Question

niobium is found to crystallise with BCC structure and found to have density of 8.55 g/cm³. Determine the atomic radius of niobium if its atomic mass is 93

Answer 1

Answer:

Given density (d) = 8.55 g/cm³

Atomic mass (M) = 93u = 93 g/mol

We know, Avogadro number, N = 6.022 × 10²³

given lattice is body - centered cubic,

so, number of atoms per unit cell (z) = 2

we know, \bf{d=\frac{zM}{a^3N}}

=> 8.55 = (2 × 93)/(a³ × 6.023 × 10²³)

=> 8.55 × a³ × 6.023 × 10²³ = 186

=> a³ = 186/(8.55 × 6.023 × 10²³)

=> a³ = 36.124 × 10^-24

=> a = 3.3057 × 10^-8 cm

for BCC unit cell, radius = √3a/4

= (√3 × 3.3057 × 10^-8)/4

= (1.732 × 3.3057)/4 × 10^-8

= 1.431 × 10^-8 cm = 1.431 × 10^-10 m

= 1.431 A°

hence, atomic radius of Niobium is 1.431A°.