Chapter 12, Problem 12.31QP

Europium crystallizes in a body-centered cubic lattice (the Eu atoms occupy only the lattice points). The density of Eu is 5.26 g/cm³. Calculate the unit cell edge length in picometers.

Interpretation Introduction

Interpretation:

The edge length of unit cell of Europium crystal 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,

                      $\text{8}\text{×}\frac{\text{1}}{\text{8}}\text{atoms}\text{in}\text{corners}\text{+}\text{1}\text{atom}\text{at}\text{the}\text{center}\text{=}\text{1}\text{+}\text{1}\text{=}\text{2}\text{atoms}$

             $\begin{array}{l}\text{ The edge length of one unit cell is given by}\\ \text{a}\text{=}\frac{\text{4R}}{\sqrt{\text{3}}}\\ \text{where a}\text{=}\text{edge length of unit cell}\\ \text{R}\text{=}\text{radius}\text{of}\text{atom}\text{.}\\ \end{array}$

Answer to Problem 12.31QP

The edge length of unit cell of Europium is calculated as $458pm.$

Explanation of Solution

Record the given data.

$\text{density}\text{of}\text{Eu}\text{=}\text{5}\text{.26}{\text{g/cm}}^{\text{3}}$

Density of Europium is given which is related to mass and volume by the following equation $\text{density}\text{=}\frac{\text{mass}}{\text{volume}}$.

Calculate the mass of unit cell of Europium.

$\begin{array}{l}\text{Average mass of one Eu atom}\text{=}\frac{\text{atomic}\text{mass}\text{of}Eu}{\text{Avogadro}\text{number}}\\ \text{=}\frac{152.0\text{g}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}25.24\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{Each}\text{unit cell contains}\text{2}\text{Eu atoms}\text{. therefore,}\\ \text{Mass}\text{of}\text{a}\text{unit}\text{cell}\text{of}\text{Europium}\text{=}\text{2}\text{×}\text{25}\text{.24}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{=}\text{50}\text{.48}\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

Each unit cell contains 2 Europium atoms.  Therefore two times the average mass of one Europium atom gives mass of a unit cell of Europium.

Calculate the volume of unit cell of Ti.

$\begin{array}{l}\text{known}\text{data:}\text{mass}\text{=}\text{50}\text{.48}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{density}\text{=}\text{5}\text{.26}{\text{g/cm}}^{\text{3}}\\ \text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{volume}\text{=}\frac{\text{mass}}{\text{density}}\\ \text{=}\frac{\text{50}\text{.48}\text{×}{\text{10}}^{\text{-23}}\text{g}}{\text{5}\text{.26}{\text{g/cm}}^{\text{3}}}\\ \text{=}\text{9}\text{.6}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\text{=}\text{96}\text{×}{\text{10}}^{\text{-24}}{\text{cm}}^{\text{3}}\\ \end{array}$

Density of $\text{Eu}$ is given and mass of the titanium is calculated in the previous step.  Substituting these two values in the equation $\text{density}\text{=}\frac{\text{mass}}{\text{volume}}$ and rearranging it, volume of the unit cell is obtained.

Determine the edge length of unit cell of $\text{Eu}$.

$\begin{array}{l}{\text{a}}^{\text{3}}\text{=}96\text{×}{\text{10}}^{\text{-24}}{\text{cm}}^{\text{3}}\\ \text{a}\text{=}{}^{\text{3}}\sqrt{96\text{×}{\text{10}}^{\text{-24}}}\\ \text{=}4.\text{58}\text{×}{\text{10}}^{\text{-8}}\text{cm}\\ \text{=}458\text{pm}\end{array}$

Volume of the bcc unit cell of Europium is calculated in the previous step. Cube root of the volume of a bcc unit cell, gives the edge length of bcc unit cell of Europium.

Conclusion

The edge length of the unit cell of Europium is determined.