Question

Calcium crystallises in FCC unit cell with edge length =0.556nm .Calculate its density if it contains 0.1% vacancy defects.

(Plz give the solution with explanation)

Answer 1

Vacancy defect is also sometimes called Schottky defect for non ionic crystals.
Here that is the case.

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

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

= 1.546x10^6 g/m3

d = 1.546 g/cm3

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

Hope this helps :)