1、1,Chapter 49 Electrical Conduction in Solids,2,49-1 Quantum Theory of Solid,Quantum theory works well in individual atoms. It works equally well in collections of atoms.,We can use quantum theory to answer the following questions, such as,Is it soft or hard for a solid ? Can it be hammered into a th
2、in sheet or drawn into a fine wire ? Is it transparent ? What kind of waves travel through it and at what speeds ? Does it conduct heat ? What are its magnetic properties ? What is its crystal structure?,3,In this chapter we will discuss the applications of quantum theory to the structure of solids
3、in order to understand their electrical conductivity.,The difference between individual atom and the solid is that in solid the conduction electrons do not belong to any particular atom. As a consequence, we may be dealing with very large numbers of electrons, so it makes no sense to try to follow t
4、he motion of an individual electron. Instead, we use statistical methods. The joining of quantum theory and statistical mechanics is called quantum statistical mechanics.,Method,4,1. Classical free electron (CFE) model,The valence electrons are free to roam through the ionic array.,The mutual repuls
5、ion between the electrons is neglected, these free electrons are treated as an ideal neutral gas.,In the absence of an electric field, these electrons are moving with the their thermal velocities. .,When an electric field is applied the free electrons acquire an average drift velocity in the directi
6、on opposite to the electric field.,5,According to these assumptions we have derived the Ohms law:,6,But there are failures of the CFE model:,(1) Specific heat,Experimentally the electronic specific heat per mole is,(2) Temperature dependence of ,Experimentally the electronic conductivity is inversel
7、y proportional to the temperature: T,7,2. Quantum-mechanical free electron model (QMFE),The electrons must be treated quantum mechanically.,The electrons must obey Paulis exclusion principle.,The electrons are free to move through the solid also. Aside from collision with the ions, the electrostatic
8、 interaction between the electrons and the lattice ions is ignored, and the interaction between the electrons is also ignored.,This will quantize the energy spectrum of the electron gas.,But:,8,49-2 Conduction Electrons in a Metal,1. Three-dimensional infinite potential well,9,The first few energy l
9、evels:,Degenerate states,10,Now we consider an electron gas made up of 34 electrons. Let us to see how do they occupied the lowest states when T=0.,We state this fact as a probability statement:,At T=0, the probability that a level below 14E0 will be occupied is 1. The probability that a level above
10、 14E0 will be occupied is 0.,Fermi energy EF: The highest occupied energy level at absolute temperature T=0.,11,New type of energy distribution function:,p(E) is the probability of the state at the energy E to be occupied by electrons.,If we faced a gas made up 1028 electrons/m3, how do we determine
11、 EF? How do we find how many electrons we can place within a certain energy range E? This means how many states in the range E to E+E?,12,2. Calculation of EF at T=0K,Construct a lattice in n-space.,In a unit of volume of this n-space there is only one state.,Although each point has a different set
12、of quantum numbers n1, n2, n3, some of the points correspond to states with the same energy (they are degenerater).,One stateone point,13,The points equidistant from the origin, they lie on a spherical shell of radius n, correspond to degenerater states. The farther away from the origin a given poin
13、t is, the higher the energy of the state represented by that point.,14,Every points inside that shell are occupied by two electrons, (why?) thus,Total number of electrons in the octant =2(one-eighth the volume of a sphere of radius nf(number of points per unit volume of this n-space ),n is number de
14、nsity of electron, V=L3 is total volume of the well.,15,This is the value of the Fermi level in a solid at T=0.,Sample problem 49-2: The Fermi level for Cu is 7.06eV. For a classical electron gas to have thermal energies this value, the temperature of the copper should be 5.50104K, about 40 times hi
15、gher than its melting temperature.,16,3. Density of the states,E=410-15eV, if L=1cm. So the discrete energy levels are very closely. Quasicontinuous!,Thus instead of asking, “what is the energy of this state?” we must ask “How many states have energies that lie in the range E to E+dE?”,Density of st
16、ates n(E): n(E) is the number of states (per unit volume) in the unit energy interval. Total number of states in the range E to E+dE is .,17,18,All states below EF are occupied and all states above EF are vacant.,Density of occupied states,19,All states below EF are occupied and all states above EF
17、are vacant.,Actually n0(E)dE is the number of electrons with energy between E and E+dE.,20,Example: Calculate the average energy per electron in a metal at T=0K.,Solution:,21,49-3 Filling the Allowed States,1. Energy distribution of electrons for T0,When the temperature is raised, only a small fract
18、ion of electrons is excited.,The probability function of occupied p(E),22,2.Fermi-Dirac distribution function,The electrons in metal wouldnt be use the Maxwell-Boltzmann distribution.,The reasons: 1. The exclusion principle. 2. The electrons are indistinguishable.,The electrons in metal would be use
19、 the Fermi-Dirac distribution.,23,Figure 49-2. Same as 49-1, but for T=1000K. Note how little the plots differ from those of Fig. 49-1.,2.Fermi-Dirac distribution function,We can see the electrons whose energies are close to the Fermi energy are the only ones that play an active role.,24,3. Specific
20、 heat,If we assume that the average energy gained per electron is kBT,25,Which is in agreement with the experiment result.,26,49-4 Electrical Conduction in Metals,The Fermi distribution of velocities in a metal,The Fermi speed vF,The electrons that contribute to the conduction are those in a small g
21、roup of velocities near vF.,27,Electrical conductivity,The electrons that contribute to the conduction have velocity vF, and vF vd, thus,The mean free path is decided by the collision with the ion core of lattice. This means is inversely proportional to the cross-section of the target.,28,Which is t
22、he observed temperature dependence of .,The resistivity of the metal to the flow of these electrons is determines by collisions made by the electrons with the ion cores of the lattice.,Ek is the total energy of a vibrating ion core of lattice. The energy of vibration comes from increasing temperatur
23、e.,29,It is surprised that a perfectly periodic lattice at the absolute zero of temperature would be totally transparent to conduction electrons. There would never be any collisions. Superconductor,Sample problem 49-5: For copper at room temperature, vF=1.57106m/s, =39nm.,No perfectly periodic latti
24、ces: vacant, impurity atoms, vibrate, etc.,Why do the values of vary over such a wide range (10-8 1022) in different types of materials?,30,49-5 Bands And Gaps,The free electron models of metals gives us a good deal of insight into several properties of metals. Yet there are many other important pro
25、perties that these models do not explain. For example why, when chemical elements crystallize to become solid, some are good conductors, some are insulators, and others are semiconductors.,We can understand these differences by extending the free electron model to take into account the interaction o
26、f the electrons with the positive ion lattice.,31,1. One-dimensional periodic potential,The Kronig-Penney model(克龙尼克潘纳模型）,32,33,34,The solution of the Schrdinger equation exists only if,35,2. Allowed and forbidden energy bands,Solve it numerically: choose a (an E), insert it into the equation and so
27、lve for k.,Interesting result: There will be ranges of for which k will be a real number. Those ranges of will be separated by other ranges for which k is imaginary. Because the momentum of a particle cannot be imaginary, so the conclusion is:,36,A particle in this periodic potential cannot have val
28、ue of for which k is imaginary; therefore, the corresponding values for E for these s are not allowed.,37,38,39,40,We conclude that the electron may possess energies within certain bands of energy but not outside of them. There are allowed and forbidden bands of energy available to electrons moving
29、in a periodic lattice.,Another conclusion is that the width of the allowed energy bands increases with increasing (increasing energy E).,3. Dispersion relation Ek,For a free electron,For an electron moving in one-dimensional array of potential wells:,41,k,42,We note that these solid lines yield the
30、values of E and the corresponding values of k for certain ranges of E. For other energy intervals the value of k is not defined by the solid lines, there are the forbidden energies.,To understand the difference between conductors, insulators and semiconductors, we must answer a question: How many en
31、ergy states are allowed within a given band?,Let us take a less quantitative, less mathematical, less rigorous method to show the physical reason for the quasicontinuous bands.,43,4. Tight-binding approximation,The first two wavefunctions for the infinite potential well.,The first two wavefunctions
32、for the finite potential well.,1,2,44,Consider the two finite wells B and C. B and C are wavefunctions associated with an electron in well B and C with the ground state energy respectively.,Suppose the electron can be found with equal probability in both wells with the ground state energy, what wave
33、function do we use to describe the electron?,B,C,45,1. The probability 2 must be symmetric with respect to a point halfway between the two wells.,2. The part of that reflects the probability that the particle be found in well B must look like the wavefunctioin when the particle is in well B. And so
34、the another part.,If C is wavefunction representing a particle in well C, (-C ) could have been used instead.,Because (- C) satisfy the Schrdinger equation for the same E and (C)2= (-C)2.,B,C,46,S: symmetric A: antisymmetric,are probability densities.,47,When the two wells are very close together, t
35、he symmetry wavefunction looks like the ground state wavefunction for a finite well of width 2a.,Thus even if the electron was initially in one well, after the two wells are brought close together, there is no way to predict exactly in which well it will be.,B,C,48,Similarly, if we had started with
36、two wells and one electron in each well, after they are brought together, there is no way of saying in which well either of the two electrons will be.,When the two wells are very close together, the antisymmetry wavefunction looks like the first excited state for a finite well of width 2a.,49,The im
37、portant conclusion: If you start with two identical s (same energy) in two identical independent system, when you bring the two together, the two degenerate s break up into two nondegenerate s.,What is the physical reason for the breakup of the degeneracy?,Either or will be appropriate to describe t
38、he behavior of the electrons in these wells, because both and are symmetric with respect to the midpoint between the wells.,50,Take example for two hydrogen atoms, each with its electron in the 1s ground state.,As the two atoms are brought together, the wavefunctions overlap, and the electrons from
39、B and C can change places.,51,Probability function is .,Thus when two atoms are brought together, two separate energy levels are formed from each level of the isolated atoms. The physical reason for this effect is the differing ways that the electrons interact with the ions in the symmetric and anti
40、symmetric states.,An electron in the state s has a lower energy than one in A .,52,Now we consider 6 hydrogen atoms with 1s level. The wavefunctions of 6 individual 1s states are 1, 2, 3, 4, 5, 6. All possible ways of adding the six individual 1s states, we can get six types of combinations having d
41、ifferent energies.,first level = 1+ 2+ 3+ 4+ 5+ 6,second level = (1+ 2+ 3) - ( 4+ 5+ 6),third level = (1+ 2) - (3+ 4) + (5+ 6),53,Six possible combinations of the ground state wavefunctions of six hydrogen atoms, each corresponding to a different energy state.,1 2 3 4 5 6,54,Suppose that N atoms are
42、 brought together to form a solid, then each of the levels of the individual isolated atoms breaks up into N discrete, closely spaced levels and becomes a band of energy levels.,What affects the width of the band is how close any two atoms are each other. The closer they are brought, the greater is
43、the separation of the energy levels. Usually band width of solid is a few electron volts. The magnitude of the energy between two bands is called gap.,55,We thus have shown that when one brings a large number N of atoms together to form a solid, the individual atomic energy levels of the atom break
44、up into a quansicontinuous energy band. Within the band there are N distinct but very close energy levels.,When the analysis is extended to multielectron atoms where we have electrons in other states, one finds that each of the individual atomic states breaks up into similar bands of quansicontinuou
45、s states.,56,We must realize that we have been using qualitative arguments and one-dimensional models for the sake of mathematical as well as conceptual simplicity. These models have educational value because they bring out the main features of band theory.,57,If we want to get theoretical results t
46、hat can be compares with those of the experiments, we must face the three-dimensional world, and our qualitative arguments must become quantitative.,But the main feature that we have found are retained.,58,49-6 Conductors, Insulators, and Semiconductors,Schematic of the occupation of the bands by el
47、ectrons in a sodium crystal of N atoms and having 11N electrons.,Valence band Conduction band,59,Occupation of the energy bands in a magnesium crystal of N atoms (12N electrons),The overlap between the 3s and 3p bands.,Valence band Conduction band,60,Filled bands,Empty bands,Conduction bands and val
48、ence bands,61,Semiconductors,At T=0k, the band structure of a semiconductor is characterized by a fully occupied valence band and a completely empty conduction band. The semiconductor ideally is an insulator. As the temperature is raised, some electrons in valence band can receive enough thermal ene
49、rgy and be excited into the conduction band because the energy gap is rather narrow. thus the semiconductor becomes conductor.,62,1) The density of charge carriers,63,Both electrons and holes are responsible for the conduction of semiconductors.,2) The resistivity ,The resistivity increases as n, th
50、e density of charge carriers, decreases.,3) The temperature coefficient of resistivity .,Definition:,(49-7),64,The resistivity of silicon (and other semiconductors) decreases with temperature (d/dT0). This happens because the density of charge carriers n increases rapidly with temperature. In the laboratory, you can identify a semiconductor by its large resistivity and especially, by its large-and negative-temperature coefficient of resistivity .,
