$A$ metal crystallizes with a $FCC$ lattice, the edge of whose unit cell is $x \text{ pm}$. The diameter of this metal atom would be $\text{pm}$.

The volume occupied by a single $CsCl$ ion pair in a crystal is $7.014 \times 10^{-23} \ cm^{3}$. The smallest $Cs^{+}-Cs^{+}$ inter-nuclear distance is equal to the length of the side of the cube corresponding to the volume of one $CsCl$ ion pair. The smallest $Cs^{+}-Cs^{+}$ inter-nuclear distance is nearly:

Calculate the volume of $fcc$ unit cell if the radius of a particle in it is $106.05 \ pm$.

In a body-centred cubic $(bcc)$ lattice of potassium, the correct relation between the atomic radius $(r)$ of potassium and the edge-length $(a)$ of the cube is:

$CsCl$ crystallises in a cubic structure that has $Cl^{-}$ at each corner and $Cs^{+}$ at the centre of the unit cell. If $r_{Cs^{+}} = 1.69 \ \mathring{A}$ and $r_{Cl^{-}} = 1.81 \ \mathring{A}$,what is the value of the edge length of the cube in $\mathring{A}$?

Calculate the volume of a $bcc$ unit cell if the radius of an atom present in it is $1.86 \times 10^{-8} \ cm$.