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

• Prof. THARWAT G. ABDEL- MALIK

(2020)

P312:SOLID STATE PHYSICS

Diffusion

• 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

 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) Important Concepts

• 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

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.

Initially

After some time

• Self-diffusion:In an elemental solid, atoms also migrate.

Self-Diffusion

Specific atom movement

A B C D

After some time

A B C D

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.

• Consider diffusion of solute atoms (B) in

solid state solution (AB) in direction x between two parallel atomic planes (separated by Δx)

• If there is no changes with time in CB at

these planes-such diffusion condition is called steady-state diffusion.Flux J

• In steady-state diffusion, the Flux is constant

with time – i.e. the rate of diffusion is independent of time.

Direction of Diffusion of Gaseous Species Gas at Pressure PB Gas at Pressure PA

Area ,A

Thin metal plate PA PB And constant

x

A-atom B-atom

B A B A

X X C C x C dx dC      

A

C

B

C

A

x

B

x

x 

C 

Distance ,x

Concentration of Diffusing Species , C

Flux Jx

Illustration of Fick’s first law

) / ( x C D Jx    

Steady-State Diffusion

Fick’s I law

dx dc DA dt dn  

• No. of atoms

crossing area A per unit time Cross-sectional area Concentration gradient

Matter transport is down the concentration gradient Diffusion coefficient/ diffusivity

A Flow direction

 As a first approximation assume D  f(t)

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

Fick’s first law

 Diffusivity (D) → f(A, B, T) D = f(c) D  f(c) C1 C2 Steady state diffusion x → Concentration →

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

Fick’s II law

Jx Jx+x x

x x x

J J

• n

Accumulati

 

 

            x x J J J

• n

Accumulati

x x

                     x x J J J x t c

x x

 

J s m Atoms m s m Atoms                    

2 3

. 1 x x J x t c                                    x c D x t c

Fick’s first law

                   x c D x t c

D  f(x)

2 2

x c D t c           

RHS is the curvature of the c vs x curve x → c → x → c → +ve curvature  c ↑ as t ↑ ve curvature  c ↓ as t ↑ LHS is the change is concentration with time

2 2

x c D t c           

2 2

x c D t c                    Dt x erf B A t x c 2 ) , (

Solution to 2o de with 2 constants determined from Boundary Conditions and Initial Condition

 

 



 



 

2

exp 2 du u Erf

• Erf () = 1
• Erf (-) = -1
• Erf (0) = 0
• Erf (-x) = -Erf (x)

u → Exp( u2) →



Area

      t x 2 exp 

 

2 exp ) ( ). (

2

       t x t h x f c 

t x

dx c d and dt dc              

2 2

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

{we find

Where 𝛍is constant . Thus we attempt the following trial solution In most cases of diffusion investigated the coordinates x and t enter into the resulting expressionas exponential of x2/t of the form

}

) 3 ( ) ( ) ( ) ( ) ( ) ( ) ( ) (

1 2 2 1 2 2

2 2 2

                             

                          

t h t h t x e t h x f t h e t x e t h x f dt dc

t x t x t x x

 

  

                         

                 

) ( 2 ) ( ) (

1

2 2

x f e t x e x f t h dx dc

t x t x t  



Assuming f(x) is finite constant then the first direvitve f1(x) =0; then the second term of the previous equation is eliminated.

) 4 ( 2 1 2 ) ( ) (

2 2 2

2

                                   

        

t x t e x f t h D dx c d D

t x t

 



) 5 ( ) ( ) ( 2 1 2

1 2 2 2

                                t h t h t x t x t D   

D t t h 4 1 ; 1 ) (   

2 / 3 1

2 1 ) ( t t h  

Equations (3) ,(4) must be the same if the suggested solution is correct therefore the following condition must be satisfied

Using The above mentioned equation is satisfied only if

t t x t t t x 2 1 2

2 2 2 / 3 2 / 1 2 2

    

                                                t x t t x t D

2 2

2 1 2 4 1 2 1 2     

t t x t x t t 2 1 2 2 4 1 2 4 1

2 2 2

            

The RHS of the equation becomes And the LHS of the equation becomes Equation (5) is satisfied under the following condition

D t t h t cons x f 4 1 ; 1 ) ( ; tan ) (     

) 6 (

4

2

        

 

Dt x

e t c 



  

dx c

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 remains finite even at t=0. If S is the diffusing material Thus

 

              

      dx e t dx c S

Dt x 4

2

 Dt x 4

2 2 



 



 



   

d e

2

D S   2 

        



Dt x

e Dt S c

4

2

2 

Put and proceed, knowing that We find that Thus The constant ∝ is evaluated using equation 1