Homework_3_solutions

# Homework_3_solutions - MAT 33 Fall 2003 Due Homework#3...

• Notes
• nizhes
• 4

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MAT 33 – Fall 2003 Homework #3 - Solutions Due: 09/15/03 1.) Calculate the number of vacancies per cubic meter in Au at 900 o C. The energy for vacancy formation is 0.98eV per atom. Futhermore, the density and atomic weight are 19.32g cm -3 and 196.9 g mol -1 respectively. Determination of the number of vacancies per cubic meter in gold at 900 ° C (1173 K) requires the utilization of Equations (4.1) and (4.2) as follows: N V = N exp - Q V kT = N A ρ Au A Au exp - Q V kT = 6.023 x 10 23 atoms/mol ( 29 19.32 g/ cm 3 ( 29 196.9 g/mol exp - 0.98 eV/ atom 8.62 x 10 - 5 eV/atom - K ( 29 (1173 K) = 3.65 x 10 18 cm -3 = 3.65 x 10 24 m -3 2.) BCC iron has a measured density of 7.87Mg m -3 . The lattice parameter of BCC iron is 0.2866nm. Calculate the percentage of vacancies in pure iron given that its atomic mass of Fe is 55.85g mol -1 . ρ = (atoms/cell) (atomic mass) (volume of unit cell) (Avagadro’s number) ρ = (atoms/cell) (55.85gmol -1 ) = 7.87 Mg m -3 (2.6866 x 10 -8 ) 3 (6.02 x 10 23 ) (atoms/cell) = (7.87) (2.866 x 10 -8 ) 3 (6.02 x 10 23 ) = 1.998 55.85 There should be 2 atoms/cell in a perfect BCC iron crystal. Thus the difference must be

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• Fall '07
• keith
• Atom, Atomic Mass, Chemical element, grain boundary energy, Cu C H O Ag Al Co Cr Fe Ni Pd Pt Zn

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