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University Physics V · Nuclear Fission and Fusion · 16.2

Induced Fission & Neutron Moderation

Two different problems wear one name here. First: why a thermal neutron splits U-235 and bounces off U-238. Then: how to carry a 2 MeV neutron down to 0.025 eV through a forest of capture resonances without losing it — and why the fastest slower-down is not the best moderator.

Open the lab ↓Practice & worked examples ↓
01

Build the model

Connect the measurement to the mechanism.

Everything here follows from one fact: the neutron and the target do not stay separate. A slow neutron entering ²³⁵U is absorbed into a compound nucleus that lives about 2.7 × 10⁻¹⁴ s — some 10⁸ times longer than a nucleon needs to cross the nucleus — so by the time it decays the state has forgotten how it was made, and the cross-section factorises into a formation term times a branching ratio Γb/Γ. That long life means a narrow width, so the cross-section is a ladder of Breit-Wigner poles rather than a smooth curve, with a 1/v tail below the lowest one.

Which nuclide those poles belong to is settled by pairing: adding a neutron to odd-N ²³⁵U makes even-even ²³⁶U and releases 6.545 MeV, above the roughly 6 MeV saddle, so the neutron needs no kinetic energy of its own; the same neutron on ²³⁸U releases only 4.806 MeV and must bring about 1 MeV with it. Fission at rest therefore means slowing the neutron by eight decades in energy — and the path down runs straight through ²³⁸U's capture resonances. Elastic kinematics fix the price: α = ((A−1)/(A+1))² is the most one collision can take, ξ = 1 + α ln α/(1−α) is what it takes on average, and the trip costs ln(E₀/E)/ξ collisions.

The cost is that no moderator is free. The nucleus that slows fastest, hydrogen, also absorbs, and the design question splits into two figures of merit that rank the candidates in opposite orders.

Simple definition
Induced fission is the decay of a compound nucleus formed by neutron capture, and moderation is the elastic slowing-down that delivers the neutron to the energy where that capture is most likely to end in fission.
Example
A 2 MeV fission neutron needs 18 elastic collisions with hydrogen, or 115 with carbon, to reach 0.0253 eV — where ²³⁵U's fission cross-section is 585 b against ²³⁸U's, which is of order 10⁻⁵ b.
Compound-nucleus factorisationσ_(n, b)(E) = σCN(E) × Γb/Γ, Γ = Σᵢ Γᵢ

One formation cross-section, shared out: capture, fission and re-emission all come from the same σCN, and fission enters only through Γf/Γ.

Γᵢ are partial widths in eV; the branching Γb/Γ is independent of how the state was formed.

Single-level Breit-Wignerσ_(n, b)(E) = π ƛ² gJ Γₙ Γb / [(E − Eᵣ)² + Γ²/4]

Peak and width are separate observables: σ(Eᵣ) = 4 π ƛ² gJ Γₙ Γb/Γ², while Γ alone fixes the compound lifetime through Γτ = ħ.

π ƛ² = 6.51 × 10⁵ b⋅eV / E; gJ = (2J+1)/[2(2I+1)]; Γ = Σ Γᵢ is the FWHM, in eV.

The 1/v tail below the first resonanceσₐ(E) ≈ π ƛ² gJ Γₙ(E) Γγ / Eᵣ² ∝ E(−1/2) ∝ 1/v

σv is constant, so one tabulated value at 0.0253 eV and 2200 m s⁻¹ reproduces the rate in any Maxwellian — up to Westcott's g-factor.

Holds for E ≪ Eᵣ, where Γₙ = Γₙ⁰ √(E/1 eV) and π ƛ² ∝ 1/E; σ in barns, E in eV.

Fissile test: excitation against the saddleE* = Sₙ + Eₙ⋅A/(A+1) vs Efδ ≈ 12 A(−1/2) MeV

Pairing alone separates fissile from threshold: ²³⁵U clears the barrier on capture, ²³⁸U must be handed about 1 MeV of kinetic energy first.

Sₙ(²³⁶U) = 6.545 MeV, Sₙ(²³⁹U) = 4.806 MeV, both saddles near 6 MeV; δ = 0.78 MeV at A = 236.

Elastic collision: the limit and the averageα = ((A−1)/(A+1))², ξ = 1 + α ln α/(1 − α)

ξ does not depend on E, so collisions to slow from E₀ to E is just ln(E₀/E)/ξ — 18 in H, 25 in D, 115 in C, 2180 in ²³⁸U.

Both dimensionless; E′ uniform on [αE, E] for isotropic s-wave CM scattering. ξ(H)=1, ξ(D)=0.725, ξ(C)=0.158.

Two figures of merit that disagreemoderating power = ξΣₛmoderating ratio = ξΣₛ/Σₐ

Power says how fast the neutron gets down; ratio says whether it arrives. Natural uranium, with η = 1.34, can only afford to optimise the second.

Σₛ, Σₐ macroscopic, in cm⁻¹. H₂O 1.28, D₂O 0.18, C 0.060 cm⁻¹; ratios near 70, 5700, 190.

01

The compound nucleus lives long enough to forget

A slow neutron does not glance off ²³⁵U; it is absorbed, and the binding energy it brings is shared among all 236 nucleons. The resulting state decays with a width Γ of order 25 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 — 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) = σCN × Γb/Γ. Capture, fission and re-emission of the neutron then become one formation cross-section shared out by branching ratios, and only Γf/Γ carries anything specifically about fission. It is an approximation with a stated domain: it fails for direct reactions, and it fails for the fast end of the fission spectrum, where the collision is over before a compound state can form.

02

A resonance has a height, a width, and a temperature

Near an isolated level, σ_(n, b) = π ƛ² gJ Γₙ Γb/[(E − Eᵣ)² + Γ²/4], with π ƛ² = 6.51 × 10⁵ b⋅eV/E and gJ = (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, Γγ = 23 meV, on a spin-zero target so gJ = 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 not its height, and that asymmetry is where the prompt negative temperature coefficient comes from.

03

Why the tail is 1/v, and why tables say 2200 m/s

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 nucleus, N σ(v) v n(v), depends only on the total neutron density 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.

04

Pairing decides which nuclide a thermal neutron can split

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. The 1.74 MeV difference is pairing. 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. The consequence at thermal energy is stark: σf = 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.

05

One collision: the worst case and the average

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 = ∫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, 0.0083 for ²³⁸U. Collisions from 2 MeV to 0.0253 eV, N = ln(E₀/E)/ξ = 18.19/ξ, are 18, 25, 88, 115 and 2180. 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.

06

Rank moderators twice and get two different orders

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. The order reverses, and that reversal is the design decision. Hydrogen's α = 0 and its 0.332 b radiative capture, n + p → d + γ, are the same strong n–p interaction seen twice; you cannot buy the first without paying the second. Natural uranium yields η = ν σf/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.

02

Change one variable at a time

Make the relationship visible.

Interactive model
12
2.0 MeV
20

Drag A from 12 to 1 and watch the line steepen: carbon spends 44 collisions inside the resonance forest on its way to thermal, hydrogen only 7 — speed that hydrogen pays for with a 0.332 b capture cross-section carbon does not have.

Interactive physics modelSlowing-down line for a neutron born at 2.0 MeV in a moderator of mass number A = 12: log energy, 10 MeV at the top to 1 meV at the bottom, against collision number. The dashed lines at 1 keV and 1 eV bracket U-238's resonance forest; the lowest marks thermal equilibrium at 0.025 eV, where elastic slowing-down stops and the line flattens. The marker is at collision 20.E₀ = 2.0 MeV moderator A = 12Eₙ = E₀ exp(−n ξ)1 MeV1 keV1 eV1 meV0.025 eV thermalU-238 resonance forestcollisions n 0 → 140after 20 collisions u = n ξ = 3.16

Α ((A−1)/(A+1))²0.716

LETHARGY GAIN ξ0.158

COLLISIONS TO 0.025 eV115 collisions

OF THOSE, IN 1 eV–1 keV44 collisions

Live interpretationΑ ((A−1)/(A+1))²: 0.716. LETHARGY GAIN ξ: 0.158. COLLISIONS TO 0.025 eV: 115 collisions. OF THOSE, IN 1 eV–1 keV: 44 collisions

03

Catch the common trap

Explain before calculating.

A natural-uranium pile moderated by graphite is rebuilt with ordinary water in its place, everything else unchanged. Water's moderating power ξΣₛ is 1.28 cm⁻¹ against graphite's 0.060 cm⁻¹, yet the pile cannot be made critical. What has been overlooked?

Choose an answer to test the model.

04

Practice & worked examples

Reason from the model, then test the result.

EasyBeryllium-9 has been proposed as a moderator. Find the largest fractional energy loss one elastic collision can produce, the mean lethargy gain ξ, and the number of collisions needed to bring a 2.0 MeV fission neutron to 0.0253 eV. Compare with graphite's 115.
  1. α = ((A−1)/(A+1))² with A = 9 gives α = (8/10)² = 0.640. A collision cannot leave the neutron below 0.640 E, so the largest possible loss in one hit is 1 − α = 36% of the incident energy.
  2. ξ = 1 + α ln α/(1 − α) = 1 + 0.640 × ln 0.640 / 0.360. With ln 0.640 = −0.4463, the second term is 0.640 × (−0.4463)/0.360 = −0.7934.
  3. So ξ = 0.2066. Each collision costs the neutron a factor e0.2066 = 1.23 in energy on average, whatever energy it starts from — that constancy is exactly why lethargy is the right variable.
  4. Total lethargy needed: u = ln(2.0 × 10⁶/0.0253) = ln(7.905 × 10⁷) = 18.19. Collisions N = u/ξ = 18.19/0.2066 = 88.
  5. Graphite (A = 12, ξ = 0.158) needs 115, so beryllium is 24% faster per neutron. It is also toxic, expensive, and (n,2n) active above about 1.7 MeV — none of which ξ can tell you.

Answerα = 0.640, ξ = 0.207, and N = 88 collisions from 2.0 MeV to 0.0253 eV — 24% fewer than graphite's 115.

MediumU-238's lowest s-wave resonance sits at Eᵣ = 6.67 eV with Γₙ = 1.5 meV, Γγ = 23 meV and J = 1/2 on a spin-zero target. Find the peak radiative-capture cross-section, then extrapolate the single-level tail down to 0.0253 eV and compare it with the tabulated thermal value of 2.68 b.
  1. π ƛ² = 6.51 × 10⁵ b⋅eV / E, so at the resonance π ƛ² = 6.51 × 10⁵/6.67 = 9.76 × 10⁴ b. The target has I = 0, so gJ = (2J+1)/[2(2I+1)] = 2/2 = 1.
  2. Total width Γ = Γₙ + Γγ = 1.5 + 23 = 24.5 meV. At E = Eᵣ the Breit-Wigner denominator collapses to Γ²/4, so σ₀ = 4 π ƛ² gJ Γₙ Γγ/Γ² = 4 × 9.76 × 10⁴ × (1.5 × 23)/24.5².
  3. = 3.904 × 10⁵ × 34.5/600.25 = 3.904 × 10⁵ × 0.05748 = 2.2 × 10⁴ b — four orders of magnitude above the geometric πR² ≈ 1.7 b. Nothing about the nuclear radius sets this; π ƛ² does.
  4. Far below the resonance, (E − Eᵣ)² + Γ²/4 → Eᵣ² = 44.5 eV², while π ƛ² grows as 1/E and the s-wave neutron width falls as √E: Γₙ(0.0253 eV) = 1.5 meV × √(0.0253/6.67) = 0.0924 meV.
  5. π ƛ²(0.0253 eV) = 6.51 × 10⁵/0.0253 = 2.573 × 10⁷ b, so σγ = 2.573 × 10⁷ × (9.24 × 10⁻⁵ × 0.023)/44.5 = 1.2 b. The two E-dependences combine to E(−1/2), which is the 1/v law.
  6. That single level supplies about 45% of the tabulated 2.68 b. The rest comes from a bound level below neutron threshold and from the ladder above — one Breit-Wigner term is never the whole cross-section.

Answerσ₀ = 2.2 × 10⁴ b at the peak; the single-level 1/v tail gives 1.2 b at 0.0253 eV, about 45% of the tabulated 2.68 b.

HardReactor graphite has density 1.60 g cm⁻³, σₛ = 4.75 b and σₐ = 0.0034 b at thermal energy. Compute its moderating power and moderating ratio, then find how much natural boron (σₐ = 767 b, molar mass 10.81 g mol⁻¹) it takes to pull the moderating ratio down to light water's 70.
  1. Atom density NC = (1.60/12.011) × 6.022 × 10²³ = 8.02 × 10²² cm⁻³. Then Σₛ = 8.02 × 10²² × 4.75 × 10⁻²⁴ = 0.381 cm⁻¹ and Σₐ = 8.02 × 10²² × 0.0034 × 10⁻²⁴ = 2.73 × 10⁻⁴ cm⁻¹.
  2. With ξ = 0.1577 the moderating power is ξΣₛ = 0.1577 × 0.381 = 0.0601 cm⁻¹ — about a twenty-first of light water's 1.28 cm⁻¹.
  3. Moderating ratio for pure carbon: ξΣₛ/Σₐ = 0.0601/2.73 × 10⁻⁴ = 220. Tables print about 190 for reactor graphite, and that gap is impurity, not physics.
  4. One part per million of boron by mass: NB = (10⁻⁶ × 1.60/10.811) × 6.022 × 10²³ = 8.91 × 10¹⁶ cm⁻³, so Σₐ(B) = 8.91 × 10¹⁶ × 767 × 10⁻²⁴ = 6.84 × 10⁻⁵ cm⁻¹ — a quarter of carbon's own absorption, from one part in a million.
  5. To reach a ratio of 70 the total absorption must be Σₐ = 0.0601/70 = 8.58 × 10⁻⁴ cm⁻¹, leaving 8.58 × 10⁻⁴ − 2.73 × 10⁻⁴ = 5.85 × 10⁻⁴ cm⁻¹ for boron.
  6. That is 5.85 × 10⁻⁴ / 6.84 × 10⁻⁵ = 8.6 ppm. Under nine parts per million of boron makes graphite no better a neutron economy than water — which is why the Chicago pile team specified graphite purity before anything else.

AnswerξΣₛ = 0.0601 cm⁻¹ and ξΣₛ/Σₐ = 220 for pure graphite; about 8.6 ppm of natural boron drops the ratio to light water's 70.