Cutting a crystal

We consider the surfaces that result when the crystal is cleaved.

The (100) plane is a quadratic lattice with edge length $a$ . At the (110) plane, we get a rectangular grid in which the long side is the face diagonals of unit cube, $\sqrt {2} a$ . In the case of the (111) plane, however, we have a hexagonal lattice consisting of equilateral triangles with the edge length $\sqrt {2} a$ .

In complete crystal, each atom has $N = 6$ nearest neighbors and hence $N$ chemical bonds. Within the surface, each atom has $k$ chemical bonds. The remaining $N - k$ bonds must therefore each be below the surface or are broken. Since the surface in the complete crystal is a mirror plane, the number of broken bonds per surface atom is just $b = \frac {1} {2} (N - k)$ . If every atom on the surface occupies an area $A$ , the number of broken bonds per surface area is $\sigma =\frac{\text{number of broken bonds}}{\text{surface area}} = \frac{b}{A} = \frac{N - k}{2 A}$ The area $A$ represents the area of the parallelogram corresponding to the unit cell of the two-dimensional lattice at the surface.

This bond density $\sigma$ results for the different cases to $\begin{aligned} \sigma_{(100)} &= \frac{6 - 4}{2 \cdot a^2} = \frac{1}{a^2} \\ \sigma_{(110)} &= \frac{6 - 2}{2 \cdot (a \cdot \sqrt{2} a) } = \frac{\sqrt{2}}{ a^2} \approx 1.41 \cdot \frac{1}{a^2} \\ \sigma_{(111)} &= \frac{6 - 0}{2 \cdot ( \sqrt{2} a \cdot \sqrt{\frac{3}{2}} a)} = \frac{\sqrt{3}}{a^2} \approx 1.73 \cdot \frac{1}{a^2} \end{aligned}$ Thus, the simplest way to cleave the crystal is along the (100) plane, since the fewest bonds must be broken here.