Niobium crystallizes in body centred cubic structure. If its density is 8.55 g `cm^(-3)`, calculate the atomic radius of niobium. (Atomic mass of Nb = 93u , `N_A = 6.02 xx 10^(23) mol^(-1)`)

Explanation

Density of niobium, d=8.55 g `cm^(-3)`, Number of atoms in bcc unit cell, Z= 2
Atomic mass of niobium, M = 93u
Density, `d = (ZM)/(N_A xx a^3)`
`a^3=(ZM)/(d xx N_A)=(2 xx 93)/(8.55 xx 6.02 xx 10^(23))`
`= 3.61 xx 10^(-23) cm^3= 36.1 xx 10^(-24) cm^3`
`a = 3.30 xx 10^(-8) cm=3.30 Å`
For a bcc unit cell, `4r = a sqrt3` or `r = (a sqrt3)/(4)=(3.30 xx sqrt3)/(4) Å`
= `(3.30 xx 1.73)/(4) = 1.427` Å= 14.27 nm