abeer alshammari,2016 1 Many conduct electricity.

There are electrons that are not bound to atoms but are able to move through the whole crystal.

Conducting solids fall into two main classes; and .

68 (RT )metals ;10 10   m And increases by the addition of small amounts of impurity. The resistivity normally decreases monotonically with decreasing temperature.

()()RTpure RT And can be reduced by the addition of small amounts of impurity. Semiconductors tend to become insulators at low T.

abeer alshammari,2016 2 Free Electron Model

Schematic model of metallic crystal, such as Na, Li, K, etc. + + + + + The equilibrium positions of the atomic cores are positioned on the crystal lattice and surrounded by a sea of conduction + + + + + electrons.

For Na, the conduction electrons are from the 3s valence + + + + + electrons of the free atoms. The atomic cores contain 10 electrons in the configuration: 1s22s2p6. + + + + + In this model, electrons are completely free to move about, free from collisions, except for a surface potential that keeps the electrons inside the metal. + + + + +

In the alkali metals, with a bcc structure, the cores take up about 15% of the volume of the crystal, but in the noble metals (Cu, Ag, Au), with an fcc structure, the atomic cores are relatively larger and maybe close to contacting or in contact with each other. PAUL Drude • resistivity ranges from 108 m (Ag) to 1020 m (polystyrene) • Drude (circa 1900) was asking why? He was working prior to the development of , so he began with a classical model: • positive cores within an electron gas that follows Maxwell- Boltzmann statistics • following the kinetic theory of gases- the electrons in the gas move in straight lines and make collisions only with the ion cores – no (1863-1906) electron-electron interactions.

• He envisioned instantaneous collisions in which electrons lose any energy gained from the electric field.

• The was approximately the inter-ionic core spacing.

• Model successfully determined the form of Ohm’s law in terms of free electrons and a relation between electrical and σ / κ , but failed to explain electron and the magnetic susceptibility of conduction electrons, and mean free path model

• The removal of the valance electrons leaves a positively charged ion. • The charge density associated the positive ion cores is spread uniformly throughout the metal so that the electrons move in a constant electrostatic potential. All the details of the crystal structure is lost when this assunption is made.

 According to FEM this potential is taken as zero and the repulsive force between conduction electrons are also ignored.

• Therefore, these conduction electrons can be considered as moving independently in a square well of finite depth and the edges of well corresponds to the edges of the sample.

abeer alshammari,2016 5 Energy Levels In One Dimension

Consider a metal with a shape of cube with edge length of L, Ψ and E can be found by solving schrödinger equation

V 퐻훹 = 퐸훹 2 2    E Since V  0 2m L/2 0 L/2

• By means of periodic boundary conditions Ψ’s are running waves. (,,)(,,)x L y  L z  L  x y z

6 abeer alshammari,2016 Energy Levels In One Dimension 22d  H     n2m dx2 n n Orbital: solution of a 1-e Schrodinger equation

Since the n  x) is a continuous function and is equal to zero beyond the length L, the boundary conditions for the wave function are 00L nn    Particle in a box n 2  n  Axsin  Axsin L   n 2 n 1,2,   L n n

2 2 n  n   2mL

These solutions correspond to standing waves with a different number of nodes within the potential well as is shown in Fig. Pauli-exclusion principle: No two electrons can occupy the same quantum state.

Quantum numbers for free electrons: (n, ms ) ms , 

where n describes the orbital  n (x) , and ms describes the projection of the momentum on a quantization axis Degeneracy: number of orbitals having the same energy.

Fermi energy εF = energy of topmost filled orbital when system is in ground state (at T=0K). For the onedimensional system of N electrons we find

2 2 nF  N N number of  F   nF  2mL 2 valence electrons In metals the value of the is of the order of 5 eV.

(a) Occupation of energy levels according to the Pauli exclusion principle, (b) The distribution function f(E), at T = 0°K and T> 0°K. Effect Of Temperature On The Fermi-Dirac Distribution 1 1 Fermi-Dirac distribution : f          e 1 kTB Fermi distribution function determines the probability of finding an electron at the energy E. μ = μ(T) is determined by N  d g  f    g = 1  At T = 0: f     for 0 

→ 0  F 1 For all T : f   2 At T = 0, occupy the lowest energy levels. Near T = 0, there is little chance that thermal agitation will kick a to an energy greater than EF.    For ε >> μ : fe 

(Boltzmann distribution)

3D e-gas Free Electron Gas In Three Dimensions 2ddd 2 2 2 Hrr  2  2  2      2m dx dy dz If the electrons are confined to a cube of edge L, the solution is the standing wave Particle in a box (fixed) boundary conditions:

n0,,yz   n Lyz ,,   n xz ,0,   n xLz ,,   n xy ,,0   n  xyL ,,  0

n  nx  y nz  Standing →  n  Asin x sin y sin  z n 1, 2, LLL i waves Periodic boundary conditions: our wavefunction is periodic in x, y, and z directions with period L,

kxyz,,,,,,,,   k xLyz    k xyLz    k  xyzL   ikr 2n  i Traveling → k  Ae ki  n 0,  1,  2, L i waves 22k   k 2m p  k → ψk is a momentum eigenstate with eigenvalue ℏk. kki k k pk v  m In the ground state a system of N electrons occupies states with lowest possible energies. Therefore all the occupied states lie inside the sphere of radius kF • The Fermi energy and the Fermi wavevector (momentum) are determined by the number of valence electrons in the system. • In order to find the relationship between N and kF, we need to count the total number of orbitals in a sphere of radius kF The surface of the Fermi sphere which should be equal to N. There are two available spin represent the boundary between states for a given set of kx, ky, and kz. occupied and unoccupied k states at absolute zero for the free • The volume in the k space which is occupies by this state is V 4 Nk2   3 electron gas. equal to (2 / L). Thus in the sphere of (4 / 3)  k the total 83 3 F number of states is 1/3 3 2 N k  22 222/3 F  kF 3 N V  F    2m 2mV which depends only of the electron concentration

2 1/3 kF 3 N Fermi velocity vF    m mV Only at temperatures above TF will the free electron gas behave like a classical gas.

Fermi (degeneracy) Temperature TF by

EFBF k T

abeer alshammari,2016 12 Density of states: •The Density of States D(E) specifies how many states exist at a given energy E. •The Fermi Function f(E) specifies how many of the existing states at energy E will be filled with electrons.

→

3/2 Vm2  22 2  V 3/2 Fermi-Dirac distribution function 3 V 2m  F Nk F  is a symmetric function; at finite 3 2 22 3  temperatures, the same number

of levels below EF is emptied and 3 N  3 N D   same number of levels above EF D F   2 FF are filled by electrons. 2  F Heat Capacity Of The Free Electron Gas

• From the diagram of N(E,T) the change in the distribution of electrons can be

resembled into triangles of height (½)g(Ef) and a base of 2kBT so (½)g(Ef)kBT

electrons increased their energy by kBT. •The electron energy levels are mostly filled up to the Fermi energy.

•So, only a small fraction of electrons, approximately T/TF, can be excited to higher levels – because there is only about kBT of thermal energy available. • N(E,T) number of free electrons per unit energy range is just the area under N(E,T) graph. N(E,T)=g(E) f(E,T) N(E,T) g(E)

T=0 • The difference in thermal energy from the value at T=0°K T>0 1 E( T ) E (0) g ( E )( k T )2 2 FB E EF • Differentiating With Respect To T Gives The Heat Capacity At Constant Volume, E C g() E k2 T vT F B 2 N E g() E Heat Capacity of 3 FF Free Flectron Gas 33NN gE() 3N F C g() E k22 T k T 22EFBF k T v F B B 2kTBF 3 T CvB Nk  2 TF Low-Temperature Heat Capacity

CCCel lat 3 CTlat  Copper

2 3 T Cel Nk B 2 TF • Total metallic heat capacity at low temperatures CTT3 Electronic Lattice Heat Heat capacity Capacity Where γ & β are constants found plotting 2 cv/T as a function of T abeer alshammari,2016 18 Transport Properties Of Conduction Electrons

• Fermi-Dirac distribution function describes the behaviour of electrons only at equilibrium. • If there is an applied field (E or B) or a temperature gradient the transport coefficient of thermal and electrical conductivities must be considered.

Transport coefficients

σ,Electrical K, conductivity • Equation of motion of an electron with an applied electric and magnetic field.

dv m  eE  ev  B e dt

• This is just newton’s law for particles of mass me and charge (-e). • The use of the classical equation of motion of a particle to describe the behaviour of electrons in plane wave states, which extend throughout the crystal. A particle-like entity can be obtained by superposing the plane wave states to form a wavepacket. • The velocity of the wavepacket is the group velocity of the waves. Thus 22k d 1 dE k p E  v     2m dk dk mm e ee pk

• So one can use equation of mdv/dt

dv v (*) me    eE  ev  B dt 

 = mean free time between collisions. An electron loses all its energy in time • In the absence of a magnetic field, the applied E results a constant acceleration but this will not cause a continuous increase in current. Since electrons suffer collisions with • • Electrons

v • The additional term me  cause the velocity v to decay  exponentially with a time constant  when the applied E

is removed. The Electrical Conductivity

• In The Presence Of DC Field Only, Eq.(*) Has The Steady State Solution

e e vE   Mobility for m e electron e me a constant of proportionality (mobility)

• Mobility Determines How Fast The Charge Carriers Move With An E. • Electrical , J e N J n() e v vE n  me V

• Where n Is The Electron Density And v Is Drift Velocity. Hence ne2 ne2 JE   Electrical conductivity me me Ohm’s law Electrical Resistivity and Resistance 1 L   R  JE   A Collisions

• In a perfect crystal; the collisions of electrons are with thermally excited lattice vibrations (scattering of an electron by a ). • This electron-phonon scattering gives a temperature

dependent collision time  ph ()T which tends to infinity as T 0.

• In real metal, the electrons also collide with impurity atoms, vacancies and other imperfections, this result in a

finite scattering time  0 even at T=0. • The total scattering rate for a slightly imperfect crystal at finite temperature;

1 1 1   ph ()T  0 Due to phonon Due to imperfections • So the total resistivity ρ,

me m e m e 2  2  2  I ()T  0 ne ne ph () T ne  0

Ideal resistivity Residual resistivity This is known as mattheisen’s rule and illustrated in following figure for sodium specimen of different purity. Residual Resistance Ratio

Residual resistance ratio = room temp. Resistivity/ residual resistivity

6 And it can be as high as 10 for highly purified single crystals.

impure pure

Temperature Collision Time

 (RT ) 2.0 x 1071 (   m )  5.3xm 1010 (   ) 1 sodium residualpureNa  can be found by taking

m 14 mme    2.6xs 10 at RT ne2 n 2.7 x 1028 m 3 7.0xs 1011 at T=0

Taking 6 ; and vF  1.1 x 10 m / s lv F l( RT ) 29 nm l( T 0) 77 m

These mean free paths are much longer than the interatomic distances, confirming that the free electrons do not collide with the atoms themselves. Thermal Conductivity, K

Due to the heat tranport by the conduction electrons > KKmetals non metals Electrons coming from a hotter region of the metal carry more thermal energy than those from a cooler region, resulting in a net flow of heat. The thermal conductivity 1 K C v l where C is the specific heat per unit volume 3 VF V

vF is the mean speed of electrons responsible for thermal conductivity

since only electron states within about kTB of  F change their occupation as the temperature varies.

1 2 l is the mean free path; lv   and Fermi energy  F mv e F F 2 2 2 2 1 1N T 2  nk T  2 2 B where  T K C v  k ()    CvB Nk  VFBF 2 T 3 3 2V TF m e 3 m e F Wiedemann-Franz Law

ne2 22nk T   K  B me 3me The ratio of the electrical and thermal conductivities is independent of the electron gas parameters;

2 2 Kk B 82 Lorentz  2.45x 10 W  K number Te3 

K L 2.23 x 1082 W  K For copper at 0 C T

• A uniform magnetic field B is applied in the y-direction. • If the charge carriers are electrons moving in the negative x-

direction with a velocity vd, they will experience an upward

magnetic force fb. • The electrons will be deflected upward, making the upper edge negatively charged and the lower edge positively charged.

abeer alshammari,2016 31 Hall Effect

Initially, v  vxxˆ  vyyˆ  vzzˆ E  Ex xˆ B  Bz zˆ   d 1     F  m  v  e(E  v  B)  dt    d 1  net force in Fx  m  vx  e(Ex  vyB) x direction  dt    d 1  net force in Fy  m  vy  e(vxB) y direction  dt   Hall Effect

As a result, electrons move in the y direction and an electric field component appears in the y direction, Ey. This will continue until the Lorentz force is equal and opposite to the electric force due to the buildup of electrons – that is, a steady condition arises. B Hall Effect

mv x  e(E  v B)  x y mv y  e(E  v B)  y x

eE v   x  v  eB x m C y   C eE m v   y  v  y m C x Hall Effect

eE v   y  v   0 y m C x Cvx  Ey  m e eB E    E   E y C x m x e v   E x m x v  E  m x x e Hall Effect

The Hall coefficient is defined as: The sign and magnitude of RH gives the sign of the charge carriers and their density. eB In most metals, the charge E Ex carriers are electrons and the y m 1 charge density determined R      from the Hall effect H 2 measurements agrees with jx B ne  ne calculated values for metals Ex B which release a single valence m electron and charge density is approximately equal to the number of valence electrons per unit volume. For copper: n = 8.47 × 1028 electrons/m3. Failure of Fermi gas model •The Hall Effect experiment suggests that a carrier can have a positive charge. •These carriers are “holes” in the electron sea - the absence of an electron acts as a net positive charge. These were first explained by Heisenberg. •We can’t explain why this would happen with our free electron theory. Some successes: 1. electrical conductivity 2. heat capacity 3. thermal conductivity Some failures: 1. physical differences between conductors, insulators, semiconductors, semi-metals abeer alshammari,2. positive2016 Hall coefficients – positive charge carriers ?? 37