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X-ray crystallography: proof

 

Question 1

Show that the triple hexagonal lattice (Vc) becomes a cubic P lattice when the perpendicular distance between two layer of lattices, h, is(IaI√6)/6.

Answer 1

The diagram below shows two layers of the triple hexagonal lattice Vc.

Focusing on just one segment of the space lattice (IVb), IaI refers to the separation between two same-layer lattice points of Vc and is also the surface diagonal of IVb. The body diagonal AB of IVb is determined using Pythagoras’ theorem as follows:

Each side of the cubic P unit cell is IaI/√2 and

(AB)^{2}=(BC)^{2}+(AC)^{2}

AB=\sqrt{\left | \textbf{\textit{a}}\right |^{2}+(\frac{\left |\textbf{\textit{a}} \right |}{\sqrt{2}})^{2}}\; =\left | \textbf{\textit{a}}\right |\sqrt{\frac{3}{2}}

Furthermore, AB is thrice the perpendicular distance between two layers of the triple hexagonal lattice Vc (see below diagram, which is IVb rotated for a better view). 

Since AB = 3h,

h=\frac{\left |\textbf{\textit{a}} \right |}{3}\sqrt{\frac{3}{2}}=\frac{\left | \textbf{\textit{a}}\right |\sqrt{6}}{6}

 

 

Question 2

Show that the triple hexagonal lattice (Vc) becomes a cubic F lattice when the perpendicular distance between two layer of lattices, h, is(IaI√6)/3.

Answer 2

Similar to the cubic P unit cell, IaI refers to the separation between two same-layer lattice points of Vc in the cubic F unit cell, e.g. the length of BC. Using Pythagoras’ theorem,

(CE)^{2}=(CH)^{2}+(HE)^{2}

By the symmetry of the cubic F unit cell, BC = CE = IaI and CH = HE. Therefore, CH = IaI/√2 and hence, AI = 2IaI/√2. Furthermore,

(AJ)^{2}=(AI)^{2}+(IJ)^{2}

Since (IJ)2 = (KI)2 + (KJ)2 = (AI)2 + (AI)2 = 2(AI)2 = 4IaI2,

AJ=\sqrt{\left ( \frac{2\left | \textbf{\textit{a}}\right |}{\sqrt{2}} \right )^{2}+4\left | \textbf{\textit{a}}\right |^{2}}=\left | \textbf{\textit{a}}\right |\sqrt{6}

Just like the cubic P unit cell, the body diagonal AJ of the cubic F unit cell (IVe) is thrice the perpendicular distance between two layers of Vc. Therefore,

h=\frac{\left | \textbf{\textit{a}}\right |\sqrt{6}}{3}

 

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