Chapter 9, Problem 53E

Interpretation Introduction

Interpretation:

        The atomic radius and density of lead cubic close packing structure has to be determined.

Concept introduction:

        In packing of atoms in a crystal structure, the atoms are imagined as spheres and closely packed in a regular pattern. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing. Cubic close packing structure has face-centered cubic (FCC) unit cell.

       In face-centered cubic unit cell, each of the six corners is occupied by every single atom. Each face of the cube is occupied by one atom.

       Each atom in the corner is shared by eight unit cells and each atom in the face is shared by two unit cells. Thus the number of atoms per unit cell in FCC unit cell is,

                     $\begin{array}{l}\text{8}\text{×}\frac{\text{1}}{\text{8}}\text{atoms}\text{in}\text{corners}\text{+}\text{6}\text{×}\frac{\text{1}}{\text{2}}\text{atoms}\text{in}\text{faces}\text{=}\text{1}\text{+}\text{3}\text{=}\text{4}\text{atoms}\\ \\ \text{ The edge length of one unit cell is given by}\\ \text{a}\text{=}\text{2R}\sqrt{\text{2}}\\ \text{where a}\text{=}\text{edge}\text{length of unit cell}\\ \text{R}\text{=}\text{radius}\text{of}\text{atom}\\ \end{array}$

Answer to Problem 53E

Answer

        The atomic radius of lead atom is 174 pm

         Density of lead is 11.57 g/cm³

Explanation of Solution

Explanation

Calculate the atomic radius of lead.

$\begin{array}{l}\text{given}\text{data:}\text{edge}\text{length,}\text{a}\text{=}\text{492}\text{pm}\text{=}\text{492}\text{×}{\text{10}}^{\text{-12}}\text{m}\\ \text{a}\text{=}\text{2R}\sqrt{\text{2}}\end{array}$

                                        $\begin{array}{l}\text{R}\text{=}\frac{\text{a}}{\text{2}\sqrt{\text{2}}}\\ \text{=}\frac{\text{492}\text{×}{\text{10}}^{\text{-12}}\text{m}}{\text{2}\text{×}\text{1}\text{.414}}\text{=}\text{174}\text{×}{\text{10}}^{\text{-12}}\text{m }\\ \text{ =}\text{174}\text{pm}\\ \end{array}$

              The edge length value for the FCC unit cell is given. The edge length value is related to the radius of the atom by the equation, $\text{a}\text{=}\text{2R}\sqrt{\text{2}}$. The given value for ‘a’ is plugged and the equation is rearranged in order to get atomic radius ‘R’ of lead.

Calculate the volume of the unit cell.

$\begin{array}{l}\text{given}\text{data}\text{:}\text{edge}\text{length}\text{,}\text{a}\text{=}\text{492}\text{pm}\text{=}\text{492}\text{×}{\text{10}}^{\text{-12}}\text{m}\\ \text{volume}\text{of}\text{unit}\text{cell}{\text{a}}^{\text{3}}\text{=}{\left(\text{492}\text{×}{\text{10}}^{\text{-12}}\text{m}\right)}^{\text{3}}\\ \text{=}\text{1}\text{.19}\text{×}{\text{10}}^{\text{-22}}{\text{cm}}^{\text{3}}\\ \text{=}\text{11}\text{.9}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\end{array}$

             The edge length value for the FCC unit cell is given. Cubic value of the edge length of the unit cell gives volume of the unit cell.

Calculate the mass of the unit cell.

                         $\begin{array}{l}\text{Average mass of one Pb atom}\text{=}\frac{\text{atomic}\text{mass}\text{of}\text{Pb}}{\text{Avogadro}\text{number}}\\ \text{=}\frac{\text{207}\text{.2}\text{g}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}\text{34}\text{.41}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{Each}\text{unit cell contains}\text{4}\text{Pb atoms}\text{. therefore,}\\ \text{Mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}\text{4}\text{×}\text{34}\text{.41}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{=}\text{137}\text{.64}\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

               Each unit cell contains 4 Pb atoms. Therefore four times the average mass of one Pb atom gives mass of a unit cell.

Calculate the density of lead.

               $\begin{array}{l}\text{known}\text{data:}\text{mass}\text{=}\text{137}\text{.64}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{volume}\text{=}\text{0}\text{.12}{\text{m}}^{\text{3}}\end{array}$

               $\begin{array}{l}\text{known}\text{data:}\text{mass}\text{=}\text{137}\text{.64}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{volume}\text{=}\text{1}\text{.19}\times {10}^{-26}{\text{cm}}^{\text{3}}\\ \text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{=}\frac{\text{137}\text{.64}\text{×}{\text{10}}^{\text{-23}}\text{g}}{\text{11}\text{.9 }\times {10}^{-23}{\text{cm}}^{\text{3}}}\\ \text{=}\text{11}\text{.57}{\text{g/cm}}^{\text{3}}\end{array}$

                  The density of lead is calculated by $\text{density}\text{=}\frac{\text{mass}}{\text{volume}}$ formula by substituting the values calculated in previous steps using given data.

Conclusion

Conclusion

                 The radius of the lead atom and its density is determined using the concept of edge length of the FCC unit cell.