###### 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(I* a*I√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), I* a*I 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 I* a*I/√2 and

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*,

###### 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(I* a*I√6)/3.

###### Answer 2

Similar to the cubic P unit cell, I** a**I 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,

By the symmetry of the cubic F unit cell, *BC = CE = *I* a*I and

*CH = HE.*Therefore,

*CH =*I

*I*

**a***/√2*and hence,

*AI = 2*I

*I*

**a***/√2.*Furthermore,

Since (*IJ*)* ^{2 }*= (

*KI*)

*+ (*

^{2 }*KJ*)

*= (*

^{2}*AI*)

*+ (*

^{2 }*AI*)

*= 2(*

^{2}*AI*)

*= 4I*

^{2 }*I*

**a***,*

^{2}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,