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:
(NEET 2018)

• 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: $Z=2,\text{Edge length,}a=\frac{4r}{\sqrt{3}}$

For fcc lattice:$Z=4,\text{Edge length,}a=2\sqrt{2}r$

$\text{Density of unit cell,}\rho =\frac{ZM}{a^3{N}_A}$
where,
Z is the number of atoms in unit cell
M is the molar mass
$N_A$ is the avogadro's number
a is the edge length of unit cell

$\therefore \frac{d_{{25}^{\circ }C}}{d_{{900}^{\circ }C}}=\frac{{\left(\frac{ZM}{N_A{a}^3}\right)}_{bcc}}{{\left(\frac{ZM}{N_A{a}^3}\right)}_{fcc}}$
$=\frac{Z_{bcc}}{Z_{fcc}}\times \frac{a_{fcc}^3}{a_{bcc}^3}$

$=\frac{2}{4}{\left(\frac{2\sqrt{2}r}{\frac{4r}{\sqrt{3}}}\right)}^3=\left(\frac{3\sqrt{3}}{4\sqrt{2}}\right)$

Hence, option (d) is correct.