The edge length of a $bcc$ unit cell is $4 \times 10^{-10} \ m$. What is the radius of the atom occupying the body center of the cubic unit cell?

$A$ certain element crystallises in a $bcc$ lattice of unit cell edge length $27 \mathring{A}$. If the same element under the same conditions crystallises in the $fcc$ lattice,the edge length of the unit cell in $\mathring{A}$ will be .........
(Round off to the Nearest Integer).
[Assume each lattice point has a single atom]
[Assume $\sqrt{3}=1.73, \sqrt{2}=1.41$]

$Fe$ crystallizes in a $bcc$ lattice. If the edge length is $286.65 \,pm$ and the density is $7.874 \,g/cm^3$, calculate the Avogadro number. $(Fe = 55.845 \,u)$

Metal $M$ crystallizes in $fcc$ lattice. If the edge length of the unit cell is $4.077 \times 10^{-8} \ cm$ and the density is $10.5 \ g \ cm^{-3}$,the atomic mass of the metal is:

Which method is used to determine the edge length of a unit cell?

The density of $Na$ is $0.613 \ g \ cm^{-3}$. If the edge length of the unit cell of $Na$ is $5 \ \mathring{A}$,the effective number of atoms of $Na$ per unit cell is (Atomic weight of $Na = 23 \ u$)