University Physics IV · Nuclear Physics · 13.8
Fission Barriers, Yields & Reactor Kinetics
Two numbers run this topic. Z²/A tells you whether a nucleus still has a fission barrier worth respecting, and β tells you whether a pile of such nuclei can be steered by hand. Everything in between — yields, cross-sections, the six factors — is the bookkeeping that connects them.
Build the model
Connect the measurement to the mechanism.
Fission is one competition run at two scales. Inside a nucleus, stretching it costs surface energy going as +ε² and refunds Coulomb energy going as −ε²; the refund wins outright once Z²/A passes about 50, and below that a barrier survives — near 6 MeV for the actinides, small enough that one captured neutron's binding energy can pay it. That is why a thermal neutron fissions 235U and not 238U: the pairing bonus in the compound nucleus, not any property of the neutron.
The same liquid drop predicts a symmetric split, and nature refuses, piling fragments near A = 95 and A = 140 because shell closures near Z = 50 and N = 82 are worth several MeV at scission — the first place the smooth model visibly fails. Then the physics runs again at reactor scale. Each fission returns about 200 MeV and 2.42 fast neutrons, and whether the chain holds is one number, keff, built from six factors that are each a ratio of things that can happen to a neutron.
What makes a reactor operable is not that number but its distance from a second one: about 0.65% of those neutrons arrive seconds late, from decaying fragments, and while excess reactivity stays under that fraction the response time is set by 13-second precursors instead of the 10⁻⁴ s prompt generation. Cross it and the excess that gave a 72 s period gives 0.07 s.
- Simple definition
- Nuclear fission is the splitting of a heavy nucleus into two fragments of comparable mass over a deformation barrier set by surface energy fighting Coulomb repulsion, releasing about 200 MeV and two or three neutrons able to drive the next fission.
- Example
- A thermal neutron captured by 235U makes 236U with 6.5 MeV of excitation against a 6.2 MeV barrier, so it fissions; the same neutron on 238U makes 239U with only 4.8 MeV against 6.6 MeV, and it does not.
Surface charges, Coulomb refunds. The bracket changes sign at Z²/A = 2aS/aC = 50.1.
aS = 17.8 MeV, aC = 0.711 MeV (Wapstra set); ε is the dimensionless spheroidal elongation
x → 1 erases the barrier. Below it the barrier only thins, and spontaneous-fission half-lives track that thinning brutally.
Dimensionless. 236U: x = 0.716; 240Pu: x = 0.735; 252Cf: x = 0.761; 209Bi: x = 0.658
236U: 6.5 ≥ 6.2, so fissile at Tₙ = 0. 239U: 4.8 < 6.6, so 238U is only fissionable, by fast neutrons.
Sₙ separation energy of the compound nucleus, Tₙ the lab kinetic energy, Eₐ the activation energy, all in MeV
Slowing a 2 MeV fission neutron to thermal lifts σf by about 500, which is what makes a natural-uranium chain possible at all.
σ₀ = 585 b for 235U fission at E₀ = 0.0253 eV, where v = 2200 m s⁻¹; 1 b = 10⁻²⁸ m²
η and ε make neutrons; f, p and the two non-leakage terms are the fractions that survive to make more.
All six factors dimensionless; ρ in Δk/k, or in dollars as ρ/β, or in pcm as 10⁵ρ
ρ = 0.0010 gives 72 s; ρ = 0.0079 gives 0.07 s. The entire margin of control sits inside one fraction.
β = 0.0065 for 235U, λ̄ = 0.0767 s⁻¹ (13.0 s mean precursor life), Λ ≈ 10⁻⁴ s in a light-water core
Stretch the drop and watch two terms fight
Deform a nucleus at fixed volume into a spheroid of elongation ε. The surface term grows, aS A²⁄³(1 + 2ε²/5); the Coulomb term relaxes, aC Z²A(−1/3)(1 − ε²/5). Their difference is ΔE = (ε²/5)[2aS A²⁄³ − aC Z²A(−1/3)], and the bracket turns negative when Z²/A > 2aS/aC = 50.1. Nothing in nature reaches that, so define the fissility x = (Z²/A)/50.1. For 236U, Z²/A = 8464/236 = 35.86, so x = 0.716 and a barrier survives. But the parabola is only the opening move: the real barrier of 236U is about 6.2 MeV, a third of a percent of its 1790 MeV binding energy, and it is exquisitely sensitive to x. Spontaneous-fission partial half-lives make the point. 236U at x = 0.716 lasts 2.4 × 10¹⁶ y; 240Pu at x = 0.735 lasts 1.2 × 10¹¹ y; 252Cf at x = 0.761 lasts 86 y. Six percent in x buys fourteen orders of magnitude of lifetime, because x sits inside the exponent of a tunnelling integral.
Whether a neutron can fission a nucleus is a pairing question
A thermal neutron brings no useful kinetic energy — 0.025 eV against a barrier of megaelectronvolts. All it delivers is its separation energy in the compound nucleus. For 235U + n → 236U that is 6.5 MeV against an activation energy of 6.2 MeV, so the compound nucleus is born above its own barrier. For 238U + n → 239U it is 4.8 MeV against 6.6 MeV, a deficit of 1.8 MeV the neutron must bring as kinetic energy; the measured threshold sits near 1 MeV once tunnelling through the real double-humped barrier is allowed for. The 1.7 MeV gap between those two separation energies is almost all pairing: 235U has an odd neutron number, so the arriving neutron pairs up and 236U is even-even, collecting δ ≈ 12/√A ≈ 0.8 MeV where 239U pays it instead. That one sign flip is the whole difference between fissile, meaning fissionable by neutrons of any energy, and merely fissionable.
The fragments are not halves, and shells say why
The liquid drop, left alone, prefers a symmetric split — that is where the Coulomb repulsion driving the fragments apart is largest. Thermal fission of 235U does the opposite. The mass-yield curve is double-humped: about 6.4% per fission near A = 95 and near A = 139, with the symmetric valley near A = 117 down at 0.01%, a peak-to-valley ratio of roughly 600. The heavy peak barely moves when the fissioning nucleus changes, staying near A = 140, while the light peak floats to take up the remainder — near A = 100 for 239Pu. The anchor is shell structure near Z = 50 and N = 82, worth several MeV at the scission point and entirely absent from the mass formula. Raise the excitation with 14 MeV neutrons and the valley fills in to within a factor of tens: shell corrections wash out with temperature, and the drop's symmetric preference reasserts itself.
Where 200 MeV goes, and why 2.42 neutrons come out
Price the split on the B/A curve. 235U sits at 7.59 MeV per nucleon and the fragments' β-stable descendants near 8.49, so repackaging some 235 nucleons releases about 210 MeV, of which 12 MeV leaves with antineutrinos and about 200 MeV is recoverable — 3.1 × 10¹⁰ fissions per second per watt. Most of it, 168 MeV, is fragment kinetic energy, and a Coulomb estimate shows where that comes from. Split 236U into 95Rb and 139Cs: EC = Z₁Z₂(1.44 MeV fm)/d with d = 1.2(95¹⁄³ + 139¹⁄³) = 11.69 fm gives 251 MeV for touching spheres. The measured 168 MeV needs d = 17.4 fm, so at scission the fragments are already elongated and half again as far apart. Charge density is roughly conserved, so the A = 139 fragment leaves as 139Cs with Z = 55, while the stable A = 139 nuclide is 139La with Z = 57. The fragments are born neutron-rich: they evaporate 2.42 prompt neutrons, then β-decay down the chain.
Six factors, one number, and no room to be sloppy
Follow one generation of neutrons. η is the number produced per neutron absorbed in fuel, ε the fast-fission bonus, p the resonance escape probability past the 238U resonances, f the thermal utilisation — the share of thermal absorptions landing in fuel — and PFNL, PTNL the fractions that fail to leak while fast and while thermal. Multiply them: keff = η f p ε PFNL PTNL. For a light-water core at 3.2% enrichment, η = 1.85, f = 0.71, p = 0.75, ε = 1.05 and the two non-leakage terms together 0.97 give keff = 1.003, a reactivity of ρ = 0.0033, or half a dollar. Notice the discipline that demands: six numbers each known to a percent or so must multiply out to within a tenth of a percent of unity. No core is designed to be exactly critical. It is built with excess reactivity for burnup and then held down by rods and soluble boron, which is what the operator actually moves.
The margin of control is 0.65 percent
Point kinetics reads dn/dt = [(ρ − β)/Λ]n + ΣᵢλᵢCᵢ. The prompt generation time Λ is about 10⁻⁴ s in a light-water core, so if ρ alone drove the exponent the period would be Λ/ρ — 0.1 s at ρ = 0.0010, hopeless for any mechanical control. It does not, because a fraction β = 0.0065 of the neutrons from 235U are emitted not at scission but seconds later, by precursors such as 87Br and 137I whose weighted mean life is 13.0 s, giving λ̄ = 0.0767 s⁻¹. While ρ < β the prompt chain alone is subcritical, every extra generation must wait on a precursor, and the stable period is T = (β − ρ)/(λ̄ρ) = 72 s at ρ = 0.0010 — a factor of 718 slower. Quote reactivity in dollars, ρ/β, and the cliff sits at exactly 1$. Above it the prompt chain closes by itself and T → Λ/(ρ − β), tens of milliseconds. 239Pu has β = 0.0021, so a plutonium core keeps under a third of that margin.
Change one variable at a time
Make the relationship visible.
Hold β at 650 pcm and walk ρ up: 100 pcm gives a 72 s period, 400 pcm gives 8.2 s, and at the dashed line, ρ = β = one dollar, it collapses to 0.46 s. Then drag β to 210 pcm, the plutonium value: the line slides left and that same 400 pcm is already prompt critical, at 0.05 s.
STABLE PERIOD T71.826 s
REACTIVITY0.154 $
POWER DOUBLING TIME49.783 s
PROMPT NEUTRONS ALONE0.1000 s
Live interpretationSTABLE PERIOD T: 71.826 s. REACTIVITY: 0.154 $. POWER DOUBLING TIME: 49.783 s. PROMPT NEUTRONS ALONE: 0.1000 s
Catch the common trap
Explain before calculating.
A thermal neutron induces fission in 235U but not in 238U. For the compound nucleus 236U the activation energy is 6.2 MeV and the neutron separation energy 6.5 MeV; for 239U they are 6.6 MeV and 4.8 MeV. Which statement gives the operative reason?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyUsing the Wapstra coefficients aS = 17.8 MeV and aC = 0.711 MeV, find the critical value of Z²/A, then the fissility parameter x for the compound nuclei 236U (Z = 92) and 240Pu (Z = 94). Which sits closer to losing its barrier?
- The second-order deformation energy is ΔE = (ε²/5)[2aS A²⁄³ − aC Z²A(−1/3)]. Divide the bracket by A²⁄³ and it becomes 2aS − aC(Z²/A), which vanishes at (Z²/A)crit = 2aS/aC.
- (Z²/A)crit = 2 × 17.8 ÷ 0.711 = 35.6 ÷ 0.711 = 50.07.
- 236U: Z² = 92² = 8464, so Z²/A = 8464/236 = 35.86 and x = 35.86/50.07 = 0.716.
- 240Pu: Z² = 94² = 8836, so Z²/A = 8836/240 = 36.82 and x = 36.82/50.07 = 0.735.
- Both are well below 1, so both keep a barrier against small deformation. 240Pu sits nearer the edge, and the tunnelling integral magnifies that gap: its spontaneous-fission partial half-life is 1.2 × 10¹¹ y against 2.4 × 10¹⁶ y for 236U.
Answer(Z²/A)crit = 50.07; x(236U) = 0.716 and x(240Pu) = 0.735, so 240Pu sits closer — and five orders of magnitude shorter in spontaneous-fission life.
Medium235U has a fission cross-section of 585 b at 0.0253 eV. (a) Use the 1/v law to find it at 0.100 eV. (b) In a thermal spectrum take σf(235U) = 585 b, σγ(235U) = 99 b and σγ(238U) = 2.68 b, with ν̄ = 2.42. Find η for natural uranium (0.72 at% 235U) and for pure 235U.
- (a) σ(E) = σ₀√(E₀/E) = 585 × √(0.0253/0.100) = 585 × √0.253 = 585 × 0.5030 = 294 b.
- (b) η = ν̄Σf/Σₐ taken over the fuel, so weight each cross-section by its atom fraction. Fission in 235U: 0.0072 × 585 = 4.212 b. Capture in 235U: 0.0072 × 99 = 0.713 b. Capture in 238U: 0.9928 × 2.68 = 2.661 b.
- Total absorption in the fuel is 4.212 + 0.713 + 2.661 = 7.586 b, of which the fission share is 4.212/7.586 = 0.5553.
- η(natural) = 2.42 × 0.5553 = 1.344.
- Pure 235U: η = 2.42 × 585/(585 + 99) = 2.42 × 0.8553 = 2.070.
- η = 1.34 leaves only 0.34 neutrons per absorption to pay for resonance capture, moderator and structural absorption, and leakage — which is why a natural-uranium core needs graphite or heavy water and cannot run on light water.
Answer(a) 294 b. (b) η = 1.34 for natural uranium and 2.07 for pure 235U.
HardA light-water reactor has Λ = 1.0 × 10⁻⁴ s, β = 0.0065 and a mean precursor decay constant λ̄ = 0.0767 s⁻¹. Rods are withdrawn to keff = 1.0010. Find (a) the reactivity in Δk/k and in dollars, (b) the stable period and doubling time, (c) the period there would be with no delayed neutrons, and (d) the period after a rod ejection to keff = 1.0080.
- (a) ρ = (k − 1)/k = 0.0010/1.0010 = 9.990 × 10⁻⁴ Δk/k, which is 99.9 pcm. In dollars, ρ/β = 9.990 × 10⁻⁴ ÷ 0.0065 = 0.154 $.
- (b) With ρ ≪ β, T = (β − ρ)/(λ̄ρ) = (0.0065 − 0.000999) ÷ (0.0767 × 0.000999) = 0.005501 ÷ (7.662 × 10⁻⁵) = 71.8 s. Doubling time = T ln2 = 71.8 × 0.6931 = 49.8 s.
- (c) Strip the precursors out and only Λ is left: T = Λ/ρ = 1.0 × 10⁻⁴ ÷ (9.990 × 10⁻⁴) = 0.100 s. The delayed neutrons stretch the response by a factor of 71.8/0.100 = 718.
- (d) ρ = 0.0080/1.0080 = 7.937 × 10⁻³ = 1.22 $. Now ρ > β, the prompt chain is critical on its own and the delayed formula no longer applies: T ≈ Λ/(ρ − β) = 1.0 × 10⁻⁴ ÷ (0.007937 − 0.0065) = 1.0 × 10⁻⁴ ÷ (1.437 × 10⁻³) = 0.070 s.
- Eight times the reactivity has shortened the period by a factor of 71.8/0.070 = 1030. That discontinuity in behaviour, not the size of Δk, is what prompt criticality means.
Answer(a) ρ = 9.99 × 10⁻⁴ Δk/k = 0.154 $. (b) T = 71.8 s, doubling in 49.8 s. (c) 0.100 s. (d) prompt critical at 1.22 $, with T ≈ 0.07 s.