Chapter 5.7, Problem 12AAP

(a)

To determine

The equilibrium concentration of vacancies per cubic meter in pure silver.

Answer to Problem 12AAP

The equilibrium concentration of vacancies per cubic meter in pure silver is $2.24\times 1{\text{0}}^{\text{23}}\text{vacancies}/{\text{m}}^{\text{3}}$.

Explanation of Solution

Express the equilibrium concentration of vacancies per cubic meter in pure copper.

 $n_v=NC{e}^{-E_v/kT}$                                                                           (I)

Here, number of vacancies per cubic meter of metal is $n_v$, total number of atom sites per cubic meter of metal is $N$, activation energy to form a vacancy is $E_v\left(\text{eV}\right)$, absolute temperature is $T$, Boltzmann’s constant is $k$ and constant is $C$.

Express total number of atom sites per cubic meter of metal.

 $N=\frac{N_0{\rho }_{\text{Ag}}}{\text{at}\text{.}\text{mass}\text{Ag}}$                                                                          (II)

Here, Avogadro’s number is $N_0$ and density of silver is ${\rho }_{\text{Ag}}$.

Conclusion:

Write the value of Avogadro’s number, atomic mass and density of silver.

 $\begin{array}{l}{N}_0=6.02\times {10}^{23}\text{atoms}\\ {\rho }_{\text{Ag}}=10.5\times {10}^6\text{g}/{\text{m}}^{\text{3}}\\ \text{at}\text{.}\text{mass}\text{Ag}=107.87\text{g}/\text{at}\text{.}\text{mass}\end{array}$

Write the value of Boltzmann’s constant.

 $k=8.62\times {10}^{-5}\text{eV}/\text{K}$

Substitute $6.02\times {10}^{23}\text{atoms}$ for $N_0$, $10.5\times {10}^6\text{g}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{Ag}}$ and $107.87\text{g}/\text{at}\text{.}\text{mass}$ for $\text{at}\text{.}\text{mass}\text{Ag}$ in Equation (II).

 $\begin{array}{c}N=\frac{\left(6.02\times {10}^{23}\text{atoms}\right)\left(10.5\times {10}^6\text{g}/{\text{m}}^{\text{3}}\right)}{107.87\text{g}/\text{at}\text{.}\text{mass}}\\ =5.86\times {10}^{28}\text{atoms}/{\text{m}}^{\text{3}}\end{array}$

Substitute $5.86\times {10}^{28}\text{atoms}/{\text{m}}^{\text{3}}$ for $N$, $8.62\times {10}^{-5}\text{eV}/\text{K}$ for $k$, $\text{750°C}$ for $T$ and $\text{1}\text{.10}\text{eV}$ for $E_v$ in Equation (I).

 $\begin{array}{c}{n}_v=\left(5.86\times {10}^{28}\text{atoms}/{\text{m}}^{\text{3}}\right)\left\{\exp \left[-\frac{\text{1}\text{.10}\text{eV}}{\left(8.62\times {10}^{-5}\text{eV}/\text{K}\right)\left(\text{750°C}\right)}\right]\right\}\\ =\left(5.86\times {10}^{28}\text{atoms}/{\text{m}}^{\text{3}}\right)\left\{\exp \left[-\frac{\text{1}\text{.10}\text{eV}}{\left(8.62\times {10}^{-5}\text{eV}/\text{K}\right)\left[\left(\text{750}+273\right)\text{K}\right]}\right]\right\}\\ =\left(5.86\times {10}^{28}\text{atoms}/{\text{m}}^{\text{3}}\right)\left\{\exp \left[-\frac{\text{1}\text{.10}\text{eV}}{\left(8.62\times {10}^{-5}\text{eV}/\text{K}\right)\left(1023\text{K}\right)}\right]\right\}\\ =2.24\times 1{\text{0}}^{\text{23}}\text{vacancies}/{\text{m}}^{\text{3}}\end{array}$

Hence, the equilibrium concentration of vacancies per cubic meter in pure silver is $2.24\times 1{\text{0}}^{\text{23}}\text{vacancies}/{\text{m}}^{\text{3}}$.

(b)

To determine

The vacancy fraction at 700°C.

Answer to Problem 12AAP

The vacancy fraction at 8700°C is $2.01\times 1{\text{0}}^{-6}\text{vacancies}/\text{atom}$.

Explanation of Solution

Express the vacancy fraction at 700°C.

 $\frac{n_v}{N}=e^{-E_v/kT}$                                                                                         (III)

Conclusion:

Substitute $8.62\times {10}^{-5}\text{eV}/\text{K}$ for $k$, $\text{700°C}$ for $T$ and $\text{1}\text{.10}\text{eV}$ for $E_v$ in Equation (III).

 $\begin{array}{c}\frac{n_v}{N}=\exp \left[-\frac{\text{1}\text{.10}\text{eV}}{\left(8.62\times {10}^{-5}\text{eV}/\text{K}\right)\left(\text{700°C}\right)}\right]\\ =\exp \left[-\frac{\text{1}\text{.10}\text{eV}}{\left(8.62\times {10}^{-5}\text{eV}/\text{K}\right)\left(\text{973}\text{K}\right)}\right]\\ =e^{-13.12}\\ =2.01\times 1{\text{0}}^{-6}\text{vacancies}/\text{atom}\end{array}$

Hence, the vacancy fraction at 700°C is $2.01\times 1{\text{0}}^{-6}\text{vacancies}/\text{atom}$.