1. EEE 3394 Electronic Materials
Chris Ferekides
Fall 2014
Week 3

2. HELP SESSIONS FRIDAY: @12:10 pm … (2 hrs) SATURDAY @ 10:05 am …(2 hrs)
In ENG 003 … Basement of Kopp Building (ENG)

3. 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
See assumptions in text ….
..molecules in constant motion .. collision time negligible compared to free motion .. collisions are elastic .. no effect from external forces etc.

4. Derivation Consider N molecules inside a cubic volume of side ay a G s t o m A r e S q u C n i F c B x
Consider N molecules inside a cubic volume of side a
The change in momentum of a molecule that collides with one of the walls is …
Force exerted by gas on a wall is equal to the rate of change in momentum …
The total pressure is equal to the total force per unit area …
Due to random motion and collisions, mean square velocity in x direction same as in y and z directions … average velocity is 1/3 of vx

5. Derivation Compare … …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

6. Heat Capacity ... Energy (U) increase per unit temperature (T)
Molar Heat Capacity Cm:
heat capacity of one mole
… for a monatomic gas
… 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)

7. Maxwell’s Principle of Equipartition of Energyz ( a ) x I y a i s o u t f p e r z =
... 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 … v x z y
Degrees of Freedom:
Monatomic gas – 3 translational…
Diatomic gas – 5 … rotational
Solid – 6 … 3 kinetic energy of vibration… +
3 potential energy of “spring” i.e. bond stretching
therefore … Cm=3R

8. Molecular Velocity and Energy Distribution
Term “average velocity” used to this point …
therefore a range of velocity values exists…
i.e. VELOCITY DISTRIBUTION
Velocities from zero (at collision) to larger values …
The Velocity Distribution is described by the Maxwell-Boltzmann distribution function . 5 1 2 S p e d ( m / s ) K 7 C 9 8 v * a r R l t i n u b o f c y k

9. Maxwell-Boltzmann Distribution for Translational Energies (monatomic gas)y , T 1 2
> A v a K t . N u m b o f s p i
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

10. Thermally Activated Processes
Arrhenius Behavior …
where the rate of change is proportional to:
The Energy EA is “characteristic” of the particular process
What are the consequences of high EA or raising the temperature?

11. Thermally Activated Processes

14. 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 eV
D(RT)=1.08x10-61cm2/s
…in 5 minutes …
L(RT)=8.04x10-26 μm
L(200C)=1.74x10-14 μm
L(800C)= μm
L(1100C)=0.134 μm

15. 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 eV
D(RT)=1.08x10-61cm2/s
…in 5 minutes …
L(RT)=8.04x10-26 μm
L(200C)=1.74x10-14 μm
L(800C)= μm
L(1100C)=0.134 μm

16. Equilibrium Concentration of Vacancies
… also a thermally activated process
nv = vacancy concentration
N = number of atoms per unit volume
Ev = vacancy formation energy

17. 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

18. 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 ??

19. Phase Diagram q d C 1 2 3 4 6 w t . % N i u r e 8 T M R ° a2 3 4 6 L I Q U D S O ( a - P H A E ) C w t . % N i X u r e 8 q d T M R
What
L0:
all liquid
L1:
nucleation begins …
what is the composition of the solid?
go to S1
what is the composition of the liquid?
go to L1 X:
both solid and liquid
what are the compositions of the solid and liquid?
go to S2 and L2
what fraction is solid and what fraction is liquid?
Use Lever Rule
S3:
“opposite” of L1; i.e. nearly all solid!
What is the composition of the solid and liquid?
go to S3 and L3
S4:
ALL solid w composition of 20% Nickel

20. Phase Diagrams – Binary Eutectic
Solvus Curve:
defines the solubility limit boundary …
Eutectic Point/Temperature:
Composition of alloy that results in the lowest melting point temperature
TWO solid phases (different compositions and Structures):
Pb-rich and Sn rich …
………… HOW DO you READ this diagram ?

21. Pb-Sn Binary Eutectic: 10% Sn
Point R:
All solid a and β; Composition of a? Composition of β
3% Sn … 98% Sn
How much is β?
And how much is a?
USE LEVER RULE …
Point Q
First nuclei of β begin to form
What are the compositions?
Point P:
All solid a;
Composition: 10% Sn
Point O:
Nearly all solid a;
What is the composition of the last “drops” of liquid?
Point M:
First solid appears – nucleation begins; (L + a); small amount of а-phase
What is the
composition of
the а-phase?
Go across to the
solvus line and
read it!
Point N:
Both L and a;
What is the
composition of
the а-phase?
Go across to the
solidus line and
read it! – 0.07
the L-phase?
liquidus line and
read it! 0.015
Point N:
What is the phase content of the alloy? i.e. what fraction is a and what fraction is L?
USE LEVER RULE
Ca=0.07
CL=0.15
CO=0.10
Point L:
All liquid … composition: 10% Sn