Calculate the unit cell edge length for an 87 wt% Ag- 13 wt% Pd alloy. All of the palladium is in solid solution, and the crystal structure for this alloy is FCC. Room temperature densities for Ag and Pd are 10.49 g/cm3 and 12.02 g/cm3, respectively, and their respective atomic weights are 107.87 g/mol and 106.4 g/mol. Report your answer in nanometers. nm

Explanation:

   Since the crystal structure is FCC , every unit cell will have 4 atoms . If a be the edge length

average density = 4 x average atomic no / volume of unit cell x N

N is avogadro no or 6.02 x 10²³

average atomic no

= 100 / (87/107.9 +13/106.4) = 107.72

average density = 100 / (87/10.49 + 13/12.02)

= 10.667

10.667 = (4 x 107.72 )/ (a³x 6.02 x 10²³)

a³ = 4 x 107.72 / (10.667 x  6.02 x 10²³)

= 6.71 x 10⁻²³

a = 4.063 x 10⁻⁸ m

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.