Point Defects. Vacancies are the most important form. Vacancies Self-interstitials

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1 Grain Boundaries 1

2 Point Defects 2

3 Point Defects A Point Defect is a crystalline defect associated with one or, at most, several atomic sites. These are defects at a single atom position. Vacancies Self-interstitials Vacancies are the most important form. 3

4 Vacancies Vacancies :a normally occupied lattice site from which an atom or ion is missing distortion of planes Vacancy Thermodynamics says that there is an equilibrium number of vacancies: Q V NV Nexp RT OR (depending on units of Q) N V N exp QV kt Why is this number of vacancies thermodynamically stable? 4

5 Thermodynamics of Vacancies (not covered in textbook) Introduction of a small number of vacancies lower the free energy. From chemistry: G = H TS G Entropy (S) is the key. A crystal only has one perfect configuration. configurational entropy is zero Vacancies (on the other hand) give rise to many configurations. add n vacancies to a lattice with N atoms so as n S G Vacancies are created until G is at a minimum (note that if too many vacancies are added, G increases because the material will then contain too many broken bonds). n* n N n 5

6 Equilibrium Vacancy Concentration Consider Cu: Q v = 96.1 kj/mol at 1000 K, X v 10-5 at 1358K ( T M ), X v 2 x 10-4 typically X v 10-4 at T M 6

7 Observing the Equilibrium Vacancy Concentration Increasing temperature causes a surface island of atoms to grow. Low energy electron microscope view of a (110) surface of NiAl. Why? The equilibrium vacancy concentration increases by atom motion from the crystal to the surface, where they join the island. Island grows/shrinks to maintain equil. vancancy conc. in the bulk. (Click to Play) Reprinted with permission from Nature (K.F. McCarty, J.A. Nobel, and N.C. Bartelt, "Vacancies in Solids and the Stability of Surface Morphology", Nature, Vol. 412, pp (2001). Image is 5.75 m by 5.75 m.) Copyright (2001) Macmillan Publishers, Ltd. 7

8 Self-Interstitials Self-interstitial: an atom from the crystal that is crowed into an interstitial site. The term interstitial refers to the empty space between atoms. Form a lot of distortion 8

9 Solute Atoms (Solid Solutions) A solute atom is defined as an atom which different from the host atoms (e.g. Cu atom in Ni). Sometimes we refer to solutes as impurities. Substitutional or Interstitial Solid solubility X sol Q sol exp kt 9

10 Factors Affecting Solubility in Substitutional Solid Solutions Atomic size factor Crystal Structure Electronegativity Valences The tale of the engineering student and the philosopher... 10

11 Solute Mobility Which kind of solute (substitutional or interstitial) will be easier to move around (diffuse) in a crystalline solid. Why? Interstitial solutes move around more easily because: They are much smaller atoms. Most of the surrounding interstitial spaces are empty. 11

12 A Note on Specifying Composition Adding solute atoms creates an alloy. We need a way to specify its composition. Typically it's done as a weight or atomic percent. e.g. steel may contain Fe with 0.4 wt% C weight percent (wt%) m 1 = weight or mass of element 1 atom percent (at%) n m1 = m 1 /A 1 m1 = mass in gram C C 1 1 m 1 m m m1 100 m i 2 n m 1 n ' 1 m i (binary) A1= atomic weight 12

13 Alloying a Surface Sn islands move along the surface and "alloy" the Cu with Sn atoms to make "bronze". Low energy electron microscope view of a (111) surface of Cu. The islands continually move into "unalloyed regions and leave tiny bronze particles in their wake. Eventually, the islands disappear. (Click to Play) Reprinted with permission from: A.K. Schmid, N.C. Bartelt, and R.Q. Hwang, "Alloying at Surfaces by the Migration of Reactive Two-Dimensional Islands", Science, Vol. 290, No. 5496, pp (2000). Field of view is 1.5 m and the temperature is 290C. 13

14 Line Defects 14

15 Dislocations- Linear Defects Dislocation: linear or onedimensional defect around which some of the atoms are misaligned Edge Dislocation: A linear crystal line defect associated with the lattice distortion produced in the vicinity of the end of an extra half-plane of atoms Burgers vector perpendicular to the dislocation line 15

16 Dislocations- Linear Defects Screw dislocations: Associated with the lattice distortion created when normally parallel planes are joined together to form a helical ramp Burgers vector is parallel to the dislocation line 16

17 Dark lines are dislocations Representation of edge, screw and mixed dislocations 17

18 Thermodynamics of Dislocations Unlike point defects, dislocations are always thermodynamically unstable Most dislocations will disappear at high enough temperatures 18

19 Dislocation Motion Dislocations move by breaking and remaking atomic bonds. The net result is a shear displacement in the crystal. (Click to Play) Courtesy P. M. Anderson 19

20 Planar Defects 20

21 Planar Defects Boundaries are two dimensions defects that separate regions of the materials that have different crystal structure or crystallographic orientations External surfaces Grain boundaries Twin boundaries (special type of grain boundary) Stacking faults Phase boundaries 21

22 Interfaces divide crystal regions External surfaces crystal edge unbonded electrons excess energy surface energy γ s (J/m 2 ) Planar Defects 22

23 External Surfaces Atoms on the free surface of the specimen are in a higher state of energy than atoms in the interior positions Surfaces have more dangling bonds To reduce this energy, materials try to minimize the total surface energy 23

24 Grain Boundaries Most solids are polycrystalline: Many grains (a single crystal region within a polycrystalline material) separated by grain boundaries grain boundaries There is atomic mismatch in a transition region where two grains meet. Boundaries (where grains come together) are called grain boundaries. 24

25 Twin Boundaries Specific mirror lattice symmetry The mirror plane is termed a twin Produced by applied mechanical shear forces or during annealing heat treatments (chapter 7) Occurs on a definite crystallographic plane and in a specific direction 25

26 Demo: Crystal model with faults Point Defects Linear Defects - Dislocations Area defects Bulk defects Model of a perfect single crystal Model of a poly-crystal with many defects 26

27 Microscopic Examination 27

28 Microstructure Structural materials can be inhomogeneous. Grain size variation Texture Metal castings Solidification involves nucleation, growth and impingement micro and macrostructure formation 28

29 Microstructure Examples Microstructure of annealed Al-3%Mg Alloy (material of the pop-can) 29

30 Al-3%Mg after large deformation and 3 hrs of annealing at 180 o C 30

31 Twins in Fe-Ni superalloy (FCC) 31

32 Solidification Nucleation small cluster (~1nm) of atoms in crystalline order Growth atoms are added to the nucleus which now grows to consume the liquid. Impingement solid grains grow until they meet one another and form a grain-boundary 32

33 Driving Force Thermodynamics G T < T M, G solid < G liquid Driving force is... ΔG sol = G solid - G liquid solid Spontaneous transformation T M liquid T Negative ΔG, ΔG<0 33

34 Nucleation Barrier When T is lower than T m, changing a volume of liquid into a solid lowers the energy of the material. If the decrease in energy per unit volume is ΔG sol then for a sphere of radius r, the decrease in energy is: 4 r 3 G sol 3 At the same time, the solid cluster which forms will have a surface and surface energy, s Again, for a sphere of radius r, the increase in energy is: 4r 2 s liquid 34

35 - Nucleation Barrier Solid cluster has a surface and surface energy s Total energy G tot =VG sol +4r s r G sol 4r s 3 Because the surface terms is larger than the volume term at small values of r, there is a nucleation barrier. Find r * from r G tot 0 35

36 Effect of temperature on ΔG* and r* 36

37 Nucleation There are 2 types of nucleation: Homogeneous nucleation A solid particle begins to form when atoms in the liquid cluster together growth of the cluster will continue if the cluster reaches the critical radius (r*) Embryo a cluster of radius < r* will shrink and redissolve Nucleus a cluster of radius r* Heterogeneous nucleation Activation energy is lowered when nuclei form on pre-existing surfaces A phase transformation will occur spontaneously when G (free energy) has a negative value G* activation free energy- energy required for the formation of a stable nucleus Free-energy vs. Radius plot for homo/heterogeneous nucleation

38 Forces Controlling Nucleation and Growth During an undercooling process (during which the temperature drops below T E ) the available reaction energy increases the rate of diffusion decreases -ΔG D T T E These are competing effects 38

39 Video of Nucleation, Growth and Impingement

40 Solidification in Metals Crystal growth into a temperature gradient leads to the formation of dendrites 40

41 Solidification Structure Three zones - chill, columnar, equiaxed. heat flow 41

42 Solidification & Continuous Casting 42

43 Micro to Macro Defects in Solids Crystals are never perfect Many scales of defect - atomic to macro Macro defects include: Pores Cracks Foreign inclusions Other phases 43

44 Grain Size Determination Grain size average grain diameter, as determined from a random cross section. N 2 n1 N Number of grains per square inch (magnified 100X) n Grain size number 44

45 Chapter 4 Practice Problems

46 Practice Problems 1. Consider a metal with a vacancy formation energy of 0.8 ev/atom. If the material is originally at 1031 K, calculate the temperature rise (in K) needed to increase the vacancy fraction by a factor of 3.5: a) 123 K b) 143 K c) 167 K d) 443 K e) 567 K

47 Practice Problems Answer: or c) / / 0.8 exp : by 3.5to get Multiply / / 0.8 exp K K T K T T atomk ev atom ev K atomk ev atom ev X K

48 Practice Problems 2. Experimental observations on the solidification of a hypothetical metal show that the liquid-solid surface energy is 0.8 J/m 2 and the driving force for solidification, ΔG sol, is -8x10 9 J/m 3. If the critical radius for homogenous nucleation is 0.2 nm, calculate the activation energy for homogenous nucleation [Hint: 1 ev =1.602x10-19 J] a) 0.74 ev b) 0.84 ev c) 0.34 ev d) 0.14 ev e) 0.24 ev

49 Practice Problems Answer: G r Gsol 4r 3 Substituteknown variablesand r G J 0.84eV or b) 0.2nm tosolve:

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