IB PHYSICS • NUCLEAR AND QUANTUM PHYSICS

Understand Fusion & Stars — Understand E.5 Fusion and stars

Discover how nuclear fusion powers the stars and forges the elements of the universe.

Historical Context & Motivation

For centuries, the source of the Sun's energy was one of the deepest puzzles in science. In the nineteenth century, physicists like Lord Kelvin estimated that if the Sun burned coal, it would exhaust its fuel in only a few thousand years — far shorter than the geological evidence suggested. Something far more powerful than chemical burning had to be at work. The quest to explain stellar energy ultimately led to nuclear fusion, the process by which light atomic nuclei combine to release enormous amounts of energy.

1905
Mass–Energy Equivalence
Albert Einstein publishes E = mc², revealing that mass can be converted into energy. This equation becomes the theoretical foundation for understanding stellar power.
1920
Eddington's Hypothesis
Arthur Eddington proposes that the Sun is powered by hydrogen nuclei fusing into helium, converting mass into energy according to Einstein's equation.
1938
The Proton–Proton Chain
Hans Bethe and Charles Critchfield work out the detailed nuclear reactions of the proton–proton chain, explaining how main-sequence stars like the Sun generate energy.
1957
Stellar Nucleosynthesis
Burbidge, Burbidge, Fowler, and Hoyle publish the B²FH paper, showing how fusion in stars creates elements heavier than helium through successive burning stages.
2022
Laboratory Ignition
The National Ignition Facility achieves fusion ignition on Earth for the first time, producing more energy from fusion reactions than the laser energy used to trigger them.

The central question this topic addresses is straightforward yet profound: How does the fusion of hydrogen into helium — and heavier elements beyond — power a star throughout its entire life cycle? Understanding fusion connects Einstein's mass–energy equivalence to the life and death of every star in the universe.

Core Principles of Nuclear Fusion

Nuclear fusion occurs when two light nuclei merge to form a single, heavier nucleus. For this to happen, the nuclei must overcome their mutual Coulomb repulsion — the electrostatic force that pushes positive charges apart. Inside stars, temperatures of tens of millions of kelvin give nuclei enough kinetic energy to get close enough for the strong nuclear force to bind them together. The resulting nucleus has slightly less mass than the sum of the original nuclei, and that difference, called the mass defect, is released as energy according to E = mc².

1

Coulomb Barrier

Positively charged nuclei repel each other. Overcoming this barrier requires extreme temperatures (≈ 10⁷ K) or quantum tunnelling, which allows nuclei to 'pass through' the barrier with a small probability.
2

Mass Defect & Binding Energy

The mass of a nucleus is less than the total mass of its individual protons and neutrons. The 'missing' mass has been converted into binding energy that holds the nucleus together.
3

Binding Energy per Nucleon

Iron-56 (⁵⁶Fe) sits at the peak of the binding energy per nucleon curve. Nuclei lighter than iron release energy by fusing; nuclei heavier than iron release energy by splitting (fission).
4

Hydrostatic Equilibrium

A star balances the inward pull of gravity against the outward radiation pressure from fusion in its core. This self-regulating balance is called hydrostatic equilibrium.
5

Stellar Nucleosynthesis

Stars create heavier elements through successive stages of fusion. The most massive stars burn through hydrogen, helium, carbon, neon, oxygen, and silicon before their cores collapse.
KEY TAKEAWAY
Think of fusion like building with LEGOs. When you snap small bricks together into a larger structure, the structure is actually slightly lighter than all the individual bricks combined. That tiny bit of 'missing weight' has been converted into the satisfying 'click' energy you feel. In a star, that 'click' is released as light and heat — and because c² is such a huge number, even a tiny mass defect produces an enormous amount of energy.

The Proton–Proton Chain Reaction

The primary fusion process in stars like our Sun is the proton–proton (pp) chain. This sequence of reactions converts four hydrogen nuclei (protons) into one helium-4 nucleus, releasing energy in the form of gamma-ray photons, positrons, and neutrinos along the way. The diagram below illustrates the three main steps of the pp-I chain, which accounts for about 85% of the Sun's energy output.

The three steps of the pp-I chain. Blue spheres represent protons (p) and purple spheres represent neutrons (n). In Step 1, two protons fuse to create deuterium (²H), releasing a positron (e⁺) and a neutrino (νₑ). In Step 2, deuterium absorbs another proton to form helium-3 (³He), releasing a gamma-ray photon (γ). In Step 3, two ³He nuclei combine to produce helium-4 (⁴He) and two free protons. The net result converts four protons into one ⁴He nucleus, releasing 26.7 MeV of energy.

Notice that each step of the chain releases energy because the products are more tightly bound than the reactants. The positrons (e⁺) quickly annihilate with electrons in the stellar plasma, converting their combined mass into additional gamma-ray energy. The neutrinos, by contrast, barely interact with matter and escape the star almost immediately — which is why detecting solar neutrinos on Earth provides direct evidence that fusion is occurring in the Sun's core right now.

Mathematical Framework

The energy released in nuclear fusion can be calculated using Einstein's mass–energy equivalence and the concept of mass defect. These equations allow you to determine how much energy a fusion reaction produces, which connects directly to how stars generate their luminosity.

MASS–ENERGY EQUIVALENCE
E = mc²
Where E is the energy released (in joules), m is the mass defect (in kg), and c is the speed of light (3.00 × 10⁸ m s⁻¹). Even a tiny mass defect produces an enormous energy because c² ≈ 9 × 10¹⁶ m² s⁻².
MASS DEFECT
Δm = (mass of reactants) − (mass of products)
The mass defect Δm is measured in atomic mass units (u), where 1 u = 1.661 × 10⁻²⁷ kg. In energy terms, 1 u corresponds to 931.5 MeV/c².
ENERGY IN MeV
E = Δm × 931.5 MeV/u
This is a shortcut: multiply the mass defect in atomic mass units directly by 931.5 to get the energy released in mega-electronvolts (MeV). This avoids converting to kilograms and using c² explicitly.
STELLAR LUMINOSITY & FUSION RATE
L = ΔE / Δt
A star's luminosity (L) is its total power output in watts. If each fusion reaction releases energy ΔE, then the number of reactions per second equals L / ΔE. The Sun's luminosity is 3.85 × 10²⁶ W, meaning roughly 3.6 × 10³⁸ pp-chain reactions occur every second.
💡 IB Exam Tip
On the IB exam, you are given a data booklet with atomic masses in unified atomic mass units (u). To find the energy released in a nuclear reaction: (1) sum the masses of reactants, (2) sum the masses of products, (3) find the mass defect Δm, and (4) convert using E = Δm × 931.5 MeV/u. Always check whether the question asks for the answer in MeV or joules.

Fusion Stages & Stellar Life Cycles

Not all stars burn their fuel in the same way. The fusion pathway a star follows and how far it can go in creating heavier elements depends critically on its mass. A star like our Sun fuses hydrogen into helium through the proton–proton chain, while more massive stars (above about 1.3 solar masses) primarily use the CNO cycle (carbon–nitrogen–oxygen cycle), which uses carbon as a catalyst. As a star exhausts its hydrogen fuel, it may begin fusing helium into carbon and oxygen through the triple-alpha process. The most massive stars continue this pattern, fusing progressively heavier elements in concentric shells until iron accumulates in the core.

The onion-shell structure of a massive pre-supernova star (≈ 25 solar masses). Each concentric shell fuses a different element: hydrogen on the outermost layer, through helium, carbon, neon, oxygen, and silicon, with an inert iron (⁵⁶Fe) core at the centre. Once iron accumulates, fusion can no longer release energy and the core collapses, triggering a supernova.

The key insight is that each successive fusion stage releases less energy per reaction and burns through its fuel faster. Hydrogen burning in a 25-solar-mass star lasts about 7 million years, but silicon burning to iron lasts only about one day. When the iron core reaches the Chandrasekhar limit (approximately 1.4 solar masses), it can no longer support itself against gravity. The core collapses in milliseconds, and the outer layers are expelled in a core-collapse supernova, which is also how elements heavier than iron (gold, uranium, etc.) are formed through rapid neutron capture (the r-process).

Successive nuclear burning stages in a massive star
Fusion StageFuel → ProductsTemperature (K)Duration (25 M☉)
Hydrogen burningH → He≈ 4 × 10⁷≈ 7 × 10⁶ years
Helium burningHe → C, O≈ 2 × 10⁸≈ 5 × 10⁵ years
Carbon burningC → Ne, Mg≈ 8 × 10⁸≈ 600 years
Neon burningNe → O, Mg≈ 1.5 × 10⁹≈ 1 year
Oxygen burningO → Si, S≈ 2 × 10⁹≈ 6 months
Silicon burningSi → Fe≈ 3.5 × 10⁹≈ 1 day

Worked Example: Energy from the pp Chain

Let's calculate the energy released when four protons fuse to form one helium-4 nucleus through the complete proton–proton chain. We will use atomic masses from the IB data booklet.

Energy Released in 4p → ⁴He + 2e⁺ + 2νₑ
1
Step 1 — Identify Given MassesFrom the data booklet: mass of a proton = 1.007276 u, mass of helium-4 nucleus = 4.002602 u, mass of an electron (= mass of positron) = 0.000549 u. Since we use nuclear masses (not atomic masses), we account for the two positrons in the products separately.
2
Step 2 — Calculate Total Reactant MassWe start with four protons: total reactant mass = 4 × 1.007276 u = 4.029104 u.
Total reactant mass = 4.029104 u
3
Step 3 — Calculate Total Product MassThe products are one ⁴He nucleus and two positrons (the neutrinos are essentially massless). Total product mass = 4.002602 + 2 × 0.000549 = 4.002602 + 0.001098 = 4.003700 u.
Total product mass = 4.003700 u
4
Step 4 — Find the Mass DefectΔm = total reactant mass − total product mass = 4.029104 − 4.003700 = 0.025404 u.
Δm = 0.025404 u
5
Step 5 — Convert to Energy (MeV)Using the conversion factor 1 u = 931.5 MeV/c²: E = 0.025404 × 931.5 = 23.66 MeV. Note that when the two positrons annihilate with two electrons in the stellar plasma, an additional 2 × 2 × 0.511 MeV = 2.044 MeV is released, bringing the total to approximately 25.7 MeV. The accepted total including all secondary effects is about 26.7 MeV.
Energy released ≈ 26.7 MeV per complete pp-chain cycle
6
Step 6 — Convert to Joules (if required)1 MeV = 1.602 × 10⁻¹³ J, so E = 26.7 × 1.602 × 10⁻¹³ = 4.28 × 10⁻¹² J. This seems tiny, but the Sun performs approximately 3.6 × 10³⁸ of these reactions every second.
E ≈ 4.28 × 10⁻¹² J per reaction

Fusion vs Fission: Strengths & Limitations

Both nuclear fusion and nuclear fission release energy by exploiting the binding energy curve, but they do so from opposite ends. Understanding their differences is essential for the IB syllabus and helps explain why we use fission in current power plants while fusion remains a goal for the future.

Comparison of fusion and fission as energy sources
FeatureFusionFission
ProcessLight nuclei combine into heavier onesHeavy nuclei split into lighter ones
FuelHydrogen isotopes (deuterium, tritium)Uranium-235, Plutonium-239
Energy per nucleonHigher (≈ 6.7 MeV per nucleon for pp chain)Lower (≈ 0.9 MeV per nucleon for ²³⁵U)
Conditions requiredExtreme temperature (> 10⁷ K) and pressureCritical mass of fissile material + neutron moderator
Radioactive wasteMinimal — helium is the main productSignificant — long-lived radioactive isotopes produced
Where it occurs naturallyStellar coresNatural uranium deposits (e.g., Oklo reactor)
Current technologyExperimental (tokamaks, laser inertial confinement)Commercial nuclear power plants operational worldwide
KEY TAKEAWAY
Both fusion and fission release energy because they move nuclei toward the peak of the binding energy per nucleon curve (iron-56). Fusion climbs the curve from the left (light nuclei getting more tightly bound), while fission descends from the right (heavy nuclei shedding excess energy by splitting). The curve's peak at iron explains why iron cannot fuel a star — there's nowhere left to go.

Connections to Advanced Topics

The physics of fusion and stars connects to several advanced topics you may encounter in HL extensions, university-level astrophysics, or the IB Option D (Astrophysics). Understanding the basics of the pp chain and the binding energy curve prepares you for deeper investigations into how the universe creates and distributes matter.

How E.5 concepts extend into advanced physics
IB E.5 ConceptAdvanced Extension
pp chain and CNO cycleDetailed nuclear reaction networks; neutrino oscillations from solar neutrino problem
Binding energy per nucleon curveSemi-empirical mass formula (SEMF / Weizsäcker formula); nuclear shell model
Hydrostatic equilibriumLane–Emden equation; Eddington luminosity limit; radiation vs convection zones
Iron core collapse → supernovaNeutron degeneracy pressure; neutron stars and pulsars; black hole formation
Stellar nucleosynthesis (up to Fe)r-process and s-process nucleosynthesis for elements beyond iron; neutron star mergers

One of the most exciting frontiers in physics is controlled fusion on Earth. Projects like ITER (International Thermonuclear Experimental Reactor) in France use powerful magnetic fields in a tokamak to confine hydrogen plasma at temperatures exceeding 150 million kelvin — about ten times hotter than the Sun's core. While the Sun can rely on its enormous gravitational pressure to sustain fusion, we must use clever engineering to replicate those conditions in a laboratory. If successful, fusion power could provide virtually limitless clean energy using deuterium extracted from seawater.

Practice Problems

PROBLEM 1CONCEPTUAL
Explain why fusion of elements heavier than iron does not release energy. Refer to the binding energy per nucleon curve in your answer.
PROBLEM 2BASIC CALCULATION
In the reaction ²H + ³H → ⁴He + n, the mass of deuterium (²H) is 2.014102 u, tritium (³H) is 3.016049 u, helium-4 (⁴He) is 4.002602 u, and the neutron is 1.008665 u. Calculate the energy released in MeV.
PROBLEM 3INTERMEDIATE
The Sun's luminosity is 3.85 × 10²⁶ W. Each complete pp-chain cycle releases approximately 26.7 MeV of energy. (a) Calculate the number of pp-chain reactions occurring per second. (b) Determine the mass of hydrogen consumed per second.
PROBLEM 4APPLIED
A tokamak fusion reactor aims to produce 500 MW of power using deuterium–tritium (D–T) fusion. Each D–T reaction releases 17.6 MeV. (a) How many fusion reactions per second are needed? (b) If the reactor runs continuously for one year, what total mass of tritium (mass = 3.016 u) is consumed?
PROBLEM 5CRITICAL THINKING
A star significantly more massive than the Sun uses the CNO cycle instead of the pp chain for hydrogen fusion. Both processes have the same net reaction (4p → ⁴He + 2e⁺ + 2νₑ) and release similar total energy. Yet the CNO cycle dominates in massive stars while the pp chain dominates in the Sun. Using your understanding of Coulomb repulsion and temperature dependence, explain why the dominant fusion pathway depends on stellar mass.

Lesson Summary

Stars are powered by nuclear fusion, which combines light nuclei into heavier ones and converts the mass defect into energy via E = mc². In the Sun, the proton–proton chain converts four protons into one helium-4 nucleus, releasing approximately 26.7 MeV per cycle. More massive stars use the CNO cycle and progress through successive burning stages (helium, carbon, neon, oxygen, silicon) until an inert iron core forms. Iron sits at the peak of the binding energy per nucleon curve, meaning no further energy can be released by fusion, and the star's core collapses.

The balance between gravitational contraction and outward radiation pressure is called hydrostatic equilibrium. When you calculate energy from fusion, find the mass defect (Δm) in atomic mass units, then multiply by 931.5 MeV/u. Fusion releases more energy per nucleon than fission, produces minimal radioactive waste, and is the ultimate goal for clean energy on Earth — though achieving the extreme temperatures needed to overcome the Coulomb barrier remains a major engineering challenge.

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