Question

Lead crystallizes in a face centered cubic unit cell structure and has a densiy of 11.40 g/cm3 . Calculate the atomic radius of lead in picometers. Molar mass of lead is 207.2 g/mol

Answer 1

Answer:

Atomic radius of lead = R = (1.75 × 10⁻¹⁰) m = 175 pm

Explanation:

For a crystalline structure,

ρ = (nA)/(V꜀Nₐ)

where

ρ = density = 11.40 g/cm³ = 11400 kg/m³ for Lead

n = number of atoms per unit cell = 4 atoms (for an FCC crystal)

A = Atomic number = 207.2 g/mol = 0.2072 kg/mol

V꜀ = Volume of the unit cell = ?

Nₐ = Avogadro's constant = (6.022 × 10²³) atoms per mol

V꜀ = (nA)/(ρNₐ)

V꜀ = (4×0.2072)/(11400×6.022×10²³)

V꜀ = (1.207 × 10⁻²⁸) m³

But, V꜀ = (edge length of the unit cell)³ = a³

a³ = (1.207 × 10⁻²⁸)

a = (4.9424058 × 10⁻¹⁰) m

But for an FCC crystal, the radius of atom is related to the edge length through,

a = 2R√2

R = a/(2√2)

R = (4.9424058 × 10⁻¹⁰)/(2√2)

R = (1.7474043 × 10⁻¹⁰) m = (1.75 × 10⁻¹⁰) m = 175 pm