research32 min studyLesson 596 in Physics

Slow neutron-capture paths and branching points

Compare capture and decay hazards, derive a local flow relation, and retain the assumptions behind neutron-density diagnostics.

On this page

What you’ll understand

  • Compute a neutron-capture rate per target and compare it with a beta mean lifetime.
  • Derive a branching fraction from two competing exponential hazards.
  • State where a stationary capture-flow approximation can and cannot predict abundance ratios.

The slow neutron-capture process builds heavier nuclei through captures interspersed with beta decays. Slow refers to the competition between the capture time and relevant weak lifetimes, rather than to an absolute clock for a whole star. Many unstable nuclei decay before capturing another neutron, keeping the path near stable isotopes. At a branching point the two hazards become comparable, and the resulting products can carry information about neutron density and the thermal environment.

Neutrons remove one barrier but not nuclear structure

A neutron has no electric charge, so there is no entrance-channel Coulomb penetration factor of the charged-particle type. Its capture cross section still depends on energy, resonances, level structure, and available gamma transitions. For thermal neutrons with a Maxwell relative distribution, the capture hazard per target is neutron density times the thermal reactivity. A Maxwellian-averaged cross section is commonly defined by dividing that reactivity by a stated thermal speed; its normalization should be checked before multiplying a tabulated value by a velocity.

In dilute targets with a maintained neutron reservoir, this hazard gives a target's exponential survival against capture alone. Beta decay supplies another hazard, obtained from its mean lifetime or half-life conversion. Stellar weak rates can differ from terrestrial values because of ionization, thermal excitation, and electron environment. A branch prediction therefore needs rates appropriate to the relevant temperature and density. Merely labeling an isotope unstable does not decide which outgoing path dominates under all astrophysical conditions.

λn=nn⟨σv⟩,λβ=ln⁡2t1/2,τn=λn−1\lambda_n=n_n\langle\sigma v\rangle,\qquad \lambda_\beta=\frac{\ln2}{t_{1/2}},\qquad \tau_n=\lambda_n^{-1}
Neutron density is number per volume, reactivity is volume per time, and both hazards are inverse time. The half-life formula assumes a specified effective exponential beta-decay law.↔ Scroll wide equations horizontally.

Competing hazards define the branch probability

Suppose a single unstable nucleus has only capture and beta-decay exits and the two hazards are constant while it waits. Survival against either is exponential with their sum. Integrating the capture hazard times that survival gives the eventual capture probability. This is not the probability of capture during an arbitrarily short exposure, because a finite exposure can leave an undecayed and uncaptured population. Counting that remaining population is essential when comparing a laboratory irradiation with an ideal eventual branch fraction.

When the hazards are comparable, changing neutron density moves the branch fraction strongly. Increasing the capture rate far above the decay rate drives the capture fraction toward one; taking neutron density toward zero drives it toward zero. These limiting checks help catch inverted ratios. More elaborate branch points can include internal isomeric states, thermally coupled levels, and several destruction routes, in which case a two-hazard formula is a reduced model whose omitted channels must be shown to be negligible.

Psurv(t)=e−(λn+λβ)t,fn=∫0∞λnPsurv(t) dt=λnλn+λβP_{\rm surv}(t)=e^{-(\lambda_n+\lambda_\beta)t},\qquad f_n=\int_0^\infty\lambda_nP_{\rm surv}(t)\,dt=\frac{\lambda_n}{\lambda_n+\lambda_\beta}
This expression assumes independent constant hazards with two exclusive exits, no replenishment, and an indefinitely continued environment. During a finite exposure the capture fraction is fn times one minus the survival probability.↔ Scroll wide equations horizontally.

A local steady-flow relation has boundaries

Away from branch points, a simple capture chain can approach a steady local flow when the neutron exposure changes slowly and source and sink boundary effects are remote. Production into one isotope then balances its capture destruction. Under a common neutron distribution this yields an approximate inverse relation between abundance and capture cross section. Nuclei near closed neutron shells often have smaller capture rates and can accumulate, but the detailed pattern still depends on the exposure history and nuclear data.

The local relation is not valid for every isotope from the start of irradiation. A finite neutron pulse, a magic-number bottleneck, a strong branch, or a changing seed inventory can prevent stationarity. Neutrons are supplied by separate reactions and can be absorbed by poisons as well as the heavy-element path. A branch ratio can constrain a model of the environment, but one observed isotope ratio does not uniquely fix all stellar mixing, neutron-source, and temperature histories without additional evidence.

nn⟨σv⟩A−1YA−1≃nn⟨σv⟩AYA⟹⟨σv⟩AYA≃constantn_n\langle\sigma v\rangle_{A-1}Y_{A-1}\simeq n_n\langle\sigma v\rangle_A Y_A\quad\Longrightarrow\quad \langle\sigma v\rangle_A Y_A\simeq\text{constant}
This is a local stationary unbranched capture-flow approximation at a common neutron distribution. Y is abundance per baryon; a corresponding cross-section relation requires the same stated thermal-speed convention.↔ Scroll wide equations horizontally.

A deliberately specified branching point

For an illustrative isotope use nn=3×108 cm−3n_n=3\times10^8\,\mathrm{cm}^{-3}, ⟨σv⟩=10−16 cm3s−1\langle\sigma v\rangle=10^{-16}\,\mathrm{cm}^3\mathrm{s}^{-1}, and effective beta hazard 10−8 s−110^{-8}\,\mathrm{s}^{-1}. Find the capture fraction, mean waiting time to either exit, and eventual expected products from 500 initially identical targets.

  1. The capture hazard is 3×108×10−16=3×10−8 s−13\times10^8\times10^{-16}=3\times10^{-8}\,\mathrm{s}^{-1}. Adding beta decay gives total hazard 4×10−8 s−14\times10^{-8}\,\mathrm{s}^{-1}.

  2. The eventual capture fraction is 3/(3+1)=3/43/(3+1)=3/4. The mean time until either event is the inverse total hazard, 2.5×107 s2.5\times10^7\,\mathrm s, rather than either individual lifetime.

  3. Multiply the branch probabilities by 500 to obtain expected counts 375 captures and 125 beta decays. These are ensemble means with the environment maintained indefinitely, not guaranteed integer outcomes in one finite neutron pulse.

The result

Three quarters eventually capture, with a mean waiting time of twenty-five million seconds and expected branch counts 375 and 125.

Explore the model

Interactive lab

Interactive lab · probability

Ensemble scatter around a capture branch fraction

Reset and set success probability to 0.75 and trials to 20. Predict mean fifteen capture-labeled outcomes and variance 3.75 under independent identical branch trials. The bars are the exact binomial distribution, and playback only highlights outcomes. This is a conditional counting model, not a neutron source, exposure-history calculation, or multi-isotope s-process network.

Ensemble scatter around a capture branch fractionThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA deterministic branch probability still produces a distribution of finite-sample counts.

Check your understanding

For two constant competing hazards, what happens to the eventual capture fraction if neutron density tends to zero while beta decay stays finite?

Try it yourself

Two neighboring unbranched isotopes share a stationary neutron distribution. Their consistently defined Maxwellian-averaged cross sections are 100 and 25 millibarns. If the first abundance is 2×10−62\times10^{-6} per baryon, find the second abundance under local steady flow and state a condition that would invalidate the inference.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The same thermal speed cancels, so cross section times abundance must match for the two isotopes.

Compare with the solution

The relation gives 100(2×10−6)=25Y2100(2\times10^{-6})=25Y_2, hence Y2=8×10−6Y_2=8\times10^{-6} per baryon. The lower-cross-section isotope has four times the abundance in this local stationary approximation. A finite pulse before stationarity, a beta branch, or a changing neutron distribution can invalidate this inferred ratio even if the tabulated cross sections are accurate.

Take it with you

  • Capture competes with weak decay through rates per target, with consistent thermal averaging.
  • The two-hazard branch formula requires a maintained environment and counts eventual exits.
  • Inverse cross-section abundance scaling is a local steady-flow approximation, not a universal yield law.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.