Chap4_4pgs - lmpsrfecll nlorllc orl$lhe In nnd...

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lmpsrfecll orl$ In lhe nlorllc nnd lonlc flrrdrlgernerlts +l calculate the number of vacancies per cm3 expected in copper at lOg@c (ust below the melting temperature). The activation "n"rgy for'vacancy formation is 20,000 cal/mol. Solution: (4 atoms/u.c.) = 8.47 x l9z atoms/cm3 n= (3.6151 x 10.8 cm)3 nu= 8.47 x lPz exp[-20,000/(1.997X1353)] = 8.47 x 1022 exp(-7 .4393) = 4.97 x lOle vacancieVcm3 L3 The density of a sample of FCC palladium is I 1.98 g/cm3 and its lattice parameter is 3.SSOZ A. Calculate (a) the fraction of the lattice poi-nts rhat contain vacanci", unJ (b) the total number of vacancies in a cubic centimeter of pd. solution: (a) ll.9g q/cm3 = ('rXl06'4 g/mol) - (3.8902 x l0-E cm;316.02 x ld3 atomVmol) .r = 3.9905 4.0
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Fl A niobium alloy is produced by introducing tungsten substitutional atoms in the BCC structure; eventually an alloy is produced that has a lattice parameter of O.32SS4 nm and a density of 11.95 g/cm3. Calculate the fraction of the atoms
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This note was uploaded on 02/01/2012 for the course MSEG 302 taught by Professor Snively during the Spring '08 term at University of Delaware.

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Chap4_4pgs - lmpsrfecll nlorllc orl$lhe In nnd...

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