Kvant Physics Problem 199

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Problem

A neutron easily passes through a layer of lead, but is stopped in an equivalent layer of paraffin, water, or another hydrogen-containing compound. Explain why.

P. L. Kapitsa

Setup and Assumptions

A neutron of mass $m$ and kinetic energy $E_0$ enters a material consisting either of heavy nuclei (lead, mass $M_{\mathrm{Pb}}$) or hydrogen-rich nuclei such as protons in paraffin or water (mass $M_{\mathrm{H}}$). The neutron interacts with nuclei through short-range nuclear forces, and the dominant process in both media is elastic scattering. The neutron moves in a stationary medium in the laboratory frame, and individual collisions are treated as isolated two-body elastic collisions. Energy losses to radiation, inelastic nuclear reactions, and collective effects are neglected except insofar as they justify using an effective scattering picture. The unknown quantity is the qualitative mechanism determining why the neutron is strongly attenuated in hydrogen-containing substances and weakly attenuated in lead.

Physical Principles

Energy and momentum conservation in an elastic collision between two particles are expressed by

$m \vec{v} + M \vec{0} = m \vec{v}' + M \vec{V},$

$\frac{1}{2} m v^2 = \frac{1}{2} m v'^2 + \frac{1}{2} M V^2,$

where $v$ and $v'$ are the initial and final neutron speeds and $V$ is the recoil speed of the nucleus.

For a head-on elastic collision, the fraction of neutron kinetic energy transferred to the nucleus is

$\eta = \frac{\Delta E}{E} = \frac{4mM}{(m+M)^2}.$

The neutron attenuation in matter is governed by repeated scattering events, where the effective slowing down depends on the average fractional energy loss per collision and the scattering cross section of nuclei.

Derivation

The behavior of a neutron passing through matter is determined by how efficiently a single collision reduces its kinetic energy. The energy transfer fraction in a head-on elastic collision is

$\eta(M) = \frac{4mM}{(m+M)^2}.$

For heavy nuclei such as lead, $M \gg m$, so the denominator expands as $(m+M)^2 \approx M^2$. Substitution into the expression gives

$\eta_{\mathrm{Pb}} \approx \frac{4mM_{\mathrm{Pb}}}{M_{\mathrm{Pb}}^2} = \frac{4m}{M_{\mathrm{Pb}}}.$

Since $M_{\mathrm{Pb}} \gg m$, the fraction $\eta_{\mathrm{Pb}}$ is very small, meaning that each collision removes only a tiny portion of the neutron energy. The neutron therefore requires a very large number of collisions to significantly slow down, and because the interaction probability per unit length in dense metals is not sufficient to enforce that many collisions before escape, the neutron effectively penetrates the lead layer.

For hydrogen-containing materials, the relevant nuclei are protons with mass $M_{\mathrm{H}} \approx m$. Substituting into the same expression yields

$\eta_{\mathrm{H}} = \frac{4m^2}{(2m)^2} = 1.$

This shows that in a head-on collision the neutron can transfer essentially all its kinetic energy to a proton. Even when averaging over scattering angles, the energy loss per collision remains a large fraction of the neutron energy. After a small number of collisions, the neutron energy decreases to thermal values $E \sim kT$, at which point it becomes susceptible to capture reactions with hydrogen or other nuclei in the medium. The repeated efficient moderation causes the neutron to lose its directed motion and be absorbed within a relatively short distance.

Result

The qualitative result is determined by the dependence of energy transfer efficiency on target mass,

$\eta(M) = \frac{4mM}{(m+M)^2}.$

For heavy nuclei such as lead,

$\eta_{\mathrm{Pb}} \approx \frac{4m}{M_{\mathrm{Pb}}} \ll 1,$

so many collisions are required for significant slowing.

For hydrogen nuclei,

$\eta_{\mathrm{H}} \approx 1,$

so a few collisions reduce the neutron energy to thermal values, after which absorption dominates.

The neutron is therefore weakly attenuated in lead and strongly attenuated in hydrogen-containing substances.

Sanity Checks

The expression $\eta(M) = \frac{4mM}{(m+M)^2}$ is dimensionless since both numerator and denominator contain squared mass factors. In the limit $M \to \infty$, the energy transfer fraction scales as $\eta \to 4m/M$, which tends to zero and corresponds to negligible recoil of an infinitely heavy nucleus. In the limit $M = m$, the fraction becomes $\eta = 1$, corresponding to complete energy transfer in a head-on collision between equal masses.

The physical behavior matches these limits: light targets efficiently slow neutrons, while heavy targets produce nearly elastic scattering with minimal energy loss. A common error arises from confusing energy transfer per collision with scattering probability; even if scattering in lead is frequent, the small value of $\eta_{\mathrm{Pb}}$ prevents effective moderation.