 Solution

For $l=2$, $Q=2.7925{\mathring{\mathrm{A}}}^{-1}$. The scattering triangle givesTo state signed sample angles without an implicit axis convention, take the incident beam along $+x$, positive angular rotation counterclockwise, and at $\omega =0$ choose $+a$ along $+x$ and $+c$ along $+y$. ThenThe positive-$l$ condition is the vector equation ${\mathbf{k}}_i-{\mathbf{k}}_f=Q\hat{\mathbf{c}}$. PutThe two detector sides then give, modulo full rotations,These pairs obey the stated same-sense sign convention and transfer momentum along $+c$. The lower-right panel below illustrates the first choice. If the initial $+c$ direction is chosen on the opposite side of $+a$, a commonly used positive-angle pair is $(55.20^{\circ },46.51^{\circ })$. Merely taking absolute values without fixing that crystal-axis convention can exchange $l=+2$ and $l=-2$. The measurable scattering angle and the vector-triangle construction are invariant.