$A$ metal crystallizes in a simple cubic lattice. The volume of one unit cell is $6.4 \times 10^7 \ pm^3$. What is the radius of the metal atom in $pm$?

An ionic compound $X^{+}Y^{-}$ has a $bcc$ structure. The distance between two nearest ions is $1.73 \, \mathring{A}$. What would be the edge length of the unit cell in $pm$?

Calculate the radius of a metal atom if it forms a $bcc$ unit cell having an edge length of $530 \ pm$. (in $pm$)

The mass of an atom present in a unit cell is $4.4 \times 10^{-23} \ g$ and the product of density and volume of the unit cell is $1.792 \times 10^{-22} \ g$. What is the type of cubic unit cell?

$A$ metal crystallizes in two phases,one as $fcc$ and another as $bcc$ with unit cell edge lengths of $3.5 \mathring{A}$ and $3.0 \mathring{A}$,respectively. The ratio of density of $fcc$ and $bcc$ phases approximately is

The edge length of the unit cell of a metal $(M_W = 24 \, g \, mol^{-1})$ having a cubic structure is $4.53 \, \mathring{A}$. If the density of the metal is $1.74 \, g \, cm^{-3}$,then the effective number of atoms in the unit cell is :- $(N_A = 6 \times 10^{23} \, mol^{-1})$