1. Relative Energy Levels of Defects Information was extracted from: Porter and Easterling, Phase Transformations in Metals and Alloys, 2nd Edition, CRC Press, 2004. Surface energy at the solid liquid interface (AKA: interfacial free energy) Rule of thumb that high angle grain boundary energies gb:  b ≈ 1/3  sv Random High Angle Grain Boundary:  > 10-15 o

2. Doping silicon with phosphorus for n-type semiconductors: Process: 3. Result: Doped semiconductor regions. silicon Processing Using Diffusion magnified image of a computer chip 0.5 mm light regions: Si atoms light regions: Al atoms 2. Heat it. 1. Deposit P rich layers on surface. silicon Adapted from chapter-opening photograph, Chapter 18, Callister 7e. Elemental Dot Map from SEM EDS 200X

3. Diffusion How do we mathematically quantify the amount or rate of mass transfer (rate of diffusion)? Measured empirically –Make thin film (membrane) of known surface area –Impose concentration gradient –Measure how fast atoms or molecules diffuse through the membrane M = mass diffused time J  slope

4. Steady-State Diffusion Fick’s first law of diffusion C1C1 C2C2 x C1C1 C2C2 x1x1 x2x2 D  diffusion coefficient [m 2 /s] “Diffusivity” Rate of diffusion independent of time (does not change with time) Flux proportional to concentration gradient = (change in concentration with position)

5. Example: Chemical Protective Clothing (CPC) Methylene chloride is a common ingredient of paint removers. Besides being an irritant, it also may be absorbed through skin. When using this paint remover, protective gloves should be worn. If butyl rubber gloves (0.04 cm thick) are used, what is the diffusive flux of methylene chloride through the glove? Data: –diffusion coefficient in butyl rubber: D = 110 x10 -8 cm 2 /s –surface concentrations: C 2 = 0.02 g/cm 3 C 1 = 0.44 g/cm 3

6. Example (cont). glove C1C1 C2C2 skin paint remover x1x1 x2x2 Solution – assuming linear conc. gradient D = 110 x 10 -8 cm 2 /s C 2 = 0.02 g/cm 3 C 1 = 0.44 g/cm 3 x 2 – x 1 = 0.04 cm Data:

7. Diffusion and Temperature Diffusion coefficient increases with increasing T. D  DoDo exp        QdQd RT = Material constant [m 2 /s] = diffusion coefficient [m 2 /s] = activation energy [J/mol or eV/atom] = gas constant [8.314 J/mol-K] = absolute temperature [K] D DoDo QdQd R T Arrhenius Equation

8. What is D o ? For Vacancy Diffusion in Cubic Crystal:  is the concentration at jump plane is the mean frequency of vibration in the jump direction or number of attempted jumps z is the number of available sites to jump to  S m is the activation entropy of migration (increase in entropy due to migration)

9. Diffusion and Temperature Adapted from Fig. 5.7, Callister 7e. (Date for Fig. 5.7 taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.) D has exponential dependence on T D interstitial >> D substitutional C in  -Fe C in  -Fe Al in Al Fe in  -Fe Fe in  -Fe 1000 K/T D (m 2 /s) C in  -Fe C in  -Fe Al in Al Fe in  -Fe Fe in  -Fe 0.51.01.5 10 -20 10 -14 10 -8 T(  C) 15001000 600 300

10. Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are D(300ºC) = 7.8 x 10 -11 m 2 /s Q d = 41.5 kJ/mol What is the diffusion coefficient at 350ºC? transform data D Temp = T ln D 1/T

11. Example (cont.) T 1 = 273 + 300 = 573 K T 2 = 273 + 350 = 623 K D 2 = 15.7 x 10 -11 m 2 /s Note that we have doubled the diffusion rate with 50 o C increase in temperature!

12. Non-steady State Diffusion The concentration of diffusing species is a function of both time and position C = C(x,t) The diffusion flux and concentration at a particular point in solid vary with time –Result in net accumulation or depletion of diffusing species Fick’s Second Law: If diffusion coefficient is independent of composition: Solutions to this expression give concentration in terms of both position and time

13. Example Adapted from Fig. 5.5, Callister 7e. B.C. at t = 0, C = C o for 0  x   at t > 0, C = C S for x = 0 (const. surf. conc.) C = C o for x =  Copper diffuses into a bar of aluminum. pre-existing conc., C o of copper atoms Surface conc., C of Cu atoms bar s C s

14. Solution: C(x,t) = Conc. at point x at time t erf (z) = error function erf(z) values are given in Table 5.1 CSCS CoCo C(x,t)C(x,t)

15. Non-steady State Diffusion Sample Problem: An FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere that gives a surface carbon concentration constant at 1.0 wt%. If after 49.5 h the concentration of carbon is 0.35 wt% at a position 4.0 mm below the surface, determine the temperature at which the treatment was carried out. Solution: use Eqn. 5.5

16. Solution (cont.): –t = 49.5 h x = 4 x 10 -3 m –C x = 0.35 wt%C s = 1.0 wt% –C o = 0.20 wt%  erf(z) = 0.8125

17. Solution (cont.): We must now determine from Table 5.1 the value of z for which the error function is 0.8125. An interpolation is necessary as follows zerf(z) 0.900.7970 z0.8125 0.950.8209 z  0.93 Now solve for D

18. To solve for the temperature at which D has above value, we use a rearranged form of Equation (5.9a); from Table 5.2, for diffusion of C in FCC Fe D o = 2.3 x 10 -5 m 2 /s Q d = 148,000 J/mol  Solution (cont.): T = 1300 K = 1027°C

19. Estimate for Diffusion Distance If you need an engineering estimate for diffusion distance, you can use