Calculate the unit cell edge length for an 79 wt% ag- 21 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.

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag,

Density of Pd,

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density,

where

 = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

Therefore, the length of the unit cell is given as:

           (1)

Average atomic weight is given as:

where

= 79 %

= 107

= 21%

= 106

Therefore,

In the similar way, average density is given as:

Therefore, edge length is given by eqn (1) as:

where are the vectors ?

the total distance traveled by the car is

30km + 49km = 79 km .