Question

When heated above $916{}^{\circ }C$, iron changes its bcc crystalline form to fcc without the change in the radius of atom. The ratio of density of the crystal before and after heating is:
Given: $\left[{\text{At. wt. of Fe = 56 g mol}}^{-1}\right]$

Answer 1

Density of the unit cell,
$\rho =\frac{Z\times M}{N_A(a^3\times {10}^{-30})}{\text{g cm}}^{-3}$
where,
$Z=\text{No. of atoms in a unit cell}M=\text{Molar mass}{N}_A=\text{Avogadro number}a=\text{Edge length in pm}$

For bcc,
$\text{Edge length of a bcc, a}=\frac{4r}{\sqrt{3}}$
$\therefore {\rho }_1=\frac{2\times 56}{N_A{\left(\frac{4r}{\sqrt{3}}\right)}^3}$

For fcc,
$\text{Edge length of a fcc, a}=\frac{4r}{\sqrt{2}}$
$\therefore {\rho }_2=\frac{4\times 56}{N_A(2\sqrt{2}r{)}^3}$
$\therefore \text{Required ratio,}=\frac{{\rho }_1}{{\rho }_2}=\frac{1}{2}\times \frac{(2\sqrt{2}{)}^3}{{\left(\frac{4}{\sqrt{3}}\right)}^3}=0.918$