Question

A metal crystallises into two cubic phases, one is face-centred cubic (fcc) and other is body-centred cubic (bcc). The edge lengths are 4 $A^{\circ }$ for fcc and 2 $A^{\circ }$ for bcc. Calculate the ratio of densities of fcc to bcc.

Answer 1

We know that,
Density of the unit cell,
$\rho =\frac{Z\times M}{N_A\times a^3}$
where,
$Z=\text{No. of atoms in a unit cell}M=\text{Molar mass}{N}_A=\text{Avagadro's number}$
$a=\text{length of the unit cell}$

$\text{Density of fcc}=\frac{Z_1\times M}{N_A\times V_1}$

$\text{Density of bcc}=\frac{Z_2\times M}{N_A\times V_2}$
where,
$V_1$ is volume of fcc unit cell
$V_2$ is volume of bcc unit cell

$\frac{{\rho }_{fcc}}{{\rho }_{bcc}}=\frac{Z_1}{Z_2}\times \frac{V_2}{V_1}$
For fcc, $Z_1=4;V_1=a^3=4^3$
For bcc, $Z_2=2;V_2=a^3=2^3$
$\frac{{\rho }_{fcc}}{{\rho }_{bcc}}=\frac{4\times 2^3}{2\times 4^3}=0.25$