Chapter 4, Problem 4.1P

Gold has $5.82\times {10}^{8}vacancies/c{m}^3$ at equilibrium at 300 K. What fraction of the atomic sites is vacant at 600 K?

To determine

The fraction of atomic sites which are vacant in gold.

Answer to Problem 4.1P

The vacant fraction of atomic sites in gold at $600\text{ K}$ is $0.00091$ .

Explanation of Solution

Given information:

At $300\text{ K}$ ,

The number of vacancies is $5.82\times {10}^8{\text{ vacancies/cm}}^3$ .

Final temperature is $600\text{ K}$ .

Concept used:

Write the expression for the number of vacancies.

  $n_v=n\times \text{exp}\left(\frac{-Q_v}{RT}\right)$ ...............(1)

Here,$Q_v$ is the energy required in $\text{cal/mol}$ , $n$ is the number of atoms per unit volume, $T$ is the temperature in Kelvin and $R$ is the universalgas constant.

Calculation:

Use FCC lattice parameter for goldas $4.065\text{ <mover accent="true"><mo>A</mo><mo>˙</mo></mover>}$ .

For $1$ cell, the number of gold atoms per ${\text{cm}}^{\text{3}}$

is calculated as follows:

  $\begin{array}{c}n=\frac{4\text{ atoms/cell}}{{\left(4.065\times {10}^{-8}\text{ cm}\right)}^3}\\ =5.95\times {10}^{22}{\text{ atoms/cm}}^3\end{array}$

The temperature is given as $300\text{ K}$ .

Universal gas constant is $1.987\frac{\text{cal}}{\text{mol}\cdot \text{K}}$ .

Calculate the activation energy for goldusing equation (1) as follows:

  $\begin{array}{c}{n}_v=n\times \text{exp}\left(\frac{-Q_v}{RT}\right)\\ 5.82\times {10}^8\frac{\text{vacancies}}{{\text{cm}}^3}=\left(5.95\times {10}^{22}\frac{\text{atoms}}{{\text{cm}}^3}\right)\exp \left(\frac{-Q_v}{\left(1.987\frac{\text{cal}}{\text{mol}\cdot \text{K}}\right)\left(300\text{ K}\right)}\right)\\ 0.978\times {10}^{-14}=\exp \left(\frac{-Q_v}{\left(1.987\right)\left(300\right)}\right)\\ Q_v=8351.159\text{ cal/mol}\end{array}$

Calculate the fraction of vacant atomic sites $\left(n_v/n\right)$ in gold at $600\text{ K}$

by using value of activation energyas follows:

  $\begin{array}{c}\frac{n_v}{v}=\text{exp}\left(\frac{-Q_v}{RT}\right)\\ =\text{exp}\left(\frac{-Q_v}{RT}\right)\\ =\exp \left(\frac{-8351.159\text{ cal/mol}}{\left(1.987\frac{\text{cal}}{\text{mol}\cdot \text{K}}\right)\left(600\text{ K}\right)}\right)\\ =0.00091\end{array}$

The vacantfraction of atomic sites in gold at $600\text{ K}$ is $0.00091$ .