The big idea: Fusion means joining light nuclei into a heavier one.
The heavier nucleus is slightly lighter than the bits that made it — and that missing mass comes out as energy. This is what powers the Sun.
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.
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.
Worked example — energy from one fusion reaction
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.)
Solution
- The energy comes from the missing mass. Write the given equation first — with the 931.5 MeV-per-u shortcut the c² is already built in, so just multiply Δm (in u) by 931.5:
- Put in the mass defect Δm = 0.026900 u:
- Work it out — keep the unit:
Final answer
E = 0.026900 × 931.5 ≈ 25.1 MeV. The tiny missing mass became the released energy.
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A balance of two pushes: A star is held together by gravity, which pulls every layer inward.
The energy from fusion makes the core blazing hot. That heat creates an outward pressure — from the radiation streaming out and the hot gas — that pushes outward.
When these two balance, the star's radius stays steady. This is stellar (hydrostatic) equilibrium.
[Diagram: phys-free-body] - Available in full study mode
It is self-correcting: The balance fixes itself:
- 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.
So a main-sequence star keeps a stable size for billions of years.
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, and shows up two ways:
- Paper 2 — calculate: given a mass defect (or the masses), find the energy released by a fusion reaction in MeV (mass defect → E = mc² → × 931.5). - Paper 2 — outline / explain: describe the conditions that make fusion possible (high temperature + density to beat Coulomb repulsion), or outline how the star stays in equilibrium (pressure out balances gravity in). - Paper 1A: a quick multiple-choice — which condition allows fusion, or what balances gravity in a star.
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.
IB-style question — energy released in a fusion reaction
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.)
Solution
- The released energy comes from the mass defect. Write the given , using the 931.5 MeV-per-u shortcut:
- Substitute the mass defect Δm = 0.025600 u:
- Work it out — keep the unit:
Final answer
E = 0.025600 × 931.5 ≈ 23.8 MeV ≈ 24 MeV.
IB-style question — outline how the star stays in equilibrium
A main-sequence star keeps a roughly constant radius for billions of years. Outline how it maintains this equilibrium. [2]
Solution
- The inward push. State what acts inward:
Gravity pulls every layer of the star inward, trying to make it collapse. - The outward push. State what balances it:
Fusion heats the core, producing an outward pressure (radiation + hot gas) that pushes outward. - The result (answer the command term). Tie them together:
These two are equal and opposite, so the net force is zero and the radius stays stable — the star is in equilibrium.
Final answer
Gravity pulls inward; the outward pressure from fusion's radiation and hot gas pushes outward; the two balance (net force zero), so the radius stays constant.