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

To find the ratio of the density of iron at room temperature to that at $900^{\circ}C$, we need to consider the change in crystal structure from body-centered cubic (bcc) to face-centered cubic (fcc) and how this affects the volume and, consequently, the density.

1. $bcc$ structure has:
   • 2 atoms per unit cell.
   • The relationship between atomic radius $r$ and the edge length $a$ of the cube is $a = \frac{4r}{\sqrt{3}}$.
2. $fcc$ structure has:
   • 4 atoms per unit cell.
   • The relationship between atomic radius $r$ and the edge length $a$ of the cube is $a = \frac{4r}{\sqrt{2}}$.
3. Calculate the density for each structure:
   • Density, $\rho = \frac{\text{Mass of atoms in unit cell}}{\text{Volume of unit cell}}$.
   • For $bcc$:
     • Mass of atoms = 2 atoms × Atomic mass, assuming molar mass $M$.
     • Volume = $a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3$.
   • For $fcc$:
     • Mass of atoms = 4 atoms × Atomic mass.
     • Volume = $a^3 = \left(\frac{4r}{\sqrt{2}}\right)^3$.
4. Determine the density ratio:
   • The density ratio is given by:
   • $\frac{\rho_{\text{bcc}}}{\rho_{\text{fcc}}} = \frac{2 \times \left(\frac{4r}{\sqrt{3}}\right)^3}{4 \times \left(\frac{4r}{\sqrt{2}}\right)^3}$.
   • Simplifying yields, $\frac{3\sqrt{3}}{4\sqrt{2}}$.

This calculation shows that the ratio of the density of iron at room temperature (bcc structure) to that at $900^{\circ}C$ (fcc structure) is $\frac{3\sqrt{3}}{4\sqrt{2}}$, aligned with the correct option.