Calculate the unit cell edge length for an 80 wt% Ag−20 wt% Pd alloy. All of the palladium is in solid solution, the crystal structure for this alloy is FCC, and the room-temperature density of Pd is 12.02 g/cm3.

λ^3 = 4.37

Explanation:

first let us to calculate the average density of the alloy

for simplicity of calculation assume a 100g alloy

80g --> Ag

20g --> Pd

ρ_avg= 100/(20/ρ_Pd+80/ρ_avg)

         = 100*10^-3/(20/11.9*10^6+80/10.44*10^6)

         = 10744.62 kg/m^3

now Ag forms FCC and Pd is the impurity in one unit cell there is 4 atoms of Ag since Pd is the impurity we can not how many atom of Pd in one unit cell let us calculate

total no of unit cell in 100g of allow = 80 g/4*107.87*1.66*10^-27

                                                          = 1.12*10^23 unit cells

mass of Pd in 1 unit cell = 20/1.12*10^23

Now,

                      ρ_avg= mass of unit cell/volume of unit cell

                      ρ_avg= (4*107.87*1.66*10^-27+20/1.12*10^23)/λ^3

                          λ^3 = 4.37

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answer:

true

explanation: