Question

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

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

Answer 1

The correct option is A $\frac{3\left(\sqrt{3}\right)}{4\left(\sqrt{2}\right)}$

For BCC, z = 2 and a = $\frac{4r}{\left(\sqrt{3}\right)}$
For FCC, z = 4 and a = 2 $\sqrt{2}$r
$\therefore \frac{d_{{25}^{\circ }C}}{d_{{900}^{\circ }C}}=\frac{{\left(\frac{ZM}{N_A{a}^3}\right)}_{BCC}}{{\left(\frac{ZM}{N_A{a}^3}\right)}_{FCC}}$
$=\frac{2}{4}{\left(\frac{2\sqrt{2}r}{\frac{4r}{\sqrt{3}}}\right)}^3$
$=\left(\frac{3\sqrt{3}}{4\sqrt{2}}\right)$