Question

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.

Answer 1

Given: It is given that the density of niobium, ρ = 8.55 g cm⁻³

Atomic mass, M = 93 g mol⁻¹

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

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

Formula: `rho = (Z xx M)/(a^3 xx N_A)`

`a = ((Z xx M)/(rho xx N_A))^(1//3)`

= `((2 xx 93)/(8.55 xx 6.022 xx 10^23))^(1//3)`

= `(186/(51.48 xx 10^23))^(1//3)`

= `(3.61 xx 10^23)^(1//3)`

= 3.3 × 10⁻⁸ cm 

= 33 nm

For body-centred cubic unit cell:

`r = (sqrt 3)/4  a`

= `1.732/4 xx 332`

= 0.433 × 33

= 14.29 nm