Calculate the unit cell edge length for an 78 wt% Ag- 22 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 unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, = 78 wt%

concentration of Pd, = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, = 107.87 g/mol

atomic weight of iron, = 106.4 g/mol

Step 1: determine the average density of the alloy

Step 2: determine the average atomic weight of the alloy

Step 3: determine unit cell volume

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

Step 4: determine the unit cell edge length

Vc = a³

= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

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specific gravity is the density of a substance divided by the density of water. since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density

so "c" is the answer

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