Question

The unit cell of a binary alloy composed of A and B metals, has a ccp structure with A atoms occupying the corners and B atoms occupying centres of each face of the cube. If during the crystallization of this alloy, in the unit cell two A atom are missed, the overall composition per unit cell is:

• A. $A{B}_6$
• B. $A{B}_4$
• C. $A{B}_8$
• D. $A_6{B}_{24}$

Answer 1

Answer: (B)

$A{B}_4$

Since the unit cell of the compound has CCP structure in which $A$ atoms are present at corners and $B$ are present at face centers. Also, $2$ of the atoms of $A$ are missing.

So, number of atoms of $A$ present at the corner $=8-2=6$

Where $8$ is the total corners of cube and $2$ is the missing.

Number of atoms of $B$ present at face centers $=6$

Now, contribution of each present atom at the corner $=\frac{1}{2}$

Contribution of each atom present at face center $=\frac{1}{2}$

So, number of atoms of $A$ present in unit cell $=\frac{6}{8}=\frac{3}{4}$

Number of atoms of $B$ present in unit cell $=\frac{6}{2}=3$

Now, $A:B$

$\frac{3}{4}:3$

So, the formula of the given compound is $A{B}_4$