Question

Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at 623 oC (this temperature is below the melting temperature (645oC)). Assume an energy for defect formation of 1.86 eV.

Answer 1

5.9 × 10^-6.

Step-by-step explanation:

In the arrangements of crystal solids there is likely going to be an imperfection or defect and one of the defect or imperfections in the arrangements of solids is known as the Schottky defects. The Schottky defects is a kind of lattice arrangements imperfection that occurs when positively charged ions and negatively charged ions leave their position.

So, let us delve right into the solution of the question. We will be making use of the formula below;

Wb/ W = e^ - c/ 2kT.

Where Wb/ W= fraction of lattice sites, c= energy for defect formation = 1.86 eV, and T = temperature= 623° C= 896 k.

So, Wb/ W = e ^ -1.86/ (2 × 896 × 8.62 × 10^ -5).

Wb/ W= 0.000005896557435956372.

Wb/ W=5.9 × 10^-6.