Nuclear fusion

Deep inside the Sun, four hydrogen nuclei are, at every instant, being crushed into one helium nucleus — and a little bit of mass simply vanishes, reappearing as light and heat. That vanishing mass is nuclear fusion: light nuclei fusing into a heavier, more tightly bound nucleus, releasing energy according to E=Δm c2E = \Delta m\,c^2. It is the engine of every star in the sky, the physics behind the hydrogen bomb, and the still-unrealised dream of a clean, near-limitless power source on Earth. The catch is that fusion demands temperatures of millions of kelvin just to get two positively charged nuclei close enough to touch — which is exactly why we have not yet tamed it, even though we understand it completely. This article works out why fusion releases energy, why it needs such extreme conditions, and how to turn measured atomic masses into a precise energy yield.

The core insight

Fusing light nuclei into a more tightly bound heavier nucleus releases energy equal to Δm c2\Delta m\,c^2, but only at temperatures of order 10610^6–107 K10^7\ \text{K} or higher, hot enough for nuclei to overcome their mutual Coulomb repulsion.

The core idea

The reaction. The syllabus's representative fusion process is the overall conversion of hydrogen into helium:

4 11H⟶ 24He+2e++2ν+energy,energy≈26.7 MeV4\,^1_1\text{H} \longrightarrow \,^4_2\text{He} + 2e^+ + 2\nu + \text{energy}, \qquad \text{energy} \approx 26.7\ \text{MeV}

Four protons (charge +4+4) become a helium-4 nucleus (charge +2+2) plus two positrons, which keeps both charge and lepton number balanced. In stars this happens through a multi-step chain (the proton–proton chain), but only this overall summary reaction is required content — not the intermediate steps.

Why energy comes out. On the binding-energy-per-nucleon curve, light nuclei like hydrogen sit low, on the steeply rising left-hand slope; helium sits noticeably higher, near the curve's early peak. When nucleons rearrange from four separate protons into one helium-4 nucleus, they end up more tightly bound on average. That increase in total binding energy is exactly the energy released, mostly as gamma photons and the kinetic energy of the products — the mirror image of fission, which extracts energy from heavy nuclei on the curve's gently falling right-hand side.

Why such extreme temperature. Every nucleus entering a fusion reaction carries positive charge, so nuclei repel each other electrostatically well before they get close enough for the short-range strong force to bind them. Only nuclei colliding with enough kinetic energy can punch through this Coulomb barrier. That requires temperatures of order 10610^6–107 K10^7\ \text{K} or more, at which matter is a fully ionised plasma and a workable fraction of collisions succeed — reactions occurring under these conditions are called thermonuclear reactions.

Why control is hard. No solid container can touch a plasma at millions of kelvin, so the plasma must be confined magnetically or by inertia (rapid compression) instead. Sustaining fusion long enough, and extracting more energy than is spent confining the plasma, has proved an extremely difficult engineering problem — which is why fusion power stays largely experimental while fission reactors are already commercial.

Where it happens. In the Sun's core, gravitational contraction maintains a temperature of about 1.5×107 K1.5\times10^7\ \text{K}, sustaining continuous proton–proton chain fusion — the source of stellar energy over billions of years. The hydrogen bomb triggers the same physics in an uncontrolled burst: a fission explosion briefly creates temperatures high enough to ignite fusion of hydrogen isotopes, releasing far more energy than fission alone.

Figure
Binding energy per nucleon versus mass number showing hydrogen low on the rising slope fusing toward helium near the peak while heavy nuclei on the falling slope undergo fission
Binding energy per nucleon — why fusion of light nuclei releases energy mass number A BE per nucleon (MeV) H (low) He (high) fusion: energy released U-235 fission (heavy nuclei): energy released too peak near iron
Derivation AMass defect to fusion energy release

The goal is to turn measured atomic masses of the reactants and products of a fusion reaction into the energy released, using nothing beyond conservation of mass-energy and the conversion factor 1 u=931.5 MeV/c21\ u = 931.5\ \text{MeV}/c^2; this is the same tool used for fission, applied here to the deuterium–deuterium reaction 12H+ 12H→ 23He+ 01n^2_1\text{H} + \,^2_1\text{H} \rightarrow \,^3_2\text{He} + \,^1_0n.

Step 1. Total mass before the reaction is the sum of the two reactant masses: with m(12H)=2.014102 um(^2_1\text{H}) = 2.014102\ u, mbefore=2×2.014102=4.028204 um_{\text{before}} = 2\times2.014102 = 4.028204\ u.

Step 2. Total mass after the reaction is the sum of the product masses: with m(23He)=3.016029 um(^3_2\text{He}) = 3.016029\ u and m(01n)=1.008665 um(^1_0n) = 1.008665\ u, mafter=3.016029+1.008665=4.024694 um_{\text{after}} = 3.016029 + 1.008665 = 4.024694\ u.

Step 3. The mass defect is what disappears in the rearrangement: Δm=mbefore−mafter=4.028204−4.024694=0.003510 u\Delta m = m_{\text{before}} - m_{\text{after}} = 4.028204 - 4.024694 = 0.003510\ u. This small deficit is the physical signature that the products are more tightly bound than the reactants.

Step 4. Convert the mass defect to energy using Einstein's relation in its practical nuclear form:

E=Δm c2=Δm (in u)×931.5 MeVE = \Delta m\,c^2 = \Delta m \,(\text{in } u)\times931.5\ \text{MeV}

where: Δm\Delta m is the mass defect in atomic mass units, and 931.5 MeV931.5\ \text{MeV} is the energy equivalent of 1 u1\ u. Substituting, E=0.003510×931.5=3.27 MeVE = 0.003510\times931.5 = 3.27\ \text{MeV}, carried away as the kinetic energy of the 23He^3_2\text{He} nucleus and the neutron.

Worked example

In the fusion reaction 12H+ 12H→ 23He+ 01n^2_1\text{H} + \,^2_1\text{H} \rightarrow \,^3_2\text{He} + \,^1_0n, calculate the energy released, using m(12H)=2.014102 um(^2_1\text{H}) = 2.014102\ u, m(23He)=3.016029 um(^3_2\text{He}) = 3.016029\ u, m(01n)=1.008665 um(^1_0n) = 1.008665\ u.

Step 1. Mass before: mbefore=2×2.014102=4.028204 um_{\text{before}} = 2\times2.014102 = 4.028204\ u.

Step 2. Mass after: mafter=3.016029+1.008665=4.024694 um_{\text{after}} = 3.016029 + 1.008665 = 4.024694\ u.

Step 3. Mass defect: Δm=4.028204−4.024694=0.003510 u\Delta m = 4.028204 - 4.024694 = 0.003510\ u.

Step 4. Energy released: E=0.003510×931.5=3.27 MeVE = 0.003510\times931.5 = 3.27\ \text{MeV}. About 3.27 MeV is released in this deuterium–deuterium fusion reaction.

Competitive stretch

Compare, per kilogram of fuel, fusing hydrogen into helium (26.7 MeV per 4 hydrogen nuclei) against fission's 200 MeV per 235 u of uranium-235.

Step 1. Mass of 4 H atoms ≈4.032 u=6.69×10−27 kg\approx 4.032\ u = 6.69\times10^{-27}\ \text{kg}, releasing 26.7×1.6×10−13=4.27×10−12 J26.7\times1.6\times10^{-13} = 4.27\times10^{-12}\ \text{J}, so energy per kg is 4.27×10−126.69×10−27=6.39×1014 J/kg\dfrac{4.27\times10^{-12}}{6.69\times10^{-27}} = 6.39\times10^{14}\ \text{J/kg}.

Step 2. Mass of one U-235 nucleus ≈235 u=3.90×10−25 kg\approx 235\ u = 3.90\times10^{-25}\ \text{kg}, releasing 200×1.6×10−13=3.2×10−11 J200\times1.6\times10^{-13} = 3.2\times10^{-11}\ \text{J}, so energy per kg is 3.2×10−113.90×10−25=8.21×1013 J/kg\dfrac{3.2\times10^{-11}}{3.90\times10^{-25}} = 8.21\times10^{13}\ \text{J/kg}. Fusion delivers roughly 6.4×10146.4\times10^{14} J/kg versus 8.2×10138.2\times10^{13} J/kg for fission — nearly 8 times more energy per kilogram of fuel.

Common traps

  • Confusing fusion (light nuclei combining, needs extreme temperature to beat Coulomb repulsion) with fission (heavy nucleus splitting, typically neutron-initiated with no repulsion barrier to overcome).
  • Thinking any high temperature suffices for fusion — 10610^6–107 K10^7\ \text{K} and above is far beyond ordinary combustion or even a fission reactor core; that gap is exactly why controlled fusion remains unsolved.
  • Trying to reproduce the detailed proton–proton chain steps — only the overall summary reaction 4 11H→ 24He4\,^1_1\text{H} \rightarrow \,^4_2\text{He} is required.
  • Mixing up the hydrogen bomb (fission-triggered, uncontrolled fusion) with a plain fission-only atomic bomb — the fusion stage is what makes it far more powerful.

Ask the tutor: Want to work through another mass-defect-to-energy fusion calculation, or discuss why magnetic confinement fusion reactors are so hard to sustain?

Why can't we just build a small fusion reactor the way we build fission reactors?

Fission reactors work because a slow neutron, which has no charge, can walk right up to a heavy nucleus and split it — no barrier to overcome. Fusion needs two positively charged light nuclei to smash together hard enough to beat their electrostatic repulsion, which only happens reliably at millions of kelvin. No solid material can contain plasma that hot, so engineers must hold it in place with magnetic fields or compress it faster than it can fly apart. Doing this while getting out more energy than is put in to sustain the plasma has been extraordinarily difficult, which is why fusion power is still experimental.

Does fusion release more energy than fission?

Per kilogram of fuel, yes — roughly eight times more, as the worked comparison shows: about 6.4×10146.4\times10^{14} J/kg for hydrogen fusing into helium versus about 8.2×10138.2\times10^{13} J/kg for uranium-235 fission. This is because the mass defect fraction in light-nucleus fusion is proportionally larger relative to fuel mass than in heavy-nucleus fission. It is also why fusion is such an attractive future energy source — small amounts of fuel could, in principle, deliver enormous energy, if the temperature and confinement problems can be solved.

Is the Sun's fusion the same reaction we see in a hydrogen bomb?

The underlying physics is the same principle — light nuclei fusing into a more tightly bound nucleus and releasing energy — but the mechanism differs. In the Sun, gravity maintains a steady core temperature of about 1.5×107 K1.5\times10^7\ \text{K} that sustains slow, continuous fusion over billions of years via the multi-step proton–proton chain. In a hydrogen bomb, a fission explosion briefly creates comparable temperatures in a tiny volume, triggering a fast, uncontrolled burst of fusion of hydrogen isotopes rather than a steady output.

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