= Lithium proton reaction energy
{title2=$^{7}\mathrm{Li}+p\longrightarrow2\alpha$}
The <nuclear reaction>
$$
{}^{7}_{3}\mathrm{Li}+{}^{1}_{1}\mathrm H\longrightarrow
{}^{4}_{2}\mathrm{He}+{}^{4}_{2}\mathrm{He}
$$
conserves <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:
$$
M({}^{7}\mathrm{Li})=7.0160034366\,\mathrm u,\quad
M({}^{1}\mathrm H)=1.00782503223\,\mathrm u,\quad
M({}^{4}\mathrm{He})=4.00260325413\,\mathrm u.
$$
These are isotope masses, not element-averaged atomic weights.<Atomic Weights and Isotopic Compositions for All Elements by National Institute of Standards and Technology>[Atomic Weights and Isotopic Compositions for All Elements, entries atomic number 1 mass number 1; atomic number 2 mass number 4; atomic number 3 mass number 7]{ref} With $\mathrm u c^2=931.49410372\,\mathrm{MeV}$,<CODATA recommended values of the fundamental physical constants: 2022 by National Institute of Standards and Technology>[CODATA recommended values of the fundamental physical constants: 2022, atomic mass constant, energy equivalent in MeV]{ref}
$$
\Delta M=0.01862196057\,\mathrm u,\qquad
Q_{\mathrm{atom}}=[M({}^{7}\mathrm{Li})+M({}^{1}\mathrm H)-2M({}^{4}\mathrm{He})]c^2
=17.34625\,\mathrm{MeV}\simeq17.3\,\mathrm{MeV}.
$$
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.
\Image[/lithium-proton-reaction-energy-and-laboratory-recoil.png]
{title=Reaction energy and the laboratory recoil of two alpha particles}
{description=Original calculation using the stated neutral-atom isotope masses, electron-mass subtraction and neglected electronic binding corrections. Chosen incident proton energy 0.500 MeV and stationary lithium-7 target; no target or detector losses. Left: energy released by the mass difference plus incident kinetic energy. Right: exact two-body relativistic transformation within these mass assumptions; equal and opposite center-of-mass momenta need not give equal laboratory kinetic energies. Curves parameterize paired events and are not an angular probability distribution or historical data.}
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