Question

Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at 573 oC (this temperature is below the melting temperature (645oC)). Assume an energy for defect formation of 1.86 eV.

Answer 1

Answer:

$2.9\times 10^{-6}$

Explanation:

$Q_s$ = Energy for defect formation = 1.86 eV

T = Temperature = $573^{\circ}\text{C}=573+273.15=846.15\ \text{K}$

k = Boltzmann constant = $8.62\times 10^{-5}\ \text{eV/K}$

The fraction of lattice sites that are Schottky defects is given by

$\dfrac{N_s}{N}=e^{-\dfrac{Q_s}{2kt}}\\\Rightarrow \dfrac{N_s}{N}=e^{-\dfrac{1.86}{2\times 8.62\times 10^{-5}\times 846.15}}\\\Rightarrow \dfrac{N_s}{N}=2.9\times 10^{-6}$

The required ratio is $2.9\times 10^{-6}$.