In a face-centered cubic (FCC) structure, the relation between the edge length \( a \) of the unit cell and the atomic radius \( r \) is given by: \[ a = \frac{4r}{\sqrt{2}} \] Here, the metallic radius \( r = \sqrt{2} \, \text{Å} \), so: \[ a = \frac{4 \times \sqrt{2}}{\sqrt{2}} = 4 \, \text{Å} \] Now, the volume of the unit cell \( V \) is given by: \[ V = a^3 \] Substitute \( a = 4 \, \text{Å} \): \[ V = (4 \times 10^{-10} \, \text{m})^3 = 6.4 \times 10^{-29} \, \text{m}^3 \] Thus, the volume of the unit cell is \( 6.4 \times 10^{-29} \, \text{m}^3 \).
The correct option is (C) : \( 6.4 \times 10^{-29}\)