advanced34 min studyLesson 588 in Physics

Neutron cross sections and resonance absorption

Connect microscopic areas to beam attenuation, spectrum-weighted rates, and the conditional scope of a 1/v law.

Before you begin

On this page

The question for today

What you’ll understand

Convert microscopic cross sections into macroscopic removal coefficients.
Distinguish absorption from removal of an uncollided beam.
Compare spectrum averages and the assumptions of low-energy 1/v capture.

A neutron cross section is an effective area for a specified target state, reaction, and neutron energy. It is not the geometric area of a hard nuclear sphere. Large resonance cross sections arise from wave dynamics and long wavelengths. To predict an experimental count or attenuation, microscopic information must be combined with target number density, the incident energy distribution, and a clear definition of which neutron histories remain in the measured beam.

Microscopic areas become probabilities per path length

For a homogeneous target containing independent scatterers, multiplying a microscopic cross section by the target number density gives a macroscopic coefficient. Its units are inverse length. The probability of a specified first interaction over a sufficiently short path is this coefficient times the path length. Adding mutually exclusive reaction channels gives the total interaction coefficient. A barn is an area unit equal to one hundred square femtometres, useful because nuclear cross sections span a wide range around that scale.

The exponential transmission law describes neutrons that have had no interaction counted as removal from a narrow uncollided beam. Elastic scattering out of its angular acceptance removes a neutron from that beam without absorbing it. An absorption-only coefficient therefore gives a different exponent unless scattering is absent or the measurement definition includes scattered particles. Following scattered neutrons through multiple later collisions requires a transport calculation, not merely substituting another number into the uncollided exponential.

1\,\mathrm b=10^{-28}\,\mathrm{m}^2=100\,\mathrm{fm}^2,\qquad\Sigma_t=n_T\sigma_t,\qquad I(x)=I(0)e^{-\Sigma_t x}

nT is target number density in m⁻³, σt is the total removal area in m² for the stated uncollided-beam geometry, and Σt is in m⁻¹. The law assumes a homogeneous stationary medium, fixed neutron energy, and independent interactions; it excludes buildup from scattered particles.↔ Scroll wide equations horizontally.

A flux average differs from a value at mean energy

A real beam can contain a distribution of energies. The absorption rate per target nucleus is the integral of cross section times differential flux. Replacing that integral by the cross section at the mean energy is generally invalid near a narrow resonance because the cross section is highly nonlinear. Beam-energy resolution and target thermal motion can spread the measured resonance profile, while isotope mixtures require a sum over their number densities and cross sections.

Thin targets sample the incoming spectrum approximately unchanged. Thick targets can selectively remove neutrons near resonance energies, changing the spectrum inside the material and reducing the effective absorption per nucleus compared with a naive uniform-flux estimate. This resonance self-shielding is a transport and energy-distribution effect, not a failure of the microscopic cross section. An average must specify its weighting, since flux weighting, particle-density weighting, and a thermal rate coefficient answer different experimental questions.

r_a=\int_0^\infty\phi(E)\sigma_a(E)\,dE,\qquad \overline\sigma_\phi=\frac{\int\phi(E)\sigma(E)dE}{\int\phi(E)dE},\qquad R_a=N_T r_a

φ(E) is differential neutron flux in m⁻² s⁻¹ per energy, ra is absorption rate per target nucleus in s⁻¹, and NT counts target nuclei in a sufficiently thin uniformly illuminated sample. The weighted cross section is an area; the total rate Ra is in s⁻¹.↔ Scroll wide equations horizontally.

The inverse-speed law is a regime, not a universal property

For low-energy s-wave neutron capture away from a nearby pole or threshold complication, the entrance partial width scales with speed while the wavelength area scales inversely with energy. Their combination can give absorption proportional to inverse speed. This produces an approximately constant product of capture cross section and speed. It does not apply through a resolved resonance peak, and other partial waves or changing exit widths can alter the dependence.

The constant product is useful only with a declared comparison. At fixed neutron number density, incident flux rises with speed, canceling the inverse-speed area in the per-target reaction rate. At fixed incident flux, faster neutrons instead give a lower rate under that same law. Report whether a reference cross section is evaluated at one speed or averaged over a spectrum, and distinguish capture into photons from other absorption channels whenever more than one is open.

Work it through

Separate transmission from the first absorption event

A toy homogeneous target has $n_T=2\times10^{28}\,\mathrm{m}^{-3}$ , absorption cross section 5 b, and scattering-removal cross section 3 b at a fixed neutron energy. For thickness $x=(\ln2)/16\,\mathrm m$ , find uncollided transmission and the probability that a first collision inside the slab is absorption. Ignore all subsequent scattered histories.

The total microscopic removal cross section is 8 b. Thus $\Sigma_t=(2\times10^{28})(8\times10^{-28})=16\,\mathrm{m}^{-1}$ , while $\Sigma_a=10\,\mathrm{m}^{-1}$ .
The specified thickness gives optical depth $\Sigma_t x=\ln2$ , so the uncollided transmission is one half. The probability of at least one first removal before exiting is also one half.
Among first removals, the absorption fraction is $\Sigma_a/\Sigma_t=10/16=5/8$ . Multiplying by the first-removal probability gives $P_{a,\rm first}=(5/8)(1/2)=5/16$ . The remaining first-scatter probability is $3/16$ .

The result

Uncollided transmission is 0.50 and first-event absorption probability is 0.3125. The 0.1875 first-scatter fraction has been removed from the beam but is not automatically absorbed.

The living notebook

Explore the model

Interactive lab

Interactive lab · probability

A cross section becomes a bounded event probability

For a separate toy thin-target experiment with first-absorption probability 0.30, Reset and set both controls to p = 0.30 and n = 10. Predict mean absorption count 3 and variance 2.1. These are independent incident neutrons and exact binomial probabilities. The lab has no target thickness, cross-section spectrum, scattering transport, or resonance controls, and playback does not generate random collision histories.

Exact probabilityHighlighted outcomeMean np

Mean · np: 5
Variance · np(1−p): 2.5
Total probability: 1

Success probability p0.5Trials n10

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 microscopic area must be combined with a target and flux model before it becomes a counting probability.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Removal from a narrow beam includes deflection outside its acceptance. The scattered neutron still exists and may later interact; absorption and uncollided-beam removal are different transport observables.

Your turn

Try it yourself

A toy capture cross section obeys 1/v and equals 4 b at speed v. At fixed incident flux $10^{10}\,\mathrm{cm}^{-2}\mathrm{s}^{-1}$ , calculate the per-target capture rate at v and at 2v. Then state what changes if neutron number density, rather than flux, is held fixed.

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

Give me a hint

One barn is 10⁻²⁴ cm². At twice the speed, the stipulated cross section halves.

Compare with the solution

At v, the rate per target is $10^{10}(4\times10^{-24})=4\times10^{-14}\,\mathrm{s}^{-1}$ . At 2v the cross section is 2 b, so the same fixed flux gives $2\times10^{-14}\,\mathrm{s}^{-1}$ . If neutron number density is fixed instead, flux doubles with speed and compensates the halved cross section, leaving the rate unchanged within the assumed inverse-speed regime. Neither result should be extrapolated through a resonance.

Keep the idea

Take it with you

A barn is an area, while a macroscopic cross section is inverse length.
Beam removal includes scattering as well as absorption.
Energy averaging and inverse-speed scaling require explicit weighting and regime assumptions.

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