Question

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

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

Answer 1

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

For bcc lattice: $Z=2,a=\frac{4r}{\sqrt{3}}$
For FCC lattice, $Z=4,a=2\sqrt{2}r$
$\frac{d_{R.T}}{d_{{900}^{o}C}}$ =$\frac{(\frac{ZM}{N_A{a}^3}{)}_{bcc}}{(\frac{ZM}{N_A{a}^3}{)}_{fcc}}$
Given, molar mass and atomic radii are constant.
$=\frac{2}{4}(\frac{2\sqrt{2}r}{\frac{4r}{\sqrt{3}}}{)}^3=\frac{3\sqrt{3}}{4\sqrt{2}}$