Question

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.

Answer 1

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, $C_{Ag}$ = 78 wt%

concentration of Pd, $C_P_d$ = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, $A_A_g$ = 107.87 g/mol

atomic weight of iron, $A_P_d$ = 106.4 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }$

$\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }$

$A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

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

$V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm$= 0.405 nm

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