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

Explanation

To solve the problem, we need to find the ratio of the density of iron at room temperature (where it has a body-centered cubic (BCC) structure) to the density of iron at 900°C (where it has a face-centered cubic (FCC) structure).

### Step-by-Step Solution:

1. **Understanding Density Formula**:
The density (\( \rho \)) of a substance can be expressed using the formula:
\[ \rho = \frac{n \cdot m}{N_A \cdot a^3} \]
where:
- \( n \) = number of atoms per unit cell
- \( m \) = molar mass of the substance
- \( N_A \) = Avogadro's number
- \( a \) = edge length of the unit cell

2. **Density at Room Temperature (BCC)**:
For BCC structure:
- Number of atoms per unit cell (\( n \)) = 2
- Relation between edge length (\( a \)) and atomic radius (\( R \)):
\[ a = \frac{4R}{\sqrt{3}} \]
Therefore, the density of BCC iron at room temperature is:
\[ \rho_{BCC} = \frac{2 \cdot m}{N_A \cdot \left(\frac{4R}{\sqrt{3}}\right)^3} \]

3. **Density at 900°C (FCC)**:
For FCC structure:
- Number of atoms per unit cell (\( n \)) = 4
- Relation between edge length (\( a \)) and atomic radius (\( R \)):
\[ a = 2\sqrt{2}R \]
Therefore, the density of FCC iron at 900°C is:
\[ \rho_{FCC} = \frac{4 \cdot m}{N_A \cdot \left(2\sqrt{2}R\right)^3} \]

4. **Calculating the Ratio of Densities**:
Now we can find the ratio of the densities:
\[ \frac{\rho_{BCC}}{\rho_{FCC}} = \frac{\frac{2 \cdot m}{N_A \cdot \left(\frac{4R}{\sqrt{3}}\right)^3}}{\frac{4 \cdot m}{N_A \cdot \left(2\sqrt{2}R\right)^3}} \]
The \( m \) and \( N_A \) cancel out:
\[ = \frac{2}{4} \cdot \frac{\left(2\sqrt{2}R\right)^3}{\left(\frac{4R}{\sqrt{3}}\right)^3} \]

5. **Simplifying the Expression**:
- Calculate \( \left(2\sqrt{2}R\right)^3 = 8\sqrt{2}R^3 \)
- Calculate \( \left(\frac{4R}{\sqrt{3}}\right)^3 = \frac{64R^3}{3\sqrt{3}} \)
- Substitute back into the ratio:
\[ = \frac{1}{2} \cdot \frac{8\sqrt{2}R^3}{\frac{64R^3}{3\sqrt{3}}} \]
- The \( R^3 \) cancels out:
\[ = \frac{1}{2} \cdot \frac{8\sqrt{2} \cdot 3\sqrt{3}}{64} \]
- Simplifying further:
\[ = \frac{1}{2} \cdot \frac{24\sqrt{6}}{64} = \frac{3\sqrt{6}}{32} \]

6. **Final Calculation**:
The final ratio of densities is:
\[ \frac{\rho_{BCC}}{\rho_{FCC}} = \frac{3\sqrt{6}}{32} \]

### Final Answer:
The ratio of the density of iron at room temperature to that at 900°C is \( \frac{3\sqrt{6}}{32} \).