Question 1.19: VACANCY CONCENTRATION IN A SEMICONDUCTOR The energy of vacan...
VACANCY CONCENTRATION IN A SEMICONDUCTOR The energy of vacancy formation in the Ge crystal is about 2.2 eV. Calculate the fractional concentration of vacancies in Ge at 938 °C, just below its melting temperature. What is the vacancy concentration given that the atomic mass Mat and density \rho of Ge are 72.64 g mol^{−1} and 5.32 g cm^{−3}, respectively? Neglect the change in the density with temperature which is small compared with other approximations in Equation 1.42.
Using Equation 1.42, the fractional concentration of vacancies at 938 °C or 1211 K is
\frac{n_{v}}{N}=\exp \left(-\frac{E_{v}}{k T}\right)=\exp \left[-\frac{(2.2 eV )\left(1.6 \times 10^{-19} J eV ^{-1}\right)}{\left(1.38 \times 10^{-23} J K ^{-1}\right)(1211 K )}\right]=7.0 \times 10^{-10}
which is orders of magnitude less than that for Al at its melting temperature in Example 1.18; vacancies in covalent crystals cost much more energy than those in metals. The number of Ge atoms per unit volume is
N=\frac{\rho N_{A}}{M_{ at }}=\frac{\left(5.32 g cm ^{-3}\right)\left(6.022 \times 10^{23} g mol ^{-1}\right)}{72.64 g mol ^{-1}}=4.41 \times 10^{22} cm ^{-3}
so that at 938 °C,
n_{v}=\left(4.4 \times 10^{22} cm ^{-3}\right)\left(7.0 \times 10^{-10}\right)=3.1 \times 10^{13} cm ^{-3}
Only 1 in 10^{9} atoms is a vacancy. A better calculation would also consider the decrease in the atomic concentration N with temperature (due to the expansion of the crystal). The final n_{v} is still about 3 \times 10^{13} cm^{−3}.