Question

Niobium Crystalizes $\left(g3\right)$ BCl lattice and has density of $8.55$ kg$/d{m}^3$. Calculate the atomic radius of Niobium.

Answer 1

Atomic mass of $Nb=93u$

Given density is $8.55kg/d{m}^3$

no. of atoms in unit cell of $BCC$ lattice, $Z=2$

We have,

$\rho =\frac{ZM}{a^3{N}_A}=\frac{2\times 93}{a^3\times 6.022\times {10}^{23}}$

$a^3=\frac{186}{8.55\times 6.022\times {10}^{23}}=\frac{186}{5.14881}\times {10}^{-24}$

$a^3=36.124\times {10}^{-24}$

$a=3.3\times {10}^{-8}dm$

For BCC,

$r=\frac{a\sqrt{3}}{4}=\frac{3.3\times {10}^{-8}\times 1.732}{4}dm$

$r=1.4289\times {10}^{-8}dm$

$r=1.43\times {10}^{-8}dm$