Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm^–3, calculate atomic radius of niobium using its atomic mass 93 u.

Given density (d) = 8.55 g cm^-3

Atomic mass (M) = 93u = 93 g mol^-1

To find: - atomic radius, r =__

We know, Avogadro Number N_A = 6.022 × 10²³ mol^-1

Since, given lattice in bcc

Therefore, Number of atoms per unit cell (z) = 2

We know that,

⇒ a³ = 36.124 × 10^-24 cm³

⇒ a = 3.3057 × 10^-8 cm

for bcc unit cell radius(r) = a/4

Therefore, r = /4 × 3.3057 × 10^-8 cm

r = 1.732/4 × 3.3057 × 10^-8 cm

r = 5.725/4 × 10^-8 cm

r = 14.31 × 10^-9

r = 14.31 nm.

Thus the atomic radius of niobium = 14.31 nm.