Question 19.6: Ionic Conduction in MgO Suppose that the electrical conducti......
Ionic Conduction in MgO
Suppose that the electrical conductivity of MgO is determined primarily by the diffusion of the Mg^{2+} ions. Estimate the mobility of the Mg^{2+} ions and calculate the electrical conductivity of MgO at 1800°C. The diffusion coefficient of Mg^{2+} ions in MgO at 1800°C is 10^{-10} cm²/s.
For MgO, Z = 2/ion, q = 1.6 × 10^{-19} C, k_{B} = 1.38 × 10^{-23} J/K, and T = 2073 K:
μ = \frac{ZqD}{k_{B}T} = \frac{(2)(1.6 × 10^{-19} \ C)(10^{-10} \ cm^2/s)}{(1.38 × 10^{-23} \ J/K)(2073 \ K)} = 1.12 × 10^{-9} C ⋅ cm^2/(J ⋅ s)
Since one coulomb is equivalent to one ampere ? second, and one joule is equivalent to one ampere ⋅ second ⋅ volt:
μ = 1.12 × 10^{-9} cm²/(V ⋅ s)
MgO has the NaCl structure with four magnesium ions per unit cell. The lattice parameter is 3.96 × 10^{-8} cm, so the number of Mg^{2+} ions per cubic centimeter is
n = \frac{(4 \ Mg^{2+} \ ions/cell)}{(3.96 × 10^{-8} \ cm)^3/cell} = 6.4 × 10^{22} \ ions/cm^3
σ = nZqμ = (6.4 × 10^{22})(2)(1.6 × 10^{-9})(1.12 × 10^{-9})
= 23 × 10^{-6} C ⋅ cm^2/(cm^3 ⋅ V ⋅ s)
Since one coulomb is equivalent to one ampere ⋅ second (A ⋅ s) and one volt is equivalent to one ampere ⋅ ohm (A ⋅ V),
σ = 2.3 × 10^{-5} \ ohm^{-1} ⋅ cm^{-1}