Every second, the Sun turns hundreds of millions of tonnes of hydrogen into helium and pours out the energy that keeps us alive. That energy comes from fusion — small nuclei being forced together to make a bigger one. In this note we’ll see what fusion is, why it releases so much energy, and the fierce conditions a star needs to make it happen.
📘 What you need to know
Nuclear fusion is the joining of two small nuclei to make one larger nucleus
It’s light nuclei (hydrogen, helium) that release energy when they fuse
In stars, four hydrogen nuclei fuse in stages to form one helium-4 nucleus — the proton–proton chain
The product nucleus has less mass than the parts that made it; that mass defect is released as energy (E = mc2)
The new nucleus has a higher binding energy per nucleon, which is why energy comes out
Nuclei only fuse if they have very high kinetic energy, to beat the electrostatic repulsion between positive charges
That means fusion needs an extremely hot, dense environment — the core of a star, or a lab reactor chasing the deuterium–tritium reaction
What Fusion Actually Is
Fusion is the opposite move to fission. Instead of a big nucleus splitting, two small nuclei join together to make a single, heavier one. The key word is small: only light nuclei give out energy when they fuse. Try to fuse heavy nuclei and you’d have to put energy in.
Nuclear fusion
two small nuclei → one larger nucleus + energy
The simplest example happens in the Sun. Two hydrogen nuclei (protons) collide and stick. As they do, one proton flips into a neutron, and the pair becomes a deuterium nucleus — a hydrogen nucleus with one proton and one neutron. The proton-to-neutron flip throws out a positron and a neutrino as well.
First step of the p–p chain11H + 11H → 21H + e+ + ν
Think of two protons as two magnets pushed north-to-north — they hate getting close. Only if you slam them together hard enough does a much stronger, very short-range “glue” (the strong nuclear force) suddenly grab them. That “slam them hard” is exactly what a star’s heat provides.
The Proton–Proton Chain
One collision isn’t the whole story. In the core of a star like the Sun, the overall result is that four hydrogen nuclei end up fused into one helium-4 nucleus, through a short series of steps called the proton–proton chain. Each step nudges the material closer to helium, and energy is released along the way.
2 protons fuse
→ one becomes a neutron →
deuterium (21H)
→ add a proton →
helium-3 (32He)
→ two He-3 fuse →
helium-4 (42He)
The tidy summary of the whole chain is that four protons go in and one helium-4 nucleus comes out, plus two positrons, two neutrinos and a burst of energy:
Overall reaction
4 11H → 42He + 2e+ + 2ν + energy
This released energy is the whole point for the star. It creates an outward radiation pressure that pushes back against the star’s own gravity, stopping it from collapsing. As long as fusion keeps running in the core, the star stays in balance.
Why Fusion Releases Energy
Here’s the surprising bit: the helium nucleus weighs less than the four protons that made it. Some mass has vanished. It hasn’t really been destroyed — it’s been converted into energy, following Einstein’s famous relation.
Mass–energy equivalenceE = Δmc2
The missing mass is called the mass defect. Multiply it by c2 (a huge number) and you get the energy set free by each reaction. Because c2 is so large, even a tiny mass defect gives an enormous energy release per kilogram of fuel — far more than any chemical reaction like burning coal.
The binding energy per nucleon view
There’s a second way to see the same thing. Binding energy per nucleon measures how tightly each proton and neutron is held inside a nucleus. When light nuclei fuse, the product sits higher on the binding-energy-per-nucleon curve — its nucleons are more tightly bound. Moving to a more tightly bound state releases the difference as energy.
Fusing light nuclei climbs the steep left side of the curve toward tighter binding — the gain in binding energy per nucleon is what’s released.
Quick recap: fusion joins light nuclei into a heavier one; the product has less mass (mass defect) and more binding energy per nucleon, and that difference escapes as energy via E = Δmc2.
The Conditions Fusion Needs
Fusion doesn’t happen easily. Both nuclei are positively charged, so they repel each other with an electrostatic (Coulomb) force that gets stronger the closer they get. The attractive strong nuclear force that would bind them only acts over an incredibly short range — the nuclei almost have to touch before it takes over.
🧭 What it takes to fuse two nuclei
Give them huge kinetic energy — fast-moving nuclei can charge through the repulsion instead of being turned away
Get them extremely close — only at tiny separations does the strong force grab hold and pull them together
Provide a very hot, very dense place — high temperature means high speeds, and high density means frequent collisions. The core of a star (around 100 million K) does exactly this
On Earth we can’t easily build a star’s core, so fusion research chases the deuterium–tritium (D–T) reaction, which ignites at more achievable conditions. A deuterium and a tritium nucleus fuse into helium-4 plus a spare neutron, releasing a large amount of energy per unit mass of fuel:
Deuterium–tritium reaction21H + 31H → 42He + 10n + energy
WE 1
Two deuterium nuclei fuse to form a helium-4 nucleus. The binding energy of helium-4 is about 28.3 MeV, and the binding energy of each deuterium nucleus is about 2.22 MeV. (a) Explain why energy is released. (b) Calculate the energy released.
Part (a)
helium-4 has a higher binding energy per nucleon than the two deuterium nuclei
its nucleons are more tightly bound, so moving to that state gives energy out
→ there is a mass defect, released as energyPart (b)
energy out = BE of product − BE of the two originals
= 28.3 − (2 × 2.22)= 28.3 − 4.44≈ 23.9 MeVthe extra binding energy is exactly the energy released
WE 2
A fusion reactor produces a power output of 25 MW using the D–T reaction, which releases 17.6 MeV per reaction. (a) Find the number of reactions per second. (b) Estimate the mass of fuel (deuterium + tritium) consumed each second. Take 1 MeV = 1.60 × 10−13 J, 1 u = 1.66 × 10−27 kg, and the total fuel mass per reaction as 5.03 u.
Part (a)
energy per reaction = 17.6 × (1.60 × 10−13)
= 2.82 × 10−12 J
number per second = power ÷ energy per reaction
= (25 × 106) ÷ (2.82 × 10−12)≈ 8.9 × 1018 per secondPart (b)
mass per reaction = 5.03 × (1.66 × 10−27) = 8.35 × 10−27 kg
fuel per second = (8.9 × 1018) × (8.35 × 10−27)
≈ 7.4 × 10−8 kg s−1a tiny mass of fuel powers a whole town — that’s the appeal of fusion
💡 Top tips
Fusion = light nuclei, fission = heavy nuclei. Only nuclei to the left of iron on the binding-energy curve release energy by fusing
Always link energy to mass defect: the product is lighter than its parts, and E = Δmc2 converts that missing mass into energy
“High temperature” is really “high kinetic energy”: heat gives the nuclei enough speed to beat the Coulomb repulsion
Balance your equations: nucleon numbers (top) and proton numbers (bottom) must match on both sides — positrons count as +1 charge, neutrons as 0
⚠ Common mistakes
Saying mass is “destroyed” — it’s converted into energy, not lost
Mixing up fusion and fission — fusion joins small nuclei, fission splits big ones
Thinking the repulsion is gravity — it’s the electrostatic force between two positive charges
Forgetting the extra particles — the p–p chain also emits positrons and neutrinos, and D–T emits a neutron
Up next: Energy Released in Fusion Reactions — we’ll put real rest masses into E = Δmc2 and work out exactly how much energy the Sun squeezes from each reaction.
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Book a free meeting and let’s work through the tricky bits together.
