Calculate the edge length of a $bcc$ unit cell if the radius of the metal atom is $227 \ pm$.

Gold has a cubic close-packed structure which can be viewed as spheres occupying $74\%$ of the total volume. What is the edge length of the unit cell if the density of gold is $19.3 \ g/cm^3$? $(Au = 197 \ amu)$

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 number of atoms present in one unit cell of it is (Given : $N_{A}=6.023 \times 10^{23} \ mol^{-1}$)

$Fe$ 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. $(Fe = 55.845 \,u)$

The density of $\beta-Fe$ is $7.6 \ g \ cm^{-3}$. It crystallizes in a cubic lattice with $a = 290 \ pm$. What is the value of $Z$? $(Fe = 56 \ g \ mol^{-1}; N_{A} = 6.022 \times 10^{23} \ mol^{-1})$

If a metal crystallises in $bcc$ structure with edge length of unit cell $4.29 \times 10^{-8} \ cm$,the radius of the metal atom is: