The big idea: Every second, the Sun turns four million tonnes of itself into sunlight — by fusing light nuclei into heavier ones deep in its core.
How fusion gives energy
- The heavier nucleus is slightly lighter than the bits that made it
- That missing mass comes out as energy
- The catch: nuclei are positive, so they repel
- Only a star's core is hot and dense enough to force them together
Before fusion
- Several light nuclei (e.g. hydrogen) flying around
- They repel each other — both are positive
- Only a super-hot, super-dense core can slam them together
After fusion
- They join into one heavier nucleus (e.g. helium)
- The product is slightly lighter than the parts added up
- That missing mass is released as energy (E = mc²)
New words, plainly: Fusion = small nuclei joining into a bigger one (the opposite of fission, which splits them).
Coulomb repulsion = the electrical push between two positive charges. Two nuclei both repel because both are positive.
Proton-proton (p-p) chain = the step-by-step set of reactions that fuses hydrogen into helium in stars like the Sun.
| The obstacle | How a star beats it |
|---|---|
| Both nuclei are positive, so they repel (Coulomb repulsion) | — |
| To touch, they must collide very fast | A very high temperature → nuclei move fast |
| Fast collisions must happen often | A very high density / pressure → nuclei packed close |
Why it needs heat AND density: High temperature gives the nuclei enough speed to slam together hard enough to beat the repulsion.
High density / pressure packs them close so those fast collisions happen often.
A star's core has both (~10⁷ K) — your kitchen does not, which is why hydrogen here just sits there.
Fusion power on Earth?: For: the fuel (hydrogen isotopes from seawater) is plentiful, and there is no long-lived radioactive waste. Against: it needs about 10⁸ K, and a plasma that hot must be held away from every wall by magnetic fields; no reactor has yet given out more energy than it takes in.
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The energy a fusion reaction gives out comes from the mass defect — how much lighter the product is than the nuclei that fused. Find that missing mass, then turn it into energy with E = mc².
- energy released by the fusion reaction (J, or MeV)
- mass defect Δm — the mass that disappears when the nuclei fuse (kg)
- speed of light, 3.00 × 10⁸ m s⁻¹ (given constant)
Two units, one idea: In joules: put Δm in kilograms (1 u = 1.661 × 10⁻²⁷ kg) and multiply by c².
In MeV (faster): keep Δm in u and multiply by 931.5 (because 1 u = 931.5 MeV c⁻²).
Both give the same energy — pick whichever the question asks for.
Hydrogen to helium: the proton–proton chain
Worked example — energy from the p–p chain
In the Sun, four protons and two electrons make one helium-4 nucleus. Proton mass = 1.007276 u, electron mass = 0.000549 u, helium-4 nucleus = 4.001506 u. Find the energy released in MeV.
Solution
- Mass in and mass out:
- Mass defect and energy:
Final answer
About 26.7 MeV for each helium nucleus made.
Given atomic masses?: Atomic masses include the electrons. If the proton numbers balance on both sides (as in ¹⁶O + ¹⁶O → ³²S), the electrons cancel: subtract the atomic masses directly.
Heavier nuclei need hotter cores: Carbon and oxygen nuclei carry more positive charge than hydrogen, so they repel far more strongly. Fusing them needs a much higher temperature, which only the cores of very massive stars reach.
In a star's core, light nuclei fuse into a heavier one. The total mass of the fusing nuclei is greater than the product by a mass defect Δm = 0.026900 u. Find the energy released, in MeV. (1 u = 931.5 MeV c⁻².)
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A balance of two pushes: Two pushes act on a star, one inward and one outward. When they balance, the star's radius stays steady. This is stellar (hydrostatic) equilibrium.
Inward: gravity
- A star is held together by gravity
- It pulls every layer inward
Outward: pressure
- The energy from fusion makes the core blazing hot
- That heat creates an outward pressure — from the radiation streaming out and the hot gas
A main-sequence star's balance: gravity pulls every layer inward, while the outward pressure from fusion's radiation and hot gas pushes back. When the two are equal the radius stays steady — this is stellar (hydrostatic) equilibrium.
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It is self-correcting: The balance fixes itself. So a main-sequence star keeps a stable size for billions of years.
If the core shrinks a little
- It gets hotter
- Fusion speeds up
- More outward pressure
- It pushes back out
If it expands a little
- It cools
- Fusion slows
- Less pressure
- Gravity pulls it back in
Equilibrium — what to say in the exam
- Gravity pulls inward on every layer of the star
- Fusion heats the core → radiation + gas pressure push outward
- The two are equal and opposite → the net force is zero → the radius is stable
- It is self-correcting: shrink → hotter → more fusion → pushes back out
How this is tested — fusion + stellar equilibrium is a classic Paper 2 topic, with a quick Paper 1A version:
Paper 1A
- A quick multiple-choice — which condition allows fusion, or what balances gravity in a star.
Paper 2
- Calculate the energy released in MeV from a mass defect (Δm → E = mc² → × 931.5).
- Outline / explain the conditions for fusion (high temperature + density to beat Coulomb repulsion), or how the star stays in equilibrium (pressure out balances gravity in).
The classic trap: Writing that fusion 'holds the star up' directly. It is the pressure produced by fusion's heat — radiation and hot gas — that balances gravity, not the reactions themselves.
Deep in a star, several light nuclei fuse into a single heavier nucleus. The combined mass of the original nuclei exceeds the mass of the product by Δm = 0.025600 u. Calculate the energy released by this reaction, in MeV. (1 u = 931.5 MeV c⁻².)
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A main-sequence star keeps a roughly constant radius for billions of years. Outline how it maintains this equilibrium.
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