When heated above `916^(@)C`, iron changes its bcc crystalline from to fcc without the change in the radius of atom . The ratio of density of the crystal before heating and after heating is :

Options

A. 1.069
B. 0.918
C. 0.725
D. 1.231

Answer

To find the ratio of the density of iron in its BCC (Body-Centered Cubic) crystalline form before heating and FCC (Face-Centered Cubic) crystalline form after heating, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Density Formula**:
The density (\( \rho \)) of a crystalline structure is given by the formula:
\[ \rho = \frac{Z \cdot m}{N_a \cdot a^3} \]
where:
- \( Z \) = number of atoms per unit cell
- \( m \) = atomic weight of the substance
- \( N_a \) = Avogadro's number (approximately \( 6.022 \times 10^{23} \))
- \( a \) = edge length of the unit cell

2. **Calculate Density for BCC Structure**:
- For BCC, \( Z = 2 \) (2 atoms per unit cell).
- The edge length \( a \) is related to the atomic radius \( R \) as:
\[ a = \frac{4R}{\sqrt{3}} \]
- The atomic weight of iron \( m = 56 \, \text{g/mol} \).

Substituting these values into the density formula for BCC:
\[ \rho_1 = \frac{2 \cdot 56}{N_a \cdot \left(\frac{4R}{\sqrt{3}}\right)^3} \]

3. **Calculate Density for FCC Structure**:
- For FCC, \( Z = 4 \) (4 atoms per unit cell).
- The edge length \( a \) is related to the atomic radius \( R \) as:
\[ a = 2\sqrt{2}R \]

Substituting these values into the density formula for FCC:
\[ \rho_2 = \frac{4 \cdot 56}{N_a \cdot (2\sqrt{2}R)^3} \]

4. **Set Up the Ratio of Densities**:
We need to find the ratio \( \frac{\rho_1}{\rho_2} \):
\[ \frac{\rho_1}{\rho_2} = \frac{\frac{2 \cdot 56}{N_a \cdot \left(\frac{4R}{\sqrt{3}}\right)^3}}{\frac{4 \cdot 56}{N_a \cdot (2\sqrt{2}R)^3}} \]

5. **Simplify the Ratio**:
Cancel out common terms:
\[ \frac{\rho_1}{\rho_2} = \frac{2}{4} \cdot \frac{(2\sqrt{2}R)^3}{\left(\frac{4R}{\sqrt{3}}\right)^3} \]
This simplifies to:
\[ = \frac{1}{2} \cdot \frac{(2\sqrt{2})^3 \cdot R^3}{\left(\frac{4}{\sqrt{3}}\right)^3 \cdot R^3} \]
Further simplification gives:
\[ = \frac{1}{2} \cdot \frac{8\sqrt{2}}{\frac{64}{3\sqrt{3}}} = \frac{1}{2} \cdot \frac{8\sqrt{2} \cdot 3\sqrt{3}}{64} \]
\[ = \frac{1}{2} \cdot \frac{24\sqrt{6}}{64} = \frac{3\sqrt{6}}{8} \]

6. **Calculate the Numerical Value**:
Approximating \( \sqrt{6} \approx 2.45 \):
\[ \frac{3 \cdot 2.45}{8} \approx \frac{7.35}{8} \approx 0.91875 \]

### Final Result:
Thus, the ratio of the density of the crystal before heating (BCC) to after heating (FCC) is approximately:
\[ \frac{\rho_1}{\rho_2} \approx 0.918 \]