Question

In a ccp crystal lattice, the edge length of the unit cell is $80pm$. Find the distance of tetrahedral void from the corner of a cube.

• A. $64.72pm$
• B. $90pm$
• C. $45.5pm$
• D. $34.64pm$

Answer 1

The correct option is D $34.64pm$

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.
$\therefore$
$\sqrt{3}a=4X$
$X=\frac{\sqrt{3}a}{4}$

$X=\frac{\sqrt{3}\times 80}{4}$
$X=34.64pm$