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 $93 \, u$.

(N/A) Given: Density $(d)$ = $8.55 \, g \, cm^{-3}$,Atomic mass $(M)$ = $93 \, g \, mol^{-1}$,$z$ (for $bcc$) = $2$,Avogadro's number $(N_{A})$ = $6.022 \times 10^{23} \, mol^{-1}$.
Using the formula for density: $d = \frac{z \cdot M}{a^{3} \cdot N_{A}}$.
Rearranging for edge length $(a)$: $a^{3} = \frac{z \cdot M}{d \cdot N_{A}} = \frac{2 \times 93}{8.55 \times 6.022 \times 10^{23}} \approx 3.612 \times 10^{-23} \, cm^{3}$.
Taking the cube root: $a \approx 3.306 \times 10^{-8} \, cm$.
For a $bcc$ structure,the relation between radius $(r)$ and edge length $(a)$ is: $r = \frac{\sqrt{3}}{4} \cdot a$.
Substituting the value of $a$: $r = \frac{1.732}{4} \times 3.306 \times 10^{-8} \, cm \approx 1.432 \times 10^{-8} \, cm$ or $0.1432 \, nm$.