Question

An element has face-centred cubic (fcc) structure with a cell edge of a. The distance between the centres of two nearest tetrahedral voids in the lattice is:

• A. $2a$
• B. $\frac{\sqrt{3}a}{2}$
• C. $a$
• D. $\frac{a}{2}$

Answer 1

The correct option is D $\frac{a}{2}$

For a fcc lattice:

2 tetrahedral voids are present at each body diagonal of a fcc lattice.

Here in the figure , there are 2 body diagonals PQ and XY containing four tetrahedral voids A,B,C,D.
The distance between two nearest tetrahedral voids in fcc lattice is the distance between centres of two adjacent minicubes.
Distance between void A and C = Nearest distance.

If a cube is divided into 8 minicubes
One tetrahedral void is present at the centre of each minicube.
Distance between void A and C:

$\text{Nearest distance is}=\frac{a}{4}+\frac{a}{4}=\frac{a}{2}$