Question

1. FCC lead has a lattice parameter of 0.4949 nm and contains one vacancy per 500 Pb atoms. Calculate (a) the density; and (b) the number of vacancies per gram of Pb.

Answer 1

Given:

The lattice parameter of FCC lead = 0.4949 nm

The number of vacancies = 1 vacancy per 500 Pb atoms

To Find:

(a) The density

(b)  The number of vacancies per gram of Pb.

Solution:

The density of lead is 11.33 g/cm³ and the number of vacancies per gram of Pb is 5.8246 X 10¹⁸

(a) Density = Mass / volume = z M / a³ Na

Here,  z = number of atoms in the unit cell

M = molar mass         (Molar mass of Pb = 207 g/mole)

a = Lattice Parameter

Na = Avogadro’s number = 6.022 X 10²³

The number of atoms in a FCC = 4

but Pb has a vacancy per 500 Pb atoms

⇒ No. of effective sites per 500 Pb atoms = 500 - 1 = 499

⇒ No. of effective sites per Pb atom = 499 / 500

The number of lead atoms per unit cell = $\frac{499}{500} X 4$ = 3.99

Substituting values,

Density (ρ) = $\frac{3.99X207}{(4.949X10^-^8)^3X 6.022X10^2^3}$

= 825.93 / 729.94 X 10⁻¹

≈ 11.331 g/cm³

(b) The number of vacancies per gram of Pb = v / a³ρ

Here, v =  Vacancies per unit = $\frac{1}{500} X 4$ = 1 / 125

Substituting,

The number of vacancies per gram of Pb = $\frac{\frac{1}{125} }{(4.949X10^-^8)^3X11.331 }$

= 5.8246 X 10¹⁸  vacancies/gram