The metallic radius is given in Å (angstroms). We need to convert it to meters: \[ 1 \, \text{Å} = 10^{-10} \, \text{m} \] \[ \text{Metallic radius} = \sqrt{3} \, \text{Å} = \sqrt{3} \times 10^{-10} \, \text{m} \]
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: \[ a = 4 \times (\sqrt{3} \times 10^{-10}) / \sqrt{3} = 4 \times 10^{-10} \, \text{m} \]
The volume \( V \) of a cubic unit cell is given by: \[ V = a^3 \] Substituting the edge length: \[ V = (4 \times 10^{-10})^3 = 64 \times 10^{-30} \, \text{m}^3 \]
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.