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

`2.9* 10^(-6)`

Step-by-step 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* 10^(-5)\ \text{eV/K}`

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

`(N_s)/(N)=e^{-(Q_s)/(2kt)}\\\Rightarrow (N_s)/(N)=e^{-(1.86)/(2* 8.62* 10^(-5)* 846.15)}\\\Rightarrow (N_s)/(N)=2.9* 10^(-6)`

The required ratio is
`2.9* 10^(-6)`.