Question

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

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

Answer 1

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

In cubic close packing, each unit cell has 8 tetrahedral voids.
Each body diagonals in a unit cell contains 2 tetrahedral voids.
Let B and C are the two tetrahedral voids and X is the distance between the centre of void B and the centre of atom at corner A as shown in figure.


We know, the distance along the body diagonal of a cube is $\sqrt{3}a$
where,
a is the edge length of the cell.

If a cube is divided into 8 minicubes, one tetrahedral void is present at the centre of each minicube.
$\therefore$
$\sqrt{3}a=4X$
$X=\frac{\sqrt{3}a}{4}$
Here two yellow tetrahedral voids are joined with each other forming a smaller cube of edge length $\frac{a}{2}$.