The big idea: In nuclear fission a large nucleus splits into two smaller ones, releasing a lot of energy and a few spare neutrons.
Those spare neutrons can hit other nuclei and split them too. One fission triggers the next, which triggers the next — a chain reaction.
Whether the chain dies out, stays steady, or grows is decided by how many neutrons are lost along the way.
What goes IN
- One slow neutron hits a large nucleus (e.g. uranium-235)
- The nucleus absorbs it and becomes unstable
- This is induced fission — the splitting was triggered by a neutron
What comes OUT
- Two smaller daughter nuclei (the fission fragments) fly apart
- A few extra neutrons are released — typically 2 or 3
- A large amount of energy (the fragments fly off fast)
New words, plainly: Induced fission = fission that is triggered by a nucleus absorbing a neutron (not happening on its own).
Chain reaction = each fission releases neutrons that go on to cause more fissions.
Self-sustaining = the chain keeps itself going without any extra neutrons being added from outside.
Dies out (subcritical)
- On average fewer than one of the released neutrons causes the next fission
- Too many neutrons escape or get absorbed without splitting anything
- The reaction fades out and stops
Steady or growing
- Exactly one per fission → steady (critical) — runs at a constant rate
- More than one per fission → growing (supercritical) — rate climbs
- A reactor is kept critical; a bomb is deliberately supercritical
Each fission releases N neutrons (about 2 or 3). The chain only stays steady if, on average, exactly one of them goes on to cause the next fission. Every other neutron is lost — it escapes the fuel, or is absorbed without causing a fission.
The rule (a counting rule — not in the data booklet): For a steady (critical) chain reaction, exactly one neutron per fission must trigger the next one.
So if N neutrons are released per fission, the number that must be lost or absorbed is N − 1.
| Neutrons released per fission | Must cause next fission (steady) | Must be lost or absorbed |
|---|---|---|
| 2 | 1 | 1 |
| 3 | 1 | 2 |
| N | 1 | N − 1 |
Read the regime off the losses
- Lose exactly N − 1 → steady (critical) — runs at a constant rate
- Lose more than N − 1 → too few left → reaction dies out (subcritical)
- Lose fewer than N − 1 → too many left → reaction grows (supercritical)
- A reactor is held at critical; the lost neutrons are mostly mopped up by control rods
Worked example — keep it steady
Each fission of a uranium-235 nucleus releases on average 3 neutrons. For a steady (self-sustaining) chain reaction, how many of those neutrons must be lost or absorbed per fission?
Solution
- State the steady-state rule. For a steady chain reaction, exactly one neutron per fission must go on to cause the next fission.
- Count the spares. Of the 3 released, 1 keeps the chain going, so the rest must be lost or absorbed:
- Read it off the rule N − 1: with N = 3, the number lost or absorbed is N − 1 = 2.
Final answer
2 neutrons per fission must be lost or absorbed (3 − 1), leaving exactly 1 to trigger the next fission and keep the chain steady.
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How this is tested: Neutron economy is a classic Paper 1A (multiple-choice) reasoning question, and the energy side turns up in Paper 2.
- Paper 1A — choose the loss: given the number of neutrons released per fission, pick the neutron-loss value that keeps the chain steady (it is always N − 1). - Paper 1A — name the regime: decide whether a chain is dying out, steady or growing from the number of neutrons that go on to fission. - Paper 2: the energy per fission from the mass defect (E = mc²), or describing the conditions for a self-sustaining reaction.
Classic trap: thinking ALL the released neutrons must continue. Only one per fission keeps it steady — the rest are lost. Losing too few makes it grow, not steady.
Steady means break-even, not zero loss: A steady chain reaction is a balance: one neutron in, one neutron out, per fission.
You are NOT trying to keep all the neutrons. You want to lose all but one — that is what keeps the rate constant.
IB-style question — which loss keeps it steady?
In a fission reactor each fission releases 2.5 neutrons on average. To run the reactor at a steady, self-sustaining rate, how many neutrons per fission (on average) must be lost or absorbed?
Solution
- State the steady-state condition. For a steady (critical) chain reaction, on average exactly one neutron per fission must cause the next fission.
- Subtract that one from the number released — the rest are lost or absorbed:
- Sanity check. Losing 1.5 leaves 1 to continue (steady). Losing more than 1.5 would let it die out; losing fewer would let it grow.
Final answer
1.5 neutrons per fission must be lost or absorbed (2.5 − 1), leaving exactly 1 to keep the chain reaction steady.