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Transcript
EEE 3394
Electronic Materials
Chris Ferekides
Fall 2014
Week 3
HELP SESSIONS
• FRIDAY: @12:10 pm … (2 hrs)
• SATURDAY @ 10:05 am …(2 hrs)
• In ENG 003 … Basement of Kopp Building
(ENG)
Kinetic Molecular Theory
What is it? What do we need it for?
• Links the “macroscopic” properties of
gases and solids to the kinetic energy
of atoms/molecules;
• Explains the pressure of gases … heat
capacity of metals … average speed of
electrons in semiconductors etc.
• Assumes that atoms/molecules of
gases, liquids, solids are in constant
motion when above absolute zero
temperature
KMT of gases
… from Newton’s 2nd Law
…dp/dt=Force
Empirical Result
N
RT
PV  
 NA 
See assumptions in text ….
..molecules in constant motion .. collision
time negligible compared to free motion ..
collisions are elastic .. no effect from
external forces etc.
Derivation
Square Container
Consider N molecules inside a cubic
volume of side a
Area A a
The change in momentum of a moleculeFace
B p  2mv
x
that collides with one of the walls is …
Force exerted by gas on a wall is equal
Face
vy
2A
Δp 2mv x
mv x
to the rate of change in momentum …
F

vx
The total pressure is equal to the total
Δt (2aGas atoms
a
)
force per unit area …
a
vx
2
2
2
2
Total force mv x1  mv x2  mv x3  ....  mv xN mNv x2
Due to random motion and collisions,
P


3
a2
a
a3
a
mean square velocity in x direction
same as in y and z directions …
average velocity is 1/3 of vx
Nmv
P
3V
2
Derivation
Compare …

Nmv 2 2 1
PV 
 N
mv 2
2
3
3

N
RT
PV  
 NA 
…where k is Boltzman’s constant
Therefore …
the mean square velocity is
proportional to T! … adding heat to a
gas … raises its temperature and total
internal energy!
Rise in internal energy per unit
temperature – HEAT CAPACITY
1
3 R
3
2
KE  mv 
T  kT
2
2 NA
2
Heat Capacity
... Energy (U) increase per unit
temperature (T)
dU
C
dT
Molar Heat Capacity Cm:
heat capacity of one mole
… for a monatomic gas
3
1
2
U  NA  mv   NAkT
2
 2
C
dU 3
3
 NAk  R
dT 2
2
… above based on constant volume … because all added energy is considered to
contribute to the temperature rise and not volume expansion (i.e. doing work to
increase volume)
Maxwell’s Principle of Equipartition of Energy
Ix= 0
... assigns 1/2kT to each “independent
way” (degrees of freedom) a
molecule can absorb energy
For example:
3 degrees of freedom …
5 degrees of freedom …
y
z
vx
x
vz
z
Iz
1

U  3 kT 
2

1

U  5 kT 
x v
y
2

Degrees of Freedom:
Monatomic gas – 3 translational…
Diatomic gas – 5 … 3 + 2 rotational
Solid – 6 … 3 kinetic energy of vibration… +
3 potential energy of “spring” i.e. bond stretching
therefore … Cm=3R
(a)
y
Iy
y axis out of paper
Molecular Velocity and Energy Distribution
R elative number of molecules
per unit velocity ( s/k m)
Term “average velocity” used to this point …
2.5
therefore a range of velocityv*values
exists…
vav
vrms
i.e. VELOCITY2DISTRIBUTION
298 K (25 °C)
v*
1.5
v
Velocities from zero (at collision) to larger av
values
…
v
rms
1
1000 K (727 °C)
The Velocity Distribution is described by the Maxwell-Boltzmann distribution function
0.5
0
0
500
1000
Speed (m/s)
3
 m  2 2
nv  4π  N
 v e
 2π  kT 
 mv 2 


 2kT 


1500
2000
Maxwell-Boltzmann Distribution for Translational
Energies (monatomic gas)

E 
 1  2 12   kT 
nE  1 N
 E e
π 2  kT 
With nE being the number of molecules per unit
volume per unit energy at an energy E!
… last term is know as the BOLTZMANN factor
Atoms have a range of energies BUT a mean
energy of 3/2kT !
And another important GENERAL relationship –
the PROBABILITY that a certain molecule in
a given system will have an energy E
nE
 Ce
N
 E 
 
 kT 
Numberof atomsperunit energy, nE
2
3
AverageKEat T1.
T1
AverageKEat T2
T2 > T1
EA
Energy, E
Thermally Activated Processes
Arrhenius Behavior …
where the rate of change is proportional to:
e
 EA 


 kT 
The Energy EA is “characteristic” of the particular
process
What are the consequences of high EA or raising the
temperature?
Thermally Activated Processes
A
A*
B
U = PE(x)
U A*
A*
EA
U A= U B
A
B
X
Displacement
Diffusion of an interstitial impurity atom in a crystal from one void
to a neighboring void. The impurity atom at position A must posses
an energy EA to push the host atoms away and move into the
neighboring void at B.
Fig 1.29
O'
After N jumps
L
Y
a
X
q = 90°
q = 0°
q = 180°
O
y
x
q = 270°
An impurity atom has four site choices for diffusion to a
neighboring interstitial vacancy. After N jumps, the impurity atom
would have been displaced from the original position at O.
Fig 1.30
Thermally Activated Processes
DIFFUSION … ??
EA for P diffusion in Si is 3.69 eV
D is the diffusion coefficient … and
DO is a constant (10.5 cm2/s)
Rms distance in t seconds is …
WATCH out for the units …
Start using eV for energy …
And K for Temperature
kT at room temp. is 0.0258 eV
D(RT)=1.08x10-61cm2/s
…in 5 minutes …
L(RT)=8.04x10-26 μm
L(200C)=1.74x10-14 μm
L(800C)=0.00171 μm
L(1100C)=0.134 μm
D  DO e
 EA 


 kT 
L  2Dt
Thermally Activated Processes
DIFFUSION … ??
EA for P diffusion in Si is 3.69 eV
D is the diffusion coefficient … and
DO is a constant (10.5 cm2/s)
Rms distance in t seconds is …
WATCH out for the units …
Start using eV for energy …
And K for Temperature
kT at room temp. is 0.0258 eV
D(RT)=1.08x10-61cm2/s
…in 5 minutes …
L(RT)=8.04x10-26 μm
L(200C)=1.74x10-14 μm
L(800C)=0.00171 μm
L(1100C)=0.134 μm
D  DO e
 EA 


 kT 
L  2Dt
Equilibrium Concentration of Vacancies
… also a thermally activated process
 Ev 
nv  N exp 
 kT 
nv = vacancy concentration
N = number of atoms per unit volume
Ev = vacancy formation energy
Phase and Phase Diagram
Phase: a HOMOGENEOUS portion of a chemical system that has same structure,
composition and properties everywhere.
Phase Diagram: A Temp vs Phase diagram in which various phases of a system are
identified by lines and regions.
100% Cu
100% Ni
Phase Diagrams – T vs. Composition
Isomorphous ??
… same morphology
everywhere
For pure Cu (or Ni) T remains
constant as liquid solidifies
(or solid melts)
Not for alloy; i.e. temperature
does not remain constant as
liquid solidifies (or solid
melts)
Initial crystal formation –
nucleation
Liquidus and Solidus lines ??
Phase Diagram
What Happens @
L(20%Ni)
L0:
Liquid
LIQUID
L0
1300
L1:
nucleation begins …
what is the composition of the solid?
go to S1
what is the composition of the liquid?
go to L1
X:
S(36%Ni)
L(20%Ni)
L1
1200
L2
S(28%Ni)
L(13%Ni)
L3
1100
TEMPERATURE(°C)
all liquid
S1
X
S3
S2
SOLID
(a-PHASE)
both solid and liquid
S4
S(20%Ni)
what are the compositions of the solid and
liquid?
1000
go to S2 and L2
0
20
40
what fraction is solid and what fraction is
wt.%Ni
liquid?
PureCu C0
Use Lever Rule
C  CO 0.28  0.20
S3:
WL  S

 53.3%
“opposite” of L1; i.e. nearly all solid!
CS  CL 0.28  0.13
What is the composition of the solid and liquid?
go to S3 and L3
C  CL 0.20  0.13
S4:
WS  O

 46.7%
C

C
0.28

0.13
ALL solid w composition of 20% Nickel
S
L
60
Phase Diagrams – Binary Eutectic
400
L
a
P
100
R'
a+L
SOLIDUS
C
183°C
19.2%
E
US
UID
LIQ
L+
SOLIDUS D
61.9%
97.5%
Q'
Q
a
SOL
VUS
200
UI
DU
S
B
SOLIDUS

SOLVUS
………… HOW DO you READ
this diagram ?
LIQ
O
S

N
ID U
• Eutectic Point/Temperature:
a
Composition of alloy that
results in the lowest melting
point temperature
a
LIQUID
M
300
SO L
a
L
• TWO solid phases (different
compositions and
Structures):
Pb-rich and Sn rich …
L
A
Temperature (oC)
• Solvus Curve:
defines the solubility limit
boundary …
R''
R
0
0
Pure Pb
20
40
60
80
Composition in wt.% Sn
100
Pure Sn
The equilibrium phase diagram of the Pb-Sn alloy. The microstructures
on the left show the observations at various points during the cooling of
a 90%Pb-10%Sn from the melt along the dashed line (the overall alloy
composition remains constant at 10 %Sn)
Pb-Sn Binary Eutectic: 10% Sn
Point Q
L:
M:
N:
O:
P:
R:
Both
What
Nearly
All
First
liquid
solid
solid
nuclei
Lisand
all
the
a;…appears
a
solid
and
a;
of
composition:
phase
ββ;a;
begin
Composition
content
– nucleation
to form
10%
of the
Sn
ofbegins;
a?
alloy?
Composition
(L
i.e.+ what
a); small
of
fraction
βamount
is a of
а-phase
and
what
fraction
is L? of the last “drops” of liquid?
What
Composition:
What
isare
thethe
composition
10%
compositions?
Sn
400
What is the
L
3% Sn … 98% Sn
L
composition of
A
What
USE
LEVER
is theRULE
LIQUID
M
the а-phase?
300
N
a
LIQ
composition
C
=0.07
of
UI
How
much is β?
a
L
DU
Go across to the
B
S
US
a+L
D
I
the
C
=0.15
а-phase?
U
And
how
much is a?
L
solidus
line and
LIQ
L+
a
E
SOLIDUS
200
SOLIDUS
SOLIDUS D
C
a
P
read
it! – 0.07
CO=0.10
183°C
97.5%
19.2%
61.9%

Q'
Q
Go
toRULE
the …
USE
Whatacross
isLEVER
the
a
100
a
composition
of
solvus line and

the L-phase?
R''
R' R
readC it! C 0.10  0.07
SOL
SOLVUS
VUS
S
read it! 0.015
ID U
WL  OC a C
 37.5%
0.98

0.10
Go across
the
βto
O 0.15
C

C

0.07
a
Wα L

 92.6%
liquidus line
Cβ and
Cα 0.98  0.03
SO L
Temperature (oC)
O
CO  CL 0.20  0.13
 40
 46.7%
20
60
80
100
C

C
0.28

0.13
Pure Pb S
Pure Sn
L Composition in wt.% Sn
0
W
0S 