= Maximum elastic recoil energy fraction
{title2=$f_{\max}=4mM/(m+M)^2$}
Let a projectile have <mass> $m$, incident <speed> $v$, and incident <kinetic energy> $E=mv^2/2$. The stationary target has <mass> $M$. In a head-on <elastic collision>, the signed outgoing projectile <velocity> $v'$ and target <velocity> $u$ obey <momentum conservation> and <conservation of energy>:
$$
mv=mv'+Mu,\qquad
\frac12mv^2=\frac12mv'^2+\frac12Mu^2.
$$
Factor the second equation and eliminate $Mu=m(v-v')$. The nontrivial collision branch has $u=v+v'$, hence
$$
v'=\frac{m-M}{m+M}v,\qquad
u_{\max}=\frac{2m}{m+M}v,\qquad
\frac{T_{\max}}{E}=\frac{\tfrac12Mu_{\max}^2}{\tfrac12mv^2}
=\frac{4mM}{(m+M)^2}.
$$
This 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> $\boldsymbol p-\boldsymbol q$ gives
$$
\frac{p^2}{2m}=\frac{p^2+q^2-2pq\cos\theta}{2m}+\frac{q^2}{2M},
\qquad
q=\frac{2Mp\cos\theta}{m+M}.
$$
Thus $T=T_{\max}\cos^2\theta$ 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> are
$$
\begin{array}{c|c|c}
M/m_p&T_{\max}/E&T_{\max}\ (\mathrm{MeV})\\
1&1&4\\
4&16/25=0.64&2.56\\
14&56/225\simeq0.2489&224/225\simeq0.9956
\end{array}
$$
These stipulated values illustrate the approximation $m_n=m_p$; they are not measurements from 1932. With $m_pc^2\simeq938\,\mathrm{MeV}$, the incident ratio $E/(m_pc^2)\simeq0.0043$ gives $v/c\simeq0.092$, 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 $(m-M)^2\geq0$, which implies $(m+M)^2\geq4mM$.
\Image[/neutron-elastic-recoil-mass-constraint.png]
{title=Recoil energy transfer and the two-target constraint on an unseen projectile's mass}
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