Induced fission, chain reactions and criticality

Induced fission begins as neutron capture. A neutron absorbed by ²³⁵U forms ²³⁶U with an excitation equal to the neutron's separation energy plus whatever kinetic energy it carried: Sₙ = [M(²³⁵U) + mₙ − M(²³⁶U)]c² = 6.55 MeV. To split, the compound nucleus must stretch past a fission barrier of roughly 6 MeV, so even a 0.025 eV neutron delivers enough. The same bookkeeping for ²³⁸U gives only 4.81 MeV for ²³⁹U, short of its barrier, so ²³⁸U fissions appreciably only when the neutron brings more than about 1 MeV of its own. The gap is pairing: the neutron captured by ²³⁵U completes an even-even nucleus and binds more tightly than one added to ²³⁸U. Absorption does not guarantee a split. For thermal neutrons on ²³⁵U the radiative-capture cross-section is 99 b against 585 b for fission, so about one absorption in seven ends as ²³⁶U and a gamma ray rather than a split.

Take one channel, n + ²³⁵U → ¹⁴¹Ba + ⁹²Kr + 3n, and balance the ledger first: charge 92 = 56 + 36, nucleon number 236 = 141 + 92 + 3. The fragments carry 92 electrons between them, as uranium does, so atomic masses subtract directly: Δm = 0.18602 u and Q = 173.3 MeV. Most of it appears as kinetic energy of the two fragments driven apart by their Coulomb repulsion. Two touching spheres at r = 11.7 fm would store Z₁Z₂e²/(4πε₀r) ≈ 250 MeV: the right order, and an overestimate, because the real scission shape is stretched and its charge centres sit further apart. The fragments inherit uranium's neutron excess, so they sit far from stability, and seven β⁻ decays carry them to stable ¹⁴¹Pr and ⁹²Zr. Atomic masses track those decays automatically, and the full difference gives 199 MeV. That whole budget is the quoted 200 MeV per fission. About 10 MeV of it leaves with antineutrinos and is never recovered, but the gamma rays emitted when the spare neutrons are captured return a similar amount, so close to 200 MeV per fission still ends up as heat. Per kilogram of ²³⁵U fissioned it is 8.2 × 10¹³ J, some 3 million times the heat from a kilogram of coal.

Fission neutrons are born at about 2 MeV, where the ²³⁵U fission cross-section is about 1.2 b. At 0.025 eV it is 585 b, roughly 500 times larger. Part of that is the 1/v law, since the reaction probability grows with the time a neutron spends near a nucleus, but extrapolating 1/v from thermal energy to 2 MeV predicts 0.07 b, twenty times too small: resonances and the fast region follow their own physics, so the advantage has to be read from measured data. It matters because the fuel competes for neutrons. At thermal energy, natural uranium (0.72% ²³⁵U) has η = 1.35 even though 35% of its absorptions happen in ²³⁸U; 3% enriched fuel reaches η = 1.84. Without moderation the picture collapses. Most fission neutrons scatter inelastically in ²³⁸U to below its roughly 1 MeV fission threshold long before they meet a ²³⁵U nucleus, so natural uranium on its own has k∞ below 1: no lump of it, however large, sustains a chain.

In an elastic collision with a nucleus of mass number A, a neutron keeps at least the fraction α = ((A − 1)/(A + 1))² of its energy, so each collision removes a roughly fixed fraction and ln E is the variable that falls evenly. Its mean drop per collision is ξ = 1 + α ln α/(1 − α): 1 for hydrogen, which can stop a neutron in one head-on hit, 0.725 for deuterium and 0.158 for carbon. Going from 2 MeV to 0.025 eV is ln(8 × 10⁷) = 18.2, so it takes about 18 collisions in hydrogen, 25 in deuterium and 115 in graphite, which is why graphite cores are large. Slowing power is only half the test. Hydrogen also captures neutrons (0.33 b), so light-water reactors need enriched fuel, while heavy water and graphite, slower to moderate but far weaker absorbers, can run on natural uranium. On the way down every neutron must also cross the capture resonances of ²³⁸U, the lowest at 6.7 eV; the chance of escaping them is one of the factors in k.

Define k as the number of fissions in one generation divided by the number in the generation before. In an infinite medium it is a product of factors, k∞ = ηfpε: η from the fuel, the share f of thermal absorptions that happen in fuel, the resonance-escape probability p and the fast-fission bonus ε. A finite core multiplies k∞ by a non-leakage probability to give k_eff. If a generation lasts ℓ, then dN/dt = (k − 1)N/ℓ and N = N₀ exp[(k − 1)t/ℓ]. With prompt neutrons alone, ℓ ≈ 10⁻⁴ s in a thermal core, and k = 1.001 would multiply the power by e¹⁰ ≈ 2 × 10⁴ every second, far faster than any rod can move. But β = 0.0065 of fission neutrons come from fragment β decays with a mean delay τ_d ≈ 13 s. Averaged over all neutrons, ℓ ≈ 10⁻⁴ s + βτ_d = 0.085 s, and the same k gives an e-folding time of 85 s. The average needs k − 1 ≪ β. A two-equation treatment gives T = τ_d(β − ρ)/ρ with ρ = (k − 1)/k, 72 s here, and T collapses towards the prompt time scale as k − 1 approaches β.

Neutrons are born throughout the volume but escape only through the surface, so the leakage fraction falls as a core grows, and some size must bring k_eff to exactly 1. Diffusion theory makes this quantitative. In a uniform bare core the flux obeys ∇²φ + B²φ = 0 with B² = (k∞ − 1)/M², the same Helmholtz equation as a standing wave, and φ must vanish at the edge. For a sphere put u = rφ: then u″ + B²u = 0, so φ = sin(Br)/r, and φ(R) = 0 requires BR = π. Criticality is the material's B fitting the geometry: R_c = πM/√(k∞ − 1). M is a transport length: a neutron ends up, in root-mean-square terms, √6 M from where it was born, and every length in the problem is a count of mean free paths. With k∞ = 1.30 and M = 7.0 cm, R_c = 40 cm; at k∞ = 1.10 it is 70 cm and the critical volume is 5.2 times larger. Shape matters too: a cube needs B² = 3(π/a)², so its critical volume is 24% larger than a sphere's.