= Neutron scattering kinematics
{title2=$\mathbf Q=\mathbf k_i-\mathbf k_f,\quad\hbar\omega=E_i-E_f$}
Treat the incident and outgoing <neutrons> as free <nonrelativistic particles> outside the specimen. Their <wave vectors>, <de Broglie wavelengths>, <momenta> and <kinetic energies> obey
$$
k_i=\frac{2\pi}{\lambda_i},\qquad k_f=\frac{2\pi}{\lambda_f},\qquad
\mathbf p_i=\hbar\mathbf k_i,\qquad \mathbf p_f=\hbar\mathbf k_f,\qquad
E_{i,f}=\frac{\hbar^2 k_{i,f}^2}{2m_n}.
$$
Here $k_i=|\mathbf k_i|$, $k_f=|\mathbf k_f|$, $m_n$ is the <neutron> <mass>, and $\hbar=h/(2\pi)$ uses the <Planck constant>. Subtracting the outgoing projectile quantities from the incoming ones gives the <momentum transfer> and <energy> delivered to the specimen:
$$
\Delta\mathbf p=\hbar\mathbf Q,\qquad
\mathbf Q=\mathbf k_i-\mathbf k_f,\qquad
\Delta E=E_i-E_f=\hbar\omega.
$$
Thus $\mathbf Q$ has units of inverse <length>, rather than <momentum>. A positive $\omega$ means the specimen gains <energy>; negative $\omega$ means it supplies <energy> to the <neutron>. If the angle between incident and outgoing propagation directions is $2\theta$, the <law of cosines> gives
$$
Q^2=k_i^2+k_f^2-2k_i k_f\cos(2\theta).
$$
Knowing the detector angle alone therefore leaves $Q$ undetermined unless the outgoing <energy> is known or an <elastic scattering> assumption is justified.
For a reproducible unit conversion choose $m_n=1.67492750056\times10^{-27}\,\mathrm{kg}$, $h=6.62607015\times10^{-34}\,\mathrm{J\,s}$ and $1\,\mathrm{eV}=1.602176634\times10^{-19}\,\mathrm J$. The first is a recommended central value; the latter two conversions are exact.<Fundamental Physical Constants — Complete Listing by National Institute of Standards and Technology and others>[2022 CODATA table: neutron mass, Planck constant and electron volt]{ref} With $1$ <ångström> equal to $10^{-10}\,\mathrm m$ and $1\,\mathrm{meV}=10^{-3}$ <electronvolts>, substitution into $E=h^2/(2m_n\lambda^2)$ gives
$$
E\,[\mathrm{meV}]=\frac{81.8042}{\bigl(\lambda\,[\mathrm{\mathring A}]\bigr)^2}
=2.07212\bigl(k\,[\mathrm{\mathring A}^{-1}]\bigr)^2.
$$
Square brackets here mean the numerical value in the indicated unit. For an original chosen example, $\lambda_i=2$ <ångströms> and $\lambda_f=2.5$ <ångströms> give $E_i\simeq20.451\,\mathrm{meV}$, $E_f\simeq13.089\,\mathrm{meV}$ and $\hbar\omega\simeq7.362\,\mathrm{meV}$. At $2\theta=60^\circ$, $Q\simeq2.879\,\mathrm{\mathring A}^{-1}$. These values stipulate an allowed projectile geometry; they do not assert that any particular specimen has an excitation there.
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