Answer
$\frac{N_{s}}{N} = 7.87 \times 10^{-6}$
Work Step by Step
Required:
Fraction of lattice sites that are Schottky defects for cesium chloride at its melting temperature (645°C), assuming an energy for defect formation of 1.86 eV.
Solution:
Using Equation 12.3 and solving for $\frac{N_{s}}{N}$ ratio:
$\frac{N_{s}}{N} = exp (-\frac{Q_{s}}{2kT}) = exp (-\frac{1.86 eV}{(2)(8.62 \times 10^{-5} eV/K)(645 + 273 K)}) = 7.87 \times 10^{-6}$
