Need help to solve lattice parameter of FCC metals Ag densit

Solution

It appears from the question that the metal is a homogeneous mixture of Ag and Au with 50 wt% Ag and 50 wt% Au.

In order to calculate the lattice parameter (length of the side of Face Centred Cubic crystal), first we need to know the average atomic weight.

Ag density = 10.49 gm/cc

Ag Atomic weight = 107.8682 gm/mole

Au density = 19.3 gm/cc

Au Atomic weight = 196.966 gm/mole

In order to calculate the average atomic weight, we need to know the mole%.

Let us consider 100 gm of metal in which 50 gm is Ag (50 wt%) and 50 gm is Au.

50 gm Ag + 50 gm Au

= (50/107.8682) mole Ag + (50/196.966) mole Au

= 0.46353 mole Ag + 0.253851 mole Au

= 0.646142 mole fraction Ag + 0.353858 mole fraction Au

Hence, average atomic weight = (0.646142x107.8682 + 0.353858x196.966) gm/mole = 139.3962 gm/mole

Mass of one atom, m = (average atomic weight/Avogadro no.) = (139.3962/6.022x10²³) gm = 2.314782x10^-22 gm

There are 4 atoms per unit cell in the FCC crystal (1/8 atom at each of 8 corners + 1/2 atom at each of 6 faces of the cube).

Density of unit cell = (4m/volume) = (4m/a³)

Therefore, a = (4m/density)^(1/3) = (4x2.314782x10^-22/14.9)^(1/3) = 3.96x10^-8 cm

Hence lattice parameter of the said FCC crystal = 3.96x10^-8 cm