Question

Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is

• A. \(\frac{\sqrt3}{\sqrt2}\)
• B. \(\frac{3\sqrt3}{4\sqrt2}\)
• C. \(\frac{4\sqrt3}{3\sqrt2}\)
• D. \(\frac{1}{2}\)

Answer 1

The Correct Option is B

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

Step 1: Understand the Structure Types

• Body-Centered Cubic (BCC): In a BCC structure, there are 2 atoms per unit cell.
• Face-Centered Cubic (FCC): In an FCC structure, there are 4 atoms per unit cell.

Step 2: Density Formula

Density (\(d\)) is given by the formula:

\(d = \frac{Z \cdot M}{a^3 \cdot N_A}\)

• \(Z\) = Number of atoms per unit cell
• \(M\) = Molar mass of iron (assumed constant)
• \(a\) = Edge length of the unit cell
• \(N_A\) = Avogadro's number

Step 3: Edge Length Relationship

• BCC: The edge length \((a_{\text{bcc}})\) is related to the atomic radius \((r)\) by \(a_{\text{bcc}} = \frac{4r}{\sqrt{3}}\).
• FCC: The edge length \((a_{\text{fcc}})\) is related to the atomic radius by \(a_{\text{fcc}} = \frac{2\sqrt{2}r}{\sqrt{2}}\).

Step 4: Calculate the Density Ratio

Let \(d_{\text{bcc}}\) and \(d_{\text{fcc}}\) be the densities in the BCC and FCC structures, respectively.

The ratio of densities is calculated as:

\(\frac{d_{\text{bcc}}}{d_{\text{fcc}}} = \frac{Z_{\text{bcc}} \cdot M/N_A}{(a_{\text{bcc}})^3} \bigg/ \frac{Z_{\text{fcc}} \cdot M/N_A}{(a_{\text{fcc}})^3}\)

Simplifying, we get:

\(\frac{d_{\text{bcc}}}{d_{\text{fcc}}} = \frac{Z_{\text{bcc}} \cdot (a_{\text{fcc}})^3}{Z_{\text{fcc}} \cdot (a_{\text{bcc}})^3}\)

Substitute the values:

• \(Z_{\text{bcc}} = 2\)
• \(Z_{\text{fcc}} = 4\)

Substitute the expressions for \(a_{\text{bcc}}\) and \(a_{\text{fcc}}\):

\(a_{\text{bcc}} = \frac{4r}{\sqrt{3}}\) and \(a_{\text{fcc}} = 2r\sqrt{\frac{2}{4}}\)

Now, we calculate:

\(\frac{d_{\text{bcc}}}{d_{\text{fcc}}} = \frac{2 \cdot (2\sqrt{2})^3}{4 \cdot \left(\frac{4}{\sqrt{3}}\right)^3}\)

Simplifying the cube and fraction:

= \frac{2 \cdot 16\sqrt{2}}{4 \cdot \frac{64}{3\sqrt{3}}}\)

= \frac{3\sqrt{3}}{4\sqrt{2}}

Thus, the correct option is \(\frac{3\sqrt3}{4\sqrt2}\).