Question

Iron exhibits bcc structure at room temperature. Above ${900}^{\circ }C$, it transforms to fcc structure. The ratio of density of iron at room temperature to that at ${900}^{\circ }C$ (assuming molar mass and atomic radii of iron remains constant with temperature) is __________.

• A. $\frac{3\sqrt{3}}{4\sqrt{2}}$
• B. $\frac{\sqrt{3}}{\sqrt{2}}$
• C. $\frac{1}{2}$
• D. $\frac{4\sqrt{3}}{3\sqrt{2}}$

Answer 1

The correct option is A $\frac{3\sqrt{3}}{4\sqrt{2}}$

We know,

Density, $d=\frac{ZM}{a^3{N}_0}$

For bcc unit cell, the number of atoms per unit cell, $Z=2$ and the edge length $a=\frac{4}{\sqrt{3}}r$.

For fcc unit cell, the number of atoms per unit cell $Z=4$ and the edge length $a=\frac{4}{\sqrt{2}}r$.

For bcc unit cell,

$d=\frac{ZM}{a^3{N}_0}$

$d=\frac{2M}{(\frac{4}{\sqrt{3}}r{)}^3{N}_0}$ ----- (1)

For fcc unit cell,

$d=\frac{ZM}{a^3{N}_0}$

$d=\frac{4M}{(\frac{4}{\sqrt{2}}r{)}^3{N}_0}$ ------ (2)

Divide equation (1) with equation (2)

$\frac{\frac{2M}{(\frac{4}{\sqrt{3}}r{)}^3{N}_0}}{\frac{4M}{(\frac{4}{\sqrt{2}}r{)}^3{N}_0}}=\frac{1}{2}\times (\frac{\frac{1}{1/\sqrt{3}}}{\frac{1}{1/\sqrt{2}}}{)}^3=\frac{1}{2}\times (\frac{\sqrt{3}}{\sqrt{2}}{)}^3=\frac{1}{2}\times \frac{3}{2}\times \frac{\sqrt{3}}{\sqrt{2}}=\frac{3\sqrt{3}}{4\sqrt{2}}$