Statistical thermodynamics of Solids: Introduction of structured solids

Transcription

1 Solid State Theory Physics 545 The lattice specific heat

2 Statistical thermodynamics of Solids: Kinetic energy Introduction of structured solids Law wof Dulong uo and Petit (Heat capacity) c 1819 Einstein Model of Crystals 1907 Born and von Karman approach 191 Debye Model of Crystals 191 Electronic energy Electronic energy Fermi level 196 Fermi-Dirac distribution

3 The crystal stores energy as: Law of Dulong and Petit - Kinetic energy of the atoms under the form of vibrations. According to the equipartition of energy, the kinetic internal energy is f. ½. k. T where f is the degree of freedom. Each atom or ion has degrees of freedom E K = / N k T - Elastic potential energy. Since the kinetic energy convert to potential energy and vice versa, the average values are equal E pot = N (½ K x ) = N x (½ k T) The stored molar energy is then: E=E E K +E pot =N N A kt= RT C=dE/dT=R =

4 Law of Dulong and Petit Within this law, the specific heat is independent of: - temperature - chemical element - crystal structure At low temperatures, all materials exhibit a decrease of their specific heat Classical harmonic oscillator Quantum + Statistical mechanics

5 Normal modes and phonons Description of lattice vibrations has so far been purely classical because we solved classical equations of motion to find the vibrational modes and dispersion relation of the lattice. In the case of a harmonic potential, the classical approach gives the same modes and dispersion relation as the quantum approach. Each mode is the mode of vibration of a quantum harmonic oscillator with wave vector k and polarisation s and quantised energy: E k, s 1 ( k ), n k, s = ω ( k ) = k, s + s / k T n ω e s 1 B 1 where n is the number of phonons in the mode k,s. A phonon is a bosonic particle with wave vector k and polarization s The more phonons in the mode, the greater the amplitude of vibration.

6 Phonon Energy The linear atom chain can only have N discrete K ω is also discrete Distance The energy of a lattice vibration mode at frequency ω was found to be u = n + 1 ω hω where ħω can be thought as the energy of a particle called phonon, as an analogue to photon n can be thought as the total number of phonons with a frequency ω, and follows the Bose-Einstein statistics: n = 1 ω exp 1 k B T Equilibrium distribution

7 Total Energy of Lattice Vibration 1 ω ( ) 1 n ω K, p + K p E l =, p K p: polarization(la,ta, LO, TO) K: wave vector Experimental observations of lattice specific heat preceded inelastic neutron scattering. Model crystal: p atoms per unit cell ) N unit cells ) pn harmonic oscillators Thermal energy (quanta) excites crystal and any number of excitations into quantized energy levels for oscillators.

8 Bose-Einstein Statistics n s 1 k = ω exp s k 1 k T B Number of excitations in particular mode, k, s In harmonic approximation, total energy density U ω ω s k = k + V s,s V k k,s ω exp s k 1 k T B s: polarization(la,ta, LO, TO) k: wave vector

9 Specific Heat (at Constant Volume) C V ω ( ) s k ω ( ) U 1 = = T V V T k k,s s exp 1 k T B s: polarization(la,ta, LO, TO) k: wave vector

10 Volume of k-space per allowed k value is k F V = F 8π ( k) ( k) Δk For Δk 0 (i.e., V ) k 1 dk Lim F V 8 π V ( k) = F( k) k 8 Δk = 8ππ V C V = s T d 8π k ωs ( k ) ω ( k) s exp 1 k T B

11 Density of Phonon States in 1D Ali linear chain of fn=10 atoms with two ends jointed a Only N wavevectors (K) are allowed(one per mobile atom): K= -8π/L -6π/L -4π/L -π/l 0 π/l 4π/L 6π/L 8π/L π/a=nπ/l/ / Only 1 K state lies within a ΔK interval of π/l # of states per unit range of K is: L/π DOS # of K-vibrational modes between ω and ω+dω : D( ω ) = L 1 π dω / dk

12 Density of States in D K x, K y, K z π 4π = 0; ± ; ± ;...; ± L L Nπ L N : # of atoms K z K y K x π/l VK 1 D ( ω) = ; V = L π dω / dk

13 Density of States Define D(ω) such that D(ω) dω is total no. of modes with frequencies in range ω to ω +dω per unit volume of crystal dω ds D( ω) dω = d δ ω ω) dk nds 8π k ( - s = = 8π 8π ω ( ) D ω = 1 8π k ds ω and for any function Q(ω s (k)) s Hence ( k ) dk k Q ( ω s ( k )) = dω D ( ω ) Q ( ω ) 8π ( k) ( k) ω CV = D( ω) dω T ω exp 1 k T B k ( k )

14 Lattice Specific Heat 1 ω ( ω ) K, p + K p E l = n, p K p: polarization(la,ta, LO, TO) K: wave vector E l = p ( ω ) ( ω ) 1 dk 1 4πK dk n K, p + K, p = n + ω ω L K, p K, p ( π L ) ( π L ) p Dispersion Relation: K = g ( ω ) Energy Density: = n ( ω ) + ω D ( ω ) d ω VK 1 D( ω ) = π d ω / dk l p Density of States (Number of K-vibrational modes between ω and ω+dω) Lattice Specific Heat: C l d l = dt = p 1 d n dt ωd ( ω ) dω

15 High-Temperature Classical Limit: ω = x «1 k B T C = k T D( ω) dω = k D( ω) dω = pnk TT V B B B which h is the same as the classical l result (Dulong and Petit law: R J/mole/K for a monatomic solid). The reason for this is because at this level of approximation the energy associated with a quantum of lattice vibration, ω, exactly cancels out and therefore it doesn't matter how big that quantum is (including zero).

16 Low Temperature Limit: Only low-frequency acoustic modes excited. ω ω =c k for each hbranch, where is initial i i s s c s = s k slope of the particular phonon dispersion curve. (Note that cs is related to the elastic constant for the mode, e.g., for [100]L elastic waves c s = v [ 100 ] = L c ρ 11 h i l i d where c 11 is an elastic constant and ρ the density

17 Low Temperature Limit: D π 1 k sinθ 1 k π 1 k 1 ω = = = 0 = c c π c π c ( ω ) ( ) dθ [ cosθ] π 0 ( π) s s s s s s s s = ω π 1 c L c c T 1 T = ω π c C C V V ω max ω ω = ω d T ω 0 exp 1 π c k T B ω max ω x = dx = dω k T k T k B 4 kbt ( ) k BT x π k B = dx = k x B T ( c) π 5 e 1 0 T c at low temperatures the specific heat is proportional to T. k B

18 Einstein Model Each molecule in the crystal lattice is supposed to vibrate isotropically about the equilibrium point in a cell delimited by the first neighbors, which are considered frozen. System of N molecules Motions in the x, y and z axis are Independent and equivalent the system can be treated as N independent one- dimensional harmonic oscillator

19 Einstein Model System of 1-Dim Harmonic Oscillator Quantized expression of the energy: ε v = hυ (v+1/) v = 0, 1,,... Partition function (without attributing 0 to the ground state) q = Σ e -(hυ(v+1/)/kt) = e -(hυ/kt) Σ e -(hυ/kt) v Considering the vibrational temperature θ = θ Ε = hυ/k e -θ/t θ/ q = 1- e -θ/t The molecular internal energy U l =-d[ln(q)] NV = kt molecular / dβ ] N, d[ln(q)] / dt ] NV N, 1 U molecular = k θ (1/ + ) e θ /T - 1

20 Einstein Model energy of the system System of N -Dim Harmonic Oscillators Q = q N U = N U molecular Intern nal energ gy / Nhυ 8000 / Nhυ + NkT U = N k θ (1/ + 1 ) e θ /T kt/hυ

21 Einstein Model the heat capacity of the system System of N -Dim Harmonic Oscillators The heat capacity of the crystal is then C = du / dt Nk 1.0 pacity Heat ca T/θ U = N k θ (1/ + 1 ) e θ /T - 1 e θ/t C = Nk(θ/T) (e θ/t 1)

22 Einstein Model Assumed model for crystal to be n harmonic oscillators, each of frequency, ω ( ) E k BθE = ωe D ω = δ ω ω n E ( ) ( ) ω ( k) V Substituting this into Equation T ω ( k ) C V = nk B ωe k BT ω exp ( ω ) C = D d ω exp 1 k T ωe θe exp exp k BT θe T = R E T θe 1 exp 1 k BT T B = R F E E θ T T» θ E, F E θ T E 1, so C V R (the classical high-temperature limit). T «θ E, F E θ T E θ T E exp θ T E C V dominated by the exponential term, which is not found experimentally at low temperatures.

23 Einstein Model comparison with experiment The value of θ Ε = 15 K was given to produce an agreement with the experiment at 1,1 1 K. θ Ε or ω Ε = kθ Ε / is the parameter that distinguishes different substances: ω Ε Α (a Ε/m) 1/ Comparison of the observed molar heat capacity of diamond (+) with Einstein s model. (After Einstein s original paper-1907) where Ei is Young s modulus dl m is atomic mass and a is the lattice parameter Einstein model gives also a qualitatively quite good agreement on term of θ Ε calculated from the elastic properties

24 Einstein Model results and limitation The Einstein Model of crystals takes into account the alteration of the heat capacity by: - temperature - chemical element - crystal structure This model explained the decrease of the heat capacity at low temperature. However: This decrease is too fast! The experimental results evolve as T Reason is that the Einstein model does not consider the collective motion and only consider one vibrational frequency.

25 Born and von Kármán approach System of N atoms possess N degrees of freedom, all expressing vibrational motion. Thus, the whole crystal has N normal modes of vibration characterized by their frequencies υ i =ω i /π THE LATTICE VIBRATIONS OF THE CRYSTAL ARE EQUIVELENT TO N INDEPENDENT OSCILLATORS N E = Σ hυ + (hυ i /kt) -1) -1 i (1/ (e )

26 Propagation of sound wave in solids notion This propagation could be solved using the classical concepts since the atomic structure (dimensions) can be ignored in comparison to the wavelength of a sound wave. The -D wave equation φ(r) + k φ(r) = 0 where: k is the magnitude of the wave vector k = π/λ Wave phase velocity v = λ υ =λ ω/π = ω/k

27 Propagation of sound wave in solids standing waves in a box The -D wave equation of motion solved in a cubic box with the side L Φ n1 n n (r) = A sin(n 1 π x / L) sin(n π y / L) sin(n π z / L) The wave vector in the Cartesian coordinates is k(πn 1 /L, πn /L, πn /L) In the k space, formed by the allowed values of k(n i = 1,,...), is composed of cubic point lattice with the separation of π/l and the volume of V u = (π/l).

28 Propagation of sound wave in solids Density of states Defining the density of states come to the determination of the number of normal modes of standing waves with the lying magnitude between k and k+dk. f(k) dk = (1/8) (4πk )dk/(π/l) = Vk dk/(π ) In term of circular frequency: f(k) dk = f(ω) dω =(Vk /π ) (dk/dω) dω = V ω dω /( v v g π ) Where v g = dω/dk is the group velocity

29 Propagation of sound wave in solids Density of states In a non dispersing medium v g =v f(k) dk = f(ω) dω = V ω dω /( v π ) The wave vector has three independent modes: 1 longitudinal and transversal modes f(ω) dω = V ω dω /( π ) (1/v L + /v T ) In an isotropic i Medium v L = v T =v m f(ω)dω =Vω ω dω /( v m π )

30 Debye Model Lattice vibrations are regarded as standing waves of the atomic planes displacement It is assumed that all normal mode frequencies satisfy the equation of fthe density of states t An upper limit for frequencies is, however, set such as ƒ ω f(ω) dω = N D f(ω) )dω = 9Nω dω/ω D Now the sum can be replaced with an integral Σ N... = ƒ ω...f(ω) )dω D

31 Debye Model Debye Approximation: Debye Density of States ω = v s K ( ω ) dg ω g D ( ω ) = = π dω π vs Frequ uency, ω ω = v s K Number of Atoms: N = 4 π K D ( π ) L K Debye cut-off Wave Vector K D = ( ) 6π η 1 0 Wave vector, K π/a Debye e Cut-off Freq. ω D = v s K D ( ) s 6π η Debye Temperature [K] ωd v Debye Temperature θ D = = B 150 NaBr 4 k k B B 1 C(dimnd) 1860 Ga 40 Si 65 NaF 49 Ge 60 NaCl 1 Al 94 NaI 164

32 Debye Model The energy of the crystal N U = Σ ε 1 i N ħω = Σ [(1/)ħω i i + ] ħω 1 e i kt -1 ω = ƒ D ħω [(1/)ħω + ] D(ω) dω ħω 0 e kt x = Nkθ D + NkT ƒ D x dx 8 x e x 0-1 D D Where θ D =ħω D /k x D = ħω D /kt x = ħω/kt

33 Debye Model The heat capacity of the crystal E = 9 8 x D 9 x Nkθ + NkT dx D x x e D 1 0 T 0 X T X 0 x D / π 4 /15 E = 9Nħω D /8 Vibrational zero-point energy E=9Nkω = D /8 + NkT C v = Nk v C v = de/dt] v = 0

34 Debye Model The heat capacity at low temperature C v =(du/dt) v T enters this expression only in x ƒ D 4 x D 0 the exponential term (β) Cv= Nk { ƒ x 4 e x dx } (e x -1) D x D ƒƒ When T<<θ x 4 e x dx = 4π 4 /15 D x (ex -1) 0 1 Nk T C V = π 4 5 θ D

35 Debye Model-Experiment The Debye Model gives good fits to the experiment; however, it is only an interpolation formula between two correct limits (T = 0 and infinite)

36 Lattice Specific Heat Energy Density Specific Heat C l l = p ω 0 D n θ D T = T 9ηk B θ D 0 1 ω π v 9ηk dω = θ D θ dx T 1 4 ( ) B ω + e x x 4 dx x ( e 1 ) 0 1 s D 0 T x e 10 7 C = ηk B = J m K x x = ω k T B 10 6 When T << θ D, T 4, C T l l (J/m -K) Specific Heat, C Quantum Regime C T Classical Regime Temperature, T (K) Diamond θ D =1860 K 10 4

37 Einstein-Debye ebye Models Lattice structure of Al Cubic Closest Packing The lattice parameter a = 0,5 nm The density ρ=,7 g/cm The wave velocity v =,4 km/s Θ elst Θ E / Θ elst = 079 0,79 Θ D / Θ elst = 0,95

38 Lattice parameter Cubic close packed, (a) Hexagonal close packed (a, c) Body centered cubic (a) Cu (.6147) Be (.856,.58) Fe (.8664) Ag (4.0857) Mg (.094, 5.105) Cr (.8846) Au (4.078) Zn (.6649, ) Mo (.1469) Al (4.0495) Cd (.9788, ) W (.1650) Ni (.540) Ti (.506, ) Ta (.06) Pd (.8907) Zr (.1, ) Ba (5.019) Pt (.99) Ru (.7058, 4.816) Pb (4.950) Os (.75, 4.191) Re (.760, 4.458)

39 Debye Temperature 9 9 x Ε = Nkθ + NkT ƒ D x dx D 8 x e x 0-1 D Where θ D =ħω D /k x D = ħω D /kt x = ħω/kt Nk C V = 1 π 4 T 5 θ D

40 The limit of the Debye Model The electronic contribution to the heat capacity was not considered

41 Electronic contribution Fermi level At absolute zero temperature, electrons pack into the lowest available energy, respecting the Pauli exclusion principle each quantum state tt can have one but btonly one particle til Electrons build up a Fermi sea, and the surface of this sea is the Fermi Level. Surface fluctuations (ripples) of this sea are induced by the electric and the thermal effects. So, the Fermi level, is the highest energetic occupied level at zero absolute 41

42 Electronic c contribution o Fermi function The Fermi function f(e), drown from the Fermi-Dirac statistics, express the probability that a given electronic state will be occupied at a given temperature. 1.0 EE E-E F < 0 ) f(e 0.5 E-E F > Temperature

43 Electronic contribution to the internal energy Orbitals are filled starting from the lowest levels, and the last filled or orbital will be characterized by the Fermi wave vector K F The total number of electron in this outer orbital is: K L F N = f ( k ) dk = k dk T π Because electrons can 0 Adopt spin orientations N T = V π k F k F = N V T π

44 Electronic contribution to the Electronic contribution to the internal energy internal energy The wavefunction of free electron is: ). ( ), ( t kx i Ae t x +ω = Ψ Its substitution in the Schrödinger equation: ) ( ), ( ) ( ) ( t x t x t x E Ψ = Ψ = Ψ ), ( ), ( x m t x t x E Ψ Ψ k E k m E =

45 Electronic contribution to the internal energy Fermi Energy N = T [ ] π m F m V E = k F F E F k Fermi Temperature T = F E F k

46 Temperature e effect ec on electrons ec Metal K Na Li Au Ag Cu NT /V (10^ cm^) KF (1/A ) EF (ev) TF (K) Only electrons near from Fermi ilevel lare affected dby the temperature.

47 Electronic contribution in the heat capacity of a metal [ ] E T C V N e e v, = [ ] T V N v, E f E T C V N i i i e v, 1 ) ( = = dn E f E C e, ) ( dn E f E T C V N i i e v, 0 ) ( = T k E m V C F e v 1/ / = k C F v π

48 Summary The nearest model describing the thermodynamic properties of crystals at low temperatures is the one where the energy is calculated considering the contribution of the lattice vibrations in the Debye approach and the contribution of the electronic motion (this is of importance when metals are studied). d) C = α. T +γ. T v v

49 Summary Terms that replaced the partition function are: Density of state (collective motion) Fermi function (electronic contribution)