Past exam of the materials science course of the University of Cambridge / 2015 / Paper 1 / 1 / a

Table of contents
- Solution a
- i a
  - Solution i
- ii a
  - Solution ii

Solution

Use fractional coordinates in a conventional cubic cell with an atom at the origin. The diamond cubic structure consists of an FCC Bravais lattice with the two-atom basis

(0, 0, 0), (\frac{1}{4}, \frac{1}{4}, \frac{1}{4}) .

(1)

The FCC translations within the conventional cell are

(0, 0, 0)

(0, \frac{1}{2}, \frac{1}{2})

(\frac{1}{2}, 0, \frac{1}{2})

and

(\frac{1}{2}, \frac{1}{2}, 0)

. Adding the basis gives eight atoms per conventional cell. Equivalently, two identical FCC sublattices are displaced by $a [111] /4$ . It is not a monatomic close-packed arrangement: every atom has four tetrahedrally arranged nearest neighbors rather than twelve.

In the projection, label each atom by its fractional height. The

(0, 0)

and

(1/2, 1/2)

columns have height0;

(0, 1/2)

and

(1/2, 0)

have height1/2. The additional four columns are

(1/4, 1/4)

and

(3/4, 3/4)

at height1/4, and

(1/4, 3/4)

and

(3/4, 1/4)

at height3/4. Boundary columns repeat periodically; they must not be counted as full extra atoms.

Figure 1.
Original fractional-coordinate projection of diamond silicon, with atom heights, vertical mirror traces and a bond-centred inversion point
. **Details**.

i

 0  0

Solution

 0  0

The vertical mirrors of the diamond cubic structure have the required orientation: their normals lie in the

(001)

plane. In the chosen atomic-origin convention, their traces satisfy

x - y = \frac{n}{2}, x + y = \frac{n}{2}, n \in Z,

(1)

with

x, y

expressed as fractions of

a

. These are the two orientations of vertical

{110}

mirrors and their translated copies. For example, the plane

x = y

interchanges the first two coordinates and leaves every diamond site in an equivalent site. Reflection in

x + y = 1/2

sends

(x, y, z)

(1/2 - y, 1/2 - x, z)

, which also preserves the full three-dimensional motif.

The projected traces alone do not test a mirror: the atom heights must also agree under reflection. In particular, a vertical

{100}

trace can appear plausible in an unlabelled projection but does not preserve all heights as a pure mirror; glide operations are different symmetry elements.

ii

 0  0

Solution

 0  0

Bond-centred inversion in diamond cubic gives one inversion centre at the midpoint of a nearest-neighbor bond:

c = (\frac{1}{8}, \frac{1}{8}, \frac{1}{8}) .

(1)

Mark its projected point

(1/8, 1/8)

and label its height $z / a = 1/8$ . Inversion about this point sends

r

(1/4, 1/4, 1/4) - r

. It exchanges the two identical FCC sublattices and therefore preserves the crystal. An individual atom at the chosen origin is not an inversion centre.