Question

An element crystallizes in fcc lattice, with a= 0.50 nm. what is the density of unit cell if it contains 0.25% schottky defect

Answer 1

The density of the crystal is given by the formula:

d = ZM/a^3NA

Given data Z for FCC = 4; M (atomic mass) = 40 g/mol; a (inter-atomic distance) = 0.556 nm or 0.556x10^-9 m, Substituting the values in the formula the density would be

 d = (4*40 g)/(0.556x10^-9m)^3*6.023x10^23

  = 160/(0.1719x10^-27m3) *6.023x10^23

  = 1.546x10^6 g/m3

  d = 1.546 g/cm3

(i) It contains 0.2% Frenkel defect.

Frenkel defect does not change the density of the crystal hence it remains the same. i.e. d = 1.546g/cm3

 

(ii) It contains 0.1% schottky defect.

Schottky defect reduces the density by 0.1%, assuming that volume remains constant.

d’= d( 1- 0.1/100)

d’= 0.999d

d’= 0.999(1.546g/cm3)

d’= 1.544g/cm3