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
