Chapter 6 Free Electron Fermi Gas (FEFG) Phys 175A Dr. Ray Kwok SJSU Classification of Solids Many ways to classify solids (a) lattice structure and (b) crystal bonding (c) measurable properties One of the easiest and earliest properties observed is the dc electrical conductivity Electrical Properties Some solids conduct current at all temperatures and, generally, the resistivity of such solids increases when temperature increases. We call them METALS Other solids stop conducting at low temperatures and their resistivity decreases with increasing temperature. We call them INSULATORS SEMI-CONDUCTORS are basically insulators with higher conductivity (due to smaller band gap). Another way to understand this is to examine the band structure of the compound (next chapter). The Free Electron Gas Model (FEG) U(x) for a 1-D crystal lattice: Simple and crude finitesquare-well model: U U=0 Can we justify this model? How can one replace the entire lattice by a constant (zero) potential? Conduction Band R 0 3p -5 V(r)˜ -1/r Atom 2 Atom 1 R Energy (eV) Discrete Electronic Energy States -10 -25 Split Energy Levels 3s Free electron band Na: 1s22s22p63s1 monovalent alkali -30 Atom 1 -35 0 Atom 2 2p 3.67 5 10 Interatomic Distance (Å) Justifications of the FEG Model 1. Metals have high electrical conductivity and no apparent activation energy, so at least some of their electrons are “free” and not bound to atoms. Measured V=IR. 2. Coulomb potential energy of positive ions U ∝ 1/r is screened by bound electrons and is weaker at large distances from nucleus. It’s further screened by mobile electrons around to minimize U ~ 0. 3. Electrons would have lowest U (highest KE) near nuclei, so they spend less time near nuclei and more time far from nuclei where U is not changing rapidly. Pauli Exclusion Principle also forbid mobile electrons to be right too close to the core. 4. Therefore model the behavior of the free electrons with U = 0 inside the volume of the metal and a finite potential step at the surface. Each atom has n0 free electrons, where n0 = chemical valence. Resistance comes from electrons interacting with lattice through occasional collisions: Particle in a box (QM - wave) Time-independent Schrödinger Equation: With U = 0: h2 2 − ∇ ψ = Eψ 2m 1. Wave functions: 2. Energies: h2 2 − ∇ ψ + U ψ = Eψ 2m ψ = Ae h2k 2 E= 2m vv i ( k ⋅r −ωt ) ∇ 2ψ = − 2mE ψ 2 h ∇ 2ψ = −k 2ψ Traveling waves (plane waves) E electronic dispersion Parabolic energy “bands” kx Particle in a 3D box 2 πx ψ n (x) = sin n L L 1-D Potential Well Here, n is the quantum number π2h 2 2 En = n 2 2mL n>0 3-D “Cubic” Potential Well (with sides length L) 2 ψ n1n 2n3 (x) = L E n1n 2n3 3/ 2 πx πy πz sin n1 sin n 2 sin n 3 L L L π2 h 2 2 2 2 2 n n n E n = + + = 1 2 3 0 2mL2 ( ) Degenerate states with n1, n2, n3. (ni > 0) Example: 34 Electrons in a box Quantum numbers (n1, n2, n3), E = n2Eo The lowest energy for this system is 3E0, which corresponds to n1 = n2 = n3 = 1 Thus only 2 (two) electrons can have this energy: one with spin ↑ and one with spin ↓ Next energy level (6E0), for which one of n’s is 2 (3 combinations) Thus total of 6 (six) electrons can have this energy Next energy level (9E0) can also accommodate 6 electrons What are the combinations of n’s for this energy level? Next energy level (11E0) also accommodates 6 electrons What are the n’s? Next energy level (12E0) is two-fold degenerate 34 Electrons in a box (cont.) So far we have placed 22 electrons, so we need to add 12 more electrons What is the next energy level? 14E0 What are n1, n2, and n3? What is degeneracy? This is the energy carried by each of the 12 electrons (what kind of energy is this?) We placed all 34 electrons!! Reading Energy Diagram Energy is plotted in terms of Eo 18 17 (3 ,2 ,2 ) 16 In this example EF = 14 Eo Configuration shown is at T = 0. Probability to find electrons with energy greater than 14Eo is 0, and below 14 Eo is 1 (at T = 0) 0 Energy (in units of E ) Values of n’s are shown on the right. 15 14 (3 ,2 ,1 ) 13 12 (2 ,2 ,2 ) 11 (3 ,1 ,1 ) 10 9 (2 ,2 ,1 ) 8 7 6 (2 ,1 ,1 ) 5 4 3 2 1 0 (1 ,1 ,1 ) Density of States: FEG By using periodic boundary conditions for a cubic solid with edge L and volume V = L3, we define the set of allowed wave vectors: 2πn x kx = L such that: ky = 2πn y L kz = 2πn z L n x , n y , n z = ±1, ± 2, ± 3, ... (No upper limit) r r r r ψ( r + L x ) = ψ( r ) = ψ( r + L y ) = ψ( r + L z ) 2π so the volume in k-space per state is: L and the density of states in k-space is: 3 1 V N(k ) = = 3 3 (2π / L ) 8π Just as in the case of phonons! Fermi Surface Since the FEG is isotropic, the surface of constant E in k-space is a sphere. Thus for a metal with N electrons we can calculate the maximum k value (kF) and the maximum energy (EF). # states V 4 (volume _ k ) → N = 2 3 πk F3 # e − = 2 8π 3 volume _ k 1/ 3 2 N k = 3π Fermi wave vector F V Fermi energy ( 2 = 3π n h 2 k F2 h 2 EF = = 3π 2 n 2m 2m ( 2 spin states per k N = # of mobile electrons ky ) 1/ 3 kF kx ) 2/3 kz Fermi sphere Fermi Parameters Fermi wavevector Fermi Energy ( k F = 3π 2ηe EF = h 2 k F2 ) 2 = 1 Fermi Energy, EF ( h 3π 2ηe 2m 2m hk h 3π 2ηe Fermi Velocity: vF = F = m m E Fermi Temp. TF = F kB ( Element Electron Density, ηe 28 -3 [10 m ] Na 2.65 Cu 8.47 Ag 5.86 Au 5.90 Fe 17.0 Al 18.1 Sn 14.8 Φ: Work Function 3 Fermi Energy EF [eV] 3.24 7.00 5.49 5.53 11.1 11.7 10.2 ) 1 ) 2 Vacuum Level Energy 3 3 Band Edge Fermi Temperature 4 TF [10 K] 3.77 8.16 6.38 6.42 13.0 13.6 11.8 Fermi Wavelength λF [Å] 6.85 4.65 5.22 5.22 2.67 3.59 3.83 Fermi Velocity 6 vF [10 m/s] 1.07 1.57 1.39 1.40 1.98 2.03 1.9 Work Function Φ [eV] 2.35 4.44 4.3 4.3 4.31 4.25 4.38 Density of States N(E) The differential number of electron states in a range of energy dE or wavevector dk is: Free electron gas dN = N ( E )dE = 2 N (k )d 3 k # of e per E density of state h2k 2 E= 2m 1/ 2 2mE k = 2 h 2 N ( E ) = 2 N (k ) N (E) = Or 2 4πk dk V k mV 2mE V 4πk = 2 3 = 2 2 = 2 2 2 dE 8π dE / dk π h k / m h π h ( ) 1/ 2 V 3 2 m E π 2 h3 V 4 V 2mE N = 2 3 πk 3 = 2 2 3π h 8π 3 1/ 2 N (E) = 1/ 2 2 N(E) 3/ 2 E 1/ 2 dN V 2mE 2m V 3 = 2 = 2 3 2m E 2 2 dE 2π h h π h ( ) EF Fermi-Dirac Distribution The probability distribution function, f(E), which describes the probability that a state with energy E is occupied for electrons is the Fermi-Dirac Distribution Function at T = 0 0, for E > E F f (E) = 1, for E ≤ E F Fermi-Dirac Distribution T > 0 Occupation Probability, f f (E) = 1 e ( E −µ ) / k BT + 1 µ is the Chemical Potential which is equal to EF at T=0. Often, µ and EF are mixed up. kBT 1 1/2 T=0K Vacuum Energy T →∞ IncreasingT 0 Electron Energy,E EF Work Function,Φ Heat Capacity of the FEG 19th century puzzle: each monatomic gas molecule in sample at temperature32 kT T has energy, so if the N free electrons in a metal make up a classical “gas” they should behave similarly. Eel = N ( 32 kT ) or expressed per mole (n): So the electronic contribution to the molar heat capacity would be expected to be Eel 3 N = 2 kT = 32 N A kT = 32 RT n n d Cel = dT Eel 3 = 2R n This is temperature independent!!! Only half of the 3R we found for the lattice heat capacity at high T. But experiments show that the total C for metals is only slightly higher than for insulators, which conflicts with the classical theory! Heat Capacity of the FEFG FEG + Pauli + Fermi-Dirac (QM) So at temperature T the total energy is: And the electronic heat capacity is: ∞ ∞ EN(E ) E el = ∫ Ef (E ) N(E ) dE = ∫ (E −µ )/ kT dE e 1 − 0 0 dE el 1 EN (E)(E − E F )e (E − E F )/ kT C el = = 2∫ dE 2 ( ) E − µ / kT dT kT 0 e −1 ∞ ( ) This integral is complicated because µ is also a function of T. Treatment of this approximation is given in advance texts. The result is: C el = π2 3 k 2 N ( E F )T Cel ∝ T Understand linear T dependence # electrons that can absorb “extra” thermal energy ≅ (2kT ) N ( EF ) f ( EF ) ≅ kT N ( EF ) Only electrons near the Fermi surface can gain energy !! total thermal energy E (T ) ≅ of electrons at T ≈2kT N(E) N(E)f(E) E0 + ( 32 kT )kT N ( EF ) E (T ) ≅ E0 + 32 k 2T 2 N ( E F ) FEG heat capacity at T Remarkably close to the exact result! Cel = dEel ≅ 3k 2 N ( EF )T dT 2 Cel = π3 k 2 N ( EF )T = γ T E EF Total C But the linear dependence is impossible to measure directly, since the heat capacity of a metal has two contributions. Now for a metal at low temperatures we can write the total heat capacity: C(T ) = Cel + Clattice = γ T + αT 3 Heat Capacity of Metals (expt) Results for simple metals (in units mJ/mol K) show that the FEG values are in reasonable agreement with experiment, but are always higher: The discrepancy is “accounted for” by defining an effective electron mass m* that is due to the neglected electron-ion interactions. i.e. phonons drag electrons & make them move slower. γexpt/ γFEG = Metal γexpt γFEG Li 1.63 0.749 2.18 Na 1.38 1.094 1.26 K 2.08 1.668 1.25 Cu 0.695 0.505 1.38 Ag 0.646 0.645 1.00 Au 0.729 0.642 1.14 Al 1.35 0.912 1.48 m*/m Ohm’s Law V E⋅L = R= = i J⋅A L 1 R=ρ = A σ E L ⋅ 1 E J A ⇒ ρ = = L σ J A 1 J =σ ⋅E = E ρ Examples of Resistivity (ρ) Ag (Silver): 1.59×10-8 Ω·m Cu (Copper): 1.68×10-8 Ω·m Graphite (C): (3 to 60)×10-5 Ω·m Diamond (C): ~1014 Ω·m Glass: ~1010 - 1014 Ω·m Pure Germanium: ~ 0.5 Ω·m Pure Silicon: ~ 2300 Ω·m At Room Temp Microscopic Model r r r vavg = vd ( E ) + vRMS (T ) v RMS = 3 k BT m ≈ 10 6 at RT me s 1 3 m e v 2 = k BT 2 2 J J = e nv d ⇒ vd = en n ~ 10 23 cm − 3 = 10 29 m − 3 ; e = 1.6 × 10 −19 C 2 v2 = -6 2 vRMS>>vd vavg ≈ vRMS (speed) i = 1.6 A and A = 1 mm (10 m ) i 1.6 −4 depends on T only, not E vd = = = 10 m/s !?? enA (1.6 × 10 −19 )(10 29 )(10 -6 ) 0.1 mm/s? Takes forever to turn on the light?? Drude Model - conductivity The FEG model was developed by Paul Drude (1900) in order to describe the electrical and thermal conductivity of metals. This work greatly influenced the course of “solid-state physics” and it introduces basic concepts we still use today. F = qE = m∆v/∆t ≈mvd/τ vd = qEτ/m J = nqvd = σE σ = nq2τ/m τ is the relaxation time = l /vavg l is the mean free path Temperature dependence of σ is just that of τ Conductivity in Reciprocal Space The Fermi sphere contains all occupied electron states in the FEG. In the absence of an electric field, there are the same number of electrons moving in the ±x, ±y, and ±z directions, so the net current is zero. But when a field E is applied along the x-direction, the Fermi sphere is shifted by an amount related to the net change in momentum of the FEG: ky Fermi surface kx kF kz h∆k x = ∆p x = mvx = −eE xτ eE τ ∆k x = ∆p x = mvx = − x h ky kx E The shift in Fermi sphere creates a net current flow since more electrons move in the –x direction than the +x direction. But the excess current carriers are only those very near the Fermi surface. So the current carriers have velocity vF. Analysis of Mean Free Path Since the velocity of current-carrying electrons is essentially independent of T, we need to examine the behavior of the mean-free-path. Electrons collide with atoms dominates… Coulomb. e- The probability of a collision in a distance ∆x is: vF P= ∆x collision cross-section = π<r2> cross-sectional area of slab = A atomic density = na total cross − sectional area for collision cross − sectional area of slab P= na A∆x π r 2 A = na ∆x π r 2 Now in a distance ∆x =l , P = 1 is true, so we can solve for Λ: 1 l= na π r2 ρ(T) Now if we assume that the collision cross-section is due to vibrations of atoms about their equilibrium positions, then we can write: r 2 = x2 + y2 And the thermal average potential energies can be written as: 1 2 Therefore Cx 2 = r2 ∝ T 1 2 Cy 2 = 12 kT and l∝ 1 1 ∝ < r2 > T σ = nq2τ/m ≈ nq2 l /mvF Finally 1 σ∝ T or ρ∝T At low T, l is limited by size or impurities, so ρ becomes constant Example : ρ(T) of K The Hall Effect This phenomenon, discovered in 1879 by American physics graduate student (!) Edwin Hall, is important because it allows us to measure the free-electron concentration n for metals (and semiconductors!) and compare to predictions of the FEG model. It also confirmed the electron-hole description of charge carriers. Hall probe FB = q v x B I Hall Coef. A hypothetical charge carrier of charge q experiences a Lorentz force in the lateral direction: FB = qvB I t w As more and more carriers are deflected, the accumulation of charge produces a “Hall field” EH that imparts a force opposite to the Lorentz force: Equilibrium is reached when these two opposing forces are equal in magnitude, which allows us to determine the drift speed: From this we can write the current density: FE = qE H qvB = qE H J = nqv = And it is customary to define the Hall coefficient in terms of the measured quantities: (EH is transverse to J) n is # of charge carriers per unit volume v= nqE H B RH = EH 1 = JB nq EH B Hall Effect Results! In the lab we actually measure the Hall voltage VH and the current I, which gives us a more useful way to write RH: EH VH / w VH t 1 I = JA = Jwt = = = = R VH = EH w H JB (I / wt )B IB nq If we calculate RH from our measurements and assume |q| = e (which Hall did not know!) we can find n. Also, the sign of VH and thus RH tells us the sign of q! (carriers) The discrepancies between the FEG predictions and expt. nearly vanish when liquid metals are compared. This reveals clearly that the source of these discrepancies lies in the electron-lattice interaction. But the results for Be and Zn are puzzling. How can we have q > 0 ??? Look at the diagram for +/- charges again. RH (10-11 m3/As) Meta l n0 solid liquid FEG value Na 1 -25 -25.5 -25.5 Cu 1 -5.5 -8.25 -8.25 Ag 1 -9.0 -12.0 -12.0 Au 1 -7.2 -11.8 -11.8 Be* 2 +24.4 -2.6 -2.53 Zn* 2 +3.3 -5 -5.1 Al 3 -3.5 -3.9 -3.9 Thermal Conductivity of Metals In metals at all but the lowest temperatures, the electronic contribution to κ far outweighs the contribution of the lattice. So we can write: The electron mean-free path can be rewritten in terms of the collision time: l = vτ Also, the electrons that can absorb extra thermal energy and therefore contribute to the heat capacity have energies very near EF, so they essentially all have velocity vF. This gives: From our earlier discussion the electronic heat capacity is: It is easy to show that N(EF) can be expressed: N(E F ) = 3N 2E F 1 κ ≅ κ el = C el vl 3 1 κ = C el v 2 τ 3 1 κ = C el v 2F τ 3 C el = π2 3 or per unit volume N ′( E F ) ≡ Which gives the heat capacity per unit volume: Cel′ ≡ Cel π 2 2 T = 2 k n V EF k 2 N ( E F )T N ( EF ) 3n = V 2 EF Wiedemann-Franz Law Now the thermal conductivity per unit volume is: or… κ′ = π 2 k 2 nτ 3m 1 π2 2 T 2 1 π2 2 T κ′ = 2 k n vFτ = 2 k n 1 2 vF2τ 3 EF 3 2 mv F T τ ~ 1/T, so κ’ ~ independent of temperature!! Now long before Drude’s time, Gustav Wiedemann and Rudolf Franz published a paper in 1853 claiming that the ratio of thermal and electrical conductivities of all metals has nearly the same value at a given T: κ σ = constant Not long after (1872) Ludwig Lorenz (not Lorentz!) realized that this ratio scaled κ linearly with temperature, and thus a Lorenz number L can be defined: σT Gustav Wiedemann ≡L very nearly constant for all metals (at room T and above) Lorenz Number We can readily compare the prediction of the FEG model to the results of experiment: π 2 k 2 nτ LFEG T 2 2 κ k π = = 3m = 2 ne τ σT 3e 2 m LFEG = 2.45 × 10 −8 T (CKJ )2 This is remarkable…it is independent of n, m, and even τ !! L = κ/σT 10-8 (J/CK)2 Metal 0 °C 100 °C Cu 2.23 2.33 Ag 2.31 2.37 Au 2.35 2.40 Zn 2.31 2.33 Cd 2.42 2.43 Mo 2.61 2.79 Pb 2.47 2.56 Agreement with experiment is quite good, although the value of L is about a factor of 10 less at temperatures near 10 K…Can you speculate about the reason? An Historical Footnote Drude of course used classical values for the electron velocity v and heat capacity Cel. By a tremendous coincidence, the error in each term was about two orders of magnitude in the opposite direction! So the classical Drude model gives the prediction: 2 −8 LDrude = 1.12 × 10 (CKJ ) [ C ~ 3/2 Nk, a factor T/TF missing, ½ mv2 = 3/2 kT instead of EF, v2 = (2kT/m) (TF/T), the factor TF/T is missing κ = 1/3 Cv2τ becomes right order!! ] But in Drude’s original paper, he inserted an erroneous factor of two, due to a mistake in the calculation of the electrical conductivity. So he originally reported: J 2 L = 2.24 × 10 −8 CK !!! ( ) So although Drude’s predicted electronic heat capacity was far too high, this prediction of L made the FEG model seem even more impressive than it really was, and led to general acceptance of the model.