Understand Fission — Understand E.4 Fission - IB Physics

SECTION 1

Historical Context & Motivation

The story of nuclear fission begins with a puzzle that captivated physicists in the 1930s. Scientists had been bombarding elements with neutrons, expecting to create heavier elements beyond uranium. Instead, they found something completely unexpected — the heavy nucleus was splitting apart into lighter fragments and releasing a staggering amount of energy. This discovery fundamentally changed our understanding of nuclear physics, and it carries enormous implications for both energy production and global security that persist to this day.

1932 • Discovery of the Neutron

James Chadwick identified the neutron, an uncharged particle in the nucleus. Because it carries no charge, the neutron can penetrate atomic nuclei without being repelled by the positive protons — making it the perfect projectile for nuclear experiments.

1934 • Fermi's Neutron Bombardment

Enrico Fermi bombarded uranium with slow neutrons, believing he had created new transuranic elements (elements heavier than uranium). His results were actually evidence of fission, but the concept was so radical that it took several more years to be correctly interpreted.

1938 • Hahn and Strassmann Split Uranium

Otto Hahn and Fritz Strassmann performed careful chemical analysis showing that bombarding uranium with neutrons produced barium — an element far lighter than uranium. This was the first experimental evidence that the uranium nucleus was actually splitting in two.

1939 • Meitner & Frisch Explain Fission

Lise Meitner and Otto Frisch provided the theoretical explanation, coining the term fission (borrowed from biology's cell division). Using Einstein's mass–energy equivalence, they calculated that the mass lost during splitting would release approximately 200 MeV per fission event.

1942 • First Controlled Chain Reaction

Enrico Fermi's team achieved the first self-sustaining chain reaction in Chicago Pile-1, a reactor built under the stands of a squash court at the University of Chicago. This proved that fission could be controlled for sustained energy release.

These discoveries raised a profound question: how does a single neutron cause an enormous nucleus to break apart, and why does this process release so much more energy than any chemical reaction? To answer this, we need to understand the forces that hold a nucleus together — and the conditions under which those forces can be overcome.

SECTION 2

Core Principles of Nuclear Fission

Nuclear fission is the process in which a heavy, unstable nucleus absorbs a neutron and splits into two or more lighter nuclei, called fission fragments, along with additional free neutrons and a tremendous amount of energy. To appreciate why this works, you need to understand several interconnected ideas about how nuclear binding energy varies with mass number.

Binding Energy per Nucleon

The binding energy per nucleon measures how tightly each proton or neutron is held inside a nucleus. Nuclei near iron-56 (A ≈ 56) have the highest binding energy per nucleon, making them the most stable. Very heavy nuclei like uranium-235 have lower binding energy per nucleon, so they can release energy by splitting into fragments closer to iron on the curve.

Mass Defect and E = mc²

The total mass of the fission products is slightly less than the mass of the original nucleus plus neutron. This missing mass, called the mass defect (Δm), has been converted into energy according to Einstein's equation E = mc². Even a tiny mass difference produces enormous energy because c² is approximately 9 × 10¹⁶ m²/s².

Induced vs. Spontaneous Fission

Most fission in reactors is induced fission — a neutron is absorbed by a heavy nucleus, which becomes excited and splits. Some very heavy isotopes also undergo spontaneous fission, splitting on their own due to quantum tunnelling, though this is much rarer.

Chain Reactions

Each fission event releases 2–3 free neutrons. If at least one of these neutrons goes on to cause another fission, a chain reaction is sustained. If exactly one neutron triggers another fission on average, the reaction is critical — a steady, controlled state. If more than one does, it becomes supercritical and can run away.

Fissile vs. Fissionable

A fissile material (like U-235 or Pu-239) can sustain a chain reaction with slow (thermal) neutrons. A fissionable material (like U-238) can undergo fission but only with fast, high-energy neutrons — it cannot easily sustain a chain reaction on its own.

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KEY TAKEAWAY

Think of a heavy nucleus like a wobbly water droplet perched on the edge of a table. A tiny nudge (the incoming neutron) is enough to make it oscillate wildly and tear itself apart into two smaller, more stable droplets. The energy released is like the splash — it comes from the fact that the two smaller droplets sit at a lower energy state than the original overstretched drop. In nuclear terms, the fission fragments have higher binding energy per nucleon, and the difference is what powers nuclear reactors.

SECTION 3

The Fission Process — Visual Explanation

The diagram shows the four stages of induced fission. A slow neutron (yellow) is absorbed by a U-235 nucleus (purple), forming an excited U-236* compound nucleus. The compound nucleus oscillates like a liquid drop until the Coulomb repulsion between the proton-rich halves overcomes the strong nuclear force, splitting into two fragments — barium-141 (cyan) and krypton-92 (pink) — along with three free neutrons and approximately 200 MeV of energy.

The diagram above illustrates the liquid drop model of fission, proposed by Niels Bohr and John Archibald Wheeler. When U-235 absorbs a thermal neutron, it momentarily becomes U-236 in a highly excited state. The nucleus begins oscillating — stretching into an elongated shape, much like a vibrating water droplet. As it stretches, the strong nuclear force (short-range, attractive) weakens across the neck region, while the Coulomb repulsion (long-range, repulsive between protons) grows dominant. Eventually the nucleus snaps apart at the neck, producing two unequal fission fragments, free neutrons, and energy.

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IB Exam Note

The IB syllabus does not require you to memorize specific fission fragment pairs. However, you must understand that fission products are asymmetric — the two fragments are usually unequal in mass — and that 2–3 neutrons are released per fission event, enabling a chain reaction.

SECTION 4

Mathematical Framework

The energy released in fission can be calculated using two related approaches: through the mass defect and Einstein's mass–energy equivalence, or through the binding energy per nucleon curve. Both methods give the same answer, and the IB expects you to be comfortable with either.

MASS–ENERGY EQUIVALENCE

E = Δm × c²

where E is the energy released (J), Δm is the mass defect — the difference between the total mass of reactants and total mass of products (kg), and c is the speed of light (3.00 × 10⁸ m/s). When using atomic mass units, 1 u of mass defect corresponds to 931.5 MeV.

ENERGY FROM BINDING ENERGY DIFFERENCE

Q = ΣBE(products) − ΣBE(reactants)

where Q is the energy released, ΣBE(products) is the total binding energy of all product nuclei, and ΣBE(reactants) is the total binding energy of the original nucleus. Since the products are more tightly bound, Q is positive — energy is released.

ENERGY PER NUCLEON ESTIMATE

Q ≈ A × (BE/A_products − BE/A_reactant)

A quick estimation method: multiply the total number of nucleons A by the difference in binding energy per nucleon between products and reactant. For U-235 fission, the BE/A increases from about 7.6 MeV to about 8.5 MeV, giving roughly 0.9 MeV × 236 ≈ 212 MeV.

The approximately 200 MeV released per fission event is distributed as follows: about 170 MeV goes into the kinetic energy of the fission fragments, approximately 5 MeV into neutron kinetic energy, about 15 MeV into gamma radiation, and the remainder into beta particles and neutrinos from the radioactive decay of fission products. In a reactor, the kinetic energy of the fragments is converted to thermal energy through collisions with surrounding atoms.

CRITICAL MASS CONDITION

k = (neutrons produced per generation) / (neutrons in previous generation)

The multiplication factor k determines the state of the chain reaction. When k = 1, the reaction is critical (steady state). When k > 1, it is supercritical (growing). When k < 1, it is subcritical (dying out). Nuclear reactors are designed to maintain k = 1 during normal operation.

SECTION 5

The Binding Energy per Nucleon Curve

The binding energy per nucleon curve is arguably the single most important graph in nuclear physics. It explains why fission of heavy nuclei and fusion of light nuclei both release energy, and why iron-56 sits at the peak of nuclear stability. For fission, the key insight is that nuclei to the right of the peak (A > 56) have lower binding energy per nucleon; splitting them into fragments closer to the peak releases the difference as kinetic energy.

The binding energy per nucleon curve peaks at iron-56 (≈ 8.8 MeV per nucleon). Uranium-235 sits at about 7.6 MeV per nucleon. When uranium fissions into fragments near A ≈ 95 and A ≈ 141, these fragments have binding energies per nucleon around 8.5 MeV. The gain of roughly 0.9 MeV per nucleon across 236 nucleons accounts for the ≈ 200 MeV released per fission.

Notice how the curve is relatively flat for medium and heavy nuclei (A > 30). This means that the energy released per nucleon in fission (about 0.9 MeV) is much less than the energy released per nucleon in fusion of very light nuclei (which can exceed 6 MeV per nucleon for hydrogen fusion). However, fission is far easier to achieve on Earth because it only requires a single neutron to trigger, whereas fusion requires overcoming immense Coulomb barriers by heating fuel to millions of degrees.

SECTION 6

Worked Example — Energy Released in Fission

Let us calculate the energy released when a U-235 nucleus undergoes fission via the following reaction:

EXAMPLE FISSION REACTION

²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n

Given atomic masses: U-235 = 235.04393 u, Ba-141 = 140.91440 u, Kr-92 = 91.92616 u, neutron = 1.00866 u. (1 u = 931.5 MeV/c²)

Calculating Fission Energy Release

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Step 1 — Identify the Reactant Masses

Total mass of reactants = mass of U-235 + mass of 1 neutron. m(reactants) = 235.04393 u + 1.00866 u = 236.05259 u

m(reactants) = 236.05259 u
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Step 2 — Identify the Product Masses

Total mass of products = mass of Ba-141 + mass of Kr-92 + mass of 3 neutrons. m(products) = 140.91440 u + 91.92616 u + 3 × 1.00866 u m(products) = 140.91440 + 91.92616 + 3.02598 = 235.86654 u

m(products) = 235.86654 u
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Step 3 — Calculate the Mass Defect

The mass defect is the difference between reactant and product masses. Δm = m(reactants) − m(products) Δm = 236.05259 − 235.86654 = 0.18605 u

Δm = 0.18605 u
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Step 4 — Convert to Energy

Using the conversion factor 1 u = 931.5 MeV/c²: E = Δm × 931.5 MeV/u E = 0.18605 × 931.5 = 173.3 MeV

E ≈ 173 MeV
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Step 5 — Interpret the Result

This 173 MeV represents the kinetic energy carried away by the fission fragments and neutrons. When we include the energy from prompt gamma rays and subsequent radioactive decay of the fission products (beta particles, neutrinos, and delayed gamma rays), the total energy per fission event comes to approximately 200 MeV. This is roughly 50 million times more energy than a single chemical combustion reaction.

Total energy per fission ≈ 200 MeV

SECTION 7

Nuclear Reactors — Components and Control

A nuclear fission reactor is an engineered system designed to sustain a controlled chain reaction at k = 1 (the critical state). The IB syllabus expects you to understand the role of each key component and how they work together to maintain safe, steady energy output. The table below summarizes the main reactor components.

Component	Material Example	Function
Fuel	Enriched uranium (3–5% U-235) or plutonium-239	Provides the fissile nuclei that undergo chain-reaction fission when they absorb neutrons.
Moderator	Water (H₂O), heavy water (D₂O), or graphite	Slows fast neutrons to thermal speeds (~0.025 eV) through elastic collisions, greatly increasing the probability that they will be absorbed by U-235.
Control Rods	Boron, cadmium, or hafnium	Absorb excess neutrons. Inserting rods deeper reduces k below 1 (subcritical); withdrawing them increases k. This is the primary method of controlling reactor power.
Coolant	Water, liquid sodium, or CO₂ gas	Transfers thermal energy from the reactor core to a heat exchanger or steam generator. In many designs, the coolant also serves as the moderator.
Shielding	Concrete, steel, lead	Absorbs gamma rays and neutrons that escape the core, protecting workers and the environment from ionizing radiation.

Key components of a nuclear fission reactor

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KEY TAKEAWAY

Think of a nuclear reactor like a campfire with careful management. The fuel rods are the logs (they provide the energy source). The moderator is like kindling that helps the fire catch (slowing neutrons so they can be captured). The control rods are like a fire blanket you can lower over the flames — they absorb the neutrons that keep the chain reaction going. The coolant carries the heat away to do useful work, like a pot of water placed over the fire to generate steam.

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Why Enrichment Matters

Natural uranium is only 0.7% U-235 and 99.3% U-238. Since U-238 tends to absorb fast neutrons without fissioning (it is fissionable but not fissile), the fuel must be enriched — the proportion of U-235 is increased to 3–5% for reactor fuel. Weapons-grade uranium requires enrichment above 90%.

SECTION 8

Fission vs. Fusion — Connections to Advanced Theory

Both fission and fusion release energy by moving nuclei toward the peak of the binding energy per nucleon curve at iron-56. However, the two processes approach this peak from opposite directions, involve very different conditions, and present distinct engineering challenges. The IB syllabus treats both in Option E, and understanding their comparison deepens your grasp of both.

Feature	Fission	Fusion
Definition	Heavy nucleus splits into lighter fragments	Light nuclei combine to form a heavier nucleus
Fuel	U-235, Pu-239 (rare, requires mining and enrichment)	Hydrogen isotopes — deuterium and tritium (abundant)
Energy per nucleon	≈ 0.9 MeV per nucleon	≈ 3–7 MeV per nucleon (much higher)
Conditions	Requires slow neutron capture; achievable at room temperature	Requires temperatures > 10⁷ K to overcome Coulomb barrier
Radioactive waste	Produces long-lived radioactive fission products (half-lives up to millions of years)	Produces minimal long-lived waste; main product (helium) is stable
Current status	Mature technology — hundreds of reactors operating worldwide	Still experimental — sustained net energy gain not yet achieved commercially
Direction on BE/A curve	Heavy nuclei move LEFT toward Fe-56 peak	Light nuclei move RIGHT toward Fe-56 peak

Comparison of nuclear fission and nuclear fusion

As you continue through Option E of the IB Physics syllabus, you will encounter fusion in greater detail (E.4 also addresses it). The key connection is that both processes are explained by the same binding energy curve. Understanding fission thoroughly provides the foundation for understanding fusion — and for evaluating the complex trade-offs involved in choosing energy sources for the future. Questions about nuclear waste management, nuclear proliferation, and safety are also explored in the IB curriculum as part of the broader societal implications of nuclear physics.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why nuclear fission releases energy by referring to the binding energy per nucleon curve. In your answer, identify where uranium-235 and its fission products sit on the curve and explain the direction of the energy change.

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Uranium-235 has a binding energy per nucleon of approximately 7.6 MeV, which is lower than the peak near iron-56 (≈ 8.8 MeV). When U-235 fissions, the two product nuclei (with mass numbers around 90–145) have binding energies per nucleon closer to 8.5 MeV. Since the products are more tightly bound per nucleon than the original uranium, the total binding energy of the system increases. This increase in binding energy corresponds to a decrease in total mass (mass defect), and this 'missing' mass is released as kinetic energy of the fragments and radiation, according to E = Δm × c². In short, fission releases energy because the products are more stable than the reactant.

Question 2: BASIC CALCULATION

A single fission event releases approximately 200 MeV of energy. Convert this energy into joules. (Use: 1 eV = 1.60 × 10⁻¹⁹ J)

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First, convert MeV to eV: 200 MeV = 200 × 10⁶ eV = 2.00 × 10⁸ eV. Then multiply by the conversion factor: E = 2.00 × 10⁸ × 1.60 × 10⁻¹⁹ J = 3.20 × 10⁻¹¹ J. This may seem like a tiny amount of energy, but remember that in 1 kg of U-235 there are approximately 2.56 × 10²⁴ atoms. If all of them fissioned, the total energy would be about 8.2 × 10¹³ J — equivalent to burning roughly 2,700 tonnes of coal.

Question 3: INTERMEDIATE

In the fission reaction ²³⁵U + ¹n → ¹⁴⁴Ba + ⁸⁹Kr + x(¹n), determine the number of neutrons x released. Then, given the following masses — U-235: 235.04393 u, Ba-144: 143.92295 u, Kr-89: 88.91763 u, neutron: 1.00866 u — calculate the energy released in MeV.

Show answer

To find x, conserve the mass number: 235 + 1 = 144 + 89 + x(1), so 236 = 233 + x, giving x = 3 neutrons. Now calculate the mass defect: m(reactants) = 235.04393 + 1.00866 = 236.05259 u. m(products) = 143.92295 + 88.91763 + 3(1.00866) = 143.92295 + 88.91763 + 3.02598 = 235.86656 u. Δm = 236.05259 − 235.86656 = 0.18603 u. Energy released = 0.18603 × 931.5 = 173.3 MeV. Including gamma and decay energy, the total is approximately 200 MeV, consistent with the standard estimate for U-235 fission.

Question 4: APPLIED

A nuclear power plant operates at 1000 MW of electrical output with a thermal efficiency of 33%. If each fission of U-235 releases 200 MeV, estimate how many kilograms of U-235 the plant consumes per day. (Avogadro's number = 6.022 × 10²³ mol⁻¹, molar mass of U-235 = 235 g/mol)

Show answer

First, find the total thermal power: P(thermal) = 1000 MW / 0.33 ≈ 3030 MW = 3.03 × 10⁹ W. Energy per day: E = 3.03 × 10⁹ × 86400 s = 2.618 × 10¹⁴ J. Energy per fission in joules: 200 MeV = 3.204 × 10⁻¹¹ J. Number of fissions per day: N = 2.618 × 10¹⁴ / 3.204 × 10⁻¹¹ ≈ 8.17 × 10²⁴ fissions. Number of moles = 8.17 × 10²⁴ / 6.022 × 10²³ ≈ 13.57 mol. Mass = 13.57 × 235 g ≈ 3189 g ≈ 3.2 kg per day. This illustrates the remarkable energy density of nuclear fuel — a 1 GW plant uses only about 3 kg of U-235 daily, compared to thousands of tonnes of coal for a similar coal plant.

Question 5: CRITICAL THINKING

A student argues: 'Since each fission of U-235 produces 2–3 neutrons, and only 1 is needed to sustain the chain reaction, the reaction should always go supercritical and explode.' Identify and explain at least three reasons why this does not happen in a nuclear reactor.

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The student's reasoning neglects several crucial factors. First, many neutrons escape the reactor core entirely — this neutron leakage depends on the geometry and size of the core, and reactors are designed so that only enough neutrons remain to sustain k = 1. Second, control rods made of neutron-absorbing materials (boron, cadmium) are inserted into the core to deliberately capture excess neutrons, precisely adjusting k. Third, many neutrons are absorbed by U-238 (which makes up the majority of the fuel) without causing fission — U-238 captures fast neutrons and becomes U-239 rather than fissioning. Fourth, the moderator slows neutrons to thermal energies, and if the moderator is lost (as in a loss-of-coolant accident in a light-water reactor), the chain reaction actually slows because fast neutrons are less likely to cause U-235 fission — this is a built-in negative feedback mechanism. Finally, delayed neutrons (emitted seconds after fission from certain fission products) provide a natural time lag that makes the reactor controllable. These factors together ensure that a properly designed reactor cannot explode like a nuclear weapon.

SUMMARY

Lesson Summary

Summary

Nuclear fission is the splitting of a heavy nucleus (such as uranium-235) into two lighter fission fragments, 2–3 free neutrons, and approximately 200 MeV of energy. The energy comes from the mass defect — the small difference in total mass between reactants and products — converted to energy via E = Δm × c². On the binding energy per nucleon curve, fission moves heavy nuclei toward the iron-56 peak, where nucleons are most tightly bound.

In a nuclear reactor, the chain reaction is sustained at a critical state (k = 1) using control rods to absorb excess neutrons, a moderator to slow neutrons to thermal speeds, and a coolant to carry thermal energy to generators. Fission differs from fusion in that it splits heavy nuclei rather than combining light ones, but both processes release energy because they move nuclei toward greater binding energy per nucleon.