Can we expand upon our discussion of metallic bonding to understand conductivity ?

1. Ch. 6-8 Question: Why do some materials conduct and others don’t? (takes a few lectures to fully answer) Can we expand upon our discussion of metallic bonding to understand conductivity? When challenging, we try to simplify. Simplistically yes, but it only works well in some cases. To understand fully why will take several lectures, but this is typically good enough for a qualifier.

2. By the end of this section you should be able to: Understand how to apply the Drude model Determine the density of conduction electrons Calculate the electrical conductivity of a metal Discuss the classical Hall effect and why the Drude model doesn’t perfectly explain the results For more on Drude Model, see Ashcroft ch. 1&3 Conductivity and the Drude ModelObjectives

3. Drude Model (1900)simple model for conductivity Za = atomic number = # of electrons isolated atom in a metal Zvalence electrons are weakly bound to the nucleus (participate in reactions) Za – Zcore 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

4. 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 all. A valance electron really belongs to the whole crystal, since it can move readily from one ion to its neighbor and so on. + + + + + + Metal (classical picture)

5. U(r) U(r) Drude Model: A gas of electrons • 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. X + + + + + +

6. U(r) U(r)=0 Drude makes the free electron approximation Neglect periodic potential & scattering between electrons Reasonable for “simple metals” (Alkali Li,Na,K,Cs,Rb) What does this remind you of?

7. Drude Assumptions • He also ignored electron-electron and electron-ion interactions (except to say electron-ion interaction were the cause of the scattering) • Electron moves in straight line between collisions. Velocity changes instantaneously, but direction random. • <v>=0, but <v2>T (gas) • We will use a trick to use this model later (future)

8. A model for electrical conduction • Drude model in 1900 • In the absence of an electric field, the conduction electrons move in random directions through the conductor with large average speeds. 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. (a) (b)

9. Drude Strengths/Weaknesses Strengths: • Very simple and intuitive model • Great at predicting strength and density of metals Weaknesses: • Predicts too high of a classical specific heat (100x larger than observed at RT) • Doesn’t explain both signs possible in Hall effect • More details in Ashcroft, Chapter 3 (magnetoresistance, sign of thermoelectric field, AC conductivity, DC conductivity anisotropy)

10. For Drude model, we will need to know theconduction 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 1023Zm /A Similar to finding number of atoms

11. Group: Predict n for Copper Density of copper = 8.96 g/cm3 n = N/V = 6.022 x 1023 Z m /A

12. “Quasi-Classical” Theory of Transport Microscopic Macroscopic Charge Current Density: J = I/A Depends on dimension  = R A /length  = 1/

13. Predicting the Conductivity Vd = drift velocity or average velocity (not speed) of conduction electrons If , what is the If we want σ or , how do we estimate vd? Current Density: J = I/A

14. Determining Velocity When Applying an Electric Field E to a Metal 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

15. Ohm’s law Calculating Conductivity 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

16.  = m / ( n e2) Group Exercise vavg = x/ Calculate the average time between collisions  for electrons in copper at 273 K. Assuming that the average speed for free electrons in copper is 1.6 x 106 m/s calculate their mean free path.

17. Electrical conductivity of materials

18. 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.

19. Resistivity vs Temperature not accounted for in Drude model 1020- Insulator Diamond Resistivity is temperature dependentmostly because of the temperature dependence of the scattering time . 1010- Semiconductor Resistivity (Ωm) Germanium 100 - Metal Copper  = m / ( n e2) 10-10- 0 100 200 300 Temperature (K) • In Metals,the resistivityincreases with increasing temperature. The scattering time decreaseswithincreasing 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.

20. The Drude Model Struggles with the Sign of the Hall Effect In a current carrying wire when in a perpendicular magnetic field, the current should be drawn to one side of the wire. As a result, the resistance will increase and a transverse voltage develops. - Lorentz force = -ev/c x H H - - - - - - - - - - - - - - - - - - ++++++++++++++++ + -

21. If a current flows (of velocity vD) in the positive x direction and a uniform magnetic field is applied in the positive z-direction, use the Lorentz force to determine the magnitude and direction of the resulting Hall field, first in terms of velocity, but then current density. • Ey= vxHz/c = jxHz/nec • Hall coefficient: induced E/current density * applied H field RH= Ey/jxH = 1/nechow big? • R is very small for metals as n is very large. Lorentz force = -ev/c x H - Hz - - - - - - - - - - - - - - - - - - ++++++++++++++++ x y

22. Ohm’s law contains e2 But for RH the sign of e is important. The Hall coefficient c 10 kG increases the resistance of bismuth 8x, while only 0.1% for silver A hole is the lack of an electron. It has the opposite charge so +e.

23. weak magnetic fields Not yet prepared to discuss other quantum versions of the Hall effect With strong magnetic fields: The integer quantum Hall effect is observed in 2D electron systems at low temperature, in which the Hall conductance undergoes quantum Hall transitions to take on quantized values The fractional quantum Hall effect: Hall conductance of 2D electrons shows precisely quantized plateaus at fractional values of e2/h from D.C. Tsui, RMP (1999) and from H.L. Stormer, RMP (1999)

24. Application: Hall Probe The measurement of large magnetic fields (~1T) is often done by the Hall effect. A sample is placed in the magnetic field and the transverse voltage is measured. For a 100-m thick Cu film, in a 1.0 T magnetic field and through which I = 0.5 A is passing, the Hall voltage is 0.737 V. Advantages: Hall effect devices are immune to dust, dirt, mud, and water, making them better for position sensing than other means such as optical and electromechanical sensing. Can also create non contact current sensors. Disadvantages: Magnetic flux from the surroundings (such as other wires) may diminish or enhance the field the Hall probe intends to detect, affecting results. Also, the output from this type of sensor cannot be used to directly drive actuators without amplification.

25. Calculate conductivity A wire (1mm in diameter by 1 m in length) of an aluminum alloy containing a partial percentage of Mn is placed in an electrical circuit as shown (not to scale). A voltage drop of 432 mV is measured across the length of the wire as it carries a 10 A current. Calculate the conductivity.

26. How Far Apart are the Conduction Electrons? density of conduction electrons in metals ~1022 – 1023 cm-3 rs = radius of a sphere whose volume is equal to the volume per electron mean inter-electron spacing in metals rs ~ 1 – 3 Å

27. Measurable quantity – Hall resistance for 3D systems More detail about Hall resistance for 2D systems n2D=n in the presence of magnetic field the resistivity and conductivity becomes tensors for 2D:

28. Another Approach: 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)

29. t/tp 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 • Also j = n e v VD(t) = VD(0)exp(-t/tp)