Sodium metal crystallizes in a body-centered cubic $(BCC)$ lattice with a unit cell edge length of $4 \ \mathring{A}$. The radius of the sodium atom is $..... \times 10^{-1} \ \mathring{A}$ (Nearest integer).

$A$ given metal crystallises out with a cubic structure having edge length of $361 \, pm$. If there are four metal atoms in one unit cell,what is the radius of one atom? .............. $pm$

Radius of cation and anion are $2.5 \ \mathring{A}$ and $2.6 \ \mathring{A}$ respectively. If a cubic crystal system is prepared by combination of above cation and anion then edge length of unit cell is ................ $\mathring{A}$ (Take $\sqrt{3} = 1.7$)

What is the relationship between the edge length $(a)$ and the radius of the atom $(r)$ in a simple cubic unit cell?

An element crystallises in $fcc$ type of unit cell. The volume of one unit cell is $24.99 \times 10^{-24} \ cm^{3}$ and density of the element is $7.2 \ g \ cm^{-3}$. Calculate the number of unit cells in $36 \ g$ of a pure sample of the element.

$X$-rays of wavelength $1 \ \mathring{A}$ are diffracted by the planes of particles in a metal. The second-order diffraction occurs at an angle of $60^{\circ}$. Calculate the distance between two parallel planes in $\mathring{A}$.