Chapter 12, Problem 12.93AP

(a)

Interpretation Introduction

Interpretation: The unit cells present in a cubic silver particle with an edge length of $25\text{nm}$ and number of silver atoms in that particle is to be calculated. The number of particles present in $1360\text{μg}$ of silver per sock is to be stated.

Concept introduction: Silver crystallizes in a fcc crystal arrangement. This crystal structure has 8 atoms at each corner of the cube and 6 atoms at each face of the cube.

To determine: The number of unit cells are present in a cubic silver particle with an edge length of $25\text{nm}$.

Answer to Problem 12.93AP

Solution

There are $\underset{\_}{62unitcells}$ present in a cubic silver particle with an edge length of $2\text{5}\text{nm}$.

Explanation of Solution

Explanation

Silver crystallizes in face centered cubic (fcc) unit cells and the unit cell edge is $0.407\text{nm}$.

$\left(\begin{array}{l}\text{The number of unit cells present in cubic silver particle }\\ \text{of edge length}\text{0}\text{.407}\text{nm}\end{array}\right)=\text{1}\text{unit}\text{cell}$

Hence,

$\left(\begin{array}{l}\text{The number of unit cells present in cubic silver particle }\\ \text{of edge length}\text{1}\text{nm}\end{array}\right)=\left(\frac{\text{1}}{\text{0}\text{.407}}\right)\text{unit}\text{cell}$

Therefore,

$\begin{array}{c}\left(\begin{array}{l}\text{The number of unit cells present in cubic silver particle }\\ \text{of edge length}25\text{nm}\end{array}\right)=\frac{\text{25}}{\text{0}\text{.407}}\text{unit}\text{cells}\\ \approx \underset{\_}{62unitcells}\text{l}\end{array}$

(b)

Interpretation Introduction

To determine: The number of silver atoms is there in a cubic silver particle.

Answer to Problem 12.93AP

Solution

There are $\underset{\_}{248}$ silver atoms in a cubic silver particle.

Explanation of Solution

Explanation

Each corner has $\frac{1}{8}$ atoms in a unit cell; therefore 8 corners will have atoms $\begin{array}{c}=\left(\frac{1}{8}\times 8\right)\\ =1\text{atom}\end{array}$

Each face has $\frac{1}{2}$ atoms in a unit cell; therefore 6 faces will have atoms $\begin{array}{l}=\frac{1}{2}\times 6\\ =3\text{atoms}\end{array}$

Therefore, total atoms per unit cell $\begin{array}{l}=1+3\\ =4\text{atoms}\end{array}$

Thus, $62\text{unit}\text{cells}$ will have atoms $\begin{array}{l}=62\times 4\\ \underset{\_}{=248atoms}\end{array}$

(c)

Interpretation Introduction

To determine: The number of particles does $1360\text{μg}$ of silver per sock corresponds to.

Answer to Problem 12.93AP

Solution

There are $\underset{\_}{75.88\times 10^{20}}$ particles present in $1360\text{μg}$ of silver per sock.

Explanation of Solution

Explanation

Number of particles in $1\text{mole}$ of silver per sock $=6.023\times {10}^{23}$

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{weight}}{\text{Molecular}\text{weight}}$

Substitute the value of the given weight and molecular weight of silver in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{weight}}{\text{Molecular}\text{weight}}\\ =\frac{1360\times {10}^{-3}\text{g}}{107.86\text{g}}\\ =12.60\times {10}^{-3}\end{array}$

Number of particles in $12.60\times {10}^{-3}\text{moles}$ of silver per sock $\begin{array}{l}=6.023\times {10}^{23}\times 12.60\times {10}^{-3}\\ \underset{\_}{=75.88\times 10^{20}}\end{array}$

Conclusion

There are $\underset{\_}{62unitcells}$ present in a cubic silver particle with an edge length of $2\text{5}\text{nm}$ and $\underset{\_}{248}$ silver atoms in a cubic silver particle.

There are $\underset{\_}{75.88\times 10^{20}}$ particles present in $1360\text{μg}$ of silver per sock.