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If the radius of a copper atom is given as 1.27 Ao, its density (in kg/m3) will be?

Density of cubic metal is given as

ρ = \(\frac{{n \times A.W}}{{{a^3}}}\) × 1.66 × 10-27 kg/m3

where

ρ = density of metal

n = effective number of atoms per unit cell

A.W = Atomic weight of the metal in amu

a = lattice parameter in meter

Given

Radius of copper = 1.27 Ao = 1.27×10-10 m

We know that the atomic weight of copper = 63.5 amu

Atom per unit cell in FCC = 4

Atomic radius r = a√2/4

Therefore, the lattice parameter of copper (a) = 4r/√2

a = (4 × 1.27×10-10)/√2 = 3.59 × 10-10 m

Therefore Density of FCC structure

\(\frac{{4 \times 63.5}}{{{{(3.59 \times {{10}^{ - 10}})}^3}}} \times {\rm{ }}1.66{\rm{ }} \times {\rm{ }}{10^{ - 27}}\;\)

ρ ≈ 8979 kg/m3