Niobium crystallises in body-centred cubic structure.

Solution:

It is given that the density of niobium, d = 8.55 g cm−3

Atomic mass, M = 93 gmol−1

As the lattice is bcc type, the number of atoms per unit cell, z = 2

We also know that, NA = 6.022 × 1023 mol−1

Applying the relation:

$d=\frac{z \mathrm{M}}{a^{3} \mathrm{~N}_{\mathrm{A}}}$

$\Rightarrow a^{3}=\frac{z \mathrm{M}}{d \mathrm{~N}_{\mathrm{A}}}$

$=\frac{2 \times 93 \mathrm{gmol}^{-1}}{8.55 \mathrm{gcm}^{-3} \times 6.022 \times 10^{23} \mathrm{~mol}^{-1}}$

= 3.612 × 10−23 cm3

So, a = 3.306 × 10−8 cm

For body-centred cubic unit cell:

$r=\frac{\sqrt{3}}{4} a$

$=\frac{\sqrt{3}}{4} \times 3.306 \times 10^{-8} \mathrm{~cm}$

= 1.432 × 10−8 cm

= 14.32 × 10−9 cm

= 14.32 nm