= Solution
For <neutron> <energy> loss, $E_f=30-20=10\,\mathrm{meV}$. Write $\mathbf Q=\mathbf k_i-\mathbf k_f$. The <neutron> dispersion $E=\hbar^2k^2/(2m_n)$, or $E[\mathrm{meV}]=2.0721\,k[\mathrm{\mathring A^{-1}}]^2$, gives
$$
k_i=3.805\,\mathrm{\mathring A^{-1}},\qquad k_f=2.197\,\mathrm{\mathring A^{-1}}.
$$
A triangle of these <wavevectors> exists if
$$
|k_i-k_f|\le Q\le k_i+k_f,
\qquad1.608\le Q\le6.002\,\mathrm{\mathring A^{-1}}.
$$
With $c=4.50\,\mathrm{\mathring A}$, the requested <momentum> transfers have $Q=(2\pi/c)|l|=1.3963|l|\,\mathrm{\mathring A^{-1}}$. Consequently
$$
\boxed{l=\pm2,\ \pm3,\ \pm4\quad\text{are kinematically accessible}.}
$$
For example $|l|=1$ is too small even for forward scattering at these unequal <neutron> energies, and $|l|=5$ exceeds the backscattering maximum. This identifies accessible <reciprocal lattice> points, not guaranteed nonzero intensity: a one-<phonon> <structure factor> contains the atomic scattering lengths, <eigenvectors> and factors $\mathbf Q\cdot\mathbf e_j$. A transverse <eigenvector> perpendicular to $\mathbf Q$ may be invisible, and <crystal motif>[basis] interference can cause extinctions. The supplied geometry and cell constants do not specify that full intensity selection rule.
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