Chapter 9, Problem 57E

Titanium metal has a body-centered cubic unit cell. The density of titanium is 4.50 g/cm³. Calculate the edge length of the unit cell and a value for the atomic radius of titanium. (Hint: In a body-centered arrangement of spheres, the spheres touch across the body diagonal.)

Interpretation Introduction

Interpretation:

        The edge length of the unit cell of Titanium has to be determined.

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.

       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,

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

Answer to Problem 57E

Answer

        The edge length of the unit cell of Titanium is 328 pm.

Explanation of Solution

Explanation

Calculate the mass of unit cell of Ti.

$\begin{array}{l}\text{Average mass of one Ti atom}\text{=}\frac{\text{atomic}\text{mass}\text{of}\text{Ti}}{\text{Avogadro}\text{number}}\\ \text{=}\frac{\text{47}\text{.9}\text{g}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}7.95\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{Each}\text{unit cell contains}\text{2}\text{Ti atoms}\text{. therefore,}\\ \text{Mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}\text{2}\text{×}7.95\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{=}15.9\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

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

Calculate the volume of unit cell of Ti.

$\begin{array}{l}\text{known}\text{data:}\text{mass}\text{=}\text{15}\text{.9}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{density}\text{=}\text{4}\text{.50}{\text{g/cm}}^{\text{3}}\\ \text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{volume}\text{=}\frac{\text{mass}}{\text{density}}\\ \text{=}\frac{\text{15}\text{.9}\text{×}{\text{10}}^{\text{-23}}\text{g}}{\text{4}\text{.50}{\text{g/cm}}^{\text{3}}}\\ \text{=}\text{35}\text{.3}\text{×}{\text{10}}^{\text{-24}}{\text{cm}}^{\text{3}}\\ \end{array}$

         Density of the Titanium 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 Ti.

$\begin{array}{l}{\text{a}}^{\text{3}}\text{=}\text{35}\text{.3}\text{×}{\text{10}}^{\text{-24}}{\text{cm}}^{\text{3}}\\ \text{a}\text{=}{}^{\text{3}}\sqrt{\text{3}\text{.53}\text{×}{\text{10}}^{\text{-24}}}\\ \text{=}\text{3}\text{.28}\text{×}{\text{10}}^{\text{-8}}\text{cm}\\ \text{=}\text{328}\text{pm}\end{array}$

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

Conclusion

Conclusion

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