Proof for distance between lattice layers

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,

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

Proof for distance between lattice layers

Question 1

Answer 1

Question 2

Answer 2

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