Chapter 12, Problem 12.46QP

Interpretation Introduction

Interpretation:

The Avogadro’s number of Barium atom in its body centered cubic lattice has to be calculated.

Concept Introduction:

In packing of atoms in a crystal structure, the atoms are imagined as spheres. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing.  Cubic close packing forms two types of lattices – body – centered lattice and face – centered lattice.

In body-centered cubic unit cell, each of the six corners is occupied by every single atom. Center of the cube is occupied by one atom.  Each atom in the corner is shared by eight unit cells and a single atom in the center of the cube remains unshared.  Thus the number of atoms per unit cell in BCC unit cell is,

$\begin{array}{l}\text{8}\text{×}\frac{\text{1}}{\text{8}}\text{atoms}\text{in}\text{corners}\text{+}\text{1}\text{atom}\text{at}\text{the}\text{center}\text{=}\text{1}\text{+}\text{1}\text{=}\text{2}\text{atoms}\\ \text{ The edge length of one unit cell is given by}\\ \text{a}\text{=}\frac{\text{4R}}{\sqrt{\text{3}}}\\ \text{where a}\text{=}\text{edge length of unit cell}\\ \text{R}\text{=}\text{radius}\text{of}\text{atom}\text{.}\\ \end{array}$

Answer to Problem 12.46QP

The Avogadro’s number of Barium atom in its body centered cubic lattice is calculated as $6.20\times 10^{23}atoms/mol$.

Explanation of Solution

One mole of Barium has mass of $\text{137}\text{.3 g}$. Density of Barium is given. Volume of one mole of Barium is obtained by the formula $\text{volume}\text{=}\frac{\text{mass}}{\text{density}}$.

Hence, calculate the volume occupied by one mole of barium atoms in its unit cell as follows –

$\begin{array}{l}\text{given}\text{data:}\text{density}\text{=}\text{3}\text{.50}{\text{g/cm}}^{\text{3}}\\ \text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{volume of 1 mole of Barium}\text{=}\frac{\text{mass}\text{of 1 mole of Barium}}{\text{density}}\\ \text{=}\frac{\text{137}\text{.3}\text{g}}{\text{3}{\text{.50g/cm}}^{\text{3}}}\text{=}\text{39}\text{.23}{\text{cm}}^{\text{3}}\end{array}$

Edge length of the cubic unit cell is given.  The cubic value of edge length gives the volume of the unit cell.  In each cell of unit cell of body centered cubic lattice, 2 Barium atoms are occupied.

$\begin{array}{l}\text{given}\text{data:}\text{edge}\text{length,}\text{a}\text{=}502\text{pm}\text{=}502\text{×}{\text{10}}^{\text{-12}}\text{m}\text{=}5.\text{02}\text{×}{\text{10}}^{\text{-10}}\text{m}\\ \text{volume,}{\text{a}}^{\text{3}}\text{=}{\text{(5}\text{.02}\text{×}{\text{10}}^{\text{-10}}\text{m)}}^{\text{3}}\\ \text{=}1.265\text{×}{\text{10}}^{\text{-22}}{\text{cm}}^{\text{3}}\\ =12.65\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\end{array}$

Dividing the volume of one mole Barium atoms by volume of one mole of Barium atoms in its unit cell gives the number of Barium atoms in one mole of Barium, which is close to Avogadro’s number.

$\begin{array}{l}\text{Number of Ba atoms in}\text{1}\text{mole of Barium =}\frac{\text{volume}\text{of}\text{1}\text{mole}\text{of}\text{Ba}}{\text{volume}\text{of}\text{1}\text{Ba}\text{in}\text{unit}\text{cell}}\\ \text{=}\frac{\text{39}\text{.23}}{\frac{\text{12}{\text{.65×10}}^{\text{-23}}}{\text{2}}}\text{= 6}\text{.20}{\text{×10}}^{\text{23}}\text{atoms/mol}\end{array}$

Conclusion

The Avogadro’s number of Barium atom in its body centered cubic lattice was calculated.