Diffusion (2020) Prof. Dr. THARWAT G. ABDEL-MALIK EMERITUS - PowerPoint PPT Presentation

P312:SOLID STATE PHYSICS Prof. THARWAT G. ABDEL- MALIK Diffusion (2020)

Prof. Dr. THARWAT G. ABDEL-MALIK EMERITUS PROFESSOR SUBJECT:-P312:SOLID STATE PHYSICS LECTURER NUMBER EIGHT (25 SLIDES) e-mail:-tharwatdr@gmail.com

Diffusion Diffusion - Mass transport by atomic motion Mechanisms • Gases & Liquids – random(Brownian) motion • Solids – vacancy diffusion or interstitial diffusion

Important Concepts  Applications of Diffusion  Activation Energy for Diffusion  Mechanisms for Diffusion  Rate of Diffusion (Fick’s First Law)  Factors Affecting Diffusion  Composition Profile (Fick’s Second Law)

Diffusion  How does diffusion occur?  Why is diffusion 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? Diffusion - Mass transport by atomic motion. Diffusion is a consequence of the constant thermal motion of atoms, molecules and particles that results in material moving from areas of high to low concentration. Mechanisms Brownian motion is the seemingly random movement of particles suspended in a liquid or gas. Solids – vacancy diffusion or interstitial diffusion.

Inter diffusion Inter diffusion (impurity diffusion): In an alloy, atoms tend to migrate from regions of high concentration to regions of low concentration. After some time Initially

Self-Diffusion • Self-diffusion:In an elemental solid, atoms also migrate. Specific atom movement After some time C C A D A D B B

Diffusion Mechanisms Atoms in solid materials are in constant motion, rapidly changing positions. For an atom to move, 2 conditions must be met: 1. There must be an empty adjacent site, and 2. The atom must have sufficient (vibrational) energy to break bonds with its neighboring atoms and then cause lattice distortion during the displacement. At a specific temperature, only a small fraction of the atoms is capable of motion by diffusion. This fraction increases with rising temperature. • There are 2 dominant models for metallic diffusion: 1. Vacancy Diffusion 2. Interstitial Diffusion

Vacancy Diffusion • atoms exchange with vacancies • applies to substitutional impurity atoms • rate depends on: -- number of vacancies -- activation energy to exchange. increasing elapsed time

Interstitial Diffusion • Interstitial diffusion – smaller atoms (H, C, O, N) can diffuse between atoms. More rapid than vacancy diffusion due to more mobile small atoms and more empty interstitial sites.

Steady-State Diffusion • Consider diffusion of solute atoms (B) in solid state solution (AB) in direction x  x between two parallel atomic planes (separated by Δx ) • If there is no changes with time in C B at A-atom these planes-such diffusion condition is called B-atom steady-state diffusion.Flux J • In steady-state diffusion, the Flux is constant     J x D ( C / x ) with time – i.e. the rate of diffusion is Diffusing Species , C C A independent of time. Concentration of Thin metal plate    C dC C C C P A  P B   A B And   dx x X X Gas at  constant x A B Pressure P B C B Direction of Flux J x Diffusion of Gas at Gaseous Species x x Pressure P A A B Distance ,x Area ,A Illustration of Fick’s first law

Fick’s I law Diffusion coefficient/ diffusivity dn dc No. of atoms   Cross-sectional area DA crossing area A dt dx per unit time Concentration gradient Matter transport is down the concentration gradient Flow direction A  As a first approximation assume D  f(t)

  / / J atoms area time concentrat ion gradient dc J  dx dc   J D dx  1 dn dc   J D A dt dx dn dc Fick’s first law   DA dt dx

 Diffusivity (D) → f(A, B, T) Steady state diffusion D  f(c) C 1 Concentration → C 2 D = f(c) x →

D  f(c) Steady state J  f(x,t) D = f(c) Diffusion D  f(c) Non-steady state J = f(x,t) D = f(c)

Fick’s II law  x   Accumulati on J J   x x x    J       Accumulati on J J x  J x J x+  x x x   x               1 Atoms Atoms c J               . m  J x J J x         x x 3 2       m s m s t x            c c c J              Fick’s first law D x x            t x x t x     2        D  f(x) c c c c     D     D     2 t x        t x x

    2 c c    D     2 t x RHS is the curvature of the c vs x curve c → c → x → x → LHS is the change is concentration with time +ve curvature  c ↑ as t ↑  ve curvature  c ↓ as t ↑

      2 c c x        D c ( x , t ) A B erf     2   t x 2 Dt Solution to 2 o de with 2 constants determined from Boundary Conditions and Initial Condition  Erf (  ) = 1  Erf (-  ) = -1      2     2  Erf (0) = 0 Erf exp u du   Erf (-x) = -Erf (x) 0 Exp(  u 2 ) → Area  0 u →

In most cases of diffusion investigated the coordinates x and t enter into the resulting expressionas exponential of x 2 /t of the form    x 2   exp   t Where 𝛍 is constant . Thus we attempt the following trial solution    2 x      c f ( x ). h ( t ) exp 2   t The solution requires that c remains finite at x=0 expect for t=0 If the solution is correct, it must satisfy the differential equation. By differentiate equation (2) and substituting in equation (1) and solving we get     2 dc d c     } and { we find     2   dt dx x t

        2 2 x x          2 dc   x           t   t    1 f ( x ) h ( t ) e e h ( t )     2 dt t   x    2 x       2 1   x h ( t )       t    f ( x ) h ( t ) e ( 3 ) 2   t h ( t )         2 2 x x             dc   2 x           t   t      1 h ( t ) f ( x ) e e f ( x )       dx t   t Assuming f(x) is finite constant then the first direvitve f 1 (x) =0; then the second term of the previous equation is eliminated.

   2 x              2 2 d c   2 2 x              t      D D h ( t ) f ( x ) e 1 ( 4 )       2     dx t  t  t Equations (3) ,(4) must be the same if the suggested solution is correct therefore the following condition must be satisfied            2 2 1 2 2 x x h ( t )               D 1 ( 5 )     2     t  t  t h ( t ) The above mentioned equation is satisfied only if 1 1    h ( t ) ; 4 D t 1   1 ( ) Using h t 3 / 2 2 t

The RHS of the equation becomes   2 1 / 2 2 x t x 1    2 3 / 2 2 t 2 t t 2 t And the LHS of the equation becomes                 2 2 2 2 x 1 2 2 x                     D 1 1              t t 4 t t         2 2 1 2 1 2 2 x x 1          2 4 4 2 t t t t t Equation (5) is satisfied under the following condition

1 1      f ( x ) cons tan t 0 ; h ( t ) ; 4 D t Thus   2 x         4 Dt   c e ( 6 ) t This solution is symmetrical for x=0. At zero time the concentration c is vanishing every where except for x=0 where it becomes infinite. Since the quantity of diffusing material is finite, the integral   c dx   remains finite even at t=0. If S is the diffusing material