Nuclear fission

On the morning of the first controlled fission chain reaction, Chicago Pile-1 sat in a squash court under a university stadium — a heap of graphite bricks and uranium pellets, nothing that looked like it could rewrite the twentieth century. When the last control rod eased out and the neutron counters began to chatter faster and faster on their own, physicists were watching a single nucleus's decision to break in two, multiplied a trillion trillion times over, turn into a self-sustaining roar of energy. That is nuclear fission: a heavy, unstable nucleus splitting into two lighter fragments, releasing energy that dwarfs anything chemistry can offer from the same mass of fuel. The idea sounds almost too simple for something this consequential. Fire a slow neutron at a $^{235}_{92}\text{U}$ nucleus, and instead of just sticking, the nucleus wobbles, elongates like a stretched water droplet, and snaps into two smaller nuclei plus a spray of extra neutrons — and about $200$ million electron-volts of energy, from a single atom you cannot even see. Multiply that by the number of atoms in a gram of uranium and you get tens of gigajoules, enough to power a city block for a day, released not by burning anything but by rearranging the nucleus itself. This article builds fission from a single curve you should already recognise — the binding-energy-per-nucleon graph — and shows why splitting a heavy nucleus is a one-way trip downhill in energy. We will write and balance a representative fission reaction, quantify how much energy comes out and why, follow the branching neutrons into a chain reaction, and see the one line of difference between a controlled nuclear reactor and an uncontrolled fission bomb. Everything here stays inside CBSE's qualitative treatment of fission — no reactor engineering, no decay laws — but by the end you should be able to read a fission equation, defend the energy number with the BE/A curve, and estimate the yield of a real mass of fuel with nothing beyond arithmetic.

Key takeaways

Fission is a heavy nucleus splitting into two lighter fragments after absorbing a slow neutron, typically also releasing two or three extra neutrons and a burst of energy.
The classic example is $^{235}_{92}\text{U} + {}^1_0n \rightarrow {}^{144}_{56}\text{Ba} + {}^{89}_{36}\text{Kr} + 3\,{}^1_0n + \text{energy}$ — but the fragment pair is not unique; many different splits occur across many events.
Energy comes from the BE/A curve. Heavy nuclei near $A \approx 235$ sit lower on the binding-energy-per-nucleon curve (about $7.6\ \text{MeV}$ per nucleon) than the mid-mass fragments produced (about $8.5\ \text{MeV}$ per nucleon); the fragments are more tightly bound, so mass converts to energy via $E = mc^2$ .
About $200\ \text{MeV}$ is released per fission event of $^{235}$ U, mostly as kinetic energy of the fragments and neutrons, with some as gamma radiation.
Extra neutrons make a chain reaction possible. Each fission releases roughly two to three neutrons, any of which can trigger another fission — the process can become self-sustaining.
One neutron per fission continuing the chain gives a steady, controlled reaction (as deliberately maintained in a nuclear reactor); more than one gives an explosively growing, uncontrolled reaction (as in a fission weapon).
Key idea: a heavy nucleus like uranium-235 splits into lighter fragments that sit higher on the binding-energy-per-nucleon curve, converting the mass difference into about 200 MeV of energy per event and releasing extra neutrons that can sustain a chain reaction.

The core insight

Nuclear fission converts mass to energy because the mid-mass fragments produced by splitting a heavy nucleus are bound more tightly per nucleon than the original nucleus was, releasing about $200\ \text{MeV}$ per event and, through the two or three neutrons also emitted, the possibility of a self-sustaining chain reaction.

Where it fits in the syllabus

Nuclear fission sits near the end of the CBSE Class 12 Physics chapter "Nuclei" (Unit IX, Atoms and Nuclei), and it is meant to be read as a direct application of the binding-energy-per-nucleon curve covered just before it. The prescribed treatment is qualitative: a representative fission reaction for $^{235}$ U, the BE/A-curve explanation for why energy is released, the idea of a chain reaction, and at most one qualitative line about a reactor keeping that chain reaction controlled — not reactor engineering, not moderator or control-rod design as an engineering topic. Radioactivity and its decay law are not part of the current 2025-26 rationalised syllabus for this chapter, so nothing here leans on half-life, decay constant, or activity; fission is treated purely as an energy-release and chain-reaction phenomenon grounded in binding energy.

The physical picture

Everything in fission traces back to one graph: binding energy per nucleon, $E_b/A$ , plotted against mass number $A$ . This curve climbs steeply for the lightest nuclei, reaches a broad, gentle peak around $A \approx 56$ – $62$ (iron and nickel territory), and then declines slowly as $A$ grows into the heavy end of the periodic table. A nucleus with a higher value of $E_b/A$ is more tightly bound — more energy would be needed to pull its nucleons apart — so nature "prefers" configurations further up the curve. Uranium-235, at $A = 235$ , sits well down the declining right-hand slope, with $E_b/A \approx 7.6\ \text{MeV}$ per nucleon.

^{235}_{92}\text{U} + {}^1_0n \;\longrightarrow\; {}^{144}_{56}\text{Ba} + {}^{89}_{36}\text{Kr} + 3\,{}^1_0n + \text{energy}

where: $^{235}_{92}\text{U}$ is the fissile uranium nucleus absorbing an incoming slow ("thermal") neutron $^1_0n$ ; the compound nucleus formed is so unstable that it almost immediately splits into two fragment nuclei — here barium-144 and krypton-89 — plus three more free neutrons, plus energy carried off as kinetic energy and gamma rays. The physical reading is a conservation statement: total mass number ( $A$ ) and total atomic number ( $Z$ ) on the left must equal the totals on the right, event by event, even though the specific fragment pair produced varies from one fission to the next. The fragments, each with mass number roughly half of $235$ , land much closer to the peak of the BE/A curve — around $E_b/A \approx 8.5\ \text{MeV}$ per nucleon. That is the entire mechanism in one sentence: the fragments are more tightly bound per nucleon than the parent nucleus was, so some rest mass has effectively disappeared and reappeared as kinetic energy and radiation, exactly as $E = mc^2$ demands.

Figure

The binding energy per nucleon curve rising steeply for light nuclei peaking near iron and nickel then declining slowly for heavy nuclei with uranium two thirty five marked low on the falling slope and the two mid mass fission fragments marked higher up near the peak with an arrow showing the energy released by moving from the low point to the higher points

Chain reaction: because each fission event releases two or three additional neutrons, and each of these can in turn be absorbed by another $^{235}$ U nucleus and trigger a further fission, the process can become self-sustaining. If, on average, more than one neutron from each fission goes on to cause a further fission, the number of fissions per generation grows explosively — this is an uncontrolled chain reaction, as in a fission weapon. If exactly one neutron per fission on average continues the chain, the reaction proceeds at a steady, controlled rate — this is the condition deliberately maintained inside a nuclear reactor, where the excess neutrons are absorbed by other materials so the chain stays just barely self-sustaining rather than runaway. That is the single qualitative point the syllabus asks for about reactors; the engineering of how that absorption is achieved is out of scope here.

Derivation AWhere the roughly 200 MeV per fission event comes from

The goal is to show, using only the binding-energy-per-nucleon values quoted above and simple bookkeeping of nucleon counts, that splitting a $^{235}$ U nucleus into two mid-mass fragments releases an energy of the right order of magnitude, about $200\ \text{MeV}$ , without needing any table of precise atomic masses — a purely BE/A-curve argument of exactly the kind CBSE expects.

Step 1. Set up the binding-energy bookkeeping. Total binding energy of a nucleus is (binding energy per nucleon) times (number of nucleons), $E_b = \dfrac{E_b}{A}\times A$ . Before fission, the $235$ nucleons of uranium are bound with $E_b/A \approx 7.6\ \text{MeV}$ each, so the total binding energy locked into the original nucleus is $E_{b,\text{before}} \approx 7.6 \times 235 \approx 1786\ \text{MeV}$ .

Step 2. Compute the total binding energy after fission. The same $235$ nucleons (roughly; three are freed as neutrons and carry negligible binding energy) are now distributed between two mid-mass fragments with $E_b/A \approx 8.5\ \text{MeV}$ each: $E_{b,\text{after}} \approx 8.5 \times 235 \approx 1998\ \text{MeV}$ .

Step 3. Read the energy release as a difference in how tightly bound the same nucleons have become. Since the fragments are more tightly bound overall, energy has been released, not absorbed — a larger total binding energy after the split means more energy is holding the nucleons together than before, and that extra binding energy corresponds to mass that has left the nuclei as kinetic energy and radiation: $\Delta E = E_{b,\text{after}} - E_{b,\text{before}} \approx 1998 - 1786 \approx 212\ \text{MeV}$ .

Step 4. Compare with the accepted figure. This back-of-envelope BE/A argument gives about $210$ – $215\ \text{MeV}$ , matching the accepted experimental value of roughly $200\ \text{MeV}$ per fission event to within the precision of the rounded BE/A values used — confirming that the entire energy release of fission is explained by the fragments sitting higher on the binding-energy-per-nucleon curve than the parent nucleus, with no separate physical mechanism needed.

Worked examples

Board level - balancing a fission reaction

Question: In the fission reaction $^{235}_{92}\text{U} + {}^1_0n \rightarrow {}^{144}_{56}\text{Ba} + {}^{X}_{36}\text{Kr} + 3\,{}^1_0n$ , find the unknown mass number $X$ of the krypton fragment, using conservation of mass number.

Step 1. Total mass number on the left side: mass number of $^{235}$ U plus the incoming neutron, $235 + 1 = 236$ .

Step 2. Total mass number on the right side must equal $236$ : barium ( $144$ ) plus krypton ( $X$ ) plus three neutrons ( $3 \times 1 = 3$ ).

Step 3. Set up the equation: $144 + X + 3 = 236$ .

Step 4. Solve: $X = 236 - 144 - 3 = 89$ .

Answer: $X = \mathbf{89}$ , so the krypton fragment is $^{89}_{36}\text{Kr}$ , consistent with the reaction quoted throughout this article. As a quick cross-check, atomic numbers must also balance: $92 + 0 = 56 + 36 + 0$ , which holds.

Competitive level - energy yield from one gram of fuel

Question: If a single fission event of $^{235}$ U releases about $200\ \text{MeV}$ of energy, estimate the total energy released, in joules, by the complete fission of $1\ \text{gram}$ of $^{235}$ U. Use $N_A = 6.022\times10^{23}\ \text{mol}^{-1}$ and molar mass of $^{235}$ U $\approx 235\ \text{g/mol}$ .

Step 1. Number of atoms in $1\ \text{g}$ of $^{235}$ U: $n = \dfrac{N_A}{235} = \dfrac{6.022\times10^{23}}{235} \approx 2.56\times10^{21}$ atoms.

Step 2. Energy released per fission event, converted to joules: $200\ \text{MeV} = 200 \times 1.6\times10^{-13}\ \text{J} = 3.2\times10^{-11}\ \text{J}$ .

Step 3. Total energy released: $E = n \times (\text{energy per event}) = 2.56\times10^{21} \times 3.2\times10^{-11}\ \text{J}$ .

Step 4. Multiply out: $E \approx 8.2\times10^{10}\ \text{J}$ .

Answer: About $\mathbf{8.2\times10^{10}\ \text{J}}$ ( $82$ gigajoules) is released by the complete fission of just $1\ \text{gram}$ of $^{235}$ U — for comparison, burning $1\ \text{gram}$ of coal releases roughly $3\times10^{4}\ \text{J}$ , so this one gram of uranium fuel is worth several tonnes of coal. Timing note: this is a two-line unit-conversion calculation once you have $N_A/235$ and the MeV-to-joule factor memorised — aim to finish it in under ninety seconds in a competitive setting, since the only place to lose time is fumbling the power-of-ten arithmetic.

Common traps

Thinking fission always produces exactly the same pair of fragments — in reality, $^{235}$ U fission produces a range of different fragment pairs across many events, though the total energy release per event is broadly similar.
Forgetting that the absorbed neutron must be included in the mass and charge balance on the left-hand side — always conserve mass number ( $A$ ) and atomic number ( $Z$ ) separately across the entire reaction, incoming neutron included.
Assuming fission releases energy simply because "the nucleus splits" — the real reason, rooted in the BE/A curve, is that the fragments are more tightly bound per nucleon than the original heavy nucleus, so mass converts to energy.
Confusing an uncontrolled chain reaction (explosive, as in a weapon) with a controlled one (steady, as in a reactor) — the underlying physics is identical; what differs is only whether, on average, more than one or exactly one neutron per fission continues the chain.
Quoting the $200\ \text{MeV}$ figure as if it were an exact universal constant — it is a typical, rounded value; different fragment pairs and different numbers of emitted neutrons give slightly different energy releases event to event.

You should be able to…

Write and balance a representative fission reaction for $^{235}$ U, conserving both mass number and atomic number across the full equation, including the incoming neutron.
Explain, using the shape of the BE/A curve, why splitting a heavy nucleus into two mid-mass fragments releases energy.
Estimate the energy released per fission event from BE/A values, and cross-check it against the accepted figure of about $200\ \text{MeV}$ .
Calculate the total energy released by the fission of a given mass of fissile material, given the energy released per event.
Describe qualitatively what a chain reaction is, and distinguish a controlled reaction (one neutron per fission continuing the chain) from an uncontrolled one (more than one).
State, in one line, why nuclear reactors deliberately absorb some neutrons — without going into reactor engineering detail.

Ask the tutor: want to see how the energy released per fission compares gram-for-gram with a chemical fuel like coal or petrol, or how the BE/A-curve argument changes when applied to fusion instead of fission — ask the AI tutor to work through either comparison.

Why does uranium-235 need a slow neutron specifically, not just any neutron, to fission

A fast neutron can miss interacting with a $^{235}$ U nucleus because it zips past too quickly to be captured; a slow, "thermal" neutron spends more time near the nucleus and has a much higher chance of being absorbed, forming the unstable compound nucleus that then splits. CBSE only expects you to know that slow neutrons are used, not the detailed physics of capture probability, so for exam purposes just remember: fission of $^{235}$ U is triggered by a slow neutron.

If fission releases 200 MeV from one atom, why does a nuclear reactor not immediately explode?

Whether the reaction runs away or stays steady depends entirely on how many of the two or three neutrons released per fission go on to cause another fission. In a bomb, that number is kept above one on average, so fissions multiply exponentially in a fraction of a second. In a reactor, materials are used to absorb the extra neutrons so that, on average, almost exactly one neutron per fission continues the chain — the same physics, but tuned to a steady rather than runaway rate.

Is the fragment pair barium and krypton the only possible outcome of uranium-235 fission?

No — that pair is simply the most commonly quoted example. In real fission events, $^{235}$ U splits into many different possible pairs of fragments, with mass numbers spread unevenly around roughly $90$ and $140$ rather than always exactly $89$ and $144$ . What stays fairly consistent from event to event is the total energy released, around $200\ \text{MeV}$ , because it always comes from the same underlying BE/A-curve mechanism.

Does this chapter also cover radioactive decay and half-life along with fission?

Not in the current CBSE rationalised syllabus — radioactivity, the decay law, half-life, and activity have been removed from the official 2025-26 unit text for this chapter, even though many older textbooks and question banks still include them. If you see a half-life question labelled "Nuclei," check whether it belongs to your specific year's syllabus before spending study time on it; this article deliberately keeps to fission and the BE/A curve, which are the parts that remain prescribed.