Rate Process in Solids: Atomic Arrangement. Activation Energy Barrier. Energy E . It Should Be Supplied To Them
Rate Process in Solids: Atomic Arrangement. Activation Energy Barrier. Energy E . It Should Be Supplied To Them
Rate Process in Solids: Atomic Arrangement. Activation Energy Barrier. Energy E . It Should Be Supplied To Them
e ( E* E ) / KT
T = Temperature in Kelvin.
4-35
Rate Process in Solids (Cont..)
4-36
Rate Process in Solids (Cont..)
4-37
• Calculate (a) the equilibrium number of vacancies per cubic
meter in pure copper at 500 degree Celsius and (b) the
vacancy fraction at 500 degree Celsius. Assume the energy of
formation of a vacancy in pure copper is 0.90eV. Given C=1,
k=8.62X 10-5 eV/K, density of copper=8.96Mg/m3, atomic
mass for copper=63.54g.
6.02 X 10 23 atoms 1 8.96 X 106 g
N
at.mass 63.54 g / at.mass m3
8.49 X 10 28 atoms / m 3
a) nv Ne Ev / kT
0.90eV
(8.49 10 28 )exp 5
(8 .62 10 eV / K )( 773 K )
(8.49 10 28 ) e 13.5
1.2 X 10 23 vacancies / m 3
b)
nv 0.90eV
exp 5
N (8 . 62 10 eV / K )( 773 K )
e 13.5 1.4 10 6
Arrhenius Equation
• The rate of chemical reaction is given by
Arrhenius equation.
Rate of reaction = Ce-Q/RT
4-38
Solidification of Metals
• Arrhenius equation can also be written as
ln (rate) = ln ( C) – Q/RT
Or Log10 (rate) = Log10 (C) – Q/2.303 RT
Which is similar to
Y = b + m X
Which is equation of a straight line
With Y intercept as ‘b’ and slope ‘m’.
Y Log10(rate)
X (1/T)
b Log10(C)
m Q/2.303R
5-2
Vacancy or Substitutional Diffusion
mechanism
• Atoms diffuse in solids if
Vacancies or other crystal
defects are present
There is enough activation
energy
• Atoms move into the vacancies present.
• More vacancies are created at higher
temperature.
• Diffusion rate is higher at high
temperatures.
5-3
Substitutional Diffusion
• Example: If atom ‘A’
has sufficient activation
energy, it moves into the
vacancy self diffusion.
5-4
Interstitial Diffusion mechanism
• Atoms move from one
interstitial site to another.
• The atoms that move must
be much smaller than the
matrix atom.
• Example:
Carbon interstitially
diffuses into BCC α or FCC
Interstitial atoms
γ iron. Matrix
atoms Figure 4.37
5-5
Steady State Diffusion
• There is no change in concentration of solute atoms at
different planes in a system, over a period of time.
• No chemical reaction occurs. Only net flow of atoms.
Distance x
Figure 4.38
5-6
Fick’s Law
• The flux or flow of atoms is given by
J = Flux or net flow of atoms.
dc
J D
D = Diffusion coefficient.
dc
dx = Concentration Gradient.
dx
5-7
Diffusivity
• Diffusivity depends upon
Type of diffusion : Whether the diffusion is
interstitial or substitutional.
Temperature: As the temperature increases
diffusivity increases.
Type of crystal structure: BCC crystal has lower
APF than FCC and hence has higher diffusivity.
Type of crystal imperfection: More open
structures (grain boundaries) increases diffusion.
The concentration of diffusing species: Higher
concentrations of diffusing solute atoms will
affect diffusivity.
5-8
Non-Steady State Diffusion
• Concentration of solute atoms at any point in metal
changes with time in this case.
• Ficks second law:- Rate of compositional change is equal to
diffusivity times the rate of change of concentration
gradient. Plane 1 Plane 2
dC x d dcx
D
dt dx dx
5-11
Carburizing
C%
5-12 (After “Metals handbook,” vol.2: “Heat Treating,” 8th ed, American Society of Metals, 1964, p.100)
Impurity Diffusion into Silicon wafer
• Impurities are made to diffuse into silicon wafer to change
its electrical characteristics.
• Used in integrated circuits.
• Silicon wafer is exposed to vapor of impurity at 11000C in a
quartz tube furnace.
• The concentration of
impurity at any point
depends on depth and
time of exposure.
Figure 4.44
5-14
Effect of Temperature on Diffusion-Example
• If diffusivity at two temperatures are determined,
two equations can be solved for Q and D0
Example:
Solution:
Effect of Temperature on Diffusion-
Example
• If diffusivity at two temperatures are determined, two
equations can be solved for Q and D0
• Example:-
The diffusivity of silver atoms in silver is 1 x 10-17 at
5000C and 7 x 10-13 at 10000C. R=8.314J/mol.K
Therefore, D1000 exp( Q / RT2 ) Q 1 1
exp
D500 exp( Q / RT1 ) R T2 T1
7 1013 Q 1 1
exp
1 1017 R 1273 773
Solving for activation energy Q
Q 183KJ / mol
5-15
Diffusivity Data for Some Metals
Solute Solvent D0 Q
(M2/S) KJ/m
ol
Carbon FCC Iron 2 x 10-5 142
5-16 (After L.H. Van Vlack. “Elements of Materials Science and Engineering.” 5 th ed., Addison-Wesley, 1985. P.137.)