Electronic Properties of Solids R.J. Nicholas Electronic Properties: • Metals • Semiconductors • Insulators • Paramagnets • Diamagnets • Ferromagnets • Superconductors Combination of : Crystal Structure Atomic Structure Free electron theory of metals • Metals are good conductors (both electrical and thermal) • Electronic heat capacity has an additional (temperature dependent) contribution from the electrons. • Why are some materials metals and others not? Simple approximation: treat electrons as free to move within the crystal Metals – HT10 – RJ Nicholas 1 Free electron theory of metals • Alkali metals (K, Na, Rb) and Noble metals (Cu, Ag, Au) have filled shell + 1 outer s-electron. • Atomic s-electrons are delocalised due to overlap of outer orbits. • Crystal looks like positive ion cores of charge +e embedded in a sea of conduction electrons • Conduction electrons can interact with each other and ion cores but these interactions are weak because: (1) Periodic crystal potential (ion cores) is orthogonal to conduction electrons - they are eigenstates of total Hamiltonian e.g. for Na conduct. electrons are 3s states, but cores are n=1 and n=2 atomic orbitals. (2) Electron-electron scattering is suppressed by Pauli exclusion principle. Assumptions: (i) ions are static - adiabatic approx. (ii) electrons are independent - do not interact. (iii) model interactions with ion cores by using an “effective mass” m* (iv) free electrons so we usually put m* = me Metals – HT10 – RJ Nicholas 2 Free Electron Model L Put free electrons into a very wide potential well the same size as the crystal i.e. they are 'de-localised' Free electron properties Free electron Hamiltonian has =22∂ ψ only kinetic energy operator: Eψ =− 2mx∂ 2 ± Free electrons are plane waves ψ = A e ikx with: Momentum: Energy: Group velocity: ∂ψ ==22∂ ψ 22k ∂ω 1 ∂E =k i= = ± =k ψ −=ψ = = ∂x 22mx∂ 2 m ∂k = ∂k m Metals – HT10 – RJ Nicholas 3 Free Electron Model – Periodic boundary conditions L L Add a second piece of crystal the same size: The properties must be the same. Density of states Calculate allowed values of k. ψ (x) = ψ (x + L) Use periodic (Born-von Karman) ∴ eikx = eik ( x + L) boundary conditions: ∴ eikL = 1 2π 4π L = size of crystal ∴ k = 0, ± , ± L L 2π ∴ δ k = L Density of allowed states in reciprocal (k-) space is: ΔK ΔV in 1− D or K in 3 − D x 2 for spin states δ k δ k 3 Metals – HT10 – RJ Nicholas 4 Density of states (2) States have energies 2 ε to ε + dε gd()εε==× gkdkkdk () 4 π2 δ k 3 dk ∴=gd()εε gkd () ε dε 22 1 ε = = k π 2 = 8212 ⎛⎞m ε 3 kd 2m ⎛⎞2π ⎜⎟2 1 ⎜⎟ ⎝⎠= ε 2 1 ⎝⎠L 2 ⎛⎞2mε 2 k = 1 ⎜⎟2 42πεmm⎛⎞ 22 1 ⎝⎠= = dε π 3 22⎜⎟ 1 1 ⎛⎞2 == 2 2 ⎜⎟ ⎝⎠ε dk= ⎛⎞21 m ⎝⎠L ⎜⎟2 1 dε = 2 3 ⎝⎠2ε 2m 2 1 =×πεε⎛⎞ 2 4 ⎜⎟2 dV ⎝⎠h Fermi Energy ∞ Ngfd= ()εεε () Electrons are Fermions ∫ FD− 0 μ at T==0() N∫ gεε d 0 3 N 8π 2mE 2 == ⎛⎞F n ⎜⎟2 Vh3 ⎝⎠ μ at T = 0 is known as the 2 3 2 = ⎛⎞3Nh Fermi Energy, EF EF ⎜⎟ ⎝⎠82πVm Metals – HT10 – RJ Nicholas 5 Typical value for EF e.g. Sodium (monatomic) crystal structure: b.c.c. crystal basis: single Na atom lattice points per conventional (cubic) unit cell: 2 conduction electrons per unit cell 2 ∴ electrons per lattice point = 1 lattice constant (cube side) = a = 0.423 nm ∴ density of electrons n = N/V= 2/a3 = 2.6 x 1028 m-3 ∴ EF = 3.2 eV ∴ Fermi Temperature TF? kBTF = EF TF = 24,000 K Finite Temperatures and Heat Capacity E-μ/k T Fermi-Dirac distribution function fF-D = 1/(e B + 1) electrons are excited by an energy ~ kBT ≈ Number of electrons is kBTg(EF) ∴Δ ≈ 2 2 E kB T g(EF) ∴ Δ Δ ≈ 2 CV = E/ T 2kB T g(EF) Metals – HT10 – RJ Nicholas 6 ∴=lnn3 ln E + const . Previously 2 3/2 dn3 dE we have n = AEF ∴= nE2 dn==3 n gE()F dE2 EF ∴=kTB = T C3nkvB 3nk B ETFF ∴ Heat Capacity is: (i) less than classical value by factor ~kBT/EF (ii) proportional to g(EF) Is this significant? Lattice Electrons Room π2 Temperature 3nat.kB /2 nkB (kBT/EF) Low 12π4/5 n k (T/Θ )3 π2/2 nk (k T/E ) Temperature at. B D B B F C/T = βT2 + γ Debye term free electron term Metals – HT10 – RJ Nicholas 7 Rigorous derivation ∞ ∂ f F −D Ugfd= εε() () εε ? ∫ FD− ∂T 0 ∞ 1 ε − μ ∂∂Uf fx==, ∴=εεgd() ε x ∂∂∫ ekT+ 1 TT0 ∂−f exx ∂ ∞ 2 x =× 2 xedx ∂∂x 2 =+gE() k T δ TT()e + 1 FB ∫ 2 − x EF ()e + 1 ∞ kTB x edxx 2 δ ∝ π 2 ≈−∞ ∫ 2 = − x + gE()FB k T EF ()e 1 3 kTB = 0(why?) Magnetic susceptibility • Susceptibility for a spin ½ particle is: μμ2 χ = B 0 / electron kT • This is much bigger than is found experimentally -Why? Metals – HT10 – RJ Nicholas 8 Pauli paramagnetism Separate density of states for spin up and spin down, ± μ shifted in energy by ½g BB (g=2) Imbalance of electron moments Δn Δ ε × μ n = ½ g( F) 2 BB giving a magnetization M μ Δ μ 2 ε M = B n = B g( F) B and a susceptibility χ μ μ 2 ε μ μ 2 ε = M/H = 0 B g( F) = 3n 0 B /2 F k-space picture and the Fermi Surface =2k 2 T=0 states filled up to E ∴ = E F 2m F Map of filled states in k-space 2mE ∴ k = F = Fermi surface F =2 π 3 = × 4 kF N 2 3 ⎛ 2π ⎞ or we can write: 3 ⎜ ⎟ ⎝ L ⎠ 3π 2 N ∴ k 3 = F V Metals – HT10 – RJ Nicholas 9 k-space picture and the Fermi Surface =2k 2 T=0 states filled up to E ∴ = E F 2m F 2mE E ∴ k = F F =2 EF k kF How big is Fermi surface/sphere compared to Brillouin Zone? Simple cubic structure volume of Brillouin Zone = (2π/a)3 electron density n = 1/a3 π 3 π3 3 volume of Fermi sphere = 4 kF /3 = 4 /a = half of one B.Z. Metals – HT10 – RJ Nicholas 10 Electron Transport - Electrical Conductivity Equation of motion: Force = rate of change of momentum ∂k = = − e (E + B × v) ∂t Apply electric field - electrons are accelerated to a steady state with a drift velocity vd - momentum is lost by scattering with an average momentum relaxation time τ mv ∴==−momentum loss d eE τ 2τ ∴==ne current j nevd E m μ is mobility with: ne2τ ∴==conductivity σ neμ vd = μE m What happens in k-space? All electrons in k-space are accelerated by electric field: = δδkFt==− eEt δ On average all electrons E eEτ shifted by: δ k =− = EF k k δk F Metals – HT10 – RJ Nicholas 11 What happens in k-space? All electrons in k-space are = δδkFt==− eEt δ accelerated by electric field: eEτ On average all electrons shifted by: δ k =− = Fermi sphere is shifted in k-space by δk << kF ∴ To relax electron momentum k must be changed by ~ kF Scattering occurs at EF ∴ we need phonons with large value of k. But phonon energy is small ε so only a small fraction of electrons kBT/ F can be scattered Scattering processes Basic Principle: Scattering occurs because of deviations from perfect crystal arrangement Electron scattering mechanisms: (i) thermal vibrations i.e. phonons (vibrations of the atoms are a deviation from perfect crystal structure) (ii) presence of impurities - charged impurities are very important - scattering is by Coulomb force i.e. Rutherford scattering. Metals – HT10 – RJ Nicholas 12 Matthiessen’s rule: Scattering rates (1/τ)add ∴=ρρρm 1 = + ne2 ∑ τ Ti Mean free path (λ): electrons are moving with Fermi velocity vF ∴ λ = τ τ vF (NOT vd ) Low temperature mean free paths can be very long as electrons are only scattered by impurities Hall Effect In a magnetic Field B the electron experiences a force perpendicular to its velocity. A current j causes a build up of charge at the edges which generates an Electric field E which balances the Lorentz force − + × = = ( e) (E v d B) y 0; E y (vd ) x Bz Metals – HT10 – RJ Nicholas 13 In a magnetic Field B the electron Hall Effect experiences a force perpendicular to its velocity. A current j causes a build up of charge at the edges which generates an Electric Balance of forces: field E which balances the Lorentz force − + × = = ( e) (E v d B) y 0; E y (vd ) x Bz The Hall coefficient RH is: = Ey RH jx Bz j = n(−e) v ⇒ R = − 1 x d H ne Negative sign is sign of the charge on the electron Metal Charge/Atom (units of electron charge e) Group Hall Expt. FE Theory Lithium -0.79 -1 I Sodium -1.13 -1 I Potassium -1.05 -1 I Copper -1.36 -1 IB Silver -1.18 -1 IB Gold -1.47 -1 IB Beryllium +0.1 -2 II Magnesium -0.88 -2 II Calcium -0.76 -2 II Zinc +0.75 -2 IIB Cadmium +1.2 -2 IIB Aluminium +1.0 -3 III Indium +1.0 -3 III Metals – HT10 – RJ Nicholas 14 Thermal conductivity In metals heat is mainly carried by the electrons Simple kinetic theory formula for thermal conductivity K: 1 λ π2 2 π2 K = /3C vF [C = /3 kB T g(EF) = /2 nkB kBT/EF] π2 λ λ τ 2 = /6 vF nkB kBT/EF [ = vF ; EF = ½mvF ] π2 2 τ = /3m n kB T Scattering processes • Low temperatures: defects, τ independent of T • Intermediate temp.