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Transcript
Metals: Drude Model and Conductivity (Covering Pages 2-11) Objectives By the end of this section you should be able to: Understand the basics of the Drude model Determine the valence of an atom Determine the density of conduction electrons Calculate the electrical conductivity of a metal Estimate and interpret the relaxation time Models of Atoms Za = atomic number = # of electrons Drude Model isolated atom in a metal Z valence electrons are weakly bound to the nucleus (participate in reactions) Za – Z core electrons are tightly bound to the nucleus (less of a role) In a metal, the core electrons remain bound to the nucleus to form the metallic ion Valence electrons wander away from their parent atoms, called conduction electrons Why mobile electrons appear in some solids and not others? According to the Drude model, the valance electrons are responsible for the conduction of electricity, thus termed conduction electrons. Na11 → 1s2 2s2 2p6 3s1 Core electrons Valance electron (loosely bound) Metallic 11Na, 12Mg and 13Al are assumed to have 1, 2 and 3 mobile electrons per atom respectively. This valance electron, which occupies the third atomic shell, is the electron which is responsible chemical properties of Na. Warning: Confusing Terminology There exist many forms of valence! By valence electrons here we mean outer electrons...the ones that could easily be pulled from the core to form an electron gas (In chemistry) Valence = Maximum number of bonds formed by atom Warning: The "oxidation state" of an atom in a molecule gives the number of valence electrons it has gained or lost through bonding. In contrast to the valency number, the oxidation state can be positive or negative. Sometimes called valence in physics. If I give you an element on the test, how would you know if metallic? I would give atomic number. Note orbital filling in Cu does not follow normal rule Metallic/insulating properties can be understood by how loose (i.e. low ionization energy) outer electrons are. Metallic or Insulating? Most metals are formed from atoms with partially filled atomic orbitals. e.g. Na, and Cu which have the electronic structure Na 1s2 2s2 2p6 3s1 Cu 1s2 2s2 2p6 3s23p63d104s1 Insulators are formed from atoms with closed (totally filled) shells e.g. Solid inert gases He 1s2 Ne 1s2 2s2 2p6 Or form close shells by covalent bonding i.e. Diamond Note orbital filling in Cu does not follow normal rule Simple picture. Metal have CORE electrons that are bound to the nuclei, and VALENCE electrons that can move through the metal. Group: Find the # of Valence Electrons for Si What is the valence of silicon, atomic Number 14? 1s2 2s2 2p6 3s2 3p2 How many electrons are in the n=3 state? Valence=4 #Valence Easy for Outer Groups Group: 1 2 18 13 Group Group Group Group Group Group Group Group 1: 1 2: 2 13: 3 14: 4 15: 5 16: 6 17: 7 18: 8 (except for helium, which has 2) Group: Find the Valence of Fe What is the valence of iron, atomic Number 26? 1s2 2s2 2p6 3s2 3p6 4s2 3d6 Number of 3d electrons = 6 What about in compounds? (SrFeO3) Except, are these the outer most? No, the 4s is farther! So valence =2! How the mobile electrons become mobile When we bring Na atoms together to form a Na metal, the orbitals overlap slightly and the valance electrons become no longer + + + attached to a particular ion, but belong to both. + + + Na metal A valance electron really belongs to the whole crystal, since it can move readily from one ion to its neighbor and so on. Drude Model: A gas of electrons X + + + + + + Drude (~1900) assumed that the charge density associated with 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 assumption is made. U(r) U(r) Drude makes the free electron approximation U(r) U(r)=0 Neglect periodic potential & scattering between electrons Reasonable for “simple metals” (Alkali Li,Na,K,Cs,Rb) What does this remind you of? Drude Assumptions He also ignored electric fields, meaning no electron-electron or electron-ion interactions If field applied, electron moves in straight line between collisions (no electron-electron collisions). Velocity changes instantaneously, but direction random. <v>=0, but <v2>T (gas) Positives: Strength and density of metals Negative: Derivations of electrical and thermal conductivity, predicts classical specific heat (100x less than observed at RT) Conduction electron Density n = N/V 6.022 x 1023 atoms per mole Multiply by number of valence electrons (Z) Convert moles to cm3 (using mass density m and atomic mass A) n = N/V = 6.022 x 1023 Z m /A Group: Predict n for Copper Density of copper = 8.96 g/cm3 n = N/V = 6.022 x 1023 Z m /A Effective Radius density of conduction electrons in metals ~1022 – 1023 cm-3 rs – measure of electronic density = radius of a sphere whose volume is equal to the volume per electron 4rs V 1 3 N n 3 1/ 3 3 rs 4n 1 ~ 1/ 3 n mean inter-electron spacing in metals rs ~ 1 – 3 Å (1 Å= 10-8 cm) 2 a0 0.529 Å – Bohr radius 2 me rs/a0 ~ 2 – 6 A model for electrical conduction (a) (b) Dr. Jie Zou Drude model in 1900 In the absence of an electric field, the conduction electrons move in random directions through the conductor with average speeds v ~ 106 m/s. The drift velocity of the free electrons is zero. There is no current in the conductor since there is no net flow of charge. When an electric field is applied, in addition to the random motion, the free electrons drift slowly (vd ~ 10-4 m/s) in a direction opposite that of the electric field. “Quasi-Classical” Theory of Transport Microscopic Macroscopic dq Current: i (Amps) dt q idt V i R Charge Current Density: J = I/A Ohm’s Law J E E where resistivity = 1/ conductivity Finding the Current Density Vd = drift velocity or average velocity of conduction electrons If Current Density: J = I/A , what is the J E E where resistivity conductivity Applying an Electric Field E to a Metal Needed from problem 1.2 F = ma = - e E, so acceleration a = - e E / m Integrating gives v = - e E t / m So the average vavg = - e E / m, where is “relaxation time” or time between collisions vavg = x/, we can find the xavg or mean free path Calculating Conductivity Ohm’s law vavg = - e E / m J = E = - n e vavg = - n e (- e E / m) Then conductivity = n e2 / m ( = 1/) And = m / ( n e2) ~10-15 – 10-14 s = m / ( n e2) Group Exercise vavg = x/ Calculate the average time between collisions for electrons in copper at 273 K. Assuming that the average E speed for free x 10 E6 m/s where resistivity electrons in copper is J1.6 calculate conductivity their mean free path. Electrical conductivity of materials How to measure the conductivity / resistivity A two-point probe can be used but the contact or wire resistance can be a problem, especially if the sample has a small resistivity. Resistivity vs Temperature not accounted for in Drude model 10 20 Resistivity is temperature dependent mostly because of the temperature dependence of the scattering time . = m / ( n e2) Resistivity (Ωm) Insulator Diamond 1010- Semiconductor Germanium 100 Metal 10-100 100 200 Temperature (K) Copper 300 • In Metals, the resistivity increases with increasing temperature. The scattering time decreases with increasing temperature T (due to more phonons and more defects), so as the temperature increases ρ increases (for the same number of conduction electrons n) • We will find the opposite trend occurs in semiconductors but we will need more physics to understand why. In Terms of Momentum p(t) Momentum p(t) = m v(t) So j = n e p(t) / m Probability of collision by dt is dt/ p(t+dt)=prob of not colliding*[p(t)+force dt] (1-dt/) = p(t)–p(t) dt/ + force dt +higher order terms in (dt) p(t+dt)-p(t) =–p(t) dt/ + force +higher order Or dp/dt = -p(t)/ + force(time) Main Results from Drude dp/dt = -p(t)/ + force(time); compare to Netwon’s second law F=dp/dt There is an effective damping term due to collisions VD(t) = VD(0)exp(-t/p ) Also j = n e v p(t)/p(t=0) 1.0 0.8 0.6 0.4 0.0 t/p 0.2 -1 0 1 2 3 We are going to use these next class to understand some specific cases 4