
Nuclear, particle & quantum-information sequence · Interactive lesson
Fission energy, decay heat and waste inventory
Shutting a reactor down stops the chain reaction, not the heat. This is the energy ledger of fission: what each 202 MeV event deposits and where, what the antineutrinos carry off, how the nuclide inventory is propagated as a matrix exponential, and why the leftover power dies far too slowly to design around casually.
UPV-16.05 · Lesson 5 of 9 in this unit. Position is not a completion score.Nuclear, particle & quantum-information sequence. These units continue into nuclear and particle physics and quantum information. Follow your university’s prerequisites, mathematical depth and assessment; this independent pathway is not a universal programme specification.
Build the model
Scram stops the chain reaction. It does not stop the heat.
The key idea
Decay heat is the thermal power a reactor keeps producing after the chain reaction has stopped, supplied by the beta and gamma decay of the fission-product and actinide inventory that its operating history built up.
Example. A 3 GW core one second after scram still delivers about 6.4% of full power — some 190 MW — and a day later about 14 MW, which is why the cooling must not stop when the chain reaction does.
The complete physical model and assumptions
Fission accounting runs on three clocks, and the topic is the discipline of keeping them apart. The first is prompt: a thermal fission of U-235 releases about 202 MeV, but that is a sum over channels landing in different places — 169 MeV of fragment kinetic energy stopped within micrometres of the event, 12 MeV of prompt neutrons and gammas spread across the lattice, 13 MeV arriving late as fission-product betas and gammas, and 8.8 MeV leaving the site as antineutrinos. Recoverable energy is therefore about 193 MeV, pushed near 200 MeV by capture gammas from (n, γ) elsewhere in the core. The second clock is the inventory: decay, capture and fission yield assemble into one linear system dN/dt = A N whose formal solution N(t) = exp(At) N(0) is the Bateman result written honestly, and whose eigenvalues span more than twenty decades, so it is evaluated by rational approximation, not by time-stepping. The third clock is decay heat. One fission's afterglow falls as roughly t⁻¹·², but a core superposes that burst response over its whole operating history, and integrating a power law raises the exponent by one: the shutdown core cools as t⁻⁰·². That is the price of the ledger — 6.4% of full power a second after the rods drop, still 0.26% a week later, and a repository timescale set not by the fission products but by the plutonium and minor actinides made along the way.
1The 202 MeV is six numbers, not one
Per thermal fission of U-235 the ledger reads 169.1 MeV of fragment kinetic energy, 4.8 MeV of prompt neutrons, 7.0 MeV of prompt gammas, 6.5 MeV of fission-product betas, 6.3 MeV of delayed gammas and 8.8 MeV of antineutrinos: 202.5 MeV in all. The split matters more than the sum, because the channels deposit in different places on different clocks. The fragments stop within a few micrometres, so essentially all 169 MeV appears inside the pellet where the fission happened — that is what sets the fuel centreline temperature. Neutrons and prompt gammas travel centimetres and heat moderator and structure. The 12.8 MeV of beta plus delayed gamma is not prompt at all: fission products emit it over seconds to years. While the core runs, that inventory sits in equilibrium, so the delayed 12.8 MeV is delivered at the same rate it is created and you cannot tell it apart — until you scram.
2The 8.8 MeV you cannot bill for
Fragments are born far on the neutron-rich side of stability, so each fission chain runs through about six β⁻ decays, and every one emits an antineutrino of mean energy near 1.5 MeV. With an interaction cross-section around 10⁻⁴³ cm², the mean free path in water is of order a hundred light-years, so all 8.8 MeV leaves the site: recoverable energy is 193.7 MeV, not 202.5. In an operating core roughly 8–9 MeV per fission comes back as capture gammas from (n, γ) on U-238, structure and control material, which is why 200 MeV per fission is the working number rather than 193. The escaping flux is not wasted information: a 1 GWₜₕ core emits about 2 × 10²⁰ antineutrinos per second, detected by inverse beta decay at KamLAND and Daya Bay, and because Pu-239 yields fewer antineutrinos above the 1.8 MeV threshold than U-235, the counting rate drifts down as the fissile vector shifts — a safeguards signal no operator can screen.
3The inventory is a matrix exponential
Write one balance equation per nuclide: dNᵢ/dt = −(λᵢ + σₐ, i φ) Nᵢ + Σⱼ (branch or yield) × (λⱼ or σⱼ φ) Nⱼ. Stack them and it is dN/dt = A N, with A the burnup matrix — a few thousand square for a real library. Hold the flux φ constant over a step and the solution is exactly N(t) = e^{At} N(0). Two facts about A govern the numerics. It is not Hermitian and generally not normal, so there is no orthonormal eigenbasis to hide behind; the object you want is the exponential itself, not the spectrum. And its eigenvalues, essentially −(λᵢ + σₐ, i φ), run from about 5 × 10⁻¹⁸ s⁻¹ for U-238 up past 10⁶ s⁻¹ for the shortest-lived precursors. A step short enough for the fastest nuclide would need of order 10¹⁴ steps to cross an 18-month cycle, so explicit integration is out. Production codes evaluate e^{At} directly, usually by CRAM, a rational approximation tuned to the negative real axis, which reaches roughly 10⁻¹⁴ relative accuracy at order 16.
4Why the curve turns from t⁻¹·² to t⁻⁰·²
Fire one fission at t′ = 0 and its β-plus-γ afterglow falls empirically as f(t) ≈ 2.66 t⁻¹·² MeV s⁻¹, with t in seconds, from about 10 s to 100 days. That power law is not fundamental; it is what a few hundred exponentials look like when the fission-product half-lives are spread roughly log-uniformly from milliseconds to decades. Now run the core at constant fission rate R for a time T and stop it. The decay power is the superposition of every burst since startup: P(t) = R ∫₀ᵀ 2.66 (t + t′)⁻¹·² dt′ = 13.3 R [t⁻⁰·² − (t + T)⁻⁰·²]. Integrating a power law raised the exponent by one. Divide by P₀ = 200 R MeV s⁻¹ and the coefficient is 13.3/200 ≈ 0.066 — the 12.8 MeV delayed share of the ledger reappearing to within 4%, since at equilibrium the delayed energy is emitted as fast as it is made. For t ≪ T the second term falls away, though only as (t/T)⁰·² relative to the first, and the core cools as t⁻⁰·²; for t ≫ T the bracket collapses to 0.2 T t⁻¹·² and the burst law returns.
5What is actually in a discharged assembly
At 48 GW⋅d per tonne of heavy metal, about 5% of the heavy-metal atoms have fissioned — nineteen in twenty leave the core intact. By mass a discharged PWR assembly is roughly 93% U-238, 0.8% U-235, 0.5% U-236, 1.2% plutonium, 0.1% minor actinides (Np-237, Am-241, Am-243, Cm-244) and 4–5% fission products. Which of those carries the heat depends entirely on when you ask. For the first century it is Cs-137 with its Ba-137m daughter (30.1 y) and Sr-90 with Y-90 (28.8 y); those two pairs set cask loading and repository drift spacing. Then the picture changes owner. Am-241 is not a fission product at all: it is the β⁻ daughter of Pu-241, whose 14.3-year half-life feeds it far faster than its own 433-year half-life removes it, so the Am-241 inventory grows after discharge and peaks near 73 years out. Design a repository on the discharge inventory and you have designed it for the wrong isotopes.
6Heat sets the engineering, mobility sets the dose
The usual figure of merit is radiotoxicity — activity weighted by ingestion dose coefficient — compared against the uranium ore the fuel came from. Fission products dominate only for the first several decades, and by about 300 years they have died with Cs-137 and Sr-90 to below the ore level; after that the curve is entirely actinide, first Am-241, then Pu-239 and Pu-240, then Np-237, and it does not fall back to the ore level for some 200 000 years. Strip out the plutonium and minor actinides and the crossover drops to centuries, which is the whole case for partitioning and transmutation. But the dose most repository performance assessments actually predict at the surface comes from none of these. It comes from long-lived fission and activation products that form anions and barely sorb on rock: Tc-99 as pertechnetate (2.1 × 10⁵ y), I-129 (1.6 × 10⁷ y), Se-79, and Cl-36 activated from chlorine impurity. They carry a negligible share of the radiotoxicity and move with the groundwater. Different nuclides set the two different constraints.
Key equations
Read the units, assumptions and limits before substituting.
Recoverable energy per fission
Erec = Eₜₒₜ − Eν̄ ≈ 202.5 − 8.8 = 193.7 MeV
Sets what a watt costs: 1 W needs 3.2 × 10¹⁰ fissions s⁻¹.
U-235 thermal fission; 1 MeV = 1.602 × 10⁻¹³ JBurnup and heavy metal consumed
B = P t / mHM1 g fissioned ≈ 0.95 MW⋅d
At 48 GW⋅d/tHM about 5% of the heavy-metal atoms have fissioned.
B in GW⋅d per tonne of initial heavy metal, at 200 MeV per fissionDepletion as a matrix exponential
dN/dt = A N ⟹ N(t) = e^{At} N(0)
One non-Hermitian operator carries decay, capture and fission yield together.
Aᵢᵢ = −(λᵢ + σₐ, i φ); Aᵢⱼ = yield or branch times λⱼ or σⱼ φBateman chain solution
Nₙ(t) = N₁(0) (∏(i<n) λᵢ) Σⱼ e(−λⱼ t) / ∏(i≠j) (λᵢ − λⱼ)
Exact, and useless when two λ nearly coincide — hence CRAM instead.
λ in s⁻¹; the eigendecomposition of A written out for a linear chainWay–Wigner decay heat
P(t)/P₀ ≈ 0.066 [ t⁻⁰·² − (t + T)⁻⁰·² ]
The t⁻⁰·² is the t⁻¹·² burst law integrated over the whole run.
t since shutdown and T operating time, both in seconds; fission products onlyAntineutrino emission rate
Rν̄ ≈ 6 P / Erec ≈ 2 × 10²⁰ s⁻¹ per GWₜₕ
The energy you cannot recover is the signal a safeguards detector reads.
about six β⁻ decays per fission chain, mean Eν̄ ≈ 1.5 MeVInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Leave the cursor at 10³ s and drag the operating-time slider from 100 s up to 3 × 10⁶ s: the knee walks right past the cursor, the power there climbs from about 1 MW to 40 MW, and the slope readout swings from −1.1 to −0.25.
Live results
- TIME AFTER SCRAM
- 1000 s
- OPERATING TIME T
- 10000 s
- DECAY POWER, 3 GW CORE
- 18.947 MW
- LOCAL SLOPE d lnP/d lnt
- -0.50
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Cursor time after scram: 3.00 log₁₀ s; Operating time before scram: 4.00 log₁₀ s
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The two exponents are not one law read twice. A single fission's afterglow really does fall as roughly t⁻¹·², and students carry that straight across to a shutdown core: it puts a reactor an hour after scram at a few parts per million of full power, when the answer is about one part in a hundred — three orders of magnitude low, and lethal as a design assumption. A core is not one burst. It is every burst since startup, so its shutdown power is the burst response integrated over the operating history, and integrating t⁻¹·² raises the exponent by one, to t⁻⁰·². The second half of the error is picturing decay heat as fissions still finishing. It is not fission at all — the chain reaction stopped when the rods went in. It is β and γ decay of nuclides already made, which no control rod and no loss of moderator can switch off.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A research reactor operates steadily at 250 kW thermal. Taking 193 MeV of recoverable energy per fission, find the fission rate and the mass of uranium consumed by fission in one day.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Energy per fission: 193 MeV × 1.602 × 10⁻¹³ J MeV⁻¹ = 3.092 × 10⁻¹¹ J.
- Fission rate: R = P / E = 2.50 × 10⁵ W ÷ 3.092 × 10⁻¹¹ J = 8.09 × 10¹⁵ fissions s⁻¹.
- In one day: N = 8.09 × 10¹⁵ s⁻¹ × 86 400 s = 6.99 × 10²⁰ fissions.
- Mass: 6.99 × 10²⁰ ÷ 6.022 × 10²³ mol⁻¹ = 1.16 × 10⁻³ mol, and × 235 g mol⁻¹ = 0.273 g. A quarter of a gram a day is why fuel is measured in years, not tanker loads.
Answer. R = 8.1 × 10¹⁵ fissions s⁻¹, and about 0.27 g of uranium fissioned per day — 1.1 g per MW⋅d.
MediumExample 2
A 3400 MW thermal PWR has run at full power for 500 days and then scrams. Use P/P₀ = 0.066[t⁻⁰·² − (t + T)⁻⁰·²], with t and T in seconds, to estimate the decay power 10 s, 1 h and 7 days after shutdown. Then name one contribution the correlation leaves out.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Operating time in seconds: T = 500 × 86 400 = 4.32 × 10⁷ s, so (t + T)⁻⁰·² ≈ 0.0297 at every t asked for here.
- At t = 10 s: 10⁻⁰·² = 0.6310, so P/P₀ = 0.066(0.6310 − 0.0297) = 0.0397, giving 0.0397 × 3400 = 135 MW.
- At t = 3600 s: 3600⁻⁰·² = 0.1944, so P/P₀ = 0.066(0.1944 − 0.0297) = 0.01087, giving 37 MW.
- At t = 6.05 × 10⁵ s (7 days): t⁻⁰·² = 0.0698 and (t + T)⁻⁰·² = 0.0296, so P/P₀ = 0.066(0.0698 − 0.0296) = 0.00265, giving 9.0 MW.
- Way–Wigner covers fission products only. It omits the actinide chain U-239 (23.5 min) → Np-239 (2.36 d) fed by capture on U-238, worth roughly 15% of the first week's heat, and it ignores neutron capture in the fission products themselves.
Answer. About 135 MW at 10 s, 37 MW at 1 h and 9.0 MW at 7 days — still a small power station a week after shutdown.
HardExample 3
A PWR assembly holds 461 kg of uranium at loading and is discharged at a burnup of 48 GW⋅d per tonne of heavy metal. Take 200 MeV of recoverable energy per fission and a molar mass of 238 g mol⁻¹ for the fresh fuel. (a) How many fissions occurred, and what fraction of the initial heavy-metal atoms is that? (b) With a cumulative Cs-137 yield of 6.2% and a half-life of 30.1 y, find the Cs-137 activity of the assembly five years after discharge.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Energy released: 48 GW⋅d/t × 0.461 t = 22.1 GW⋅d = 22.1 × 10⁹ W × 86 400 s = 1.91 × 10¹⁵ J.
- Fissions: 200 MeV = 3.204 × 10⁻¹¹ J, so Nfis = 1.91 × 10¹⁵ ÷ 3.204 × 10⁻¹¹ = 5.97 × 10²⁵.
- Initial atoms: 4.61 × 10⁵ g ÷ 238 g mol⁻¹ = 1937 mol = 1.17 × 10²⁷ atoms, so FIMA = 5.97 × 10²⁵ ÷ 1.17 × 10²⁷ = 5.1%. Nineteen heavy-metal atoms in twenty leave the core unfissioned.
- Cs-137 made: 0.062 × 5.97 × 10²⁵ = 3.70 × 10²⁴ atoms, neglecting decay during the roughly 4.5-year irradiation, which overstates it by about 5%.
- Decay for 5 y: λ = ln2 / 30.1 y = 0.02303 y⁻¹, so N = 3.70 × 10²⁴ × e(−0.1151) = 3.30 × 10²⁴ atoms.
- Activity: λ = 0.693 ÷ (30.1 × 3.156 × 10⁷ s) = 7.30 × 10⁻¹⁰ s⁻¹, so A = 7.30 × 10⁻¹⁰ × 3.30 × 10²⁴ = 2.4 × 10¹⁵ Bq — 2.4 PBq, or 65 kCi, from one nuclide in one assembly.
Answer. 5.97 × 10²⁵ fissions and FIMA ≈ 5.1%; the Cs-137 activity is ≈ 2.4 PBq (65 kCi) five years after discharge.
Quick check
Select an answer and explain it before checking.
Review & continue
Decay heat is the thermal power a reactor keeps producing after the chain reaction has stopped, supplied by the beta and gamma decay of the fission-product and actinide inventory that its operating history built up.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.