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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