Question

Niobium crystallises in body-centred cubic structure. If density is $8.55gc{m}^{-3}$, calculate atomic radius of niobium using its atomic mass $93U$.

Answer 1

Density $\left(D\right)$, number of atoms per unit cell $\left(Z\right)$, atomic mass $\left(M\right)$, Avogadro's number $\left(N_0\right)$ and edge length ($a)$ are related as below:
$D=\frac{ZM}{N_0{a}^3}$

Given $D=8.55$ g/cm${}^3$ and $M=93$ u.

Edge length $=a=(\frac{ZM}{D{N}_0}{)}^{1/3}=(\frac{2\times 93}{8.55\times 6.023\times {10}^{23}}{)}^{1/3}=3.306\times {10}^{-8}$ cm

Atomic radius $=r=\frac{\sqrt{3}}{4}a=\frac{\sqrt{3}}{4}\times 3.306\times {10}^{-8}=1.432\times {10}^{-8}cm=14.32$ nm