Nuclear Fusion

Nuclear fusion is the process in which two light nuclei combine to form a heavier nucleus, releasing energy in the process. It is the power source of every star in the universe, including our Sun, and it has the potential to provide virtually unlimited clean energy here on Earth — if we can solve the enormous engineering challenges involved.

Why Fusion Releases Energy

Looking at the binding energy per nucleon curve, light nuclei (low A) have low binding energy per nucleon. When two light nuclei fuse, the product has a higher binding energy per nucleon, meaning the nucleons are more tightly bound in the product. The increase in total binding energy corresponds to a decrease in mass (via E = mc²), and this mass difference is released as energy.

For example, deuterium (²₁H) has a binding energy per nucleon of about 1.11 MeV, while helium-4 has about 7.08 MeV per nucleon. Fusion reactions involving hydrogen isotopes therefore release enormous amounts of energy per nucleon.

Worked Example: Energy from D-T Fusion

Problem: Calculate the energy released in the D-T fusion reaction using binding energies.

²₁H + ³₁H → ⁴₂He + ¹₀n

Binding energies: D = 2.22 MeV (total), T = 8.48 MeV (total), He-4 = 28.3 MeV (total), neutron = 0 (free)

Total BE of reactants = 2.22 + 8.48 = 10.70 MeV Total BE of products = 28.3 + 0 = 28.3 MeV

Energy released = 28.3 − 10.7 = 17.6 MeV ✔

This agrees with the experimentally measured value.

Worked Example: Energy from D-T Fusion Using Masses

Masses: D = 2.01410 u, T = 3.01605 u, He-4 = 4.00260 u, n = 1.00866 u

Mass of reactants = 2.01410 + 3.01605 = 5.03015 u Mass of products = 4.00260 + 1.00866 = 5.01126 u Δm = 5.03015 − 5.01126 = 0.01889 u

Energy = 0.01889 × 931.5 = 17.6 MeV ✔

Both methods give the same answer, as expected.

Conditions for Fusion

Fusion requires bringing two positively charged nuclei close enough together (within ~10⁻¹⁵ m) for the strong nuclear force to take over and bind them. But both nuclei are positively charged, so there is a strong Coulomb (electrostatic) barrier that repels them.

To overcome this barrier, the nuclei must approach each other with sufficient kinetic energy. This requires:

1. Extremely High Temperature

At temperatures of around 10⁷ to 10⁸ K (tens to hundreds of millions of kelvin), the particles in a gas have enough thermal kinetic energy to overcome the Coulomb barrier. At these temperatures, matter exists as a plasma — a fully ionised gas of bare nuclei and free electrons.

Even at these temperatures, most collisions between nuclei do not result in fusion. Quantum tunnelling plays a crucial role: nuclei can "tunnel through" the Coulomb barrier even when their classical energy is insufficient, but the probability of tunnelling increases sharply with kinetic energy.

2. High Particle Density (Pressure)

High density increases the rate of collisions between nuclei. More collisions per second means more fusion reactions per second, which is necessary to sustain the process and produce useful amounts of energy.

3. Sufficient Confinement Time

The hot plasma must be held together long enough for enough fusion reactions to occur. If the plasma disperses too quickly, the energy released is less than the energy required to heat and confine it.

The product of density, temperature, and confinement time must exceed a certain threshold — the Lawson criterion — for the fusion process to produce net energy.

Estimating the Coulomb Barrier Height

Problem: Estimate the energy required for two protons to approach to within 1.0 fm of each other.

E = kq₁q₂/r = (8.99 × 10⁹)(1.6 × 10⁻¹⁹)² / (1.0 × 10⁻¹⁵)

E = (8.99 × 10⁹)(2.56 × 10⁻³⁸) / (1.0 × 10⁻¹⁵)

E = 2.3 × 10⁻¹³ J = 1.44 MeV

At temperature T, the average kinetic energy is (3/2)kT. Setting this equal to 1.44 MeV:

T = 2E/3k = 2 × 2.3 × 10⁻¹³ / (3 × 1.38 × 10⁻²³) = 1.1 × 10¹⁰ K

This is far higher than the Sun’s core temperature (1.5 × 10⁷ K), which demonstrates that quantum tunnelling is essential — without it, the Sun could not shine.

Fusion in Stars

Stars are natural fusion reactors. The enormous gravitational pressure in a star’s core provides the necessary density and temperature. The core of our Sun reaches about 1.5 × 10⁷ K and a density of about 150,000 kg m⁻³.

The Proton-Proton Chain

In stars like the Sun (and smaller), the dominant fusion process is the proton-proton (pp) chain, which converts hydrogen into helium through a sequence of steps:

Step 1: Two protons fuse to form deuterium, emitting a positron and an electron neutrino: ¹₁H + ¹₁H → ²₁H + ⁰₊₁β + νₑ

This is the slowest step (limited by the weak force) and is the bottleneck of the entire process. On average, a given proton in the Sun’s core waits about 10 billion years before undergoing this reaction.

Step 2: The deuterium fuses with another proton to form helium-3, emitting a gamma ray: ²₁H + ¹₁H → ³₂He + γ

Step 3: Two helium-3 nuclei fuse to form helium-4, releasing two protons: ³₂He + ³₂He → ⁴₂He + 2¹₁H

Overall: 4¹₁H → ⁴₂He + 2⁰₊₁β + 2νₑ + γ + 26.7 MeV

The Sun converts about 600 million tonnes of hydrogen into helium every second, losing about 4 million tonnes of mass as energy (P = Δm × c² / t ≈ 3.8 × 10²⁶ W).

The CNO Cycle

In heavier, hotter stars (core temperatures above about 1.7 × 10⁷ K), the carbon-nitrogen-oxygen (CNO) cycle becomes the dominant fusion pathway. Carbon, nitrogen, and oxygen nuclei act as catalysts in a cycle that has the same net effect — converting four protons into one helium-4 nucleus — but through a different sequence of reactions. The carbon acts as a catalyst: it is consumed and then regenerated during the cycle.

The CNO cycle is more temperature-sensitive than the pp chain (its rate depends roughly on T¹⁶ compared to T⁴ for the pp chain), which is why it dominates only in hotter stars.

Fusion on Earth

Reproducing the conditions inside a star on Earth is extraordinarily challenging. The two most promising fusion fuel reactions are:

Deuterium-Tritium (D-T) fusion: ²₁H + ³₁H → ⁴₂He + ¹₀n + 17.6 MeV

This is the easiest fusion reaction to achieve because it has the lowest Coulomb barrier for a useful energy yield. It is the reaction targeted by most current fusion experiments.

Deuterium-Deuterium (D-D) fusion: ²₁H + ²₁H → ³₂He + ¹₀n + 3.27 MeV or ²₁H + ²₁H → ³₁H + ¹₁H + 4.03 MeV

Comparison of Fusion Reactions

Reaction	Energy Released	Coulomb Barrier	Fuel Availability	Notes
D-T	17.6 MeV	Lowest	Tritium must be bred	Currently most promising
D-D	3.27 or 4.03 MeV	Higher	Abundant (seawater)	Harder to achieve
D-³He	18.3 MeV	Higher still	³He very rare on Earth	Produces no neutrons — ideal but impractical
p-p	26.7 MeV (chain)	Very high	Abundant	Far too slow without stellar conditions

Why Fusion Is Difficult on Earth

Temperature: plasma must reach ~10⁸ K (hotter than the Sun’s core, because we cannot achieve the Sun’s extreme density and gravitational confinement)
Confinement: no physical container can hold plasma at 100 million kelvin — it would instantly melt or vaporise any material
Instabilities: hot plasma is extremely difficult to control — turbulence and instabilities cause it to lose energy rapidly
Net energy: to date, no fusion experiment has achieved sustained net energy output (energy out > energy in) in a commercially viable way

Tokamak Reactors

The most advanced approach to magnetic confinement fusion uses a tokamak — a doughnut-shaped (toroidal) chamber that uses powerful magnetic fields to confine the plasma and keep it away from the walls.

Key features:

Toroidal magnetic field — wraps around the doughnut shape
Poloidal magnetic field — wraps the short way around, created by a current flowing through the plasma
The combined fields create helical field lines that confine the charged plasma particles
Superconducting magnets are used to generate the enormous fields required
The plasma is heated by a combination of ohmic heating, neutral beam injection, and radio-frequency heating

Major tokamak projects include JET (Joint European Torus, which achieved the first significant D-T fusion in 1997) and ITER (International Thermonuclear Experimental Reactor, under construction in France, aiming to demonstrate net energy gain).

Fusion vs Fission: Energy Output

Aspect	Fusion	Fission
Energy per reaction	~17.6 MeV (D-T)	~200 MeV (U-235)
Energy per unit mass of fuel	Much higher (fuel is lighter)	Lower
Energy per nucleon	~3.5 MeV	~0.9 MeV
Fuel availability	Deuterium from seawater (virtually unlimited)	Uranium (limited reserves)
Radioactive waste	Short-lived (activation products, tritium)	Long-lived (fission products, actinides)
Risk of runaway reaction	No — plasma cools and fusion stops	Yes — requires active control

Common Exam Mistakes

Saying fusion does not produce radioactive waste. It does — the neutrons from D-T fusion activate the reactor walls, creating radioactive isotopes. However, these are much shorter-lived than fission waste.
Confusing the Sun’s core temperature with the temperature needed on Earth. Earth-based reactors need ~10⁸ K (higher than the Sun) because they cannot achieve the Sun’s density and confinement.
Saying the Sun uses D-T fusion. The Sun uses the proton-proton chain, which starts with p-p fusion. D-T fusion is the reaction targeted for Earth-based reactors because it has the lowest barrier.
Forgetting quantum tunnelling. Without tunnelling, fusion in the Sun would be impossible at its actual core temperature. Many students calculate the classical barrier height and then forget to mention tunnelling.

Summary

Fusion releases energy because the product has higher binding energy per nucleon than the reactants. It requires extreme temperatures (~10⁸ K), high density, and long confinement times to overcome the Coulomb barrier. Stars achieve this through gravitational confinement; on Earth, magnetic confinement in tokamaks is the leading approach. The proton-proton chain powers the Sun, while heavier stars use the CNO cycle. Despite decades of research, achieving sustained net energy from fusion on Earth remains one of the greatest engineering challenges.

A-Level Deep Dive: Nuclear Fusion

Spec mapping

Edexcel 9PH0 specification, Topic 8 — Nuclear and Particle Physics, with extension into Topic 11 — Astrophysics and Cosmology covers the fusion of light nuclei, the role of binding energy per nucleon, and the conditions required for fusion to proceed in stars and in laboratory devices (refer to the official specification document for exact wording). Although fusion is introduced inside the Topic 8 nuclear-physics block, candidates are expected to draw heavily on Topic 5 (Thermodynamics and ideal gases) for the kinetic-theory link $\tfrac{3}{2}k_BT$ that justifies the temperatures required, on Topic 9 (Thermal energy) for confinement-time arguments, and on Topic 11 for stellar-fusion pathways including the proton–proton chain and the CNO cycle. The Edexcel formula booklet provides $E = \Delta m c^2$ , the unified atomic mass unit conversion ( $1\,\text{u} = 931.5\,\text{MeV}/c^2$ ), and $\tfrac{3}{2}k_BT$ , but does not provide reaction equations or barrier heights — these must be reasoned from first principles.

Worked example with full mark scheme

Question (8 marks):

The deuterium–tritium (D–T) fusion reaction is the most accessible route to terrestrial fusion energy:

$^2_1\text{H} + {}^3_1\text{H} \rightarrow {}^4_2\text{He} + {}^1_0\text{n} + Q$

Given atomic masses: $m(^2_1\text{H}) = 2.01410\,\text{u}$ , $m(^3_1\text{H}) = 3.01605\,\text{u}$ , $m(^4_2\text{He}) = 4.00260\,\text{u}$ , $m(^1_0\text{n}) = 1.00867\,\text{u}$ . ( $1\,\text{u} = 931.5\,\text{MeV}/c^2$ .)

(a) Show that the equation is balanced for nucleon and proton number. (2)

(b) Calculate the energy released, $Q$ , in MeV. (4)

(c) Explain why this reaction proceeds only at temperatures of order $10^8\,\text{K}$ , despite being energetically favourable. (2)

Solution with mark scheme:

(a) Nucleon number on the LHS: $2 + 3 = 5$ . Nucleon number on the RHS: $4 + 1 = 5$ . B1 — nucleon balance shown explicitly.

Proton number on the LHS: $1 + 1 = 2$ . Proton number on the RHS: $2 + 0 = 2$ . B1 — proton balance shown explicitly.

(b) Step 1 — total mass of reactants.

$m_\text{R} = 2.01410 + 3.01605 = 5.03015\,\text{u}$

M1 — sum of reactant masses written down.

Step 2 — total mass of products.

$m_\text{P} = 4.00260 + 1.00867 = 5.01127\,\text{u}$

M1 — sum of product masses written down.

Step 3 — mass defect.

$\Delta m = m_\text{R} - m_\text{P} = 5.03015 - 5.01127 = 0.01888\,\text{u}$

A1 — mass defect to at least three significant figures.

Step 4 — convert to energy using $1\,\text{u} \equiv 931.5\,\text{MeV}/c^2$ .

$Q = 0.01888 \times 931.5 \approx 17.6\,\text{MeV}$

A1 — final answer $Q \approx 17.6\,\text{MeV}$ with units. Note: roughly $3.5\,\text{MeV}$ is carried by the helium-4 nucleus and $14.1\,\text{MeV}$ by the neutron, by conservation of momentum, but the question asks only for $Q$ .

(c) Both nuclei are positively charged (D has $Z = 1$ , T has $Z = 1$ ), so they repel each other electrostatically. To approach within the range of the strong nuclear force ( $\sim 1\text{–}3\,\text{fm}$ ), they must overcome the Coulomb barrier. The thermal kinetic energy per particle scales as $\tfrac{3}{2}k_BT$ , so reaching the barrier requires temperatures of order $10^8\,\text{K}$ (with quantum tunnelling allowing the reaction to proceed at somewhat lower temperatures than the classical estimate predicts). B1 — Coulomb barrier identified as the reason. B1 — link to thermal kinetic energy and the resulting temperature scale.

Total: 8 marks (B2 M2 A2 B2 as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A future fusion reactor is proposed to fuse deuterium with helium-3:

$^2_1\text{H} + {}^3_2\text{He} \rightarrow {}^4_2\text{He} + {}^1_1\text{p} + Q$

(a) State, with a reason, whether the Coulomb barrier for D– $^3$ He fusion is higher or lower than that for D–T fusion. (2)

(b) Explain, with reference to the binding-energy-per-nucleon curve, why the reaction is exothermic. (2)

(c) Suggest one advantage and one disadvantage of D– $^3$ He fusion compared with D–T fusion as the basis of a power-station reactor. (2)

Mark scheme decomposition by AO:

(a)

B1 (AO1.1) — barrier higher than D–T.
B1 (AO2.1) — because $Z = 2$ for helium-3 versus $Z = 1$ for tritium, and the Coulomb barrier scales as $Z_1 Z_2 / r$ , so doubling one charge doubles the barrier.

(b)

B1 (AO1.2) — helium-4 lies higher on the binding-energy-per-nucleon curve than the lighter reactants.
B1 (AO2.1) — therefore mass defect is positive and energy is released as the system moves toward the more strongly bound product.

(c)

B1 (AO1.2) — advantage: the products of D– $^3$ He are charged particles only (no neutron), reducing neutron-induced activation of the reactor structure and simplifying shielding.
B1 (AO3.1a) — disadvantage: $^3$ He is extremely rare on Earth (it is produced almost entirely from the decay of tritium or extracted in trace amounts) so fuel supply is the limiting factor.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. This is a typical Edexcel synoptic Paper 2 structure where AO1 recall (charge values, BE/N curve location) is rewarded but the marks above C-grade demand AO2 reasoning that links scaling laws and physical consequences.

Synoptic links

Connects to:

Topic 8 — Binding energy curve: the master diagram of nuclear physics. Fusion releases energy when light nuclei (left of iron-56) combine to form a more strongly bound product. The curve rises steeply from hydrogen to helium-4 (the source of the $\sim 17\,\text{MeV}$ released in D–T) and continues to rise more slowly to iron-56, the most strongly bound nucleus. Beyond iron, fission rather than fusion releases energy.
Topic 11 — Stellar nucleosynthesis: stars are gravitationally confined fusion reactors. The Sun fuses hydrogen via the proton–proton chain ( $4\,{}^1_1\text{H} \rightarrow {}^4_2\text{He} + 2\text{e}^+ + 2\nu_e$ ), releasing about $26.7\,\text{MeV}$ per helium produced. Heavier main-sequence stars use the CNO cycle, in which carbon, nitrogen and oxygen catalyse hydrogen burning. Beyond core hydrogen exhaustion, stars climb the BE/N curve through helium burning, carbon burning and so on, up to iron-56.