Sodium metal crystallises in a body centred cubic lattice with unit cell edge length of $4 \mathrm{\AA }$. The radius of sodium atom is $\_\_\_\times {10}^{-1}\mathrm{\AA }$. (Nearest integer)

Edge length of the unit cell $\left(\mathrm{a}\right) = 4 \mathrm{x} {10}^{-8}\mathrm{cm}$

Radius of sodium atom $\left(\mathrm{r}\right)=?$

Since, sodium crystallises in body centered cubic arrangement.

The radius of the atom and the edge length of the cubic unit cell are related as follows,

For bcc structure, $4\mathrm{r}=\sqrt{3}\mathrm{a}\Rightarrow \mathrm{r}=\frac{\sqrt{3}}{4}\mathrm{a}$

$$\begin{gathered} \mathrm{r}=\frac{1.732\times 4\times {10}^{-8}}{4} \\ \mathrm{r}=1.732\times {10}^{-8} \mathrm{cm} \\ \mathrm{r}=17.3\mathrm{\AA } \end{gathered}$$.