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Transcript
ELECTRONIC CONDUCTION IN METALS 1 Free Electron Theory • A metal is defined as a solid with valence electrons that are not tightly bound to the atoms but are easily able to move through the whole crystal. • They can be called free electrons or conduction electrons. • So according to Free Electron Model (FEM) the valence electrons are responsible for the conduction of electricity, & for this reason these electrons are called “Conduction Electrons”. 2 • As an example, consider Sodium (Na). • The electron configuration of the Free Na Atom is: 1s2 2s2 2p6 3s1 Valence Electron Core Electrons (loosely bound) • The outer electron in the third atomic shell (n = 3, ℓ = 0) is the electron which is responsible for the chemical properties of Na. 3 • When we bring Na atoms together to form a Na metal, we see the picture as Na metal • Outer orbitals of Na atoms overlap slightly. From this observation it follows that a valance electron is no longer attached to a particular ion, but belongs to both neighbouring ions at the same time. 4 a valance electron belongs to the whole crystal, since it can move readily from one ion to its neighbour, and then the neighbour’s neighbour, and so on. This mobile electron becomes a conduction electron in a solid. + + + + + + – The removal of the valance electrons leaves a positively charged ion. – The charge density associated with the positive ion cores is spread uniformly throughout the metal – The electrons move in a constant electrostatic potential. – Electric field due to lattice ions is uniform so neglected – Repulsive force b/w conducting electron ignored – Attractive force b/w lattice ions and elections ignored 5 •These free electrons in a metal behaves like a gas molecule with random motion •Why electric current is zero in the absence of electric field??? Randomness of electrons due to thermal energy??? + + + + + + average kinetic energy 1 3 me vth2 kT 2 2 thermal velocity of electron vth 3kT me Applied Electric Field Drift velocity of the electrons Electrical Conductivity 6 Electrical Conductivity: Electric field leads to a potential difference which creates a force on electron F eE This force accelerates the electrons and give velocity away from the field F eE me a m d Ee dt m d dt Ee vd m Drift velocity: Mean collision time: average time b/w two collision during its random motion r Mobility : magnitude of drift velocity achieved by electron in unit electric field e me vd E If applied field E is removed???? The drift velocity will decays exponentially to zero The relaxation time: r 1 cos r v0 vd e t / r vd at t r , v0 e the drift velocity reduces to 1/e to its peak value after switch off the E Mean Free Path (λ): average distance travelled by electrons b/w two collision During random motion Electrical Conductivity: A conductor of Length L, cross-section area A Current flow I, potential drop V, Electric field E I V / R, J I V LV / L A AR AR I E A Current density J nedv L AR where dv e E m ne 2 J E m Electrical Conductivity ne 2 But J E E m ne 2 me Effect of temperature and impurity on conductivity????? Electrical Conductivity Electrical Conductivity ne 2 me Mean free path vth , where vth is the thermal velocity / vth me 3kT ne 2 ne 2 me 3me kT thermal velocity of electron vth 3kT me 3me kT ne 2 Failure of Classical Free Electron Theory: • resistivity of metal vary as T1/2 But experimentally is varies with T •It gives almost seven times value at t=300K of exp. •Conductivity depends on free electron density n so dia or trivalent metal should have higher conductivity Electrical Conductivity The kinetic energy of the electron is 1 3 2 mvth kT 2 2 3kT and vth m At room temperature , the drift velocity imparted to the electrons by the applied electric field is very much smaller than the average thermal velocity. The average distance travelled by an electron between two successive collisions in the presence of applied field is known as free mean path . The time taken by an electron between two successive collisions is known as mean collision time of the electron c. Hence the time taken by the electrons in traversing the distance will be decided mainly by rms velocity. m Now c vth 3kT Mobility • Mobility of the electron μ is defined as the steady state drift velocity<vd> per unit electric field. d e E m ne 2 e ne. m m ne 1 m m 2 ne ne Where (resistivit y ) ne ne 2 m m •The electrical conductivity σ depends on two factors ,the charge density n and their mobility. These two quantities depend on temperature. •In metals n is constant and μ decreases slightly with temperature and hence with increase of temperature ,the conductivity decreases. •In semiconductors the exponential increase of n with temperature is responsible for increase of conductivity with temperature. • In insulator n remains constant and above certain temperature μ increase exponentially resulting in dielectric breakdown. For the thermal conductivity of a metal in terms of temperature. Relaxation Time Relaxation Time can be defined as the time taken for the drift velocity to decay to 1/e of its initial value Let assume that the applied field is cut off after the drift velocity of the electron has reached its steady value. Drift velocity after this instant is governed by d d m dt d d dt m d d (t ) d (0) exp( t / ) at t (0) d (t ) d e Differenti ating equation d d (t ) d dt Vd(0)is the steady state drift velocity Collision time vd 0 the change in average velocity on collision is opposite. Hence the rate of change of average velocity is given by Q.1/Tut 9 The relaxation time and root mean square velocity of the electron at room temperature are 2.5x10-14 s and 1x105 m/s. Calculate the value of mean free path of the electron. Q.2/Tut 9 The resistivity of a metal at temperature 20°C is 1.69 x 10-8 ohm m and concentration of the free electrons in metal, ne = 8.5x1028/m3. Calculate root mean square velocity (c), relaxation time (τ), mean free path (λ), mobility of electrons (μ) and value of electrical conductivity (σ) on the basis of classical free electron theory. • Therefore, these conduction electrons can be considered as moving independently in a square well of finite depth & the edges of the well correspond to the edges of the crystal. • Consider a metal with a cubic shape with edge length L: Ψ and E can be found by solving the Schrödinger equation: 2 Since V E 2 2m V 0 Use periodic boundary conditions & get the Ψ’s as travelling plane waves. -L/2 0 L/2 ( x L, y L, z L) ( x, y, z ) 19 • The solutions to the Schrödinger equation are plane waves, 1 ik r 1 i ( kx x k y y kz z ) ( x, y , z ) e e V V Normalization constant kL 2n 2 where , k So the wave vector must satisfy 2 2 2 k x nx k y n y k z nz L L L nx , n y , nz taking any integer values; +ve, -ve or zero. 20 • The wave function Ψ(x,y,z) corresponds to an energy 2 2 k2 E 2m E 2m (k x 2 k y 2 k z 2 ) • The corresponding momentum is: p (k x , k y , k z ) • The energy is completely kinetic: 2 2 1 2 k mv 2 2m mv 2 2 2 k 2 22 2 2 2 k x nx k y n y k z nz L L L One unit cell corresponding to one wave vector has sides kx 2 2 2 ky kz L L L So volume of unit cell will be 2 2 2 8 3 8 3 kxk ykz 3 L L L L V V is the volume of the crystal. The number of k values in the range K and k+dk is 4k 2 dk 4k 2 dk g (k )dk volume of unit cell 8 3 / V 23 • So the number of allowed k values inside a spherical shell of k-space of radius k is: Vk 2 g (k )dk 2 dk , 2 • Here g(k) is the density of states per unit magnitude of k. 24 The number of allowed states per unit energy range? • Each k state represents two possible electron states, one for spin up, the other is spin down. dk g ( E ) 2 g (k ) dE g ( E )dE 2 g (k )dk 2 2 2 dE k dk m k E 2m g ( E ) 2 g (k ) k 2mE 2 2Vk 2 m Vk m V m m 2 2 2 2 2 2 k 2 2 2 2 2 2 k g (E) V 2 2 2mE 2 3/ 2 1/ 2 (2 m ) E 3 25 Ground State of the Free Electron Gas • Electrons are Fermions (s = ± ½) and obey the Pauli exclusion principle; each state can accommodate only one electron. • The lowest-energy state of N free electrons is therefore obtained by filling the N states of lowest energy. 26 • Thus all states are filled up to an energy EF, known as The Fermi energy, obtained by integrating the density of states between 0 and EF, The result should equal N. Remember that g (E) N EF 0 g ( E )dE EF V 2 2 V 2 3/ 2 1/ 2 (2 m ) E 3 (2m) 3 2 3/ 2 E dE 1/ 2 0 V 3 2 3/ 2 (2 mE ) F 3 • Solving for EF (Fermi energy); 3 N EF 2m V 2 2 2/3 27 • The occupied states are inside the Fermi sphere in k-space as shown below; the radius is Fermi wave number kF. EF kz 3 N 2m V 2 Fermi Surface E = EF kF 2/3 2 kF 2 EF 2me From these two equations, kF can be found as, 1/ 3 3 N kF V 2 ky kx 2 The surface of the Fermi sphere represents the boundary between occupied & unoccupied k states at absolute zero for the free electron gas. 28 • Typical values may be obtained by using monovalent potassium metal (K) as an example; for potassium the atomic density and hence the valence electron density N/V is 1.402x1028 m-3 so that EF 3.40 1019 J 2.12eV k F 0.746 A1 EF kBTF • The Fermi (degeneracy) Temperature TF is given by EF 4 TF kB 2.46 10 K 29 • It is only at a temperature of this order that the particles in a classical gas can attain (gain) kinetic energies as high as EF . • Only at temperatures above TF will the free electron gas behave like a classical gas. • Fermi momentum PF meVF PF kF PF VF 0.86 106 ms 1 me • These are the momentum & velocity values of the electrons at the states on the Fermi surface of the Fermi sphere. • So, the Fermi Sphere plays an important role in the behavior of metals. 30 Typical values for monovalent potassium metal; 3 N EF 2m V 2 2 2.12eV 1/ 3 3 N kF V 2 2/3 0.746 A1 PF VF 0.86 106 ms 1 me EF TF 2.46 104 K kB 31 Free Electron Gas at Finite Temperature • At temperature T the probability of occupation of an electron state of energy E is given by the Fermi distribution function f FD 1 1 e ( E EF ) / k B T • The Fermi distribution function determines the probability of finding an electron at energy E. 32 • The Density of States g(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. • I • It is a probability distribution function. 33 33 Fermi-Dirac Statistics The Fermi Energy EF is essentially the same as the Chemical Potential μ. EF is called The Fermi Energy. Note the following: • When E = EF, the exponential term = 1 and FFD = (½). • In the limit as T → 0: L L • At T = 0, Fermions occupy the lowest energy levels. • Near T = 0, there is little chance that thermal agitation will kick a Fermion to an energy greater than EF. 34 In the limit as T → 0 For E < EF : 1 f ( E EF ) 1 1 exp () For E > EF : 1 f ( E EF ) 0 1 exp () E A step function! EF 0 1 f(E) 35 Fermi Function at T=0 & at a Finite Temperature f FD 1 1 e If E = EF then f(EF) = ½ ( E EF ) / k B T fFD(E,T) 0.5 E E<EF EF E>EF 36