Docstoc

Free electron theory of metals

Document Sample
Free electron theory of metals Powered By Docstoc
					                                 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

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:19
posted:11/19/2011
language:English
pages:17