IB Physics SLTopic 5 — Fusion & StarsPaper 1 & 2E = Δmc2~8 min read
Energy from Fusion
Last page we saw why fusion releases energy: the product nucleus is lighter than the parts that made it. Now we’ll actually put numbers on it. Give me the rest masses of the nuclei involved and one famous equation, and we can work out exactly how much energy pours out of every single reaction — and why a pinch of fusion fuel outclasses a truckload of coal.
📘 What you need to know
The energy released comes from the mass defect — the difference between the mass of the reactants and the mass of the products
Convert mass into energy with E = Δmc2
Masses are usually given in unified atomic mass units (u); convert to kg using 1 u = 1.66 × 10−27 kg
Energy can be quoted in joules or in MeV, where 1 MeV = 1.60 × 10−13 J
To find power output: energy per reaction × number of reactions per second
To find fuel used per second: reactions per second × mass of fuel per reaction
Fusion gives a very large energy per unit mass of fuel — millions of times more than burning fossil fuels
Where the Energy Comes From
When light nuclei fuse, add up the masses on each side of the reaction and they don’t match. The products are always a little lighter than the reactants. That missing mass is the mass defect, and it hasn’t disappeared — it has turned into energy.
Mass defectΔm = (total mass of reactants) − (total mass of products)
Einstein’s mass–energy relation tells us how much energy that lost mass is worth. Because the speed of light squared is such a colossal number, even a mass defect of a fraction of a percent produces a huge amount of energy.
Energy releasedE = Δmc2
A helium-4 nucleus weighs about 0.7% less than the four protons that formed it. That sounds tiny — but multiply by c2 (nine followed by sixteen zeros) and that 0.7% is why the Sun has shone for billions of years.
The Units You’ll Need
The masses in exam questions come in unified atomic mass units (u), so you almost always convert before plugging into E = Δmc2. Answers can then be left in joules or turned into MeV, whichever the question asks for.
mass in u
× 1.66×10−27 →
mass in kg
× c2 →
energy in J
÷ 1.60×10−13 →
energy in MeV
🧭 How to solve an energy-from-fusion problem
Write the balanced reaction — check nucleon numbers (top) and proton numbers (bottom) match on both sides
Find the mass defect in u: total reactant mass − total product mass
Convert to kilograms — multiply the mass defect by 1.66 × 10−27
Apply E = Δmc2 using c = 3.0 × 108 m s−1 to get energy in joules
Convert to MeV if needed — divide by 1.60 × 10−13
WE 1
Two deuterium nuclei fuse to form a helium-3 nucleus and a neutron: 21H + 21H → 32He + 10n. Using the rest masses below, calculate the energy released, in MeV. Take m(21H) = 2.014102 u, m(32He) = 3.016029 u, m(n) = 1.008665 u, 1 u = 1.66 × 10−27 kg, 1 MeV = 1.60 × 10−13 J.
Step 1 — mass defect in u
Δm = 2(2.014102) − (3.016029 + 1.008665)
= 4.028204 − 4.024694= 0.003510 uStep 2 — convert to kg
Δm = 0.003510 × (1.66 × 10−27) = 5.83 × 10−30 kg
Step 3 — E = Δmc²E = (5.83 × 10−30) × (3.0 × 108)²= 5.25 × 10−13 JStep 4 — convert to MeV= (5.25 × 10−13) ÷ (1.60 × 10−13)≈ 3.3 MeV
From One Reaction to a Power Output
A single reaction releases a minuscule amount of energy. The reason fusion matters is the sheer number of reactions happening every second. Link them together and you can go from energy-per-reaction to the power output of a whole reactor — or a whole star.
Power from many reactions
power = energy per reaction × reactions per second
Turn that around and you can also find how fast the fuel is being used up. Each reaction consumes a fixed mass of fuel, so:
Fuel consumed each second
fuel mass per second = reactions per second × fuel mass per reaction
WE 2
A small star radiates 8.0 × 1025 W, produced entirely by the overall reaction 4 11H → 42He. (a) Find the energy released per reaction. (b) Find the number of reactions per second. (c) Estimate the mass of hydrogen fused each second. Take m(11H) = 1.007825 u, m(42He) = 4.002603 u, 1 u = 1.66 × 10−27 kg.
Part (a) — energy per reaction
Δm = 4(1.007825) − 4.002603 = 0.028697 u
Δm = 0.028697 × (1.66 × 10−27) = 4.76 × 10−29 kg
E = (4.76 × 10−29) × (3.0 × 108)²≈ 4.29 × 10−12 JPart (b) — reactions per second
number = power ÷ energy per reaction
= (8.0 × 1025) ÷ (4.29 × 10−12)≈ 1.9 × 1037 per secondPart (c) — hydrogen per second
each reaction fuses 4 H nuclei, so mass = 4 × 1.007825 u
= (1.9 × 1037) × 4 × 1.007825 × (1.66 × 10−27)≈ 1.3 × 1011 kg s−1that’s over a hundred billion kg of hydrogen every second — and the star barely notices
Why Fusion Fuel Is So Powerful
The headline feature of fusion is its energy per unit mass of fuel. The deuterium–tritium reaction releases about 17.6 MeV from roughly 5 u of fuel — work that out per kilogram and it comes to around 3 × 1014 J. Compare that with burning coal, which yields about 3 × 107 J per kilogram: fusion is on the order of ten million times more energy-dense.
Per kilogram of fuel, fusion outstrips coal by roughly seven orders of magnitude — note the vertical axis is logarithmic.
Quick recap: the released energy is the mass defect times c2; convert u→kg (× 1.66×10−27) and J→MeV (÷ 1.60×10−13); scale up with reactions per second to get power or fuel use.
💡 Top tips
Convert units first. The classic mark-loser is plugging a mass in u straight into E = Δmc2 — always go to kg first
Keep the full mass-defect value through the calculation; rounding 0.0287 u to 0.03 u too early wrecks your final answer
Watch what “per second” attaches to: power ÷ energy-per-reaction gives reactions per second, not the fuel mass directly
Remember the neutrinos. In a real star a small percentage of the energy escapes as neutrinos, so not all of it reaches us as light
⚠ Common mistakes
Using mass in u without converting to kg — your energy comes out 1027 times too big
Forgetting to square the speed of light in E = Δmc2
Doing products minus reactants — the mass defect is reactants minus products (and must be positive for energy to be released)
Multiplying by the wrong count — if four nuclei fuse per reaction, the fuel mass per reaction uses four of them
Up next: Star Formation — we leave the equations behind for a while and follow a cold cloud of gas as gravity squeezes it hot enough for the fusion we’ve just been calculating to switch on.
Want this to actually click before the exam?
Book a free meeting and let’s work through the tricky bits together.
