Calculate the radius of a silver atom in cm, given that ag has an fcc crystal structure, a density of 10.5 g/cm3, and an atomic weight of 107.87 g/mol.

For solid cubic crystals, the equation for density is:

ρ = nA/VcNa

where

n is 4 atoms per unit cell of FCC structure

A is atomic weight (107.87 g/mol)

Vc is the volume of the cubic cell

Na is Avogadro's number (6.022*10²³ atoms/mol)

10.5 g/cm³ = [ (4) (107.87 g/mol) ]/[Vc (6.022*10²³ atoms/mol) ]

Solving for Vc,

Vc = 6.823707*10⁻²³ cm³

The volume is equal to (2r√2) ³. Thus,

6.823707*10⁻²³ cm³ = (2r√2) ³

Solving for r,

r = 1.445*10⁻⁸ cm