Chapter 12, Problem 12.138P

(a)

Interpretation Introduction

Interpretation:

The unit-cell edge length of the densest diamond is to be calculated.

Concept introduction:

Crystal structure or lattice is the three-dimensional representation of atoms and molecules arranged in a particular manner. The unit cell is the smallest part of the lattice that is repeated in all directions to yield the crystal lattice. There are 3 types of cubic unit cells as follows:

1. The simple cubic unit cell

2. Body-centered unit cell

3. Face-centered unit cell

In the cubic unit cell, atom at the corner is shared by 8 adjacent cells so the contribution of an atom at the corner is $\frac{1}{8}$, atom at the face is shared by 2 adjacent cells so the contribution of an atom at the face is $\frac{1}{2}$ and atom at the body center is shared by 1 cell so the contribution of an atom at the body center is $1$. The number of atoms present in a simple cubic, a body-centered cubic, and a face-centered cubic unit cell are 1, 2 and 4 respectively.

Answer to Problem 12.138P

The unit-cell edge length of the densest diamond is $3.57\times {10}^{-8}\text{cm}$.

Explanation of Solution

The formula to calculate the volume of one mole of carbon atom with a higher density is as follows:

$\text{Volume of one mole of C atom}=\left(\frac{\text{molar mass of C}}{\text{Density of C}}\right)$ (1)

Substitute the value of $12.01\text{g/mol}$ for molar mass of $\text{C}$ atom and $3.52{\text{ g/cm}}^{\text{3}}$ for the density of $\text{C}$ in the equation (1).

$\begin{array}{c}\text{Volume of one mole of C atom}=\left(\frac{12.01\text{g/mol}}{3.52{\text{ g/cm}}^{\text{3}}}\right)\\ =3.411931{\text{ cm}}^{\text{3}}\text{/mol}\end{array}$

Diamond has a face-centered cubic unit cell. Face-centered cubic unit cell, 8 atoms are present at the corners of the cell and 6 atoms at the face of the cell. The contribution of an atom present at the corner is $\frac{1}{8}$ and the contribution of an atom at the face of a cell is $\frac{1}{2}$. Therefore the number of an atom in the face-centered cubic unit cell is calculated as follows:

$\begin{array}{c}\text{Number of atoms}=\left(\frac{1}{8}\text{atom per cell}\right)\left(\text{8}\text{atoms}\right)+\left(\frac{1}{2}\text{atom per cell}\right)\left(6\text{atom}\right)\\ =1\text{atom}+3\text{atoms}\\ =4\text{atoms}\end{array}$

There are 4 atoms at the tetrahedral voids and therefore, the diamond has 8 carbon atoms in the lattice.

The formula to calculate the volume of the unit cell is as follows:

$\text{Volume of unit cell}=\left[\begin{array}{l}\left(\text{Volume of one mole of}\text{C atom}\right)\\ \left(\frac{1\text{mol}\text{C}}{6.022\times {10}^{23}\text{C}\text{atoms}}\right)\left(\frac{8\text{C atom}}{1\text{unit cell}}\right)\end{array}\right]$ (2)

Substitute the value of $3.411931{\text{ cm}}^{\text{3}}\text{/mol}$ for volume of one mole of $\text{C}$ atom in the equation (2).

$\begin{array}{c}\text{Volume of unit cell}=\left[\begin{array}{l}\left(3.411931{\text{ cm}}^{\text{3}}\text{/mol}\right)\\ \left(\frac{1\text{mol}\text{C}}{6.022\times {10}^{23}\text{C}\text{atoms}}\right)\left(\frac{8\text{C atom}}{1\text{unit cell}}\right)\end{array}\right]\\ =4.56326228\times {10}^{-23}{\text{cm}}^3\end{array}$

The formula to calculate edge length of the cube is as follows:

$\text{Edge length of the cube}=\sqrt[3]{\text{Volume of unit cell}}$ (3)

Substitute the value of $4.56326228\times {10}^{-23}{\text{cm}}^3$ for volume of the unit cell in the equation (3).

$\begin{array}{c}\text{Edge length of the cube}=\sqrt[3]{4.56326228\times {10}^{-23}{\text{cm}}^3}\\ =3.5654679\times {10}^{-8}\text{cm}\\ \approx 3.57\times {10}^{-8}\text{cm}\end{array}$

Conclusion

The unit-cell edge length of the densest diamond is $3.57\times {10}^{-8}\text{cm}$.

(b)

Interpretation Introduction

The number of C atoms present diamond unit cell with the lowest density is to be calculated.

Concept introduction:

Crystal structure or lattice is the three-dimensional representation of atoms and molecules arranged in a particular manner. The unit cell is the smallest part of the lattice that is repeated in all directions to yield the crystal lattice. There are 3 types of cubic unit cells as follows:

1. The simple cubic unit cell

2. Body-centered unit cell

3. Face-centered unit cell

In the cubic unit cell, atom at the corner is shared by 8 adjacent cells so the contribution of an atom at the corner is $\frac{1}{8}$, atom at the face is shared by 2 adjacent cells so the contribution of an atom at the face is $\frac{1}{2}$ and atom at the body center is shared by 1 cell so the contribution of an atom at the body center is $1$. The number of atoms present in a simple cubic, a body-centered cubic, and a face-centered cubic unit cell are 1, 2 and 4 respectively.

Answer to Problem 12.138P

The number of C atoms present diamond unit cell with the lowest density is $6.84\text{C}\text{atoms}$.

Explanation of Solution

The formula to calculate the volume of one mole of carbon atom with lower density is as follows:

$\text{Volume of one mole of C atom}=\left(\frac{\text{molar mass of C}}{\text{Density of C}}\right)$ (4)

Substitute the value of $12.01\text{g/mol}$ for molar mass of $\text{C}$ atom and $3.01{\text{ g/cm}}^{\text{3}}$ for the density of $\text{C}$ in the equation (4).

$\begin{array}{c}\text{Volume of one mole of C atom}=\left(\frac{12.01\text{g/mol}}{3.01{\text{ g/cm}}^{\text{3}}}\right)\\ =3.9900{\text{ cm}}^{\text{3}}\text{/mol}\end{array}$

The formula to calculate the volume of the unit cell is as follows:

$\text{Volume of unit cell}={\left(A\right)}^3$ (5)

Substitute $3.3058\stackrel{\text{o}}{\text{A}}$ for $A$ in the equation (5).

$\begin{array}{c}\text{Volume of unit cell}={\left(3.5654679\times {10}^{-8}\text{cm}\right)}^3\\ =12.712561\times {10}^{-24}{\text{cm}}^{\text{3}}\end{array}$

The formula to calculate the number of C atoms with lower density is as follows:

$\text{Number of C atom}=\left[\begin{array}{l}\left(\frac{\text{Volume of unit cell}}{\text{Volume of one mole of C atom}}\right)\\ \left(\frac{6.022\times {10}^{23}\text{C}\text{atoms}}{1\text{mol}\text{C}}\right)\end{array}\right]$ (6)

Substitute $12.712561\times {10}^{-24}{\text{cm}}^{\text{3}}$ for the volume of unit cell and $3.9900{\text{ cm}}^{\text{3}}\text{/mol}$ for the volume of one mole of carbon atom in the equation (6).

$\begin{array}{c}\text{Number of C atom}=\left[\begin{array}{l}\left(\frac{12.712561\times {10}^{-24}{\text{cm}}^{\text{3}}}{3.9900{\text{ cm}}^{\text{3}}\text{/mol}}\right)\\ \left(\frac{6.022\times {10}^{23}\text{C}\text{atoms}}{1\text{mol}\text{C}}\right)\end{array}\right]\\ =6.8409\text{C}\text{atoms}\\ \approx 6.84\text{C}\text{atoms}\end{array}$

Conclusion

The number of C atoms present diamond unit cell with the lowest density is $6.84\text{C}\text{atoms}$.