Question

Niobium crystallizes in the body-centred cubic structure. If the density is $8.55{\text{g c}}{{\text{m}}^3}$, calculate atomic radius of niobium using its atomic mass $93{\text{ u}}$.

Answer 1

To answer this question, knowledge on Density and Volume of the unit cell is required which is defined as the mass per volume of the cell.
Formula used:
${\text{density of the unit cell = }}\dfrac{{{\text{mass of the unit cell}}}}{{{\text{volume of the unit cell}}}}$
$\Rightarrow {\text{d = }}\dfrac{{{\text{zM}}}}{{{{\text{a}}^{\text{3}}}{{\text{N}}_{\text{A}}}}}$
Where, z is the number of atoms present in the unit cell, M is the molar mass of the substance, a is the edge length of the solid, and ${{\text{N}}_{\text{A}}}$ is the Avogadro’s number = $6.023 \times {10^{23}}$

Complete step by step answer:
Here the atomic radius of niobium has been asked for which is equal to $\dfrac{{\sqrt 3 }}{4}$ the edge length of the unit cell of the metal in the body-centred cubic structure.
Therefore, let the edge length be ‘a’ and the volume of the cell, V = ${{\text{a}}^3}$
From the above equation we have, density of the cell, ${\text{d = }}\dfrac{{{\text{zM}}}}{{{{\text{a}}^{\text{3}}}{{\text{N}}_{\text{A}}}}}$
$\Rightarrow {{\text{a}}^{\text{3}}}{\text{ = }}\dfrac{{{\text{z}} \times {\text{M}}}}{{{\text{d}} \times {{\text{N}}_{\text{A}}}}}$
For the body-centred cubic lattice, z = 2, from the question we have, d = $8.55{\text{g c}}{{\text{m}}^3}$, Molar mass = $93{\text{ u}}$ . So putting the values of the parameters in the above equation we get,
${{\text{a}}^{\text{3}}}{\text{ = }}\dfrac{{2 \times 93}}{{8.55 \times {\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}}}$
$\Rightarrow {{\text{a}}^3} = 3.6 \times {10^{ - 23}}{\text{c}}{{\text{m}}^{\text{3}}}$
So the volume of the unit cell is, $3.6 \times {10^{ - 23}}{\text{c}}{{\text{m}}^{\text{3}}}$ , and the atomic radius of Niobium is,
${\text{a}} = \sqrt[3]{{3.6 \times {{10}^{ - 23}}{\text{c}}{{\text{m}}^{\text{3}}}}}$
$\Rightarrow {\text{a}} = 3.3 \times {10^{ - 8}}$
For Body-Centred Cubic Cell, the atomic radius, r = $\dfrac{{\sqrt 3 }}{4}{\text{a}}$
$\Rightarrow \dfrac{{\sqrt 3 }}{4} \times 3.3 \times {10^{ - 8}}$ = $1.43 \times {10^{ - 8}}$ cm.

So the atomic radius of niobium is 14.3nm

Note:
A unit cell is the smallest portion of a crystal lattice which when repeated in different directions, generates the entire lattice.
There are four types of unit cells based on the arrangement of atoms in the lattice and among them, the Body Centred Cubic Cell contains one constituent particle present at body centre position besides the corner of the cells.