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

ANSWER

Density(D), number of atoms per unit cell(Z), atomic mass(M), Avogadro's number(N0​)and edge length (a)are related asbelow:
D= ZM / N₀a³

​GivenD=8.55g/cm³ and M=93u.

Edge length= a = (ZM / DN₀)^1/3
= ( 2 * 93 / 8.55 * 6.023*10²³ )
= 3.306×10^-8cm
Atomic radius= r = (3)^1/2* a / 4
= 1.432×10^-8
=14.32nm

Thanks