Neutron detection by proton recoil

In neutron detection by proton recoil, a neutron strikes a hydrogen atomic nucleus, namely a proton, and the moving proton produces detectable ionization. Paraffin wax provides many hydrogen targets. There are two distinct questions: how much kinetic energy can a collision transfer, and how often does it occur? Elastic collision kinematics answers the first. It does not determine a collision cross-section or a detector's efficiency.

For the following derivations the target is initially at rest in the laboratory but is free to recoil. Its mass is finite. In particular a proton must not be treated as an immovable target. All massive particles are nonrelativistic particles, their translational kinetic energies are $m{v}^2/2$, and the collision changes no internal state. Target binding and thermal motion are neglected relative to the incident kinetic energy. These assumptions suit an introductory description of fast-neutron elastic collisions; they do not cover every neutron capture or other nuclear reaction.

Let a projectile have mass $m$, incident speed $v$, and incident kinetic energy $E=m{v}^2/2$. The stationary target has mass $M$. In a head-on elastic collision, the signed outgoing projectile velocity $v^{\prime }$ and target velocity $u$ obey momentum conservation and conservation of energy:Factor the second equation and eliminate $Mu=m(v-v^{\prime })$. The nontrivial collision branch has $u=v+v^{\prime }$, henceThis is the maximum over scattering angles, not the energy of every recoil. To see the bound without assuming a head-on event, set the recoil momentum magnitude to $q$ and its laboratory angle to the incident direction to $\theta$. Conservation of energy with outgoing projectile momentum $\mathbf{p}-\mathbf{q}$ givesThus $T=T_{\max }{\cos }^2\theta$ for a nonzero recoil, and the endpoint occurs at $\theta =0$.

For an original numerical example, stipulate $E=4\mathrm{MeV}$ and $m=m_p$, with $m_p$ the proton mass. Approximate hydrogen, helium and nitrogen target masses by $M=m_p$, $4m_p$ and $14m_p$ respectively. Their maximum fractions and recoil energies areThese stipulated values illustrate the approximation $m_n=m_p$; they are not measurements from 1932. With $m_p{c}^2\simeq 938\mathrm{MeV}$, the incident ratio $E/(m_p{c}^2)\simeq 0.0043$ gives $v/c\simeq 0.092$, consistent with the nonrelativistic limit used here. A million electronvolts is an energy unit, whereas a mass expressed in $\mathrm{MeV}/c^2$ requires the factor $c^2$ when inserted into an energy equation.

The hydrogen endpoint permits essentially complete transfer because the projectile and target masses match. For a very heavy target, $M\gg m$, the fraction approaches $4m/M$, which is small: the target can take substantial momentum with little kinetic energy. The exact inequality follows from $(m-M{)}^2\ge 0$, which implies $(m+M{)}^2\ge 4mM$.

A single recoil endpoint leaves a tradeoff between incident speed and projectile mass. Two targets can remove the unknown incident speed. Assume both runs sample the same maximum projectile speed $v$ and that single elastic collisions reach the endpoint. For initially stationary target masses $M_1<M_2$, let their maximum recoil speeds be $u_1,u_2$. The maximum elastic recoil energy fraction calculation givesIf energies rather than speeds are inferred, use $R=\sqrt{T_{1,\max }{M}_2/(T_{2,\max }{M}_1)}$. The energy ratio itself is not the speed ratio.

For the stipulated $4\mathrm{MeV}$ example, hydrogen and nitrogen giveThe inferred mass then determines $v=u_1(m+M_1)/(2m)$. Changing the assumed mass to improve one recoil endpoint changes the prediction for the other: this is the constraining power of the comparison.

James Chadwick's June paper instead used approximate endpoints $u_p=3.3\times 10^{7}m/s$ and $u_N=4.7\times 10^{6}m/s$, obtaining $m/m_p\simeq 1.15$; it allowed roughly ten percent uncertainty in $u_N$. This was a rough mass comparison.^[ref]

The inverse calculation has useful built-in checks. A positive finite mass requires $1<R<M_2/M_1$. As $R$ approaches $1$, distinguishing large projectile masses becomes difficult. Differentiation givesUncertain recoil endpoints therefore produce an uncertain inferred mass; the collision model cannot turn imprecise range measurements into exact spectroscopy. If different target runs select different incident energy groups, the common-$v$ cancellation also fails.

For comparison, Compton scattering of a photon of energy $E_{\gamma }$ from a stationary free target of mass $M$ has the exact relativistic endpointThis follows by combining momentum conservation with the target's relativistic energy, maximizing transfer when the photon reverses direction. Unlike the massive-projectile calculation, this equation does not approximate the incident photon as a nonrelativistic particle.

As a separate, explicitly hypothetical test, take $m_p{c}^2=938\mathrm{MeV}$ and target rest energies $938\mathrm{MeV}$ and $14\times 938\mathrm{MeV}$. To reproduce the stipulated hydrogen recoil endpoint $T=4\mathrm{MeV}$, the photon would need $E_{\gamma }\simeq 45.36\mathrm{MeV}$. The same photon would give the heavier target at most $0.311\mathrm{MeV}$, whereas the stipulated massive projectile gave $0.996\mathrm{MeV}$. This is the structure of a two-target discrimination, not a transcription of a 1932 measurement. Actual radiation can contain more than one energy group or both photons and neutrons; a useful model must account for the observed endpoints and their origin together.