Question

A metal crystallises in a body centered cubic lattice with the metallic radius \(\sqrt3 Å\). The volume of the unit cell in m³ is

• A. 64 × 10^-29
• B. 4 × 10^-29
• C. 6.4 × 10^-29
• D. 4 × 10^-10

Answer 1

The Correct Option is C

Step 1: Convert the Metallic Radius to Meters

The metallic radius is given in Å (angstroms). We need to convert it to meters: \[ 1 \, \text{Å} = 10^{-10} \, \text{m} \] \[ \text{Metallic radius} = \sqrt{3} \, \text{Å} = \sqrt{3} \times 10^{-10} \, \text{m} \]

Step 2: Calculate the Edge Length of the Unit Cell

In a BCC lattice, the edge length \( a \) of the unit cell can be related to the atomic radius \( r \) by the formula: \[ a = 4r / \sqrt{3} \] Substituting the given radius: \[ a = 4 \times (\sqrt{3} \times 10^{-10}) / \sqrt{3} = 4 \times 10^{-10} \, \text{m} \]

Step 3: Calculate the Volume of the Unit Cell

The volume \( V \) of a cubic unit cell is given by: \[ V = a^3 \] Substituting the edge length: \[ V = (4 \times 10^{-10})^3 = 64 \times 10^{-30} \, \text{m}^3 \]

Conclusion

The volume of the unit cell is \( 64 \times 10^{-30} \, \text{m}^3 \). The closest option to this value is: \[ \boxed{6.4 \times 10^{-29} \, \text{m}^3} \] Therefore, the correct answer is (C) 6.4 × 10^-29.

Answer 2

In a body-centered cubic (BCC) lattice, the relationship between the edge length \( a \) of the unit cell and the metallic radius \( r \) is given by: \[ \text{Diagonal of the cube} = 4r \] The diagonal of the cube is also equal to \( \sqrt{3}a \), so: \[ \sqrt{3}a = 4r \] Thus, \[ a = \frac{4r}{\sqrt{3}} \] Now, we calculate the volume of the unit cell, which is \( a^3 \). Substituting \( a = \frac{4r}{\sqrt{3}} \), the volume of the unit cell is: \[ V = \left( \frac{4r}{\sqrt{3}} \right)^3 = \frac{64r^3}{3\sqrt{3}} \] Given that the metallic radius \( r = \sqrt{3} \, \text{Å} = \sqrt{3} \times 10^{-10} \, \text{m} \), we substitute this into the equation for volume: \[ V = \frac{64 \times (\sqrt{3} \times 10^{-10})^3}{3\sqrt{3}} = 6.4 \times 10^{-29} \, \text{m}^3 \] Thus, the volume of the unit cell is \( 6.4 \times 10^{-29} \, \text{m}^3 \).