The Sun has been shining for 4.6 billion years, and it does it by fusing an almost unbelievable 600 million tonnes of hydrogen every second. Where does all that energy come from? The same place as fission energy: a tiny mass defect turned into energy by E = mc2. On this page we’ll follow the exact step-by-step reaction — the proton–proton chain — and calculate just how much hydrogen the Sun burns to stay alight.
📚 What you need to know
Fusion releases energy because the product has a higher binding energy per nucleon
The energy comes from the mass defect via ΔE = Δmc2
In the Sun, hydrogen fuses to helium through the proton–proton (p–p) chain
The overall reaction is four hydrogen-1 nuclei → one helium-4 nucleus
Each overall reaction releases about 27 MeV of energy
Some energy is carried away by neutrinos and doesn’t contribute to luminosity
To find fusion rate: divide the star’s power output by the energy per reaction
Why fusion releases energy
When two small nuclei fuse, the larger nucleus produced has a higher binding energy per nucleon than the originals. That means the nucleons end up more tightly bound, and the extra binding energy is released. There’s a mass defect between the original nuclei and the new one — the product weighs slightly less — and that missing mass becomes energy:
Energy from the mass defectΔE = Δmc2
For example, when two deuterium nuclei fuse into helium-4, the binding energy jumps from about 4 MeV (the two deuteriums) to about 28 MeV (the helium), releasing 24 MeV. The bigger the rise in binding energy, the more energy comes out.
The proton–proton chain
In the hot core of a star like the Sun, fusion doesn’t happen in one giant leap. It goes through a series of smaller steps called the proton–proton chain, which builds a helium-4 nucleus out of hydrogen. The net effect is that four protons become one helium-4 nucleus, plus positrons, neutrinos and gamma photons.
The chain builds helium-4 from hydrogen in three main steps. Steps 1 and 2 each happen twice to make the two helium-3 nuclei that step 3 needs.
Adding it all up, the overall reaction and its released particles are:
Don’t try to memorise every particle in every step — examiners rarely want that. What they do want is the overall picture: four hydrogen nuclei become one helium-4, and the mass defect gives the energy. If you can write the overall reaction and calculate its energy from the masses, you’ve got the marks. The individual steps are just the “how”.
Calculating the energy released
To find the energy from one overall reaction, we use the same method as always: find the mass defect between the four hydrogen nuclei and the helium-4 nucleus, then convert it to energy.
WE 1
In the overall p–p reaction, four hydrogen-1 nuclei fuse into one helium-4 nucleus. Using rest masses hydrogen-1 = 1.007825 u and helium-4 = 4.002603 u, calculate the energy released per reaction, in joules. (1 u = 1.66 × 10−27 kg, c = 3 × 108 m s−1)
Step 1 — mass defect in uΔm = 4(1.007825) − 4.002603Δm = 0.028697 uStep 2 — convert to kgΔm = 0.028697 × 1.66×10⁻²⁷ = 4.76×10⁻²⁹ kgStep 3 — energyE = Δmc² = 4.76×10⁻²⁹ × (3×10⁸)²≈ 4.29 × 10−12 JThat’s about 27 MeV per reaction. Keep the mass in u until the last moment, then convert — it keeps the arithmetic clean.
From reactions to the Sun’s power
Once we know the energy per reaction, we can work backwards from the Sun’s total power output (its luminosity) to find how many reactions happen each second — and therefore how much hydrogen it burns. One subtlety: a small fraction of the energy is carried away by neutrinos, which escape the Sun entirely and don’t contribute to its light.
Number of reactions per secondnumber of reactions = power output ÷ energy released per reaction
WE 2
The Sun’s luminosity is 3.85 × 1026 W. Suppose 80% of this comes from the p–p chain, each overall reaction releases 4.29 × 10−12 J, and neutrinos carry away 2% of that energy. Estimate the mass of hydrogen fused each second by this process. (1 u = 1.66 × 10−27 kg)
Step 1 — usable energy per reaction (after neutrinos)E = 0.98 × 4.29×10⁻¹² = 4.20×10⁻¹² JStep 2 — power from this processP = 0.8 × 3.85×10²⁶ = 3.08×10²⁶ WStep 3 — reactions per secondN = 3.08×10²⁶ ÷ 4.20×10⁻¹² = 7.33×10³⁷ s⁻¹Step 4 — mass of hydrogen (4 nuclei per reaction)m = 4 × 1.007825 × 1.66×10⁻²⁷ × 7.33×10³⁷≈ 4.9 × 1011 kg s−1Around 490 million tonnes of hydrogen a second — and the Sun has enough to keep this up for billions of years. Work in order: usable energy → power → reactions → mass.
⚛ Fusion energy & rate questions
Mass defect: total reactant mass − product mass.
Energy per reaction: Δmc² (or × 931.5 for MeV).
Subtract neutrino energy if asked (they escape).
Reactions per second: power ÷ usable energy per reaction.
Mass of fuel per second: nuclei per reaction × nucleus mass × reaction rate.
💡 Top tips
Learn the overall reaction (4 H → He-4), not every sub-step.
Energy comes from the mass defect, via Δmc².
Neutrinos carry off some energy — subtract it before using luminosity.
Keep masses in u until you convert to kg.
Rate = power ÷ energy per reaction.
⚠ Common mistakes
Forgetting neutrinos carry away energy that never becomes light
Using 3 or 2 hydrogen nuclei instead of 4 in the overall reaction
Forgetting to square c in Δmc²
Mixing up energy per reaction with the star’s total power
Forgetting there are 4 nuclei per reaction when finding fuel mass
Quick recap: Fusion energy comes from the mass defect (ΔE = Δmc2) as light nuclei climb the binding-energy curve. In the Sun, the proton–proton chain fuses 4 hydrogen → helium-4, releasing about 27 MeV per reaction, with neutrinos carrying off a little. Divide the star’s power by the energy per reaction to get the reaction rate and the huge mass of fuel burned each second.
We keep saying fusion “holds the star up against gravity” — but where do stars even come from, and how does that balance get set up in the first place? Every star begins as a cold cloud of gas that collapses until its core is hot enough to ignite. Next page: How Stars Form.
Fusion energy calculations still tricky?
Book a free meeting and we’ll drill the mass-defect method, the neutrino correction, and the power-to-fuel-rate chain the Sun questions always test.
