Answer:
[tex]2.9\times 10^{-6}[/tex]
Explanation:
[tex]Q_s[/tex] = Energy for defect formation = 1.86 eV
T = Temperature = [tex]573^{\circ}\text{C}=573+273.15=846.15\ \text{K}[/tex]
k = Boltzmann constant = [tex]8.62\times 10^{-5}\ \text{eV/K}[/tex]
The fraction of lattice sites that are Schottky defects is given by
[tex]\dfrac{N_s}{N}=e^{-\dfrac{Q_s}{2kt}}\\\Rightarrow \dfrac{N_s}{N}=e^{-\dfrac{1.86}{2\times 8.62\times 10^{-5}\times 846.15}}\\\Rightarrow \dfrac{N_s}{N}=2.9\times 10^{-6}[/tex]
The required ratio is [tex]2.9\times 10^{-6}[/tex].
