#### 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.

#### Solution

It is given that the density of niobium, *d* = 8.55 g cm^{−3}

Atomic mass, M = 93 gmol^{−1}

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

We also know that, N_{A} = 6.022 × 10^{23} mol^{−1}

Applying the relation:

`d = (zM)/(a^3N_A)`

`=>a^3= (zM)/(dN_A)`

= `(2xx93 gmol^(-1))/(8.55 "gcm"^(-3)xx6.022xx10^(23) mol^(-1))`

= 3.612 × 10^{−23} cm^{3}

So, *a* = 3.306 × 10^{−8} cm

For body-centred cubic unit cell:

`r = sqrt3/4a`

=`sqrt3/4xx3.306xx10^(-8) cm`

= 1.432 × 10^{−8} cm

= 14.32 × 10^{−9} cm

= 14.32 nm..

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Solution 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. Concept: Packing Efficiency - Efficiency of Packing in Body-centred Cubic Structures.