Calculate the edge length of a unit cell for a metal with an atomic radius of $128 \ pm$ that forms an $fcc$ unit cell structure.

In a $bcc$ lattice having the edge length of $200 \ pm$, the cation has the radius of $70 \ pm$. The radius ratio of $\frac{r^{+}}{r^{-}}$ is (Given, $\sqrt{2}=1.4, \sqrt{3}=1.7$ and $\sqrt{6}=2.4$ )

Calculate the number of unit cells in $1 \ cm^3$ volume of an element if unit cell edge length is $4.0 \times 10^{-8} \ cm$.

If the edge length of a $bcc$ unit cell is $386 \ pm$, then the atomic radius will be ........... $pm$.

In a body-centered cubic $(BCC)$ unit cell of a metal,the edge length is $4 \times 10^{-10} \ m$. The diameter of the atom will be:

Ferrous oxide has a cubic structure and each edge of the unit cell is $5.0 \, \mathring{A}$. Assuming the density of the oxide is $4.0 \, g \, cm^{-3}$,the number of $Fe^{2+}$ and $O^{2-}$ ions present in each unit cell will be ($M_w$ of $FeO = 72$)