Question

If an element with atomic mass=50 crystalizes in FCC lattice, with edge length of unit cell as 0.5nm. What is the density of unit cell if it contains 0.25% Schottky defects

(a) 2 g/cc (b)2.66 g/cc (c)3.06 g/cc (d)none of these

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) = 0.5 nm or 0.556x10^-9 m, Substituting the values in the formula the density would be
d = (4*40 g)/(0.5x10^-9m)^3*6.023x10^23
= 160/(0.1719x10^-27m3) *6.023x10^23
= 2.68x10^6 g/m3
d = 2.68 g/cm3

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(2.68g/cm3)
d’= 2.66g/cm3