What is the edge length of the unit cell of a body-centred cubic $(BCC)$ crystal of an element whose atomic radius is $75 \ pm$ (in $pm$)?

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

Potassium has a $bcc$ structure with a nearest neighbour distance of $4.52 \ \mathring{A}$. Its atomic weight is $39$. Its density (in $kg \ m^{-3}$) will be:

Calculate the radius of an atom of an element in $pm$ if it forms $bcc$ unit cell structure having edge length $4.3 \times 10^{-8} \ cm$.

$A$ 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).

Bragg's law is given by the equation