Chapter 12, Problem 12.99P

(a)

Interpretation Introduction

Interpretation:

The number of atoms 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 12.99P

The number of atoms in each unit cell is 4.

Explanation of Solution

Atom adopts face-centered cubic unit arrangement.

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 silver atoms in the face-centered cubic unit cell is calculated as follows:

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

Conclusion

The number of atoms in each unit cell is 4.

(b)

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.

Answer to Problem 12.99P

The volume of a unit cell is $9.23\times {10}^{-23}{\text{cm}}^3$.

Explanation of Solution

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

$\text{Volume}\text{of}\text{unit}\text{cell}={\left(\text{edge length of unit cell}\right)}^3$ (1)

Substitute $4.52\times {10}^{-8}\text{cm}$ for edge length of the unit cell in the equation (1).

$\begin{array}{c}\text{Volume}\text{of}\text{unit}\text{cell}={\left(4.52\times {10}^{-8}\text{cm}\right)}^3\\ =9.23454\times {10}^{-23}{\text{cm}}^3\\ \approx 9.23\times {10}^{-23}{\text{cm}}^3\end{array}$

Conclusion

The volume of a unit cell is $9.23\times {10}^{-23}{\text{cm}}^3$.

(c)

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 12.99P

The mass of a unit cell is $1.34\times {10}^{-22}\text{g}$.

Explanation of Solution

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

$\text{Mass of}\text{unit cell}=\left(\text{Volume of unit cell}\right)\left(\text{Density}\text{of solid}\right)$ (3)

Substitute $9.23454\times {10}^{-23}{\text{cm}}^3$ for the volume of the unit cell and $1.45{\text{g/cm}}^{\text{3}}$ for the density of solid in the equation (3).

$\begin{array}{c}\text{Mass of}\text{unit cell}=\left(9.23454\times {10}^{-23}{\text{cm}}^3\right)\left(1.45{\text{g/cm}}^{\text{3}}\right)\\ =1.3390\times {10}^{-22}\text{g}\\ \approx 1.34\times {10}^{-22}\text{g}\end{array}$

Conclusion

The mass of a unit cell is $1.34\times {10}^{-22}\text{g}$.

(d)

Interpretation Introduction

The approximate atomic mass for the element 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 12.99P

The approximate atomic mass for the element is $20.2\text{amu/atom}$.

Explanation of Solution

One unit cell consists of 4 atoms.

The formula to calculate the mass of an atom is as follows:

$\text{Mass}\text{of}\text{atom}=\left(\text{mass}\text{of}\text{unit}\text{cell}\right)\left(\frac{\text{1}\text{unit}\text{cell}}{\text{4}\text{atoms}}\right)$ (4)

Substitute $1.3390\times {10}^{-22}\text{g}$ for the mass of the unit cell in the equation (4).

$\begin{array}{c}\text{Mass}\text{of}\text{atom}=\left(\frac{1.3390\times {10}^{-22}\text{g}}{\text{1}\text{unit}\text{cell}}\right)\left(\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\right)\left(\frac{1.66054\times {10}^{-24}\text{ g}}{\text{1}\text{amu}}\right)\left(\frac{\text{1}\text{unit}\text{cell}}{\text{4}\text{atoms}}\right)\\ =20.1592\text{amu/atom}\\ \approx 20.2\text{amu/atom}\end{array}$

Conclusion

The approximate atomic mass for the element is $20.2\text{amu/atom}$.