Answer :
Answer:
The edge length is 0.4036 nm
Solution:
As per the question:
Density of Ag, [tex]\rho = 10.49 g/cm^{3}[/tex]
Density of Pd, [tex]\rho = 12.02 g/cm^{3}[/tex]
Atomic weight of Ag, A = 107.87 g/mol
Atomic weight of Pd, A' = 106.4 g/mol
Now,
The average density, [tex]\rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}[/tex]
where
[tex]V_{c} = a^{3}[/tex] = Volume of crystal lattice
a = edge length
n = 4 = no. of atoms in FCC
Therefore,
[tex]\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}[/tex]
Therefore, the length of the unit cell is given as:
[tex]a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}[/tex] (1)
Average atomic weight is given as:
[tex]A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}[/tex]
where
[tex]C_{Ag}[/tex] = 79 %
[tex]A_{Ag}[/tex] = 107
[tex]C_{Pd}[/tex] = 21%
[tex]A_{Pd}[/tex] = 106
Therefore,
[tex]A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78[/tex]
In the similar way, average density is given as:
[tex]\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}[/tex]
[tex]\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}[/tex]
Therefore, edge length is given by eqn (1) as:
[tex]a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm[/tex]