Iron exhibits `bcc` structure at roomj temperature. Above `900^(@)C`, it transformers to `fcc` structure. The ratio of density of iron at room temperature to that at `900^(@)C` (assuming molar mass and atomic radius of iron remains constant with temperature) is

Options

A. `(3sqrt(3))/(4sqrt(2))`
B. `(4sqrt(3))/(3sqrt(2))`
C. `(sqrt(3))/(sqrt(2))`
D. `(1)/(2)`

Explanation

To find the ratio of the density of iron at room temperature (where it exhibits a BCC structure) to that at 900°C (where it transforms to an FCC structure), we can follow these steps:

### Step 1: Understand the structures and their parameters
- **BCC (Body-Centered Cubic)**:
- Number of atoms per unit cell (Z) = 2
- Edge length (A) = \( \frac{4R}{\sqrt{3}} \)

- **FCC (Face-Centered Cubic)**:
- Number of atoms per unit cell (Z) = 4
- Edge length (A) = \( 2\sqrt{2}R \)

### Step 2: Write the formula for density
The density (\( \rho \)) of a crystal structure can be calculated using the formula:
\[ \rho = \frac{Z \cdot M}{N_a \cdot A^3} \]
Where:
- \( Z \) = number of atoms per unit cell
- \( M \) = molar mass
- \( N_a \) = Avogadro's number
- \( A \) = edge length of the unit cell

### Step 3: Calculate the density for BCC and FCC
- **Density at room temperature (BCC)**:
\[ \rho_{BCC} = \frac{Z_{BCC} \cdot M}{N_a \cdot A_{BCC}^3} = \frac{2 \cdot M}{N_a \cdot \left(\frac{4R}{\sqrt{3}}\right)^3} \]

- **Density at 900°C (FCC)**:
\[ \rho_{FCC} = \frac{Z_{FCC} \cdot M}{N_a \cdot A_{FCC}^3} = \frac{4 \cdot M}{N_a \cdot \left(2\sqrt{2}R\right)^3} \]

### Step 4: Set up the ratio of densities
We need to find the ratio \( \frac{\rho_{BCC}}{\rho_{FCC}} \):
\[ \frac{\rho_{BCC}}{\rho_{FCC}} = \frac{\frac{2 \cdot M}{N_a \cdot \left(\frac{4R}{\sqrt{3}}\right)^3}}{\frac{4 \cdot M}{N_a \cdot \left(2\sqrt{2}R\right)^3}} \]

### Step 5: Simplify the ratio
Since the molar mass \( M \) and Avogadro's number \( N_a \) cancel out, we have:
\[ \frac{\rho_{BCC}}{\rho_{FCC}} = \frac{2}{4} \cdot \frac{\left(2\sqrt{2}R\right)^3}{\left(\frac{4R}{\sqrt{3}}\right)^3} \]
\[ = \frac{1}{2} \cdot \frac{(2\sqrt{2})^3 \cdot R^3}{\left(\frac{64R^3}{3\sqrt{3}}\right)} \]
\[ = \frac{1}{2} \cdot \frac{16\sqrt{2}R^3 \cdot 3\sqrt{3}}{64R^3} \]
\[ = \frac{1}{2} \cdot \frac{48\sqrt{6}}{64} \]
\[ = \frac{48\sqrt{6}}{128} = \frac{3\sqrt{6}}{8} \]

### Step 6: Final Ratio
Thus, the final ratio of the density of iron at room temperature to that at 900°C is:
\[ \frac{3\sqrt{6}}{8} \]