Niobium crystallises in a body-centred cubic $(bcc)$ structure. If the density is $8.55 \ g \ cm^{-3}$, then the atomic radius of niobium is (atomic mass of niobium $= 93 \ u$) (in $pm$)

In a cubic unit cell of an ionic compound, the eight corners are occupied by anions and cations are at the center of the cube. Calculate the volume of the unit cell if the density of the unit cell is $4 \text{ g cm}^{-3}$. [Molar mass of compound $= 168.6 \text{ g mol}^{-1}$ and $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$]

$A$ substance has a density of $2 \ g \ cm^{-3}$. It crystallizes in the $fcc$ crystal with an edge length of $600 \ pm$. The molar mass of the substance (in $g \ mol^{-1}$) is
$(N_{A} = 6 \times 10^{23} \ mol^{-1})$

For a crystal, the angle of diffraction $(2 \theta)$ is $90^{\circ}$ and the second order line has a $d$ value of $2.28 \ \text{Å}$. The wavelength (in $\text{Å}$) of $X$-rays used for Bragg's diffraction is

$A$ 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 element with a molar mass $27 \text{ g mol}^{-1}$ forms a cubic unit cell. Calculate the number of atoms present in a unit cell if the density of the metal is $2.7 \text{ g cm}^{-3}$. Given: $[a^3 \times N_A = 40 \text{ cm}^3 \text{ mol}^{-1}]$