Sodium metal crystallizes in a $bcc$ lattice with a unit cell edge length of $a = 4.29 \ \mathring{A}$. What is the radius of the sodium atom in $\mathring{A}$?

Calculate the number of atoms in $1 \text{ g}$ of a metal that forms a $bcc$ crystal structure. Given that the product of density and unit cell volume is $\rho \times a^3 = 6.6 \times 10^{-22} \text{ g}$.

An element with molar mass $2.7 \times 10^{-2} \ kg \ mol^{-1}$ forms a cubic unit cell with edge length of $405 \ pm$. If its density is $2.7 \times 10^3 \ kg \ m^{-3}$, the nature of the cubic unit cell is: $(N_{A} = 6.02 \times 10^{23} \ mol^{-1})$

$KBr$ has a rock salt type structure and has a density of $3.70 \ g/cm^3$. The edge length of the unit cell is approximately [molecular weight of $KBr = 120 \ g/mol$]:

The crystal structure of an element has an $fcc$ lattice. If the edge length of the crystal is $4 \ \mathring{A}$,what is the atomic weight (in $g \ mol^{-1}$) of the element,if the density of the crystal is $11.21 \ g \ cm^{-3}$ $(N_{A} = 6.023 \times 10^{23} \ mol^{-1})$?

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