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

It's a