Physics › Nuclear physics › Fission and fusion
Fission and fusion
Uranium-235 absorbs a thermal neutron and splits, releasing about 200 MeV and the two or three neutrons that sustain a chain reaction. Below the critical mass too many neutrons escape through the surface for the chain to continue. Fusion of light nuclei instead needs temperatures near 10⁷ K to beat the electrostatic repulsion.
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Builds on Mass-energy and binding energy and Radioactive decay and the decay constant and Coulomb's law and electric field strength.
IN THIS TOPIC
- Describe induced fission by thermal neutrons and balance a fission equation.
- Explain the chain reaction and the meaning of critical mass.
- Calculate the energy released in fission and fusion reactions from nuclear masses.
- State the conditions fusion needs, and why the electrostatic repulsion of two positive nuclei sets them.
COMMON MISCONCEPTION
Splitting any atom releases energy.
Fission only releases energy for nuclei heavier than iron, the most tightly bound nucleus. Splitting lighter nuclei would absorb energy; for them fusion is the energy-releasing process.
Induced fission of uranium-235
A uranium-235 nucleus can absorb a slow neutron and form an unstable uranium-236 nucleus, which may split into two smaller nuclei and additional neutrons. The neutron has to be a thermal neutron, one moving at everyday molecular speeds, since uranium-235 captures those far more readily than fast ones. Only nuclei on the heavy side of iron sit low enough on the binding-energy curve for their fragments to end up better bound, so only they release energy by splitting; splitting a light nucleus would require an input of energy.
A typical split turns uranium-235 plus a neutron into barium-141 and krypton-92 plus three neutrons. Check that both nucleon number and proton number are conserved: proton number gives 92 = 56 + 36, and nucleon number gives 236 = 141 + 92 + 3. The products have a greater binding energy per nucleon, so their total mass is lower, and the mass difference is released mainly as kinetic energy. A typical fission releases about 200 MeV. Those fragments carry too many neutrons for their new size and sit above the stable band, so they decay by β⁻ in chains, a fact whose consequences fill the next lesson.
Energy from the masses
The energy calculation follows one method every time. Total nuclear mass before, minus total after, times 931.5. The products of any release reaction weigh less than the ingredients, and the difference leaves as kinetic energy of the products. For fission the sum uses the fuel nucleus, the incoming neutron, both fragments and the freed neutrons. The question supplies the masses.
GUIDED PRACTICE
A fission yield in familiar units
A uranium-235 fission event has a total mass difference of 0.186 u. Find the energy released in MeV and in joules.
Show the working
Multiply by 931.5, and 0.186 × 931.5 = 173 MeV.
In SI that becomes 173 × 106 × 1.60 × 10−19 = 2.8 × 10−11 J. Tiny per event, but the events come by the countless billion, as the next example shows.
The chain and the critical mass
Each fission's spare neutrons can induce further fissions, which is the door to a chain reaction. Left alone in a large enough mass of fuel, one fission becomes three, becomes nine, growing by powers. Held so that exactly one neutron per fission goes on to cause another, the chain ticks over at a steady rate, which is what a reactor is built to do.

Describe the drawing
Left, uncontrolled: each fission frees three neutrons and each finds a nucleus, so one becomes three becomes nine becomes twenty-seven, the frame filling faster each generation. Right, controlled: absorber rods eat two of every three, and the chain ticks over at exactly one, steady power instead of a bomb. The difference between a reactor and a weapon is that one number.
Whether a chain can sustain itself at all is a question of size. In a small lump, too many neutrons reach the surface and escape before meeting a nucleus; the chain fizzles. The critical mass is the minimum amount of fissile material in which, on average, one neutron per fission is retained to fission again, and it is not a constant of the material alone: shape, density, enrichment and any surrounding reflector all move it. Below it a chain is impossible; at and above it, a chain can be sustained.
WORKED EXAMPLE
Counting a power station's fissions
Each fission releases about 200 MeV. How many fissions per second sustain a reactor producing 1.0 GW of thermal power?
One fission in joules is 200 × 106 × 1.60 × 10−19 = 3.2 × 10−11 J.
Rate = power/energy per event = 1.0 × 109/(3.2 × 10−11) = 3.1 × 1019 fissions per second.
Thirty billion billion fissions a second, each releasing 3.2 × 10−11 J. The output is large because the number of events is large, not because any one event is.
Fusion of light nuclei
At the curve's other end, light nuclei release energy by fusion, and the steep left slope makes each fused nucleon worth more than a fissioned one, so kilogram for kilogram fusion yields several times more. The standard reaction fuses deuterium and tritium into helium-4 plus a neutron, releasing 17.6 MeV.
The condition is the approach distance. Two nuclei have to come within a few femtometres for the strong force to act, and both of them are positive, so this is the closest-approach problem from earlier in the unit, now working against us. Temperatures of order 107 K raise the collision energy and so cut the distance of closest approach, and in a star the crushing pressure keeps the density, and with it the collision rate, enormous. That much is the answer an exam question is after.
The arithmetic does not close there, though, and it is worth seeing why. At 1.5 × 107 K the average nucleus carries only about 2 keV, while the electrical hill between two protons runs to several hundred keV. Practically nothing in a star has the energy to get over it, so a purely classical treatment predicts a fusion rate hopelessly below what stars actually deliver. What rescues it is quantum tunnelling, the same barrier-penetration effect that lets alpha particles leave a nucleus.
So the full picture combines the two. The fastest nuclei in the tail of the speed distribution get close, and tunnelling carries them the rest of the way through a barrier they were never going to climb. Stars manage it, but not easily: the Sun's core releases only a few hundred watts per cubic metre, and it is sheer size that makes the total colossal. Building a reactor that sustains it remains one of engineering's great open problems.
ASSESSMENT FOCUS
- Thermal means slow: neutrons at speeds comparable to ordinary molecular motion, which uranium-235 captures far more readily than fast ones. Use the word thermal, since “neutron” alone does not state the condition for capture.
- Balance fission equations twice, nucleon numbers along the top and proton numbers along the bottom, and count the freed neutrons; two or three per fission is the expected answer.
- Energy calculations are before minus after, in u, times 931.5 for MeV. Show the subtraction explicitly: reversing it gives a negative energy for a reaction that releases energy.
- Critical mass answers need the surface argument. Below it, too many neutrons escape through the surface before they cause a fission, so the chain cannot sustain itself.
- For why fusion needs extreme temperature, name the electrostatic repulsion between positive nuclei and the need to approach within strong-force range. That is the required answer. Do not write that the nuclei have enough energy to get right over the barrier, because at stellar temperatures they do not, and tunnelling does the rest.
- E = mc² is the only equation the booklet prints for this topic, with 931.5 MeV per u sitting among the constants. Everything else here is method, so the fission and fusion recipes have to come from memory.
CHECK YOURSELF
Deuterium (2.01355 u) and tritium (3.01550 u) fuse to form helium-4 (4.00150 u) and a neutron (1.00867 u), masses given as nuclear masses. Find the energy released.
Show a hint
Total mass before, total after, difference times 931.5.
Show the answer
Before, 2.01355 + 3.01550 = 5.02905 u. After, 4.00150 + 1.00867 = 5.01017 u.
Δm = 5.02905 − 5.01017 = 0.01888 u.
E = 0.01888 × 931.5 = 17.6 MeV, carried off as kinetic energy of the helium nucleus and, mostly, the neutron.
Fission of heavy nuclei and fusion of light nuclei both release energy because their products have greater binding energy per nucleon and lie closer to the peak of the binding-energy curve.
Sum the masses before and after; the missing u, times 931.5, is the MeV set free.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the fission and fusion questions page.
CHECK YOUR PROGRESS
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- Describe induced fission by thermal neutrons and balance a fission equation.
- Explain the chain reaction and the meaning of critical mass.
- Calculate the energy released in fission and fusion reactions from nuclear masses.
- State the conditions fusion needs, and why the electrostatic repulsion of two positive nuclei sets them.
Open the full revision checklist to track your progress across the whole unit.
