清华大学：《材料科学基础》课程教学资源（PPT课件讲稿）Chapter 5.1 Introduction

Chapter Imperfections in Crystals Lesson seventeen

§5.1 ntroduction 1. Classification Ideal or perfect crystal real or defect crystal According to dimension point defect: vacancies, interstitial atoms line defect: dislocation planar defect: stacking fault o volume defect: mosaic structure

2. Effects physical properties(point defect) mechanical properties(dislocation) metallurgical properties(all of the defects)

§5.2 Point defects I Classification: 1. Schottky defect 2. Frenkel defect only vacancies pairs of vacancy and interstitial atom Schottky defect Frenkel defect

only vacancies pairs of vacancy and interstitial atom

Equi libr ium concentration of point defects Equilibrium concentration of point defects in metal △G,=△H1-7(△S,+△Sm) Suppose we introduce n vacancies into a crystal containing N atom sites Determine: C N

( ) v v v m G  H T S  S N n Cv 

△G,=△H-TAs≈△U+P△-7(△S,+△Sm) △S Entropy of mixing △U Energy of formation of vacancy N! △S=klno=klnC inkIn n! (N-n) =-NH[C,nC1+(1-C,)n(1-C,) △G,=CN△U+M[C,lnC,+(1-C,)n(1-C

P ( ) v v m G  H TS  U  V T S  S Sm —— Entropy of mixing [ ln (1 )ln(1 )] !( )! ! ln ln ln v v v v n m N Nk C C C C n N n N S k k C k              [ ln (1 )ln(1 )] v v TN Cv Cv Cv Cv G  C NU  k    U —— Energy of formation of vacancy

S S)一

G,=Gn+△G G N△U+MT[nC+-1n(1-C)+=]=0 aC △U △H-T△S,) RT RT m—Nm—N △U/RT G or Cy Ae( or Fcc a≈1) △U/RT NN e △U/Rr T△S n,:the number of vacancies per cm N: the number of atoms per cm3

] 0 1 1 [ln ln(1 ) 0                C C C C C N U NkT C C G Gv G Gv  RT H T S RT U C C v v v ( ) 1 ln          G G0 C C m TS U RT v e N n C    (for FCC A≈1) H kT v Ae N n C  or   U RT n Ne    U RT v v n N e    Ø Ø Ø nv : the number of vacancies per cm3 . Nv : the number of atoms per cm3

For ionic crystals 正离子空位 ]:负离子空位 正离子间隙 负离子间隙 △U1+△Ur Schottky △U +1-1=eXp( exp( RT RT Ci+ Ci ep AU+AUX=eXP(-AUF Frenkel RT RT

For ionic crystals:  : 正离子空位 : 负离子空位 : 正离子间隙 : 负离子间隙   i i exp( ) exp( ) [ ] [ ] S [ ] [ ] RT U RT U U C C             exp( ) exp( ) [ ] F [ ] RT U RT U U C C i i             Schottky Frenkel

Example 1. Determine the number of vacancies needed for a BCC iron crystal to have a density of 7.87g/cm3. The lattice parameter of the iron is 2. X 10-8cm Solution Xatoms/cell)(55.847g/mol) 7.8814g/cm (2.866×10cm)3(602×1023 atoms/mo) X atoms/cell(7.87)2866×108)(602×1 1.9971 5.847 o. there shou ald be 2.00-1.9971=0.0029 vacancies per unit cell 0.0029 vacancies/cell Vacancies/cm 1.23×10 (2.866×108cm)3

1. Determine the number of vacancies needed for a BCC iron crystal to have a density of 7.87g/cm3. The lattice parameter of the iron is 2.866×10-8cm. 3 8 3 23 7.8814g / cm (2.866 10 cm) (6.02 10 atoms/ mol) ( atoms/ cell)(55.847g / mol)      X  1.9971 55.847 (7.87)(2.866 10 ) (6.02 10 ) atoms/ cell 8 3 23       X So, there should be 2.00－1.9971=0.0029 vacancies per unit cell. 20 8 3 3 1.23 10 (2.866 10 cm) 0.0029 vacancies/ cell Vacancies/ cm      