Chapter 12.4, Problem 5PPC

Given that the diameter and average mass of a billiard ball are 5.72 cm and 165 g. respectively, determine the density of a billiard ball. Assuming that they can be packed like atoms in a metal, determine the density of a collection of billiard balls packed with a simple cubic unit cell, and those packed with a face-centered unit cell. Explain why the three densities are different despite all referring to the same objects.

Interpretation Introduction

Interpretation:

• Given the diameter and mass of a billiard ball its Density has to be determined.
• By assuming the billiard balls are packed in simple cubic lattice, density of the billiard ball in its simple cubic unit cell has to be determined.
• By assuming the billiard balls are packed in face centered cubic lattice, density of the billiard ball in its face centered cubic unit cell has to be determined.

Concept Introduction:

In packing of atoms or molecules of a solid, the atoms/molecules 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 structure has face-centered cubic (FCC) unit cell.

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,

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

In simple cubic unit cell, each atom in the corner is shared by eight unit cells. Thus, number of atoms per simple cubic unit cell is,

$\text{8}\text{×}\frac{\text{1}}{\text{8}}\text{atoms}\text{in}\text{corners}\text{=}\text{1}\text{atom}$

The edge length of simple cubic unit cell is represented by the formula   “ $\text{l}\text{=}\text{2r}$”.

Answer to Problem 5PPC

Density of billiard ball is determined as $1.68g/c{m}^3.$

Density of billiard ball in its simple cubic unit cell is determined as $0.882g/c{m}^3.$

Density of billiard ball in its face-centered cubic unit cell is determined as $1.25g/c{m}^3.$

Explanation of Solution

To calculate: Density of billiard ball.

Billiard ball is spherical is shape.

Thus its volume is,

$\begin{array}{l}Volumeofbilliardball\text{ = }\frac{4}{3}\pi r^3\\ where,\text{ r = radius of the billiard ball}\end{array}$

We know,

$diameter\text{ of the billiard ball = 5}\text{.72 cm}$

And hence radius of the billiard ball is half of the diameter. Therefore,

$radius\text{ = }\frac{5.72cm}{2}\text{ = 2}\text{.86 cm}$

Therefore volume of the billiard ball is,

$\frac{4}{3}\times 3.14\times {\left(2.86\text{ cm}\right)}^3=\text{ 97}{\text{.94 cm}}^3$

Also, mass of the billiard ball is given as $165\text{ g}\text{.}$ hence the density of billiard ball is calculated as follows –

$\begin{array}{l}density\text{ = }\frac{mass}{volume}\\ \text{ = }\frac{\text{165 g}}{97.94c{m}^3}\text{ = 1}{\text{.68 g/cm}}^3\end{array}$

To calculate:

Density of the billiard ball in its simple cubic unit cell.

Edge length of the simple cubic unit cell is given as,

$\begin{array}{l}l\text{ = 2r}\\ \text{l = edge length; r = radius of the atom inside unit cell}\end{array}$

Cubic value of the edge length of the unit cell gives the volume of the component in its unit cell.  Therefore, Volume of the billiard ball in its simple cubic unit cell is,

$\begin{array}{l}volume=l^3\\ \text{ = }{\left(2r\right)}^3\text{ = }{\left(2\times 2.86\text{ cm}\right)}^3\\ \text{ = 187}{\text{.15 cm}}^3\end{array}$

Also, mass of the billiard ball is given as $165\text{ g}\text{.}$

Each simple cubic unit cell has one billiard ball.  Therefore,  Mass of the billiard ball in its unit cell is $\text{165 g}\text{.}$

Hence the density of billiard ball in simple cubic unit cell is calculated as follows –

$\begin{array}{l}density\text{ = }\frac{mass}{volume}\\ \text{ = }\frac{\text{165 g}}{\text{187}\text{.15 }c{m}^3}\text{ = 0}{\text{.882 g/cm}}^3\end{array}$

To calculate:

Density of the billiard ball in its face centered cubic unit cell.

Edge length of the face centered cubic unit cell is given as,

$\begin{array}{l}\text{l = 2r}\sqrt{\text{2}}\\ \text{l = edge length; r = radius of the atom inside unit cell}\end{array}$

Cubic value of the edge length of the unit cell gives the volume of the component in its unit cell.  Therefore, Volume of the billiard ball in its simple cubic unit cell is,

$\begin{array}{l}volume={\left(2r\sqrt{2}\right)}^3\\ \text{ = }{\left(2\times 2.86\text{ cm }\times \text{1}\text{.414}\right)}^3\\ \text{ = 529}{\text{.08 cm}}^3\end{array}$

Also, mass of the billiard ball is given as $165\text{ g}\text{.}$

Each face centered cubic unit cell has four billiard balls.  Therefore, Mass of the billiard ball in its unit cell is $\text{165 g }\times \text{ 4 = 660 g}\text{.}$

Hence the density of billiard ball in face centered cubic unit cell is calculated as follows –

$\begin{array}{l}density\text{ = }\frac{mass}{volume}\\ \text{ = }\frac{\text{660 g}}{\text{529}\text{.08 }c{m}^3}\text{ = 1}{\text{.25 g/cm}}^3\end{array}$

To explain:

Density of the billiard ball is different in all these three cases.

Density of billiard ball when it is not intact with other billiard ball is different from those when close packed in certain fashion.  The billiard ball is spherical in shape and efficiency in packing of them is not always $100\%$.  The arrangement and close packing fashion differs in simple cubic and face-centered cubic lattice, so does the density of the component.

Conclusion

• Given the diameter and mass of a billiard ball its Density has been determined.
• By assuming the billiard balls are packed in simple cubic lattice, density of the billiard ball in its simple cubic unit cell has been determined.
• By assuming the billiard balls are packed in face centered cubic lattice, density of the billiard ball in its face centered cubic unit cell has been determined.