University Physics V · Nuclear Fission and Fusion · 16.3
Chain Reactions & Critical Size
Criticality is not a mass you look up. It is one scalar equation: a buckling that the composition fixes, set equal to a buckling that the shape fixes. Learn to compute both sides, and to say which of your numbers came from a table and which from an eigenvalue solve.
Build the model
Connect the measurement to the mechanism.
A chain reaction is a generation-to-generation ratio, and this model splits that ratio into two halves that can be computed independently. The first half is composition. Walk one generation round the life cycle — fast fission before slowing down (ε), escape past the U-238 capture resonances (p), absorption in fuel rather than in moderator or structure (f), neutrons released per thermal absorption in fuel (η) — and the product k∞ = ε p f η is the multiplication of an infinite medium, a number with no length in it.
The second half is where the neutrons go. One-speed diffusion writes the steady balance as D∇²φ − Σₐφ + k∞Σₐφ = 0, which rearranges into a Helmholtz eigenvalue problem, ∇²φ + B²φ = 0, on the core's domain with φ = 0 at an extrapolated surface. The composition names the eigenvalue it demands, Bₘ² = (k∞ − 1)/L²; the geometry offers a whole spectrum of eigenvalues, and non-negativity of a neutron flux selects the lowest one, Bg².
Criticality is the statement that these two separately computed scalars are equal, and every design trade — enrichment against size, sphere against slab, bare against reflected — moves one side of that equality. The cost is the approximation: Fick's law is a near-isotropic expansion that fails within a few transport mean free paths of any boundary, and the extrapolated boundary is a patch borrowed from transport theory.
- Simple definition
- A core is critical when the buckling its composition demands, Bₘ² = (k∞ − 1)/L², equals the lowest eigenvalue Bg² of −∇² on that core's geometry with the flux vanishing at the extrapolated boundary.
- Example
- With k∞ = 1.20 and L² = 40 cm², Bₘ² = 5.00 × 10⁻³ cm⁻². A bare sphere offers Bg² = (π/R̃)², so R̃ = π/0.0707 = 44.4 cm, and subtracting d = 2.13 D = 1.79 cm leaves a critical radius of 42.6 cm.
Splits one multiplication into four knobs a designer can move separately.
All four dimensionless: ε fast fission, p resonance escape, f thermal utilisation, η neutrons per thermal absorption in fuel
η = 2.08 for pure U-235 but only 1.35 for natural uranium — U-238 capture, not geometry, is the first obstacle.
Macroscopic Σ in cm⁻¹, flux-weighted over the cell; ν = 2.42 for thermal U-235
Every geometric fact about the core enters through the single scalar Bg².
L² = D/Σₐ in cm²; in two groups replace L² by the migration area M² = L² + τ
Two numbers computed in different places, set equal — that is the whole design equation.
Bₘ² from composition alone, Bg² from shape and size alone; both in cm⁻²
Positivity picks the lowest eigenvalue: every higher mode has a node, and flux cannot change sign.
R̃ = R + 2.13 D; sphere (π/R̃)², cube 3(π/ã)², cylinder (2.405/R̃)² + (π/H̃)²
The same eigenfunction answers a power question: a bare core's centre runs 3.29 times its average.
r and R in cm, φ in cm⁻² s⁻¹, ρ in g cm⁻³; peak-to-average flux π²/3 = 3.29
Count one generation, then split the count
k∞ is a ratio of successive generations in a medium so large that nothing escapes, and the four-factor formula is that ratio walked round the neutron life cycle in order. Start with the thermal neutrons absorbed in fuel in one generation. Each releases η = νΣfF/ΣₐF fast neutrons; ν = 2.42 for thermal U-235, and with σf = 585 b against σₐ = 681 b that gives η = 2.42 × 585/681 = 2.08. In natural uranium the same arithmetic runs over the mixture, 2.42 × 0.0072 × 585 against 0.0072 × 681 + 0.9928 × 2.68, and η collapses to 1.35. The fast neutrons get one bonus, ε, from the few fissions they cause in U-238 before slowing down. They then run the resonance gauntlet, surviving with probability p. Those that reach thermal energy are shared out, and the fraction landing in fuel rather than moderator, cladding or coolant is f. Multiply: k∞ = ε p f η. A natural-uranium graphite lattice with ε = 1.03, p = 0.90, f = 0.89 and η = 1.34 gives k∞ = 1.106. No length has appeared yet.
Self-shielding is why fuel is lumped, and it hides inside p
U-238 has enormous capture resonances on the way down — the lowest at 6.67 eV peaks near 2 × 10⁴ barns against a scattering background of about 10 b — and p is the probability of getting past all of them. In a homogeneous mixture every U-238 nucleus sees the same flux, and the infinitely dilute resonance integral of roughly 280 b applies to all of it; that is why a homogeneous natural-uranium and graphite mix has never been made critical. Gather the fuel into rods and two things shield the interior. Energetically, absorption digs a dip in the flux at each resonance energy. Spatially, the outer millimetre of the rod absorbs so hard at resonance that the inside is screened. The effective resonance integral then depends chiefly on surface-to-mass ratio, Ieff = a + b√(S/M), and drops by more than an order of magnitude. Lumping costs a little f — thermal flux is depressed inside the rod, so the fuel competes less well — but it buys far more p. That trade is why reactors are lattices.
One-speed diffusion makes it an eigenvalue problem
Now let neutrons leave. Fick's law J = −D∇φ makes the net leakage per unit volume ∇⋅J = −D∇²φ, and in steady state production balances absorption plus leakage. In an infinite medium each absorption yields k∞ new neutrons, so the source is k∞Σₐφ and the balance reads D∇²φ − Σₐφ + k∞Σₐφ = 0. Divide by D and use the diffusion area L² = D/Σₐ: ∇²φ + [(k∞ − 1)/L²]φ = 0. That is Helmholtz. The operator is the Laplacian on the core's domain, the basis is its eigenfunctions there, and the boundary condition is φ = 0 on the extrapolated surface R̃ = R + 2.13 D. The bracket is the material buckling Bₘ², built from composition only. Notice what has happened: a neutron-balance statement has become the demand that the flux be an eigenfunction of −∇² with a prescribed eigenvalue.
Geometry supplies one number, and positivity picks it
The Laplacian on a bounded domain with a Dirichlet condition has a discrete spectrum. On a sphere of extrapolated radius R̃ the eigenfunctions are sin(nπr/R̃)/r with eigenvalues (nπ/R̃)². Only n = 1 is positive throughout; every higher mode has an interior node and would ask the flux to be negative somewhere, which no neutron density can be. So the geometric buckling Bg² is the lowest eigenvalue, and for a bare sphere it is (π/R̃)². Other shapes: a slab (π/ã)², a cube 3(π/ã)², a finite cylinder (2.405/R̃)² + (π/H̃)², whose volume is least at H/R = 1.85. At fixed volume the sphere has the smallest buckling of any shape, which is the eigenvalue version of saying it leaks least. Recast a critical sphere as a cube of the same volume and ã = (4π/3)¹⁄³ R̃ = 1.612 R̃, so Bg² rises by the factor 3/1.612² = 1.155 and the core goes subcritical.
Critical radius, critical mass, and what a reflector buys
Set Bg² = Bₘ² and solve for size. With k∞ = 1.20, D = 0.84 cm and Σₐ = 0.021 cm⁻¹, L² = 40.0 cm² and Bₘ² = 5.00 × 10⁻³ cm⁻², so R̃ = π/0.0707 = 44.43 cm and R = 44.43 − 1.79 = 42.64 cm, a critical volume of 3.25 × 10⁵ cm³. Multiply by density for the critical mass. A reflector cannot be folded into this one-region solve, because the surrounding material has no fission source: it needs a second region, φᵣ ∝ sinh((R + T − r)/Lᵣ)/r, matched to the core solution by continuity of φ and of the current J = −D dφ/dr at r = R. That match is a transcendental equation in R, solved numerically. Wrapping the same core in 10 cm of light water moves the critical radius from 42.6 cm to 29.4 cm — a reflector saving of 13.2 cm and a 67 per cent cut in critical volume, bought entirely by scattering escaping neutrons back.
Where one-speed diffusion stops being true
Fick's law is the first term of a near-isotropic expansion of the transport equation. It needs weak absorption compared with scattering, a flux that varies slowly on the scale of a transport mean free path, and distance from boundaries, voids and strong absorbers — two or three mean free paths at least. The extrapolated boundary is the honest admission of this: pure diffusion theory would put the flux zero at 2D = 0.667 λₜᵣ beyond the surface, and the value actually used, 0.7104 λₜᵣ, comes from solving the transport Milne problem and grafting the answer on. One energy group is a separate fiction, since fast and thermal neutrons have quite different D and Σₐ; two-group theory repairs it by replacing L² with the migration area M² = L² + τ. And for a small fast metal assembly the model simply fails: Godiva, a bare sphere of 94 per cent enriched uranium metal, is critical at radius 8.7 cm and about 52 kg, and only transport theory gets that number.
Change one variable at a time
Make the relationship visible.
Set k∞ = 1.20 and L² = 40 cm²: the solid marker lands at Rc = 42.6 cm. Drag R through it and watch keff cross 1 while the flux shape hardly changes. Then raise L² to 60 cm² and Rc moves out to 52.6 cm — a longer diffusion length lets more neutrons walk out.
GEOMETRIC BUCKLING Bg²25.9 ×10⁻⁴ cm⁻²
MULTIPLICATION keff1.088
CRITICAL RADIUS Rc42.6 cm
LEAKAGE9.4 %
Live interpretationGEOMETRIC BUCKLING Bg²: 25.9 ×10⁻⁴ cm⁻². MULTIPLICATION keff: 1.088. CRITICAL RADIUS Rc: 42.6 cm. LEAKAGE: 9.4 %
Catch the common trap
Explain before calculating.
A bare spherical core is exactly critical. Without changing its composition at all, it is recast as a cube of the same volume. What happens, and why?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA natural-uranium graphite lattice has ε = 1.03, p = 0.90, f = 0.89 and η = 1.34. Find k∞, and state the largest fraction of its neutrons a finished core of this material may lose to leakage and still run steadily.
- Take the four factors in life-cycle order: fast fission before slowing down (ε), survival past the U-238 resonances (p), thermal absorption in fuel rather than elsewhere (f), and neutrons released per thermal absorption in fuel (η).
- ε p = 1.03 × 0.90 = 0.927. Then × f: 0.927 × 0.89 = 0.8250. Then × η: 0.8250 × 1.34 = 1.1055.
- So k∞ = 1.106 — an infinite lattice of this material multiplies by 10.6 per cent per generation. Nothing in that number refers to size.
- A finite core is critical when keff = k∞ PNL = 1, so the non-leakage probability must be PNL = 1/1.1055 = 0.9045.
- Leakage fraction = 1 − 0.9045 = 0.0955.
Answerk∞ = 1.106, and the core may leak at most 9.5 per cent of its neutrons. That 9.5 per cent is the entire budget the geometry has to spend, which is why natural-uranium piles are metres across.
MediumA homogeneous thermal assembly has k∞ = 1.20, diffusion coefficient D = 0.84 cm and macroscopic absorption Σₐ = 0.021 cm⁻¹. Find L², the material buckling, the physical radius of a bare critical sphere, and its critical volume.
- Diffusion area L² = D/Σₐ = 0.84/0.021 = 40.0 cm², so L = 6.32 cm and the root-mean-square straight-line distance from thermalisation to absorption is √(6L²) = 15.5 cm.
- Material buckling Bₘ² = (k∞ − 1)/L² = 0.200/40.0 = 5.00 × 10⁻³ cm⁻², so Bₘ = 0.07071 cm⁻¹. This uses composition only.
- Criticality demands Bg² = Bₘ². The lowest Dirichlet eigenvalue of −∇² on a sphere is (π/R̃)², so R̃ = π/0.07071 = 44.43 cm.
- Subtract the extrapolation distance d = 2.13 D = 2.13 × 0.84 = 1.79 cm: the physical radius is R = 44.43 − 1.79 = 42.64 cm.
- V = (4/3)πR³ = 4.18879 × 42.64³ = 4.18879 × 7.752 × 10⁴ = 3.25 × 10⁵ cm³, or 325 litres.
AnswerL² = 40.0 cm², Bₘ² = 5.00 × 10⁻³ cm⁻², R = 42.6 cm, V = 3.25 × 10⁵ cm³. The flux is φ(r) ∝ sin(πr/44.43)/r, still 4.2 per cent of its central value at the physical surface.
HardWrap that core in 10 cm of light water, Dᵣ = 0.16 cm and Lᵣ = 2.85 cm. Set up the two-region matching condition, solve it for the reflected critical radius, and compare the saving with the thick-reflector slab estimate.
- The core keeps B = Bₘ = 0.07071 cm⁻¹ and φc = A sin(Br)/r. The reflector has no fission source, so φᵣ = C sinh((R + T − r)/Lᵣ)/r, which vanishes at the extrapolated outer edge R + T.
- Match φ and the current J = −D dφ/dr at r = R, i.e. match D(1/φ)(dφ/dr): Dc[B cot(BR) − 1/R] = −Dᵣ[(1/Lᵣ)coth(T/Lᵣ) + 1/R].
- Insert numbers. Dc B = 0.84 × 0.07071 = 0.05940; Dc − Dᵣ = 0.68; coth(10/2.85) = 1.0018, so (Dᵣ/Lᵣ)coth = 0.05624. The condition is 0.05940 cot(0.07071 R) = 0.68/R − 0.05624.
- Root-find it (brentq on 20–40 cm): R = 29.41 cm. The bare sphere needed 42.64 cm, so the reflector saving is δ = 13.2 cm.
- Core volume goes as R³: (29.41/42.64)³ = 0.328, so V falls from 3.25 × 10⁵ to 1.07 × 10⁵ cm³ — a 67 per cent cut for 10 cm of water.
- Compare the slab estimate δ ≈ (Dc/Dᵣ)Lᵣ tanh(T/Lᵣ) = 5.25 × 2.85 × 0.998 = 14.9 cm. It overshoots by 1.7 cm, 13 per cent, because it is a plane-geometry result used on a curved surface.
AnswerR = 29.4 cm, δ = 13.2 cm, critical volume down 67 per cent to 1.07 × 10⁵ cm³. The slab formula gives 14.9 cm, about 13 per cent too generous.