Question

We have a metal alloy (bcc) that

Metal A is FCC, 30 atom% (40.244 wt%), r=0.2 nm, A = 55 g/mol

and, Metal B is BCC, 70 atom% (59.756 wt%), r=0.24 nm, A = 35 g/mol

1. Find the volume per atom for metal A and metal B by dividing the unit cell volume by the number of atoms in the unit cell. Put your units in cm3/atom.

2. Use the atom% values and the answers to (a) to find the average volume per atom

Answer 1

Metal A is fcc 30 atoms. 40.244 wts. Sol Given that V=0-2nm. A-55 g/mol Metal B is BCC to atormy do 59.756 WE% 70-24nm 4= 35 g/mol for fcc Unit Cell 2a=48 a = h . Cell length = a) unit cell volume for A = ( 4 2/3 (init cell volume = 23) = (4 x 0.2 x 1073 = 1.81810 22 cm Volume per atom for A = 1.8181522 (No of atoms in fccunit cell=4) 4 -4.52x1ő ciklatom for B, BCC unit cell 13 a =48 ५ a= 4 Unit cell volume for metal B = (4 x0.24) - X0 24 262 -22 = 1.70x102 cm3 -22 volume per atom for B = 1.70x10 2 = 8.51x10 23 cm latom

for second problem ht 100 atoms in the Metal alloy * Them atoms in A = 30 [30% atom] atoms in 8=70 [70% atom) Next volume of A in alloy = 30x, volane Por atoms of = 30X4.52X2623 Simillarly = 1+35x12²cm B = 70x8.5jx1623 = 5.96X1021 cm3 Tout volume =(1-35x162) +5.96X164) 7.32x102 cm3 finally average volume paratom = 732X182/ Ци 3 مادگس 2 7032X1023