Question 1.18: VACANCY CONCENTRATION IN A METAL The energy of formation of ...
VACANCY CONCENTRATION IN A METAL The energy of formation of a vacancy in the aluminum crystal is about 0.70 eV. Calculate the fractional concentration of vacancies in Al at room temperature, 300 K, and very close to its melting temperature 660 °C. What is the vacancy concentration at 660 °C given that the atomic concentration in Al is about 6.0 \times 10^{22} cm^{−3}?
Using Equation 1.42, the fractional concentration of vacancies are as follows: At 300 K,
\begin{aligned} \frac{n_{v}}{N}=\exp \left(-\frac{E_{v}}{k T}\right) &=\exp \left[-\frac{(0.70 eV )\left(1.6 \times 10^{-19} J eV ^{-1}\right)}{\left(1.38 \times 10^{-23} J K ^{-1}\right)(300 K )}\right] \\ &=1.7 \times 10^{-12} \end{aligned}
At 660 °C or 933 K,
\begin{aligned} \frac{n_{v}}{N}=\exp \left(-\frac{E_{v}}{k T}\right) &=\exp \left[-\frac{(0.70 eV )\left(1.6 \times 10^{-19} J eV ^{-1}\right)}{\left(1.38 \times 10^{-23} J K ^{-1}\right)(933 K )}\right] \\ &=1.7 \times 10^{-4} \end{aligned}
That is, almost 1 in 6000 atomic sites is a vacancy. The atomic concentration N in Al is about 6.0 \times 10^{22} cm^{−3}, which means that the vacancy concentration n_{v} at 660 °C is
n_{v}=\left(6.0 \times 10^{22} cm ^{-3}\right)\left(1.7 \times 10^{-4}\right)=1.0 \times 10^{19} cm ^{-3}
The mean vacancy separation (on the order of n_{V}^{-1 / 3}) at 660 °C is therefore roughly 5 nm. The mean atomic separation in Al is ∼0.3 nm ( \sim N^{-1 / 3}), so the mean separation between vacancies is only about 20 atomic separations! (A more accurate version of Equation 1.42, with an entropy term, shows that the vacancy concentration is even higher than the estimate in this example.) The increase in the linear thermal expansion coefficient of a metal with temperature near its melting temperature, as shown for Mo in Figure 1.20, has been attributed to the generation of vacancies in the crystal.
