Question

Iron exhibits $bcc$ structure at room temperature. Above $900^{\circ}C$, it transforms to fee 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. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{\sqrt{2}}$
• C. $\frac{3\sqrt{3}}{4\sqrt{2}}$
• D. $\frac{4\sqrt{3}}{3\sqrt{2}}$

Answer 1

The Correct Option is C

For $BCC$ lattice $: Z =2, a =\frac{4 r }{\sqrt{3}}$ For $FCC$ lattice $: Z =4, a =2 \sqrt{2} r$ $\therefore \frac{d_{25^{\circ} C }}{d_{900^{\circ} C }}=\frac{\left(\frac{ ZM }{ N _{ A } a ^{3}}\right)_{ BCC }}{\left(\frac{ ZM }{ N _{ A } a ^{3}}\right)_{ FCC }}=\frac{2}{4}\left(\frac{2 \sqrt{2} r }{\frac{4 r}{\sqrt{3}}}\right)^{3}=\left(\frac{3 \sqrt{3}}{4 \sqrt{2}}\right)$