Extra:Hexagonal peaks

Consider a hexagonal lattice viewed end-on using scattering. For instance, a hexagonally-packed-cylinder block-copolymer moprhology that is oriented horizontally ('laying down'; i.e. cylinder long axes parallel to the substrate plane). There will be a peak along the specular at $(q_{x},q_{z})=(0,q_{l})$, where $q_{l}$ corresponds to the cylinder layering distance in realspace (i.e. the distance between subsequent rows of cylinders). Since:

$\cos 30^{\circ }={\frac {\sqrt {3}}{2}}$

$\sin 30^{\circ }={\frac {1}{2}}$

The full set of (first order) peaks is:

• $(q_{x},q_{z})=(0,+q_{l})$
• $(q_{x},q_{z})=\left(-{\frac {\sqrt {3}}{2}}q_{l},+{\frac {q_{l}}{2}}\right)$ and $(q_{x},q_{z})=\left(+{\frac {\sqrt {3}}{2}}q_{l},+{\frac {q_{l}}{2}}\right)$
• $(q_{x},q_{z})=\left(-{\frac {\sqrt {3}}{2}}q_{l},-{\frac {q_{l}}{2}}\right)$ and $(q_{x},q_{z})=\left(+{\frac {\sqrt {3}}{2}}q_{l},-{\frac {q_{l}}{2}}\right)$
• $(q_{x},q_{z})=(0,-q_{l})$

The realspace layering distance is:

$d_{l}={\frac {2\pi }{q_{l}}}$

And the realspace cylinder-cylinder distance is:

${\begin{alignedat}{2}d_{cc}&={\frac {d_{l}}{{\sqrt {3}}/2}}\\&={\frac {2\pi /q_{l}}{{\sqrt {3}}/2}}\\&={\frac {2\pi }{+({\sqrt {3}}/2)q_{l}}}\end{alignedat}}$

Thus, the $q_{x}$ position of the away-from-specular peaks gives the cylinder-cylinder distance:

${\begin{alignedat}{2}d_{cc}&={\frac {2\pi }{q_{x}}}\end{alignedat}}$