Iron crystallizes in several modifications. At about `910^(@)C`, the body centred cubic `alpha` - form undergoes a transition to the face centred cubic `gamma`- form. Assuming that the distance between nearest neighbours is same in the two forms at the transition temperature. Calculate the ratio of density of `gamma` - iron to that of `alpha` - iron at the transition temperature.

Answer

To solve the problem of calculating the ratio of the density of gamma iron to that of alpha iron at the transition temperature, we can follow these steps:

### Step 1: Understand the Crystal Structures
- **Alpha Iron (α-Fe)**: It crystallizes in a body-centered cubic (BCC) structure.
- **Gamma Iron (γ-Fe)**: It crystallizes in a face-centered cubic (FCC) structure.

### Step 2: Determine the Number of Atoms per Unit Cell
- For BCC, the number of atoms per unit cell (z) is 2.
- For FCC, the number of atoms per unit cell (z) is 4.

### Step 3: Relate Atomic Radius to Lattice Parameter
- In BCC, the relationship between the atomic radius (r) and the lattice parameter (a) is given by:
\[ a_{\alpha} = \frac{4r}{\sqrt{3}} \]
- In FCC, the relationship is:
\[ a_{\gamma} = \frac{2\sqrt{2}r}{1} \]

### Step 4: Calculate the Density of Alpha Iron
The density (ρ) of a substance can be calculated using the formula:
\[ \rho = \frac{z \times M}{N_a \times a^3} \]
Where:
- \( z \) = number of atoms per unit cell
- \( M \) = molar mass of iron (approximately 56 g/mol)
- \( N_a \) = Avogadro's number (approximately \( 6.022 \times 10^{23} \) mol\(^{-1}\))
- \( a \) = lattice parameter

For alpha iron:
\[ \rho_{\alpha} = \frac{2 \times M}{N_a \times a_{\alpha}^3} \]
Substituting \( a_{\alpha} = \frac{4r}{\sqrt{3}} \):
\[ \rho_{\alpha} = \frac{2 \times M}{N_a \times \left(\frac{4r}{\sqrt{3}}\right)^3} \]
This simplifies to:
\[ \rho_{\alpha} = \frac{2 \times M \times 3\sqrt{3}}{N_a \times 64r^3} \]

### Step 5: Calculate the Density of Gamma Iron
For gamma iron:
\[ \rho_{\gamma} = \frac{4 \times M}{N_a \times a_{\gamma}^3} \]
Substituting \( a_{\gamma} = 2\sqrt{2}r \):
\[ \rho_{\gamma} = \frac{4 \times M}{N_a \times (2\sqrt{2}r)^3} \]
This simplifies to:
\[ \rho_{\gamma} = \frac{4 \times M}{N_a \times 16\sqrt{2}r^3} \]

### Step 6: Calculate the Ratio of Densities
Now, we can find the ratio of the densities:
\[ \frac{\rho_{\gamma}}{\rho_{\alpha}} = \frac{\frac{4 \times M}{N_a \times 16\sqrt{2}r^3}}{\frac{2 \times M \times 3\sqrt{3}}{N_a \times 64r^3}} \]
This simplifies to:
\[ \frac{\rho_{\gamma}}{\rho_{\alpha}} = \frac{4 \times 64}{2 \times 16\sqrt{2} \times 3\sqrt{3}} = \frac{128}{96\sqrt{2}\sqrt{3}} = \frac{128}{96\sqrt{6}} \]
Calculating this gives approximately:
\[ \frac{\rho_{\gamma}}{\rho_{\alpha}} \approx 1.088 \]

### Final Answer
Thus, the ratio of the density of gamma iron to that of alpha iron at the transition temperature is approximately **1.088**.

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