{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chap4_4pgs

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

This preview shows pages 1–3. Sign up to view the full content.

lmpsrfecll orl\$ In lhe nlorllc nnd lonlc flrrdrlgernerlts +l calculate the number of vacancies per cm3 expected in copper at [email protected] (ust belowthe meltingtemperature). 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 - 3.990s tiacuon=-J-=0.002375 31

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Fl A niobium alloy is produced by introducing tungsten substitutional atoms in the BCC structure; eventually an alloy is produced that has a latticeparameter of O.32SS4 nm and a density of 11.95 g/cm3. Calculate the fraction of the atoms in the alloy that are tungsten.
This is the end of the preview. Sign up to access the rest of the document.
• Spring '08
• snively
• Trigraph, ions substituteor magnesium, cm3expectedn copperat lOg, rgy for'vacancy ormation, ndits lattice arameter, slip planesn FCC

{[ snackBarMessage ]}