Question

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.

Answer 1

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag,
`\rho = 10.49 g/cm^(3)`

Density of Pd,
`\rho = 12.02 g/cm^(3)`

Atomic weight of Ag, A = 107.87 g/mol

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

Now,

The average density,
`\rho_(a) = \frac{n A_(avg)} {V_(c)* N_(A)}`

where

`V_(c) = a^(3)` = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

`\rho_(a) = = \frac{n A_(avg)} {V_(c)* N_(A)}`

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

`a = ((nA_(avg))/(\rho_(a)* N_(a)))^(1/3)` (1)

Average atomic weight is given as:

`A_(avg) = (100)/((C_(Ag))/(A_(Ag)) + (C_(Pd))/(A_(Pd)))`

where

`C_(Ag)` = 79 %

`A_(Ag)` = 107

`C_(Pd)` = 21%

`A_(Pd)` = 106

Therefore,

`A_(avg) = (100)/((79)/(107) + (21)/(106)) = 106.78`

In the similar way, average density is given as:

`\rho_(a) = (100)/((C_(Ag))/(\rho_(Ag)) + (C_(Pd))/(\rho_(Pd)))`

`\rho_(a) = (100)/((79)/(10.49) + (21)/(12.02)) = 10.78 g/cm^(3)`

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

`a = ((4* 106.78)/(10.78* 6.023 X 10^23))^(1/3) = 4.036* 10^(- 8) cm = 0.4036* 10^(- 9) m = 0.4036 nm`