Question

Calcium crystallizes in a face centred cubic unit cell with a $\text{=} \, \text{0} \text{.560} \, \text{nm} \text{.}$ The density of the metal if it contains 0.1% schottky defects would be:

• A. $\text{1} \text{.51\, gcm}^{\text{-3}}$
• B. $\text{2} \text{.51\,gcm}^{\text{-3}}$
• C. $\text{15} \text{.1\, gcm}^{\text{-3}}$
• D. $\text{0} \text{.151\, gcm}^{\text{-3}}$

Answer 1

Answer: (A)

A face centred cubic unit cell contains four atoms. Schottky defect is basically a vacancy defect in ionic solids. Due to Schottky defects, the vacant spaces will increase, resulting in decrease in number of atoms per unit cell.
Thus, number of atoms will reduce to:
$Z=\left(4-\frac{4 \times 0.1}{100}\right)=3.996$
The density of the metal can be calculated by:
$\rho=\frac{ Z \times M }{ N _{0} \times a ^{3}}$
Where, $M$ is the molecular mass of calcium $=40\, g \,mol ^{-1}$
$N _{0}$ is the Avogadro number $=6.022 \times 10^{23} mol ^{-1}$
$a$ is the edge length $=0.560 \,nm =0.56 \times 10^{-7} cm$
Putting the values in the formula above:
$=\frac{3.996 \times\left(40 g mol ^{-1}\right)}{\left(0.560 \times 10^{-7} cm \right)^{3} \times\left(6.022 \times 10^{23} mol ^{-1}\right)}$
$=1.51\, g\, cm ^{-3}$