Chapter 9, Problem 52E

Interpretation Introduction

Interpretation:

        The density of nickel having face-centered cubic unit cell is given and its atomic radius 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,

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

Answer to Problem 52E

Answer

         The radius of nickel atom is $1\text{36}\stackrel{\text{°}}{\text{A}}$.

Explanation of Solution

Explanation

Calculate the mass of a unit cell.

        $\begin{array}{l}\text{Average mass of one Ni atom}\text{=}\frac{\text{atomic}\text{mass}\text{of}\text{Ni}}{\text{Avogadro}\text{number}}\\ \text{=}\frac{\text{58}\text{.69}\text{g}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}\text{9}\text{.746}\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

$\begin{array}{l}\text{Each}\text{unit}\text{cell}\text{has}\text{4}\text{Ni}\text{atoms}\text{.}\text{Therefore,}\\ \text{mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}\text{4}\text{×}\text{average}\text{mass}\text{of}\text{one}\text{Ca}\text{atom}\\ \text{=}\text{4}\text{×}\text{9}\text{.746}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{=}\text{390}\text{×}{\text{10}}^{-24}\text{g}\\ \end{array}$

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

Calculate the volume and edge length of the unit cell.

    $\begin{array}{l}\text{given}\text{data}\text{:}\text{density}\text{=}\text{6}\text{.84}{\text{g/cm}}^{\text{3}}\\ \text{calculated}\text{data}\text{from}\text{previous}\text{step}\text{:}\text{mass}\text{=}\text{390}\text{×}{\text{10}}^{\text{-24}}\text{g}\\ \text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{volume}\text{of}\text{a}\text{unit}\text{cell,}{\text{a}}^{\text{3}}\text{=}\frac{\text{mass}}{\text{density}}\\ {\text{a}}^{\text{3}}\text{=}\frac{\text{390}\text{×}{\text{10}}^{\text{-24}}\text{g}}{\text{6}\text{.84}{\text{g/cm}}^{\text{3}}}\\ \text{=}\text{57}\text{.02}\text{×}{\text{10}}^{\text{-24}}{\text{cm}}^{\text{3}}\\ \text{a}\text{=}{\left(\text{57}\text{.02}\text{×}{\text{10}}^{\text{-24}}{\text{cm}}^{\text{3}}\right)}^{\text{1/3}}\\ \text{=}\text{3}\text{.85}\text{×}{\text{10}}^{\text{-8}}\text{cm}\end{array}$

          The volume ‘a³’ of the unit cell is calculated using $\text{density}\text{=}\frac{\text{mass}}{\text{volume}}$ formula. Volume of fcc unit cell is cube of its edge length. Hence, Cube root of the volume gives the edge length of the unit cell.

Calculate the radius of Ni atom.

$\begin{array}{l}\text{The edge length of one unit cell is given by}\\ \text{a}\text{=}\text{2R}\sqrt{\text{2}}\\ \text{where a}\text{=}\text{edge length of unit cell}\\ \text{R}\text{=}\text{radius}\text{of}\text{atom}\end{array}$

$\begin{array}{l}\text{a}\text{=}\text{2R}\sqrt{\text{2}}\\ \text{R}\text{=}\frac{\text{a}}{\text{2}\sqrt{\text{2}}}\\ \text{=}\frac{\text{3}\text{.85}\text{×}{\text{10}}^{\text{-8}}\text{cm}}{\text{2}\text{×}\text{1}\text{.414}}\\ =1.\text{36}\text{×}\text{10}{}^{\text{-8}}\text{cm}\\ \text{=}1\text{36}\stackrel{\text{°}}{\text{A}}\end{array}$

             The side length of fcc unit cell is given as $\text{a}\text{=}\text{2R}\sqrt{\text{2}}$ where ‘a’ is side length and ‘R’ is radius of the atom. By rearranging the above expression and substituting the value for side length calculated in the previous step, the radius value ‘R’ for the nickel atom is determined.

Conclusion

Conclusion

               The radius of the nickel atom is determined using the concept of side length of the FCC unit cell and its relation with density given.