= Lithium reaction laboratory recoil
Equal product <masses> imply equal <kinetic energies> and opposite <momenta> in the <center of mass> frame. The laboratory frame has nonzero initial <momentum> because the <proton> is moving. <Conservation of momentum> therefore forbids the two laboratory <momenta> from always being equal and opposite. The products are equal in laboratory <kinetic energy> only for particular emission directions, not for every event.
An exact two-body calculation makes this distinction quantitative. Write $a=m_pc^2$, $b=m_{\mathrm{Li}}c^2$, $h=m_\alpha c^2$, and $T$ for incident <kinetic energy>. Express every quantity below in the same <energy> units; $\mathcal P=pc$ is <momentum> times the <speed of light>. For a stationary target,
$$
\mathcal P=\sqrt{T(T+2a)},\quad
W=\sqrt{(a+b)^2+2bT},\quad
\beta=\frac{\mathcal P}{a+b+T},\quad
\gamma=\frac{a+b+T}{W}.
$$
Here $W$ is the <center-of-mass energy>, and $\beta$ is the center-of-mass <speed> divided by the <speed of light>. Each outgoing <alpha particle> has center-of-mass total <energy> $E_*=W/2$ and momentum-times-$c$ magnitude $\mathcal P_*=\sqrt{E_*^2-h^2}$. A <Lorentz transformation> gives the paired laboratory <kinetic energies>, with $\theta_*$ measured from the incident beam in the <center of mass> frame:
$$
K_1=\gamma(E_*+\beta\mathcal P_*\cos\theta_*)-h,\qquad
K_2=\gamma(E_*-\beta\mathcal P_*\cos\theta_*)-h.
$$
Their sum is independent of $\theta_*$ and equals $Q+T$. Their difference is $2\gamma\beta\mathcal P_*\cos\theta_*$; the symmetry belongs to the product interchange and the <center of mass> frame.
For the chosen $T=0.500\,\mathrm{MeV}$ model, subtract $Z$ electron rest <masses> from each neutral-atom mass, neglect electronic <binding energies>, and use $m_ec^2=0.51099895069\,\mathrm{MeV}$.<CODATA recommended values of the fundamental physical constants: 2022 by National Institute of Standards and Technology>[CODATA 2022 wall chart, electron mass, energy equivalent in MeV]{ref} At $\theta_*=0$, the paired laboratory <kinetic energies> are approximately $9.979$ and $7.867\,\mathrm{MeV}$; at $\theta_*=90^\circ$ both are approximately $8.923\,\mathrm{MeV}$. In the latter case both products have a forward laboratory <momentum> component, so even equal laboratory <kinetic energies> do not imply back-to-back laboratory trajectories. A real thick target and finite detector acceptance would broaden or select these ideal kinematic possibilities.
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