Answer
$N_{fr} = 8.24 \times 10^{23} defects/m^{3}$
Work Step by Step
Required:
Number of Frenkel defects per cubic meter in silver chloride at 350°C. The energy for defect formation is 1.1 eV, whereas the density for AgCl is 5.50 g/cm3 at 350°C.
Solution:
Using Equation 12.2 but determining first the number of lattice sites per cubic meter (N) using the modified form of Equation 4.2,
$N = \frac{N_{A} ρ}{A_{Ag} + A_{Cl}} = \frac{(6.022 \times 10^{23} atoms/mol) (5.50 g/cm^{3} (10^{6} \frac{cm^{3}}{m^{3}})}{107.87 g/mol + 35.45g/mol} = 2.31 \times 10^{28} lattice sites/m^{3}$
$N_{fr} = e^{ (-\frac{Q_{fr}}{2kT}) }= (2.31 \times 10^{28} lattice sites/m^{3})e^{ (-\frac{1.1 eV}{2(8.62 \times 10^{-5} eV/K) (350 +273 K)}) }\\= 8.24 \times 10^{23} defects/m^{3}$
