Question

Niobium crystallises in body - centred cubic structure. If density is $8.55gc{m}^{-3}$. Calculate atomic radius of niobium using its atomic mass 93 u.

Answer 1

Step I Calculation of edge length of unit cell(a)

Atomic mass of the element $\left(M\right)=93gmo{l}^{-1}$

Number of particles in bcc type unit cell (Z) = 2

Mass of the unit cell $=\frac{Z\times M}{N_A}=\frac{2\times \left(93gmo{l}^{-1}\right)}{(6.022\times {10}^{23}mo{l}^{-1})}$

$=30.89\times {10}^{-23}g$

Density of unit cell (d) $=8.55gc{m}^{-3}$

Volume of unit cell $\left(a^3\right)=\frac{Massofunitcell}{Densityofunitcell}$

$=\frac{(30.89\times {10}^{-23}g)}{\left(8.55gc{m}^{-3}\right)}$

$=36.16\times {10}^{-24}c{m}^3$

Edge length of unit cell (a) $=(36.13\times {10}^{-24}c{m}^3{)}^{\frac{1}{3}}$

$=3.31\times {10}^{-8}cm$

Step II Calculation of radius of unit cell (r)

For bcc structure, $r=\frac{\sqrt{3}a}{4}$

$=\frac{\sqrt{3}\times (3.31\times {10}^{-8}cm)}{4}$

$=1.43\times {10}^{-8}cm$

$=143pm$