Given Values-- Atomic Radius (nm) 0.18 CC Metal = Silver BCC Metal: = Sodium Tem

Question

Given Values-- Atomic Radius (nm) 0.18 CC Metal = Silver BCC Metal: = Sodium Temperature (C) 1127 Metal A Zinc Equilibrium Number of vacancies (m 3) = 7.42E+23 Temperature for Metal A 247 Metal B Calcium If the atomic radius of a metal is the value shown above and it ha the face-centered cubic crystal structure, calculate the volume of its unit cell in nm 3? What is the atomic packing factor for the BCC crystal structure? Find the theoretical density for the FCC Element shown above in g/cm^3: Calculate the atomic radius, in nm, of the BCC Metal above utilizing the density and the atomic weight provided y examination booklet 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 Repeat the calculation in question 5 at room temperature (25 C): Calculate the energy (in eV/atom) for vacancy formation for the Metal A and the equilibrium number of vacancies at the temperature provided above Calculate the number of atoms per cubic meter in Metal B (units atoms/m^3): What is the composition, in atom percent, an alloy that contains a 36 g Metal A and b) 47 g Metal B? Composition for Metal A (%) What is the composition, in atom percent of an alloy that contains a) 36 g Metal A and 47 Metal B? Composition for Metal B(%)

Explanation / Answer

1) Given that atomic radius = 0.18 nm

For FCC,   atomic radius = a /(8)1/2   where a is edge length of the cube

                 a = atomic radius x (8)1/2

                   = 0.18 nm x (8)1/2

                   = 0.51 nm

Therefore,

volume of the unit cell = a3

                              = (0.51 nm)3

                                = 0.13 nm3

2)

Atmic packing factor for BCC crystal structure = 0.68

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