Diffusion in Solids ISSUES TO ADDRESS... How does diffusion occur? Why is it an important part of processing? How can the rate of diffusion be predicted for some simple cases? How does diffusion depend on structure and temperature? 1
Review: Types of solid solutions Cu-Ni Fe-Fe 3 C What are the atomic structures of the Cu-Ni solid solution? What about the Fe-Fe 3 C solid solution (austenite)? Are these the same?
Solute atoms in Alloys Two outcomes if impurity (B) added to host (A): Solid solution of B in A (i.e., random dist. of point defects) OR Substitutional solid soln. (e.g., Cu in Ni) Interstitial solid soln. (e.g., C in Fe) Solid solution of B in A plus particles of a new phase (usually for a larger amount of B) Second phase particle --different composition --often different structure. 3
Diffusion Diffusion - Mass transport by atomic motion Mechanisms Gases & Liquids random (Brownian) motion Solids vacancy diffusion or interstitial diffusion 4
Interdiffusion: In an alloy, atoms tend to migrate from regions of high concentration to regions of low concentration Initially Diffusion After some time Adapted from Figs. 5.1 and 5., Callister 7e. 5
Self-diffusion: In an elemental solid, atoms also migrate. Label some atoms D C A B Diffusion After some time C A B D 6
Vacancy Diffusion: Diffusion Mechanisms atoms exchange with vacancies applies to substitutional impurities atoms rate depends on: --number of vacancies --activation energy to exchange. increasing elapsed time 7
Diffusion Simulation Simulation of interdiffusion across an interface: Rate of substitutional diffusion depends on: --vacancy concentration --frequency of jumping. (Courtesy P.M. Anderson) 8
Diffusion Mechanisms Interstitial diffusion smaller atoms can diffuse between atoms. Adapted from Fig. 5.3 (b), Callister 7e. More rapid than vacancy diffusion 9
Processing Using Diffusion Case Hardening: --Diffuse carbon atoms into the host iron atoms at the surface. --Example of interstitial diffusion is a case hardened gear. Adapted from chapter-opening photograph, Chapter 5, Callister 7e. (Courtesy of Surface Division, Midland-Ross.) Result: The presence of C atoms makes iron (steel) harder. Surface hardness makes it harder to initiate cracks better fatigue resistance 10
Doping silicon with phosphorus for n-type semiconductors: Process: 0.5 mm 1. Deposit P rich layers on surface. Processing Using Diffusion. Heat it. silicon 3. Result: Doped semiconductor regions. magnified image of a computer chip light regions: Si atoms silicon light regions: Al atoms Adapted from chapter-opening photograph, Chapter 18, Callister 7e. 11
Diffusion How do we quantify the amount or rate of diffusion? moles(or mass)diffusing mol kg J Flux or surfacearea time cm s m Measured empirically Make thin film (membrane) of known surface area Impose concentration gradient Measure how fast atoms or molecules diffuse through the membrane s J M At l A dm dt M = mass diffused time J slope 1
Steady-State Diffusion Rate of diffusion independent of time Flux proportional to concentration gradient = dc dx C 1 C 1 Fick s first law of diffusion C C J D dc dx if linear dc dx x 1 x x C x C x C x 1 1 D diffusion coefficient 13
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 /s surface concentrations: C 1 = 0.44 g/cm 3 C = 0.0 g/cm 3 14
Example (cont). Solution assuming linear conc. gradient C 1 paint remover glove x 1 x t b 6D C skin J Data: - D dc dx C D x D = 110 x 10-8 cm /s C 1 = 0.44 g/cm 3 C = 0.0 g/cm 3 x x 1 = 0.04 cm C x 1 1 J (110 x 10-8 cm (0.0 g/cm 0.44 g/cm /s) (0.04 cm) 3 3 ) 1.16 x 10-5 g cm s 15
Diffusion and Temperature Diffusion coefficient increases with increasing T. D D 0 Q d exp or D D0 RT exp D = diffusion coefficient [m /s] D 0 = pre-exponential [m /s] Q d = activation energy [J/mol or ev/atom] R = gas constant [8.314 J/mol-K] k B = Boltzmann s constant = R/N A = 8.617x10-5 ev/k T = absolute temperature [K] Qd k T B 17
1500 1000 600 300 Diffusion and Temperature D has exponential dependence on 1/T 10-8 T(C) D (m /s) D interstitial >> D substitutional 10-14 C in a-fe C in g-fe Al in Al Fe in a-fe Fe in g-fe 10-0 0.5 1.0 1.5 1000 K/T 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, 199.) 18
Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are D(300ºC) = 7.8 x 10-11 m /s Q d = 41.5 kj/mol What is the diffusion coefficient at 350ºC? D transform data ln D Temp = T 1/T lnd lnd lnd 0 lnd Q R 1 d 1 T D ln D 1 and Q R d lnd 1 1 1 T T1 lnd 0 Q R d 1 T1 19
Example (cont.) D D 1 exp Q R d 1 T 1 T 1 T 1 = 73 + 300 = 573 K T = 73 + 350 = 63 K D (7.8 x 10 11 m 41,500 J/mol /s) exp 8.314 J/mol-K 1 63K 1 573K D = 15.7 x 10-11 m /s 0
Non-steady State Diffusion The concentration of diffucing species is a function of both time and position C = C(x,t) In this case Fick s Second Law is used Fick s Second Law C t D C x 1
Non-steady State Diffusion Copper diffuses into a bar of aluminum. Surface conc., C s of Cu atoms bar pre-existing conc., C o of copper atoms C s 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 =
Solution: C x,t C C C s o o 1 erf x Dt C(x,t) = Conc. at point x at time t erf (z) = error function z e y 0 dy erf(z) values are given in Table 5.1 3
Solution: C x, t C C C s o o 1 erf x Dt erfc x Dt C(x,t) = Conc. at point x at time t erf (z) = error function z e y 0 dy erf(z) values are given in Table 5.1 C S C(x,t) C o 4
Non-steady State Diffusion Sample Problem: An FCC iron-carbon alloy initially containing 0.0 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. C( x, t) Co x Solution: use Eqn. 5.5 1 erf Cs Co Dt 5
Solution (cont.): C( x,t ) C C C s o o 1 erf x Dt t = 49.5 h x = 4 x 10-3 m C x = 0.35 wt% C s = 1.0 wt% C o = 0.0 wt% C( x, t) C C C s o o 0.35 0.0 1.0 0.0 1 erf x Dt 1 erf( z) erf(z) = 0.815 6
Solution (cont.): We must now determine from Table 5.1 the value of z for which the error function is 0.815. An interpolation is necessary as follows z erf(z) 0.90 0.7970 z 0.815 0.95 0.809 z 0.90 0.815 0.7970 0.95 0.90 0.809 0.7970 z 0.93 Now solve for D z x Dt D x 4z t D x 4z t (4 x 10 (4)(0.93) 3 m) (49.5 h) 1h 3600s.6 x 10 11 m /s 7
Solution (cont.): To solve for the temperature at which D has above value, we use a rearranged form of Equation (5.9a); from Table 5., for diffusion of C in FCC Fe D o =.3 x 10-5 m /s Q d = 148,000 J/mol T Qd R( lnd lnd) o T (8.314 J/mol- K)(ln.3x10 148,000 J/mol 5 11 m /s ln.6x10 m /s) T = 1300 K = 107 C 8
Summary Diffusion FASTER for... open crystal structures materials w/secondary bonding smaller diffusing atoms lower density materials Diffusion SLOWER for... close-packed structures materials w/covalent bonding larger diffusing atoms higher density materials 31