Niobium crystallizes in a $bcc$ structure. If its density is $8.55 \, g/cm^3$, calculate the atomic radius of niobium. [Atomic mass of $Nb = 93 \, u$] (in $pm$)

If $a$ stands for the edge length of the cubic systems: simple cubic,body-centred cubic,and face-centred cubic,then the ratio of radii of the spheres in these systems will be respectively:

$A$ metal $M$ crystallizes into two lattices: face-centered cubic $(fcc)$ and body-centered cubic $(bcc)$ with unit cell edge lengths of $2.0 \ \mathring{A}$ and $2.5 \ \mathring{A}$ respectively. The ratio of densities of the $fcc$ lattice to the $bcc$ lattice for the metal $M$ is $...........$ (Nearest integer).

Calculate the density of an element having molar mass $63 \ g \ mol^{-1}$ that forms $FCC$ structure $\left[a^3 \times N_{A} = 28 \ cm^3 \ mol^{-1}\right]$ (in $g \ cm^{-3}$)

What is the relation between edge length $a$ and the total volume occupied by atoms in a $BCC$ unit cell?

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: