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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 ∂ 2ψ 2 only kinetic energy operator: Eψ = − 2m ∂x 2 Free electrons are plane waves ψ = A e ± ikx with: Momentum: Energy: Group velocity: ∂ψ 2 ∂ 2ψ 2 2 k ∂ω 1 ∂E k i = ± kψ − = ψ = = ∂x 2m ∂x 2 2m ∂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) ∴ e ikx = e ik ( x + L ) boundary conditions: ∴ e ikL = 1 2π 4π L = size of crystal ∴ k = 0, ± , ± L L 2π ∴δk = L Density of allowed states in reciprocal (k-) space is: ΔK ΔVK in 1 − D or in 3 − D x 2 for spin states δ k δ k3 Metals – HT10 – RJ Nicholas 4 Density of states (2) States have energies ε to ε + dε 2 g (ε )d ε = g (k )dk = 4π k 2 dk × δ k3 dk ∴ g (ε )d ε = g (k ) dε dε 2 k2 8π 1 ε = ⎛ 2m ⎞ 2 1 2m = 3 k 2 ⎜ 2 ⎟ 1 dε ⎛ 2π ⎞ ⎜ ⎟ ⎝ ⎠ 2ε 2 1 ⎝ L ⎠ ⎛ 2mε ⎞ 2 k =⎜ 2 ⎟ 1 ⎝ ⎠ 4π 2mε ⎛ 2m ⎞ 2 1 = 3 2 ⎜ 2 ⎟ 1 dε 1 ⎛ 2π ⎞ ⎜ ⎟ ⎝ ⎠ ε 2 dk ⎛ 2m ⎞ 2 1 ⎝ L ⎠ =⎜ ⎟ dε ⎝ 2 ⎠ 2ε 1 2 ⎛ 2m ⎞ 3 2 1 = 4π ⎜ 2 ⎟ ε 2 dε × V ⎝h ⎠ Fermi Energy ∞ Electrons are Fermions N = ∫ g (ε ) f 0 F −D (ε ) d ε μ at T = 0 N = ∫ g (ε ) d ε 0 3 N 8π ⎛ 2mEF ⎞ 2 n= = ⎜ ⎟ V 3 ⎝ h2 ⎠ μ at T = 0 is known as the 2 ⎛ 3N ⎞ 3 h2 Fermi Energy, EF EF = ⎜ ⎟ ⎝ 8π V ⎠ 2m 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 Fermi-Dirac distribution function fF-D = 1/(eE-μ/kBT + 1) electrons are excited by an energy ~ kBT Number of electrons is ≈ kBT g(EF) ∴ ΔE ≈ kB2T2 g(EF) ∴ CV = ΔE/ ΔT ≈ 2kB2T g(EF) Metals – HT10 – RJ Nicholas 6 ∴ ln n = 3 ln E + const. Previously 2 we have n = AEF3/2 dn 3 dE ∴ = n 2 E dn 3 n = = g ( EF ) dE 2 EF k BT T ∴ C v = 3nk B = 3nk B EF TF ∴ Heat Capacity is: (i) less than classical value by factor ~kBT/EF (ii) proportional to g(EF) Is this significant? Lattice Electrons Room Temperature 3nat.kB π2/2 nkB (kBT/EF) Low 12π4/5 nat.kB (T/ΘD)3 π2/2 nkB (kBT/EF) Temperature C/T = βT2 + γ Debye term free electron term Metals – HT10 – RJ Nicholas 7 Rigorous derivation ∞ ∂f F − D U = ∫ ε g (ε ) f F − D (ε ) d ε ? ∂T 0 ∂U ∞ ∂f 1 ε − μ ∴ = ∫ ε g (ε ) dε f = x , x= e + 1 kT ∂T 0 ∂T ∂f − ex ∂x ∞ x e dx 2 x = × ∫ ( ) 2 = g ( EF ) k B T 2 + δ ∂T ex + 1 ∂T (e ) 2 x − EF + 1 k BT ∞ x e x dx π2 ≈−∞ δ ∝ ∫ (e ) 2 2 = g ( EF ) k B T − EF x + 1 3 k BT = 0 (why?) Magnetic susceptibility • Susceptibility for a spin ½ particle is: μ B μ0 2 χ= / 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, shifted in energy by ± ½gμBB (g=2) Imbalance of electron moments Δn Δn = ½ g(εF) × 2μBB giving a magnetization M M = μB Δn = μB2 g(εF) B and a susceptibility χ = M/H = μ0 μB2 g(εF) = 3nμ0 μB2 /2εF k-space picture and the Fermi Surface 2 k2 T=0 states filled up to EF ∴ = EF 2m Map of filled states in k-space 2mEF = Fermi surface ∴ kF = 2 4π k F 3 N = 2× or we can write: 3 ⎛ 2π ⎞ 3 ⎜ ⎟ ⎝ L ⎠ 3 3π 2 N ∴ kF = V Metals – HT10 – RJ Nicholas 9 k-space picture and the Fermi Surface 2 k2 T=0 states filled up to EF ∴ = EF 2m E 2mEF ∴ kF = 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 volume of Fermi sphere = 4πkF3/3 = 4π3/a3 = 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 mvd ∴ momentum loss = = −eE τ ne 2τ ∴ current j = nevd = E m μ is mobility with: ne 2τ vd = μE ∴ conductivity σ = = neμ m What happens in k-space? All electrons in k-space are accelerated by electric field: δ k = Fδ t = − eE δ t On average all electrons E shifted by: δ k = − eEτ EF k kF δk Metals – HT10 – RJ Nicholas 11 What happens in k-space? All electrons in k-space are δ k = Fδ t = − eE δ t accelerated by electric field: eEτ On average all electrons shifted by: δk = − 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 so only a small fraction of electrons kBT/εF 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 m ∴ρ = ne 2 ∑ 1τ = ρ T + ρ i Mean free path (λ): electrons are moving with Fermi velocity vF ∴ λ = v F τ ( NOT v d τ ) 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 B z 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 B z The Hall coefficient RH is: RH = Ey j x Bz j x = n ( − e) vd ⇒ RH = − 1 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: K = 1/3CλvF [C = π2/3 kB2T g(EF) = π2/2 nkB kBT/EF] = π2/6 λvF nkB kBT/EF [λ = vFτ ; EF = ½mvF2] = π2/3m n kB2 τ T Scattering processes • Low temperatures: defects, τ independent of T • Intermediate temp. : Low Temp phonons - Debye model τ ∝ T-3 • High temperatures: ‘classical’ phonons τ ∝ T-1 Wiedeman-Franz ratio Electrical and Thermal conductivities of electrons are both proportional to the relaxation time τ Taking the ratio of the two should make this cancel so if we define the Lorenz number as L = K/(σT) we have the L = K π 2kB 2 Wiedeman-Franz Law: σT = 3e 2 Predicted value is absolute and the same for all metals. Works well at high and low temps, - breaks down in ‘Debye’ region where energy and charge scattering are different Metals – HT10 – RJ Nicholas 15 Successes and Failures of Free Electron Model Successes: • Temperature dependence of Heat Capacity • paramagnetic (Pauli) susceptibility • Ratio of thermal and electrical conductivities (Lorentz number) • Magnitudes of heat capacities and Hall effect in simple metals Failures: • Heat capacities and Hall effect of many metals are wrong • Hall effect can be positive • Does not explain why mean free paths can be so long • Does not explain why some materials are metals, some insulators and some are semiconductors Nearly Free Electron Approximation Use a travelling wavefunction for an electron, e ikx, with kinetic energy 2k2/2m Assume that this is Bragg scattered by the wavevector G=2π/a to give a second wave e i(k-G)x with energy 2(k-G)2/2m Crystal potential is periodic in real space. Therefore we can Fourier Transform the potential so that: V ( x) = ∑ G VG exp (iGx ) For a schematic solution we calculate what happens for a single Fourier component VG so V(x) = VG(eiGx + e-iGx) Metals – HT10 – RJ Nicholas 16 Nearly Free Electron Approximation Use a travelling wavefunction for an electron, e ikx, with kinetic energy 2k2/2m Bragg scattering θ d Δk = 2|k|sinθ 2dsinθ = λ = 4π/λ sinθ = 2π/d Δk = G = ha* + kb* +lc* a 2θ With d = h2 + k 2 + l 2 Formally what we are doing is to solve the Hamiltonian form of Schrödinger equation Hψ = Eψ where ψ are the two travelling wave solutions. Expanding gives: ⎛ H11 − λ H12 ⎞ ⎛ eikx ⎞ ⎛ eikx ⎞ ⎜ ⎟⎜ ⎟ = (E − λ ) ⎜ i ( k −G ) x ⎟ ⎜ H 21 H 22 − λ ⎟ ⎜ ei (k −G ) x ⎟ ⎜e ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ H11 = ψ1 * − ∂2 2 ψ1 = k22 , H 22 = 2 (k − G )2 2m ∂x 2 2m 2m H12 = ψ 1 * V ( x) ψ 2 = VG Metals – HT10 – RJ Nicholas 17

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