Chapter 4, Problem 4.1P

Interpretation Introduction

Interpretation:

The fraction of atomic sites which are vacant in gold at given conditions is to be determined.

Concept introduction:

The number of vacancies per ${\text{cm}}^3$ is calculated by the formula;

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

Here, $n$ is the number of atoms per ${\text{cm}}^3$, $Q_v$ is the energy required to deliver one mole of vacancy in $\text{cal/mol}$, $T$ is the temperature in Kelvin, and $R$ is the universalgas constant.

Answer to Problem 4.1P

The fraction of atomic sites which are vacant in gold at given conditions is $0.00091$.

Explanation of Solution

Given information:

At $300\text{ K}$, the number of vacancies in gold at equilibrium is $5.82\times {10}^8{\text{ vacancies/cm}}^3$. Fraction of vacant atomic sites are to be calculated at $600\text{ K}$.

The lattice parameter for gold with Face-centered cubic crystal structure is $4.065\text{ <mover accent="true"><mo>A</mo><mo>˙</mo></mover>}$.

For $1$ cell, the number of gold atoms per ${\text{cm}}^{\text{3}}$ is calculated as shown below if there are $4$ gold atoms per cell is:

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

The temperature is given as $300\text{ K}$ .Universal gas constant to be used is $1.987\frac{\text{cal}}{\text{mol}\cdot \text{K}}$.

Using equation (1) to calculate the activation energy for gold as:

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

Using this calculated value of activation energy, calculate the fraction of vacant atomic sites $\left(n_v/n\right)$ in gold at $600\text{ K}$ as:

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

Conclusion

The fraction of atomic sites which are vacant in gold at $600\text{ K}$ is calculated to be $0.00091$.