.602\times10^{-13}\ \text{J}$, find the energy released if all the nuclei in $1\ \text{g}$ of uranium-235 undergo fission, and compare it with the $\sim\!33\ \text{kJ}$ from burning a gram of coal.

Step 1. Number of nuclei in $1\ \text{g}$: one mole is $235\ \text{g}$, so $N = (1/235)\times 6.022\times10^{23} = 2.56\times10^{21}$ nuclei.

Step 2. Energy per nucleus in joules: $200\ \text{MeV} = 200\times1.602\times10^{-13} = 3.20\times10^{-11}\ \text{J}$.

Step 3. Total energy: $E = N \times 3.20\times10^{-11} = 2.56\times10^{21}\times3.20\times10^{-11} = \boxed{8.2\times10^{10}\ \text{J}}$.

Step 4. Compare with coal: $8.2\times10^{10} / 3.3\times10^{4} \approx 2.5\times10^{6}$ — a gram of uranium yields about 2.5 million times the energy of a gram of coal. Equivalently $8.2\times10^{10}\ \text{J} \div 3.6\times10^{6}\ \text{J/kWh} \approx 2.3\times10^{4}\ \text{kWh}$, enough to run a typical home for a couple of years.

Board level — completing a fission reaction and counting neutrons (CBSE and A-Level pattern)

A uranium-235 nucleus absorbs a neutron and splits as ${}^{235}_{\ 92}\text{U} + {}^{1}_{0}\text{n} \to {}^{144}_{\ 56}\text{Ba} + {}^{A}_{Z}\text{Kr} + x\,{}^{1}_{0}\text{n}$. Find the atomic number $Z$ and mass number $A$ of the krypton fragment and the number $x$ of neutrons released.

Step 1. Conserve charge (protons): $92 = 56 + Z$, so $Z = \boxed{36}$ — krypton, as its symbol demands.

Step 2. Conserve nucleon number: $235 + 1 = 144 + A + x$, i.e. $236 = 144 + A + x$, so $A + x = 92$.

Step 3. A krypton isotope of mass number $A = 89$ is the physically produced fragment here, giving $x = 92 - 89 = \boxed{3}$ neutrons. The general lesson: fix $Z$ from charge conservation, then trade off $A$ against the neutron count $x$ using nucleon conservation. Every fission equation must balance top numbers (nucleons) and bottom numbers (protons) separately.

Competitive level — daily fuel burn of a 1000 MW reactor (JEE and NEET pattern)

A nuclear reactor generates thermal power $P = 1000\ \text{MW} = 1.0\times10^{9}\ \text{W}$. Each fission releases $200\ \text{MeV}$. Find (a) the number of fissions per second, and (b) the mass of uranium-235 consumed per day. Take $1\ \text{MeV} = 1.602\times10^{-13}\ \text{J}$, $N_A = 6.022\times10^{23}$.

Step 1. Energy per fission: $200\ \text{MeV} = 3.20\times10^{-11}\ \text{J}$. Fissions per second: $R = P/E = (1.0\times10^{9})/(3.20\times10^{-11}) = \boxed{3.1\times10^{19}\ \text{s}^{-1}}$.

Step 2. Fissions per day: $3.1\times10^{19}\times 86400 = 2.7\times10^{24}$.

Step 3. Mass consumed: each fissioned nucleus removes one atom, so mass $= (2.7\times10^{24}/6.022\times10^{23})\times 235\ \text{g} = 4.48\times235 \approx \boxed{1.05\ \text{kg per day}}$.

Step 4. Sanity check against a famous rule of thumb: fissioning $1\ \text{g}$ of uranium-235 releases about one megawatt-day of heat ($8.2\times10^{10}\ \text{J}$ versus $1\ \text{MW}\times1\ \text{day} = 8.64\times10^{10}\ \text{J}$). A $1000\ \text{MW}$ reactor therefore burns about $1000\ \text{g} = 1\ \text{kg}$ of fuel a day — matching Step 3. Timing note: about two minutes. The whole item is $R=P/E$ then multiply by seconds in a day and by $235/N_A$. Carry $200\ \text{MeV} = 3.2\times10^{-11}\ \text{J}$ as a memorised conversion and the arithmetic is two multiplications. The "$1\ \text{g} \approx 1\ \text{MW-day}$" rule lets you write the kilogram-per-day answer down almost by inspection — a huge time saver under exam pressure.

Competitive level — mass defect and the fraction of mass converted (JEE Advanced pattern)

In a fission releasing $200\ \text{MeV}$, what mass is converted to energy, and what fraction is this of the original $\sim\!236\ \text{u}$? Take $1\ \text{u} = 931.5\ \text{MeV}/c^2$.

Step 1. Mass converted: $\Delta m = Q/931.5 = 200/931.5 = \boxed{0.215\ \text{u}}$, about $3.6\times10^{-28}\ \text{kg}$.

Step 2. Fraction of the total mass: $\Delta m / A = 0.215/236 = 9.1\times10^{-4} \approx \boxed{0.09\%}$.

Step 3. Interpretation: less than one part in a thousand of the uranium's mass becomes energy, yet that sliver is $200\ \text{MeV}$ — because $c^2$ is enormous. For comparison, chemical burning converts only about one part in a billion of the fuel's mass, which is exactly why nuclear fission is roughly a million times more energy-dense than combustion. Timing note: under a minute if you remember that $Q$ in MeV divided by $931.5$ is directly the mass defect in atomic mass units — no need to detour through kilograms and joules. The fraction $0.09\%$ is worth memorising as the headline "mass-to-energy" number of nuclear fission.

Saying the energy comes from the neutron. The trigger neutron carries about $0.025\ \text{eV}$; the fission releases about $200\ \text{MeV}$. The energy is the binding-energy jump of the rearranged nucleons, banked as a mass defect. Writing "the fast neutron smashes the nucleus and its kinetic energy is released" loses the marks on CBSE, NEET and A-Level alike.

Confusing fission with fusion. Both release energy by climbing the binding-energy curve toward iron, but from opposite ends: fission splits heavy nuclei (right of the peak), fusion joins light ones (left of it). A stock exam sentence — "fusion is the splitting of a heavy nucleus" — is simply wrong. Anchor it: heavy → split → fission; light → join → fusion.

Getting the isotope wrong: U-235 versus U-238. Only uranium-235 (and plutonium-239) fissions readily with slow neutrons; uranium-238, $99.3\%$ of natural uranium, mostly absorbs fast neutrons. Questions that ask why natural uranium needs enrichment or a moderator are testing exactly this distinction.

Forgetting to balance a fission equation on both counts. Nucleon number (top) and proton number (bottom) must balance separately. Students routinely balance the mass numbers and forget the neutrons carry no charge, or vice versa. Fix $Z$ from the bottom row first, then the neutron count from the top.

Misreading the multiplication factor. $k=1$ is the steady critical state of an operating reactor, not "off." $k<1$ is subcritical (dying), $k>1$ supercritical (growing). And "critical mass" is about neutron leakage geometry, not a chemical or temperature threshold.

Board scope. Fission, chain reactions, critical mass, the moderator and control rods, and reactor-versus-bomb are core on CBSE, ICSE (ISC), JEE Main, JEE Advanced, NEET, AP Physics 2, IB SL and HL, and A-Level 9702. IGCSE 0625 carries the qualitative story only — what fission is, the chain reaction, the uses — but not the $200\ \text{MeV}$ mass-defect calculation, the $Q$-value, or the exponential $k^g$ growth. Give an IGCSE answer in words, not in mass-energy sums.

Texts. NCERT Physics Class 12, Chapter 13 ("Nuclei"), the fission and nuclear-energy sections · H. C. Verma, Concepts of Physics Vol. 2, "The Nucleus" · Krane, Introductory Nuclear Physics, for the liquid-drop and fission-barrier picture · Serway/Jewett, Physics for Scientists and Engineers, "Nuclear Energy."

Provenance. Standard nuclear masses and constants. Recompute: representative split ${}^{235}\text{U}+\text{n}\to{}^{141}\text{Ba}+{}^{92}\text{Kr}+3\text{n}$ mass defect $0.186\ \text{u}$, prompt $Q = 173\ \text{MeV}$; binding-energy estimate $236\times0.85 \approx 200\ \text{MeV}$; energy tally $168+5+7+8+7+12 = 207\ \text{MeV} \to \sim\!200\ \text{MeV}$; mass defect per fission $200/931.5 = 0.215\ \text{u} = 0.09\%$ of $236\ \text{u}$; $1\ \text{g}$ of U-235 gives $N = 6.022\times10^{23}/235 = 2.56\times10^{21}$ nuclei $\times\,3.20\times10^{-11}\ \text{J} = 8.2\times10^{10}\ \text{J}$; coal ratio $8.2\times10^{10}/3.3\times10^{4} = 2.5\times10^{6}$; kWh $= 8.2\times10^{10}/3.6\times10^{6} = 2.3\times10^{4}$; $1000\ \text{MW}$ reactor $R = 10^{9}/3.20\times10^{-11} = 3.1\times10^{19}\ \text{s}^{-1}$, $1.05\ \text{kg/day}$; $1\ \text{g}\approx1\ \text{MW-day}$ ($8.64\times10^{10}\ \text{J}$); $1\ \text{kg}$ has $2.56\times10^{24}$ atoms, $\log_2 = 81$ generations, $2^{80}=1.21\times10^{24}$, $\times10\ \text{ns}\approx0.8\ \mu\text{s}$; thermal neutron $0.025\ \text{eV}$, fission neutron $\sim\!2\ \text{MeV}$; natural U $0.7\%$ / $99.3\%$; critical mass U-235 $\sim\!52\ \text{kg}$ bare, $\sim\!15\ \text{kg}$ reflected; Krypton fragment $Z=36$, $A=89$, $x=3$.

Prerequisites. The Binding-Energy Curve — the single graph that explains why fission (heavy, right of the iron peak) and fusion (light, left of it) both release energy. That article's peak at iron-56 and uranium's $7.6\ \text{MeV}$ per nucleon are the foundation of everything here. Leads to. Nuclear Fusion and Stellar Energy, the mirror-image process at the light end of the curve, powering the Sun and offered as fission's cleaner successor. For enrichment beyond this article, the university seam on nuclear reactor physics and neutron transport treats the four-factor formula and critical geometry properly (go-deeper only, never an exam dependency here).