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The Faraday constant is the electric charge associated with one mole of elementary charges, $F=N_{\mathrm{A}}e\approx 96485.33212\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$. The product of the Avogadro constant and elementary charge is exact under the SI definitions; the displayed decimal is rounded. For ions with signed charge number $z$, the molar electrical contribution to an electrochemical potential is $zF\psi$, where $\psi$ is electric potential. ^[ref].

The nuclear reactionconserves electric charge and total nucleon count: $3+1=2+2$ and $7+1=4+4$. Its nuclear reaction Q-value is the initial minus final rest energy. It is additional to the incoming beam kinetic energy; the accelerator triggers a rearrangement whose products can carry much more kinetic energy than the incident proton.

Use one consistent set of neutral-atom isotope masses, in unified atomic mass units, from the National Institute of Standards and Technology:These are isotope masses, not element-averaged atomic weights.^[ref] With $\mathrm{u}{c}^2=931.49410372\mathrm{MeV}$,^[ref]The four electron rest masses cancel between sides. Differences of total electronic binding energies do not cancel exactly: this atomic-mass result approximates the bare-nucleus nuclear reaction Q-value at the displayed $0.001\mathrm{MeV}$ scale, not at arbitrary precision. No mixture of rounded nuclear binding energies is needed. At chosen incident kinetic energy $T=0.500\mathrm{MeV}$ on a stationary target, the two outgoing alpha particles together carry approximately $Q+T=17.846\mathrm{MeV}$ before losses in the target and detector window.

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 $a=m_p{c}^2$, $b=m_{\mathrm{Li}}{c}^2$, $h=m_{\alpha }{c}^2$, 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,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 $E_{\ast }=W/2$ and momentum-times-$c$ magnitude ${\mathcal{P}}_{\ast }=\sqrt{E_{\ast }^2-h^2}$. A Lorentz transformation gives the paired laboratory kinetic energies, with ${\theta }_{\ast }$ measured from the incident beam in the center of mass frame:Their sum is independent of ${\theta }_{\ast }$ and equals $Q+T$. Their difference is $2\gamma \beta {\mathcal{P}}_{\ast }\cos {\theta }_{\ast }$; 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 $m_e{c}^2=0.51099895069\mathrm{MeV}$.^[ref] At ${\theta }_{\ast }=0$, the paired laboratory kinetic energies are approximately $9.979$ and $7.867\mathrm{MeV}$; at ${\theta }_{\ast }=90^{\circ }$ 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.