IB Physics HLTopic 5 — The Atomic & Nuclear WorldPaper 1 & 2tiny mass, huge energy~15 min read
Energy from Fission
When a uranium nucleus splits, it releases an astonishing amount of energy from an almost unimaginably small amount of mass. Here’s the headline: one kilogram of uranium fuel can release as much energy as nearly two million kilograms of petrol. In this page we’ll work out exactly where that energy comes from, how to calculate it, and why nuclear fuel is so incredibly energy-dense compared to anything you can burn.
📚 What you need to know
Fission energy comes from the mass defect: the products weigh slightly less than the reactants
That lost mass becomes energy via ΔE = Δmc2
Each U-235 fission releases roughly 200 MeV of energy
Most of that energy appears as kinetic energy of the fast-moving fragments
Nuclear fuel is about a million times more energy-dense than chemical fuels
The energy released can be found from the binding energy per nucleon jump
To find total energy: multiply energy per fission by the number of nuclei that fission
Where does the energy come from?
When a heavy nucleus splits, the two daughter nuclei sit higher on the binding-energy-per-nucleon curve than the original — they’re held together more tightly. The extra binding energy has to go somewhere, and it appears as the kinetic energy of the flying fragments and neutrons, plus some gamma radiation.
Crucially, this is the same idea as mass defect. The total mass of everything after the split is slightly less than the total mass before. That missing mass is converted straight into energy by Einstein’s equation:
Energy from mass defectΔE = Δmc2Δm = mass lost (kg) • c = 3.0 × 108 m s−1
Don’t say the energy comes from “breaking bonds” or “splitting the atom releases stored energy” — examiners hate that. The correct story is always: the products have a tiny bit less mass than the reactants, and that mass difference becomes energy through ΔE = Δmc². Keep coming back to mass defect and you’ll never lose the mark.
How much energy per fission?
A single fission of uranium-235 releases about 200 MeV. That might sound small — and for one nucleus it is — but remember there are a colossal number of nuclei in even a gram of uranium. When you add them all up, the total is enormous. Compare it to a chemical reaction: burning one molecule of fuel releases only a few eV. So fission is roughly tens of millions of times more energetic per event than burning.
On a log scale, uranium fission towers over chemical fuels — roughly a million times more energy from each kilogram.
WE 1
Each fission of U-235 releases about 200 MeV. Calculate the energy released by the complete fission of 1.0 kg of U-235, and compare it to petrol (46 MJ per kg). (NA = 6.02 × 1023 mol−1, molar mass 235 g mol−1, 1 MeV = 1.6 × 10−13 J)
Step 1 — number of nuclei in 1.0 kgN = (1000 ÷ 235) × 6.02×10²³ = 2.56×10²⁴Step 2 — energy per fission in joules200 MeV = 200×10⁶ × 1.6×10⁻¹⁹ = 3.2×10⁻¹¹ JStep 3 — total energyE = 2.56×10²⁴ × 3.2×10⁻¹¹ ≈ 8.2×10¹³ J≈ 8.2 × 1013 J per kgThat’s about 1.8 million times the 46 MJ you get from a kilogram of petrol. A handful of uranium holds the energy of a fuel tanker — this is the whole appeal of nuclear power.
From one nucleus to a power station
To scale up from a single fission to a working power plant, we just multiply. The total energy is simply the energy per fission times the number of nuclei that split:
Total energy releasedEtotal = N × Eper fissionN = number of nuclei that fission
Real reactors don’t turn all that heat into electricity, though. Like any heat engine they’re limited by efficiency — typically only about a third of the thermal energy becomes electrical energy, with the rest lost as waste heat.
WE 2
A nuclear power station generates 1.0 GW of electrical power at an efficiency of 33%. If each kilogram of U-235 releases 8.2 × 1013 J, estimate the mass of U-235 used per day.
Step 1 — thermal power neededPthermal = 1.0×10⁹ ÷ 0.33 = 3.03×10⁹ WStep 2 — thermal energy in one dayE = 3.03×10⁹ × 86400 = 2.62×10¹⁴ JStep 3 — mass of fuelm = 2.62×10¹⁴ ÷ 8.2×10¹³≈ 3.2 kg per dayJust over 3 kg of uranium a day powers a city. A coal station of the same output burns thousands of tonnes daily — that difference is the energy density of fission in action.
⚛ Finding the energy from fission
Energy per fission: from Δm (via Δmc² or 931.5 MeV/u), or use ~200 MeV for U-235.
Count the nuclei: N = (mass ÷ molar mass) × NA.
Total energy: multiply the two together.
Convert units carefully (MeV → J using 1.6×10−13).
Efficiency? Multiply by the efficiency for electrical output.
💡 Top tips
Energy comes from the mass defect, not “broken bonds”.
Most fission energy is kinetic energy of the fragments.
Use ~200 MeV per fission for U-235 when no masses are given.
Total energy = energy per fission × number of nuclei.
Only about a third of the heat becomes electricity in a real plant.
⚠ Common mistakes
Saying energy comes from “breaking bonds” — it’s the mass defect
Forgetting to convert MeV to joules before multiplying
Using grams instead of the molar mass in grams when counting nuclei
Forgetting the efficiency step when asked for electrical output
Confusing energy per fission with energy per kilogram
Quick recap: Fission energy comes from the mass defect (ΔE = Δmc2), about 200 MeV per fission, appearing mostly as kinetic energy of the fragments. Total energy = energy per fission × number of nuclei, making nuclear fuel roughly a million times more energy-dense than chemical fuel. Real reactors convert only about a third of the heat into electricity.
We’ve seen that one fission releases about 200 MeV — but a single split on its own would be useless. The magic of a reactor is that each fission throws out spare neutrons that trigger more fissions, keeping the whole thing going. How we control that self-sustaining process is the key to nuclear power. Next page: Chain Reactions.
Fission energy calculations tripping you up?
Book a free meeting and we’ll drill Δmc², counting nuclei, the MeV-to-joule conversion and the efficiency step examiners always sneak in.
