Question

# --Given Values-- Atomic Radius (nm) = 0.116 FCC Metal = Gold BCC Metal: = Sodium Temperature...

--Given Values--
FCC Metal = Gold
BCC Metal: = Sodium
Temperature ( C ) = 1017
Metal A = Tin
Equilibrium Number of Vacancies (m-3) = 6.02E+23
Temperature for Metal A = 369
Metal B = Gallium

1) If the atomic radius of a metal is the value shown above and it has the face-centered cubic crystal structure, calculate the volume of its unit cell in nm3? Write your answers in Engineering Notation.
2) What is the atomic packing factor for the BCC crystal structure?
3) Find the theoretical density for the FCC Element shown above in g/cm3. Write your answer to the ten thousandths place (0.0000):
4) Calculate the atomic radius, in nm, of the BCC Metal above utilizing the density and the atomic weight provided by examination booklet - Write your answer with 4 significant figures:
5) Calculate the fraction of atom sites that are vacant for copper (Cu) at the temperature provided above. Assume an energy for vacancy formation of 0.90 eV/atom:
6) Repeat the calculation in question 5 at room temperature (25 C):
7) Calculate the energy (in eV/atom) for vacancy formation for the Metal A and the equilibrium number of vacancies at the temperature provided above - Write your answer with 4 significant figures:
8)Calculate the number of atoms per cubic meter in Metal B (units atoms/m3). Write your answer with 4 significant figures :
9) What is the composition, in atom percent, of an alloy that contains a) 36 g Metal A and b) 47 g Metal B? Composition for Metal A (%):
10) What is the composition, in atom percent, of an alloy that contains a) 36 g Metal A and b) 47 g Metal B? Composition for Metal B (%):
11) What is the composition of Metal A in atom percent, if the alloy consists of 4.5 wt% Metal A and 95.5 wt% of Metal B?
12) What is the composition of Metal B in atom percent, if the alloy consists of 4.5 wt% Metal A and 95.5 wt% of Metal B?

Releation between edge length and radius of atom for FCC = a = r√8

a = edge length

a = 0.116 nm x √8

volume = a3 = (0.116 nm x √8 )3 = 0.03532 nm3 = 3.53 x 10-2 nm3

2.

atomic packing factor = volume of atoms in unit cell / total voulme of unit cell

relation between edge length and radius for BCC = a = 4r/√3

volume of unit cell = (4r/√3)3

total no. of atoms in BCC inside unit cell = 2

volume of atoms inside unit cell = 2 x 4/3 x pi x r3

atomic packing factor = 2 x 4/3 x pi x r3 / (4r/√3)3 = 0.6798 ~ = 0.68

3.

theoretical density = mass of atoms inside cell / volume of unit cell

in FCC there are 4 atoms inside unit cell = 4 x 197g / 6.022 x 1023 = 130.85 x 10-23 g

volume calculated in part 1 = 3.53 x 10-2 nm3 = 3.53 x 10-2 x 10-21 cm3

density = mass / volume = 130.85 x 10-23 g / 3.53 x 10-2 x 10-21 cm3 = 37.06899 g/cm3

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