Chapter 12, Problem 12.98P

(a)

Interpretation Introduction

Interpretation:

The number of $\text{Zn}$ and $\text{Se}$ ions in each unit cell 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 12.98P

The number of $\text{Zn}$ ions is 4 and $\text{Se}$ ions in each unit cell are 4.

Explanation of Solution

The structure of zinc selenide $\left(\text{ZnSe}\right)$ is similar to the zinc blende structure. Therefore selenide ions are present at the corners of the cell and the faces 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 selenide ions is calculated as follows:

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

The charge ratio between the ions in $\text{ZnSe}$ is $2:2$ and therefore, the number of zinc ions is $4$. Zinc ions are present in half of the tetrahedral voids.

Conclusion

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 $1$.

(b)

Interpretation Introduction

The mass of a unit cell 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 12.98P

The mass of a unit cell is $577.48\text{amu}$.

Explanation of Solution

The number of $\text{Zn}$ ions is 4 and $\text{Se}$ ions in each unit cell are 4.

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

$\text{Mass}\text{of}\text{unit}\text{cell}=\left(4\right)\left(\text{mass}\text{of}\text{Zn}\text{atom}\right)+\left(4\right)\left(\text{mass}\text{of}\text{Se}\text{atom}\right)$ (1)

Substitute $65.41\text{ amu}$ for the mass of $\text{Zn}$ atom and $78.96\text{ amu}$ for the mass of $\text{Se}$ atom in the equation (1).

$\begin{array}{c}\text{Mass}\text{of}\text{unit}\text{cell}=\left(4\right)\left(65.41\text{ amu}\right)+\left(4\right)\left(78.96\text{ amu}\right)\\ =577.48\text{amu}\end{array}$

Conclusion

The mass of a unit cell is $577.48\text{amu}$.

(c)

Interpretation Introduction

The volume of a unit cell 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.

The conversion factor to convert $\text{amu}$ to $\text{g}$ is as follows:

$\text{1}\text{amu}=1.66054\times {10}^{-24}\text{ g}$

Answer to Problem 12.98P

The volume of a unit cell is $1.77\times {10}^{-22}{\text{cm}}^{\text{3}}$.

Explanation of Solution

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

$\text{Volume of unit cell}=\left(\frac{\text{mass of}\text{unit cell}}{\text{Density}\text{of solid}}\right)$ (2)

Substitute $577.48\text{amu}$ for the mass of unit cell and $5.42{\text{g/cm}}^{\text{3}}$ for the density of solid in the equation (2).

$\begin{array}{c}\text{Volume of unit cell}=\left(\frac{577.48\text{amu}}{5.42{\text{g/cm}}^{\text{3}}}\right)\left(\frac{1.66054\times {10}^{-24}\text{ g}}{\text{1}\text{amu}}\right)\\ =1.76924\times {10}^{-22}{\text{cm}}^{\text{3}}\\ \approx 1.77\times {10}^{-22}{\text{cm}}^{\text{3}}\end{array}$

Conclusion

The volume of a unit cell is $1.77\times {10}^{-22}{\text{cm}}^{\text{3}}$.

(d)

Interpretation Introduction

The edge length of a unit cell 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 12.98P

The edge length of a unit cell is $5.61\times {10}^{-8}\text{cm}$.

Explanation of Solution

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

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

Rearrange the equation (3) t calculate edge length as follows:

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

Substitute $1.76924\times {10}^{-22}{\text{cm}}^{\text{3}}$ for the volume of the unit cell in the equation (3).

$\begin{array}{c}\text{Edge length of cube}=\sqrt[3]{1.76924\times {10}^{-22}{\text{cm}}^{\text{3}}}\\ =5.6139\times {10}^{-8}\text{cm}\\ \approx 5.61\times {10}^{-8}\text{cm}\end{array}$

Conclusion

The edge length of a unit cell is $5.61\times {10}^{-8}\text{cm}$.