Question

If an element (atomic weight $=50$) crystallizes in FCC lattice, with $a=0.50\text{nm}$, the density $\left({\text{in g/cm}}^3\right)$ of unit cell if it contains $0.25\%$ Schottkey defects is: (use $N_A=6\times {10}^{23}$)

Answer 1

Density of crystal decreases due to Schottkey defect. Effective number of atoms per unit cell in given lattice is
$=4\times (1-\frac{0.25}{100})=3.99$
So, the density is
$\rho =\frac{50\times 3.99}{6\times {10}^{23}(0.5\times {10}^{-7}{)}^3}\Rightarrow \rho =\frac{50\times 3.99}{6\times 125\times 0.1}\Rightarrow \rho =2.66{\text{g/cm}}^3$