Chapter 12, Problem 12.94AP

(a)

Interpretation Introduction

Interpretation: The diameter and the length of the nonorod of gold is given. The approximate number of atoms of gold present in each nanorod; the number of unit cells corresponding to each nanorod and the mass of each nanorod is to be calculated.

Concept introduction: The volume of the cylinder is calculated by the formula,

$\text{Volume}=\frac{\text{π}\times {\left(\text{Diameter}\right)}^2\times \text{Height}}{4}$

The volume of the sphere is calculated by the formula,

$\text{Volume}=\frac{\text{4π}\times {\left(\text{Radius}\right)}^3}{3}$

To determine: The approximate number of atoms of gold present in each nanorod.

Answer to Problem 12.94AP

Solution:

The approximate number of atoms of gold present in each nanorodis $\underset{\_}{2.835\times 10^6}$.

Explanation of Solution

Given

The diameter of the nanorodis $32\text{nm}$.

The height of the nanorodis $67\text{nm}$.

The volume of the cylinder is calculated by the formula,

$\text{Volume}=\frac{\text{π}\times {\left(\text{Diameter}\right)}^2\times \text{Height}}{4}$

Substitute the value of diameter and height in the above equation.

$\begin{array}{c}\text{Volume}=\frac{\text{3}\text{.14}\times {\left(32\text{nm}\right)}^2\times 67\text{nm}}{4}\\ =53.86\times {10}^3{\text{nm}}^3\end{array}$

The radius of the gold atom is $0.166\text{nm}$.

The volume of the sphere is calculated by the formula,

$\text{Volume}=\frac{\text{4π}\times {\left(\text{Radius}\right)}^3}{3}$

Substitute the value of the radius of gold sphere in the above equation.

$\begin{array}{c}\text{Volume}=\frac{\text{4}\times \text{3}\text{.14}\times {\left(0.166\text{nm}\right)}^3}{3}\\ =0.019{\text{nm}}^3\end{array}$

Therefore, the number of gold atoms present in each nanorod is,

$\frac{53.86\times {10}^3{\text{nm}}^3}{0.019{\text{nm}}^3}=\underset{\_}{2.835\times 10^6}$

(b)

Interpretation Introduction

To determine: The number of unit cell of gold corresponding to each nanorod.

Answer to Problem 12.94AP

Solution:

The number of unit cell of gold corresponding to each nanorodis $\underset{\_}{7.09\times 10^5}$.

Explanation of Solution

The unit cell of gold is face centered.

Therefore, there are four atoms of gold present in each unit cell.

The number of unit cell of gold corresponding to each nanorodis $2.835\times {10}^6$.

Therefore, the number of unit cell corresponding to each unit cell is,

$\frac{2.835\times {10}^6}{4}=\underset{\_}{7.09\times 10^5}$

(c)

Interpretation Introduction

To determine: The mass of each nanorod.

Answer to Problem 12.94AP

Solution:

The mass of each nanorodis $\underset{\_}{9.27\times 10^{-16}g}$.

Explanation of Solution

The mass of one mole of gold is $196.9666\text{g}$.

The mass of each atom is calculated by the formula,

$\text{Mass}\text{of}\text{atom}=\frac{\text{Mass}\text{of}\text{one}\text{mole}}{\text{Avogadro's}\text{Number}}$

Substitute the value of mass of one mole of gold and Avogadro’s number as $6.023\times {10}^{23}$ in the mass of atom formula.

$\begin{array}{c}\text{Mass}\text{of}\text{atom}=\frac{196.9666\text{g}}{6.023\times {10}^{23}}\\ =32.7\times {10}^{-23}\text{g}\end{array}$

The number of unit cell of gold corresponding to each nanorodis $2.835\times {10}^6$.

Therefore, the mass of each nanorod is,

$32.7\times {10}^{-23}\text{g}\times 2.835\times {10}^6=\underset{\_}{9.27\times 10^{-16}g}$

Conclusion:

The approximate number of atoms of gold present in each nanorodis $\underset{\_}{2.835\times 10^6}$.

The number of unit cell of gold corresponding to each nanorodis $\underset{\_}{7.09\times 10^5}$.

The mass of each nanorodis $\underset{\_}{9.27\times 10^{-16}g}$.