--Given Values-- Atomic Radius (nm) = 0.116 FCC Metal = Gold BCC Metal: = Sodium
ID: 701157 • Letter: #
Question
--Given Values--
Atomic Radius (nm) = 0.116
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.
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2) What is the atomic packing factor for the BCC crystal structure?
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3) Find the theoretical density for the FCC Element shown above in g/cm3. Write your answer to the ten thousandths place (0.0000):
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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:
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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:
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6) Repeat the calculation in question 5 at room temperature (25 C):
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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:
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8)Calculate the number of atoms per cubic meter in Metal B (units atoms/m3). Write your answer with 4 significant figures :
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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 (%):
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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 (%):
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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?
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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?
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Explanation / Answer
Releation between edge length and radius of atom for FCC = a = r8
a = edge length
r = radius of atom
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
please ask rest of them seperately thanks!
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