Induced fission and neutron moderation

A slow neutron does not glance off a heavy nucleus; it is absorbed, and the binding energy it brings is shared among all the nucleons. At ²³⁸U's 6.67 eV level the ²³⁹U* state decays with a total width Γ = Γₙ + Γ_γ = 24.5 meV, so it lives τ = ħ/Γ = 6.58 × 10⁻¹⁶ eV⋅s ÷ 0.0245 eV = 2.7 × 10⁻¹⁴ s. A nucleon crosses a nucleus of that size in about 2 × 10⁻²² s, so the compound nucleus survives some 10⁸ crossings; the ²³⁶U* levels reached from ²³⁵U are broader, because a fission width adds to a radiative width of about 35 meV, but they too last millions of crossings — long enough for every memory of the entrance channel except energy, angular momentum and parity to be washed out. That is Bohr's independence hypothesis, and it buys the factorisation the rest of the topic rests on: σ_{(n, b)} = σ_CN × Γ_b/Γ. Capture, fission and re-emission of the neutron then become one formation cross-section shared out by branching ratios, and only Γ_f/Γ carries anything specifically about fission. It is an approximation with a stated domain: it fails for direct reactions, and it weakens toward the fast end of the fission spectrum, where direct and pre-equilibrium emission take a growing share of collisions before a compound state has formed.

Near an isolated level, σ_{(n, b)} = π ƛ² g_J Γₙ Γ_b/[(E − Eᵣ)² + Γ²/4], with π ƛ² = 6.51 × 10⁵ b⋅eV/E and g_J = (2J+1)/[2(2I+1)] counting the fraction of entrance spin states that can reach the level. Take ²³⁸U's lowest resonance, Eᵣ = 6.67 eV, Γₙ = 1.5 meV, Γ_γ = 23 meV, on a spin-zero target so g_J = 1. At the peak the denominator is Γ²/4, giving σ₀ = 4 × (6.51 × 10⁵/6.67) × (1.5 × 23)/24.5² = 2.2 × 10⁴ b. That is ten thousand times the geometric πR² ≈ 1.7 b, because the scale in the formula is the neutron's reduced wavelength, not the nuclear radius. What a spectrometer actually records is broader and lower: thermal motion of the target smears the level over a Doppler width Δ = √(4Eᵣ kT/A) = √(4 × 6.67 × 0.0259/238) = 54 meV, more than twice the natural Γ. Heating the fuel conserves the area under the peak but lowers its height and widens its wings; in a fuel lump the flux is already depressed at the peak and not in the wings, so the broadened line captures more neutrons, and that is where the prompt negative temperature coefficient comes from.

Below the lowest resonance, two energy dependences meet. The geometric factor π ƛ² = πħ²/(2mE) falls as 1/E. The s-wave neutron width rises as Γₙ = Γₙ⁰√(E/1 eV), because a zero-angular-momentum neutron faces no centrifugal barrier and its penetrability goes as k. The denominator, meanwhile, is frozen at Eᵣ². Multiply the three: σₐ ∝ E^(−1/2) ∝ 1/v. So σv is a constant, and the reaction rate per unit volume, N ∫σ(v) v n(v) dv = N σv n, depends only on the total neutron density n and not on how those neutrons are spread over speed. That is why cross-sections are tabulated at a single point, E = 0.0253 eV, v = 2200 m s⁻¹: for a 1/v absorber that one number reproduces the rate in any Maxwellian spectrum, at any moderator temperature. Real nuclides are not perfectly 1/v, and the correction is Westcott's g-factor — 0.976 for ²³⁵U fission at 20 °C, so a 2.4% error if you ignore it, and much larger for ²³⁹Pu, whose 0.3 eV resonance sits on the shoulder of the thermal peak.

Capture on ²³⁵U makes ²³⁶U* with excitation E* = Sₙ + Eₙ⋅A/(A+1). From atomic masses, Sₙ(²³⁶U) = (235.043930 + 1.008665 − 236.045568) u × 931.494 MeV/u = 6.545 MeV, and the saddle-point barrier of ²³⁶U sits near 6 MeV. The neutron therefore clears the barrier bringing no kinetic energy at all: ²³⁵U is fissile. Repeat on ²³⁸U and Sₙ(²³⁹U) = 4.806 MeV, roughly 1.4 MeV short of ²³⁹U's saddle, so ²³⁸U is fissionable but only above a threshold near 1 MeV. Most of the 1.74 MeV difference is pairing, counted twice. ²³⁵U has 143 neutrons, odd; the captured neutron pairs off and the even-even product keeps the pairing energy δ ≈ 12 A^(−1/2) = 0.78 MeV at A = 236. ²³⁸U's 146th neutron is already paired, and the 147th arrives with no partner, so ²³⁹U forgoes the same δ. Gaining δ in one case and forgoing it in the other gives 2δ ≈ 1.56 MeV; the smooth fall of Sₙ with neutron excess supplies the remaining 0.18 MeV. The consequence at thermal energy is stark: σ_f = 585 b for ²³⁵U against order 10⁻⁵ b for ²³⁸U. Get a neutron down to 0.0253 eV and the 0.72% minority isotope does essentially all the work.

Elastic scattering in the centre of mass leaves the neutron's speed unchanged and only turns it, so back in the lab the outgoing energy runs from E down to αE with α = ((A−1)/(A+1))². For hydrogen α = 0 and a single collision can take everything; for carbon α = 0.716, so no collision can take more than 28%; for ²³⁸U itself α = 0.983. Below a few hundred keV, s-wave scattering is isotropic in the centre of mass, which makes E′ uniform on [αE, E]. Averaging the lethargy gain over that flat distribution, ξ = ∫ from αE to E of ln(E/E′) dE′/[(1−α)E] = 1 + α ln α/(1−α), gives a number that does not depend on E — the reason lethargy rather than energy is the natural variable, and the reason the slowing-down line is straight. The values: ξ = 1 for H, 0.725 for D, 0.207 for Be, 0.158 for C, 0.00838 for ²³⁸U. Collisions from 2 MeV to 0.0253 eV, N = ln(E₀/E)/ξ = 18.19/ξ, are 18, 25, 88, 115 and 2170. Count only the 1 eV–1 keV band, of lethargy width 6.91, and hydrogen spends 7 collisions inside ²³⁸U's resonance forest where carbon spends 44.

Moderating power is ξΣₛ, the lethargy removed per centimetre travelled: 1.28 cm⁻¹ for light water, 0.18 for heavy water, 0.060 for graphite. On that measure water wins by a factor of 21. Moderating ratio is ξΣₛ/Σₐ, the same slowing-down weighed against parasitic absorption: about 70 for water, 190 for reactor graphite, 5700 for 99.75% heavy water. Water drops from first to last, and that reversal is the design decision. Hydrogen's α = 0 is pure equal-mass kinematics, but every hydrogen nucleus that offers it also offers 0.332 b of radiative capture, n + p → d + γ; you cannot buy the first without paying the second. Natural uranium yields η = ν σ_f/σₐ = 2.42 × 4.21/7.59 = 1.34 neutrons per absorption in fuel, so barely a third of a neutron of margin has to cover leakage, structure and every other loss. Heavy water fits inside that margin and light water does not — which is why CANDU runs unenriched fuel while every light-water reactor enriches to 3–5% ²³⁵U, and why graphite piles compensate with wide lattice pitch and lumped fuel that self-shields the resonances.