Chapter 12, Problem 12.120QP

Calcium metal crystallizes in a face-centered cubic unit cell with a cell edge length of 558.84 pm. Calculate (a) the radius of a calcium atom in angstroms (A) and (b) the density of calcium metal in g/crn³.

Interpretation Introduction

Interpretation:

Edge length of unit cell of Calcium is given. Its density and atomic radius have to be determined.

Concept Introduction:

In crystalline solids, the components are packed in regular pattern and neatly stacked. The components are imagined as spheres and closely packed. This phenomenon is called “close packing” in crystals. 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,

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

Answer to Problem 12.120QP

Atomic radius of Calcium is $\text{1}\text{.976}\stackrel{\text{o}}{\text{A}}$

Density of Calcium is $1.52g/c{m}^3.$

Explanation of Solution

The edge length of the unit cell of Calcium is related to the atomic radius of the Calcium as $\text{a = 2R}\sqrt{2}$ from which atomic radius of Calcium is obtained as shown below.

$\begin{array}{l}\text{given data: edge length, a = 558}\text{.84 pm}\text{.}\\ \text{ a = 2R}\sqrt{2}\\ \text{ R}\text{= }\frac{\text{a}}{2\sqrt{2}}\\ \text{ = }\frac{558.84\times {\text{10}}^{-12}\text{m}}{2\times 1.414}\\ \text{ =}\text{ 197}\text{.6 pm = 1}\text{.976}\stackrel{°}{\text{A}}\end{array}$

The atomic radius of the Calcium is calculated in the previous step. The formula for the edge length of the cubic unit cell is known as $\text{a}\text{=}\text{2R}\sqrt{\text{2}}$ in which ‘a’ corresponds to edge length of the unit cell and ‘R’ corresponds to radius of the atom in the unit cell. The value is substituted and ‘a³’value is calculated. This value corresponds to the volume of the unit cell.

$\text{known}\text{data:}\text{R}\text{=}\text{197}\text{.6}\text{pm}\text{=}\text{197}\text{.6}\text{×}{\text{10}}^{\text{-12}}\text{m}$

$\text{volume}\text{of}\text{unit}\text{cell}\text{=}{\text{a}}^{\text{3}}$

$\begin{array}{l}\text{a}\text{=}\text{2R}\sqrt{\text{2}}\\ {\text{a}}^{\text{3}}\text{=}{\left(\text{2R}\sqrt{\text{2}}\right)}^{\text{3}}\\ \text{=}{\left(\text{2}\text{×}\text{197}\text{.6}\text{×}{\text{10}}^{\text{-12}}\text{m}\text{×}\text{1}\text{.414}\right)}^{\text{3}}\\ \text{=}\text{1}\text{.745}\text{×}{\text{10}}^{\text{-22}}{\text{cm}}^{\text{3}}\end{array}$

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

$\begin{array}{l}\text{Average mass of one Ca atom}\text{=}\frac{\text{atomic}\text{mass}\text{of}\text{Ca}}{\text{Avogadro}\text{number}}\\ \text{=}\frac{\text{40}\text{.08}\text{g}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}\text{6}\text{.6423}\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

$\text{Each}\text{unit}\text{cell}\text{has}\text{4}\text{Ca}\text{atoms}\text{.}\text{Therefore,}$

$\begin{array}{l}\text{mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}\text{4}\text{×}\text{average}\text{mass}\text{of}\text{one}\text{Ca}\text{atom}\\ \text{=}\text{4}\text{×}\text{6}\text{.6423}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{=}\text{2}\text{.66}\text{×}{\text{10}}^{\text{-22}}\text{g}\end{array}$

$\begin{array}{l}\text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{=}\frac{\text{2}\text{.66}\text{×}{\text{10}}^{\text{-22}}\text{g}}{\text{1}\text{.745}\text{×}{\text{10}}^{\text{-22}}{\text{cm}}^{\text{3}}}\\ \text{=}\text{1}\text{.52}{\text{g/cm}}^{\text{3}}\end{array}$

Mass and volume of the unit cell is calculated in the previous steps. By substituting the values in the formula, $\text{density}\text{=}\frac{\text{mass}}{\text{volume}}$, the density of the calcium atom is determined.

Conclusion

The atomic radius and density of Calcium have been determined.