Chapter 4, Problem 1QAP

To determine

The number of vacancies per cubic meter in Gold.

Answer to Problem 1QAP

The number of vacancies: $N_V=14.1\times {10}^{23}\text{vacancies}/{\text{m}}^{\text{3}}$.

Explanation of Solution

Given Information:

The equilibrium fraction of lattice sites at ${800}^{\circ }\text{C}$

  $=2.5\times {10}^{-5}$ .

The density of gold: ${\rho }_{Fe}=18.45\text{g}/{\text{cm}}^{\text{3}}$ .

The temperature: $T={800}^{°}\text{C}$ .

From the table, find the atomic weight of Gold i.e. $A_{Au}=196.97\text{g}/\text{mol}$ .

First, calculate the total number of lattice sites in gold using the relation:

  $\begin{array}{l}{N}_{Au}=\frac{N_A{\rho }_{Au}}{A_{Au}}\\ \text{Where,}\\ N_A\to \text{Avogadro's}\text{number = }6.022\times {10}^{23}\text{atoms/mol}\end{array}$

Calculation:

  $\begin{array}{l}{N}_{Au}=\frac{N_A{\rho }_{Au}}{A_{Au}}\\ \text{Where,}\\ N_A\to \text{Avogadro's}\text{number = }6.022\times {10}^{23}\text{atoms/mol}\\ \text{Substituting}\text{values;}\\ N_{Au}=\frac{\left(6.022\times {10}^{23}\text{atoms/mol}\right)\left(18.45g/c{m}^3\right)}{196.97\text{g}/\text{mol}}\\ N_{Au}=5.64\times {10}^{22}\text{atoms}/{\text{cm}}^{\text{3}}\\ or,\\ N_{Au}=5.64\times {10}^{28}\text{atoms}/{\text{m}}^{\text{3}}\end{array}$

Therefore, the number of vacancies in gold will be determined as:

  $\begin{array}{l}{N}_V=\left(2.5\times {10}^{-5}\right)N_{Au}\\ or,\\ N_V=\left(2.5\times {10}^{-5}\right)\left(5.64\times {10}^{28}\text{atoms}/{\text{m}}^{\text{3}}\right)\\ \Rightarrow N_V=14.1\times {10}^{23}\text{vacancies}/{\text{m}}^{\text{3}}\end{array}$