Average energy Eav of an electron in a metal is deter­mined by the Fermi-Dirac statistics and the density of states. It increases with the Fermi energy and also with the temperature.

Boltzmann statistics describes the behavior of a collection of particles (e. g., gas atoms) in terms of their energy distribution. It specifies the number of particles N(E) with given energy, through N(E) oc exp (-E/kT), where k is the Boltzmann constant. The description is nonquantum mechanical in that there is no restriction on the number of particles that can have the same state (the same wavefunction) with an energy E. Also, it applies when there are only a few particles compared to the number of possible states, so the likelihood of two particles having the same state becomes negligible. This is generally the case for thermally excited electrons in the conduction band of a semiconductor, where there are many more states than electrons. The kinetic energy distribution of gas molecules in a tank obeys the Boltzmann statistics.

Cathode is a negative electrode. It emits electrons or attracts positive charges, that is, cations.

Debye frequency is the maximum frequency of lat­tice vibrations that can exist in a particular crystal. It is the cut-off frequency for lattice vibrations.

Debye temperature is a characteristic temperature of a particular crystal above which nearly all the atoms are vibrating in accordance with the kinetic molecular theory, that is, each atom has an average energy (potential + kinetic) of 3kT due to atomic vi­brations, and the heat capacity is determined by the Dulong-Petit rule.

Density of states g(E) is the number of electron states [e. g., wavefunctions, ^(n, I, mt, ms)] per unit energy per unit volume. Thus, g(E)dE is the number of states in the energy range E to (E + dE) per unit volume.

Density of vibrational states is the number of lattice vibrational modes per unit angular frequency range.

Dispersion relation relates the angular frequency co and the wavevector K of a wave. In a crystal lattice, the coupling of atomic oscillations leads to a particular relationship between co and K which determines the allowed lattice waves and their group velocities. The dispersion relation is specific to the crystal structure, that is, it depends on the lattice, basis, and bonding.

Effective electron mass m*e represents the inertial re­sistance of an electron inside a crystal against an accel­eration imposed by an external force, such as the ap­plied electric field. If Fext = eEx is the external applied force due to the applied field <EX, then the effective mass m* determines the acceleration a of the electron by eEx = m*a. This takes into account the effect of the internal fields on the motion of the elec­tron. In vacuum where there are no internal fields, m*e is the mass in vacuum me.

Fermi-Dirac statistics determines the probability of an electron occupying a state at an energy level E. This takes into account that a collection of electrons must obey the Pauli exclusion principle. The Fermi-Dirac function quantifies this probability via f(E) = 1/{1 + exp[(Ј — EF)/kT]}, where EF is the Fermi energy.

Fermi energy is the maximum energy of the electrons in a metal at 0 K.

Field emission is the tunneling of an electron from the surface of a metal into vacuum, due to the application of a strong electric field (typically Ј > 109Vm_1).

Group velocity is the velocity at which traveling waves carry energy. If co is the angular frequency and K is the wavevector of a wave, then the group velocity vg = dco/dK.

Harmonic oscillator is an oscillating system, for ex­ample, two masses joined by a spring, that can be de­scribed by simple harmonic motion. In quantum me­chanics, the energy of a harmonic oscillator is quantized and can only increase or decrease by a dis­crete amount fico. The minimum energy of a harmonic oscillator is not zero but hco (see zero-point energy).

Lattice wave is a wave in a crystal due to coupled os­cillations of the atoms. Lattice waves may be traveling or stationary waves.

Linear combination of atomic orbitals (LCAO) is a

Method for obtaining the electron wavefunction in the molecule from a linear combination of individual atomic wavefunctions. For example, when two H atoms A and B come together, the electron wavefunctions, based on LCAO, are

Fa = fs(A) + fs(B) fb = fls(A) ~ fs(B)

Where j/s(A) and fs(B) are atomic wavefunctions centered around the H atoms A and B, respectively. The l/a and f/b represent molecular orbital wavefunctions for the electron; they reflect the behavior of the elec­tron, or its probability distribution, in the molecule.

Mode or state of lattice vibration is a distinct, inde­pendent way in which a crystal lattice can vibrate with its own particular frequency co and wavevector K. There are only a finite number of vibrational modes in a crystal.

Molecular orbital wavefunction, or simply molecu­lar orbital, is a wavefunction for an electron within a system of two or more nuclei (e. g., molecule). A mo­lecular orbital determines the probability distribution of the electron within the molecule, just as the atomic orbital determines the electron’s probability distribu­tion within the atom. A molecular orbital can take two electrons with opposite spins.

Orbital is a region of space in an atom or molecule where an electron with a given energy may be found. An orbit, which is a well-defined path for^a-electron, cannot be used to describe the whereabouts of the elec­tron in an atom or molecule because the electron has a probability distribution. Orbitals are generally repre­sented by a surface within which the total probability is high, for example, 90 percent.

Orbital wavefunction, or simply orbital, describes the spatial dependence of the electron. The orbital is f (r, 0, 0), which depends on n, t, and mt, and the spin dependence ms is excluded.

Phonon is a quantum of lattice vibrational energy of magnitude tico, where co is the vibrational angular fre­quency. A phonon has a momentum h K where K is the wavevector of the lattice wave.

Seebeck effect is the development of a built-in poten­tial difference across a material as a result of a temper­ature gradient. If dV is the built-in potential across a temperature difference dT, then the Seebeck coeffi­cient S is defined as S = dV/dT. The coefficient gauges the magnitude of the Seebeck effect. Only the net Seebeck voltage difference between different met­als can be measured. The principle of the thermocouple is based on the Seebeck effect.

State is a possible wavefunction for the electron that defines its spatial (orbital) and spin properties, for example, /s(n, t, mt, ms) is a state of the elec­tron. From the Schrцdinger equation, each state cor­responds to a certain electron energy E. We thus speak of a state with energy E, state of energy E, or even an energy state. Generally there may be more than one state jr with the same energy E.

Thermionic emission is the emission of electrons from the surface of a heated metal.

Work function is the minimum energy needed to free an electron from the metal at a temperature of absolute zero. It is the energy separation of the Fermi level from the vacuum level.

Zero-point energy is the minimum energy of a har­monic oscillator fico. Even at 0 K, an oscillator in quantum mechanics will have a finite amount of en­ergy which is its zero-point energy. Heisenberg’s un­certainty principle does not allow a harmonic oscillator to have zero energy because that would mean no un­certainty in the momentum and consequently an infi­nite uncertainty in space (Apx Ax > fi).

Questions and problems

Phase of an atomic orbital

What is the functional form of a Is wavefunction i/^(r)? Sketch schematically the atomic wave­function Vrb(‘*) as a function of distance from the nucleus.

What is the total wavefunction (r, f) ?

What is meant by two wavefunctions (A) and (B) that are out of phase?

Sketch schematically the two wavefunctions (A) and Vs (B) at one instant.

Molecular orbitals and atomic orbitals Consider a linear chain of four identical atoms representing a hypothetical molecule. Suppose that each atomic wavefunction is a Is wavefunction. This system of identical atoms has a center of symmetry C with respect to the center of the molecule (midway between the second and the third atom), and all molecular wavefunctions must be either symmetric or antisym­metric about C.

Using the LCAO principle, sketch the possible molecular orbitals.

Sketch the probability distributions | yjr |2.

If more nodes in the wavefunction lead to greater energies, order the energies of the molecular orbitals. Note: The electron wavefunctions, and the related probability distributions, in a simple potential energy well that are shown in Figure 3.15 can be used as a rough guide toward finding the appropriate njotectr* lar wavefunctions in the four-atom symmetric molecule. For example, if we were to smooth the electron potential energy in the four-atom molecule into a constant potential energy, that is, generate a potential energy well, we should be able to modify or distort, without flipping, the molecular orbitals to somewhat resemble yjr to ^4 sketched in Figure 3.15. Consider also that the number of nodes increases from none for V^i to three for in Figure 3.15.

Diamond and tin Germanium, silicon, and diamond have the same crystal structure, that of diamond. Bonding in each case involves sp3 hybridization. The bonding energy decreases as we go from C to Si to Ge, as noted in Table 4.7.

What would you expect for the bandgap of diamond? How does it compare with the experimental value of 5.5 eV?

Tin has a tetragonal crystal structure, which makes it different than its group members, diamond, silicon, and germanium.

Is it a metal or a semiconductor?

What experiments do you think would expose its semiconductor properties?

Table 4.7






Melting temperature, °C





Covalent radius, nm





Bond energy, eV





First ionization energy, eV





Bandgap, eV





Compound III-V Semiconductors Indium as an element is a metal. It has a valency of III. Sb as an element is a metal and has a valency of V. InSb is a semiconductor, with each atom bonding to four neighbors, just like in silicon. Explain how this is possible and why InSb is a semiconductor and not a metal alloy. (Consider the electronic structure and sp3 hybridization for each atom.)

Compound II-VI semiconductors CdTe is a semiconductor, with each atom bonding to four neigh­bors, just like in silicon. In terms of covalent bonding and the positions of Cd and Те in the Periodic Table, explain how this is possible. Would you expect the bonding in CdTe to have more ionic character than that in III-V semiconductors?

*4.6 Density of states for a two-dimensional electron gas Consider a two-dimensional electron gas in

Which the electrons are restricted to move freely within a square area a2 in the xy plane. Following the procedure in Section 4.5, show that the density of states g(E) is constant (independent of energy).

Fermi energy of Cu The Fermi energy of electrons in copper at room temperature is 7.0 eV. The elec­tron drift mobility in copper, from Hall effect measurements, is 33 cm2 V-1 s~].

What is the speed vp of conduction electrons with energies around Ep in copper? By how many times is this larger than the average thermal speed ^thermal of electrons, if they behaved like an ideal gas (Maxwell-Boltzmann statistics)? Why is vF much larger than ^thermal?

What is the De Broglie wavelength of these electrons? Will the electrons get diffracted by the lat­tice planes in copper, given that interplanar separation in Cu = 2.09 A? (Solution guide: Diffrac­tion of waves occurs when 2d sin# = k, which is the Bragg condition. Find the relationship be­tween к and d that results in sin# > 1 and hence no diffraction.)

Calculate the mean free path of electrons at Ep and comment.

Free electron model, Fermi energy, and density of states Na and Au both are valency I metals; that is, each atom donates one electron to the sea of conduction electrons. Calculate the Fermi energy (in eV) of each at 300 К and 0 K. Calculate the mean speed of all the conduction electrons and also the speed of electrons at Ep for each metal. Calculate the density of states as states per eV cm-3 at the Fermi energy and also at the center of the band, to be taken at (Ep + Ф)/2. (See Table 4.1 for Ф.)

Fermi energy and electron concentration Consider the metals in Table 4.8 from Groups I, II, and III in the Periodic Table. Calculate the Fermi energies at absolute zero, and compare the values with the ex­perimental values. What is your conclusion?

Table 4.8

Ef (eV) Erie V)

Metal Group Mat Density (g cm-3) [Calculated] [Experiment]

TOC o "1-5" h z Cu I 63.55 8.96 — 6.5

Zn II 65.38 7.14 — 11.0

A1 III 27 2.70 — 11.8

Temperature dependence of the Fermi energy

Given that the Fermi energy for Cu is 7.0 eV at absolute zero, calculate the Ef at 300 K. What is

The percentage change in Ј> and what is your conclusion?

Given the Fermi energy for Cu at absolute zero, calculate the average energy and mean speed per conduction electron at absolute zero and 300 K, and comment.

X-ray emission spectrum from sodium Structure of the Na atom is [Ne]3s1. Figure 4.59a shows the formation of the 3,s and 3p energy bands in Na as a function of intemuclear separation. Figure 4.59b shows the X-ray emission spectrum (called the L-band) from crystalline sodium in the soft X-ray range as explained in Example 4.6.

From Figure 4.59a, estimate the nearest neighbor equilibrium separation between Na atoms in the

Crystal if some electrons in the 3s band spill over into the states in the 3p band.

Explain the origin of the X-ray emission band in Figure 4.59b and the reason for calling it the


What is the Fermi energy of the electrons in Na from Figure 4.59b?

Taking the valency of Na to be I, what is the expected Fermi energy and how does it compare with


That in part (c)?







0.5 1

Intemuclear distance (nm)

Figure 4.59

Energy band formation in sodium.

/.-emission band of X-rays from sodium.

I SOURCE: (b) Data extracted from W. M. Cadt and D. H. Tomboulian, Phys. Rev., 59, 1941, p. 381.

Conductivity of metals in the free electron model Consider the general expression for the conduc­tivity of metals in terms of the density of states g(Ep) at Ep given by

A = e2v2Fxg{EF)

Show that within the free electron theory, this reduces to a — e2nz/me, the Drude expression.


Mean free path and conductivity in the free electron model

подпись: mean free path and conductivity in the free electron modelMean free path of conduction electrons in a metal Show that within the free electron theory, the mean free path t and conductivity a are related by

87 x 1<


Calculate i for Cu and Au, given each metal’s resistivity of 17 nf2 m and 22 nЈ2 m, respectively, and that each has a valency of I. We are used to seeing a <x n. Can you explain why a <x n2/3?

*4.14 Low-temperature heat capacity of metals The heat capacity of conduction electrons in a metal is proportional to the temperature. The overall heat capacity of a metal is determined by the lattice heat ca­pacity, except at the lowest temperatures. If 8E{ is the increase in the total energy of the conduction elec­trons (per unit volume) and ST is the increase in the temperature of the metal as a result of heat addition, Et has been calculated as follows:



Where Et (0)is the total energy per unit volume at 0 K, n is the concentration of conduction electrons, and Efo is the Fermi energy at 0 K. Show that the heat capacity per unit volume due to conduction electrons in the free electron model of metals is



подпись: [4.84]Heat capacity of



Where y = {n2/2){nk2/Efo)’ Calculate Ce for Cu, and then using the Debye equation for the lattice heat capacity, find Cv for Cu at 10 K. Compare the two values and comment. What is the comparison at room temperature? (Note: Cvoiume = Cmoiar(p/Mat), where p is the density in g cm-3, CvoiUme is in J K-1 cm“3, and Mat is the atomic mass in g mol-1.)

Secondary emission and photomultiplier tubes When an energetic (high velocity) projectile elec­tron collides with a material with a low work function, it can cause electron emission from the surface. This phenomenon is called secondary emission. It is fruitfully utilized in photomultiplier tubes as il­lustrated in Figure 4.60. The tube is evacuated and has a photocathode for receiving photons as a signal. An incoming photon causes photoemission of an electron from the photocathode material. The electron is then accelerated by a positive voltage applied to an electrode called a dynode which has a work func­tion that easily allows secondary emission. When the accelerated electron strikes dynode D, it can release several electrons. All these electrons, the original and the secondary electrons, are then acceler­ated by the more positive voltage applied to dynode D2. On impact with Ј>2, further electrons are re­leased by secondary emission. The secondary emission process continues at each dynode stage until the final electrode, called the anode, is reached whereupon all the electrons are collected which results in a signal. Typical applications for photomultiplier tubes are in X-ray and nuclear medical instruments






Photomultiplier tubes.

I SOURCE: Courtesy of Hamamatsu.




Vacuum — tube








(X-ray CT scanner, positron CT scanner, gamma camera, etc.), radiation measuring instruments (e. g., radon counter), X-ray diffractometers, and radiation measurement in high-energy physics research.

A particular photomultiplier tube has the following properties. The photocathode is made of a semiconductor-type material with Eg ~ 1 eV, an electron affinity x of 0.4 eV, and a quantum efficiency of 20 percent at 400 nm. Quantum efficiency is defined as the number of photoemitted electrons per absorbed photon. The diameter of the photocathode is 18 mm. There are 10 dynode electrodes and an ap­plied voltage of 1250 V between the photocathode and anode. Assume that this voltage is equally dis­tributed among all the electrodes.

What is the longest threshold wavelength for the phototube?

What is the maximum kinetic energy of the emitted electron if the photocathode is illuminated with a 400 nm radiation?

What is the emission current from the photocathode at 400 nm illumination?

What is the KE of the electron as it strikes the first dynode electrode?

It has been found that the tube has a gain of 106 electrons per incident photon. What is the average

Number of secondary electrons released at each dynode?

Thermoelectric effects and Ep Consider a thermocouple pair that consists of gold and aluminum. One junction is at 100 °C and the other is at 0 °C. A voltmeter (with a very large input resistance) is in­serted into the aluminum wire. Use the properties of Au and A1 in Table 4.3 to estimate the emf regis­tered by the voltmeter and identify the positive end.

The thermocouple equation Although inputting the measured emf for V in the thermocouple equa­tion V — aAT + b(AT)2 leads to a quadratic equation, which in principle can be solved for A7 in general AT is related to the measured emf via

AT = aV + a2V2 + <23 V3 H

With the coefficients a, 02* etc., determined for each pair of TCs. By carrying out a Taylor’s expansion of the TC equation, find the first two coefficients a and «2- Using an emf table for the K-type thermo­couple or Figure 4.33, evaluate a and <22­

Thermionic emission A vacuum tube is required to have a cathode operating at 800 °C and providing an emission (saturation) current of 10 A. What should be the surface area of the cathode for the two ma­terials in Table 4.9? What should be the operating temperature for the Th on W cathode, if it is to have the same surface area as the oxide-coated cathode?

Table 4.9

Be (Am 2 K-2)


Th on W

3 x 104


Oxide coating



Field-assisted emission in MOS devices Metal-oxide-semiconductor (MOS) transistors in micro­electronics have a metal gate on an Si02 insulating layer on the surface of a doped Si crystal. Consider this as a parallel plate capacitor. Suppose the gate is an A1 electrode of area 50 fim x 50 fim and has a voltage of 10 V with respect to the Si crystal. Consider two thicknesses for the Si02, (a) 100 A and (b) 40 A, where (1 A = 10“10 m). The work function of A1 is 4.2 eV, but this refers to electron emission into vacuum, whereas in this case, the electron is emitted into the oxide. The potential energy barrier be­tween A1 and Si02 is about 3.1 eV, and the field-emission current density is given by Equation 4.46a and

Calculate the field-emission current for the two cases. For simplicity, take me to be the electron mass in free space. What is your conclusion?


подпись: 4.20CNTs and field emission The electric field at the tip of a sharp emitter is much greater than the “applied field,” *E0. The applied field is simply defined as Vg Id where d is the distance from the cathode tip to the gate or the grid; it represents the average nearly uniform field that would exist if the tip were replaced by a flat surface so that the cathode and the gate would almost constitute a parallel plate capacitor. The tip ex­periences an effective field Ј that is much greater than which is expressed by a field enhancement fac­tor /3 that depends on the geometry of the cathode-gate emitter, and the shape of the emitter; Ј = ^fE(). Further, we can take 4>^2 O % <Ј>3//2 in Equation 4.46. The final expression for the field-emission current density then becomes

6.44 x 107<1>3/2

W„ )

, 1.5×10« ,2 /10.4 (

J = – —- exp(^jexp(-

Fowler – Nordheim field emission current






подпись: 4.21Where <Ј> is in eV. For a particular CNT emitter, <Ј> = 4.9 eV. Estimate the applied field required to achieve a field-emission current density of 100 mA cm2 in the absence of field enhancement (ft = 1) and with a field enhancement of p = 800 (typical value for a CNT emitter).

Nordheim-Fowler field emission in an FED Table 4.10 shows the results of I-V measurements on a Motorola FED microemitter. By a suitable plot show that the I-V follows the Nordheim-Fowler emis­sion characteristics. Can you estimate <I>?

Table 4.10 Tests on a Motorola FED micro field emitter

VG 40.0 42 44 46 48 50 52 53.8 56.2 58.2 60.4

/emission 0.40 2.14 9.40 20.4 34.1 61 93.8 142.5 202 279 367

*4.25 Overlapping bands Consider Cu and Ni with their density of states as schematically sketched in Fig­ure 4.61. Both have overlapping 3d and 4s bands, but the 3d band is very narrow compared to the 4s

Band. In the case of Cu the band is full, whereas in Ni, it is only partially filled.

In Cu, do the electrons in the 3d band contribute to electrical conduction? Explain.

In Ni, do electrons in both bands contribute to conduction? Explain.

Do electrons have the same effective mass in the two bands? Explain.

Can an electron in the 4s band with energy around Ep become scattered into the 3d band as a re­

Sult of a scattering process? Consider both metals.

> E

подпись: > e DEFINING TERMSScattering of electrons from the 4s band to the 3d band and vice versa can be viewed as an additional scattering process. How would you expect the resistivity of Ni to compare with that of Cu, even though Ni has two valence electrons and nearly the same density as Cu? In which case would you ex­pect a stronger temperature dependence for the resistivity?


> E

Figure 4.61 Density of states and electron filling in Cu and Ni.

*4.26 Overlapping bands at Ep and higher resistivity Figure 4.61 shows the density of states for Cu (or

Ag) and Ni (or Pd). The d band in Cu is filled, and only electrons at Ef in the s band make a contribu­tion to the conductivity. In Ni, on the other hand, there are electrons at Ep both in the s and d bands. The d band is narrow compared with the s band, and the electron’s effective mass in this d band is large; for simplicity, we will assume m* is “infinite” in this band. Consequently, the d-band electrons cannot be accelerated by the field (infinite m*), have a negligible drift mobility, and make no contribution to the conductivity. Electrons in the s band can become scattered by phonons into the d band, and hence be­come relatively immobile until they are scattered back into the s band when they can drift again. Con­sider Ni and one particular conduction electron at Ep starting in the s band. Sketch schematically the magnitude of the velocity gained vx — ux from the field Ex as a function of time for 10 scattering events; vx and ux are the instantaneous and initial velocities, and | vx — ux | increases linearly with time, as the electron accelerates in the s band and then drops to zero upon scattering. If rss is the mean time for s to 5-band scattering, rsd is for 5-band to d-band scattering, rds is for d-band to 5-band scattering, assume the following sequence of 10 events in your sketch: r^, rss, ?sd> 4s. *ss, *sd,

What would a similar sketch look like for Cu? Suppose that we wish to apply Equation 4.27. What does g(Ep) and r represent? What is the most important factor that makes Ni more resistive than Cu? Con­sider Matthiessen’s rule. (Note: There are also electron spin related effects on the resistivity of Ni, but for simplicity these have been neglected.)

4.27 Grtineisen’s law A1 and Cu both have metallic bonding and the same crystal structure. Assuming that the Griineisen’s parameter у for A1 is the same as that for Cu, у = 0.23, estimate the linear expansion coefficient X of Al, given that its bulk modulus К = 75 GPa, cs = 900 J K~1 kg”1, and p = 2.7 g cm-3. Compare your estimate with the experimental value of 23.5 x 10-6 K-1.


First point-contact transistor invented at Bell Labs. I SOURCE: Courtesy of Bell Labs.


The three inventors of the transistor: William Shockley (seated), John Bardeen (left), and Walter Brattain (right) in 1948; the three inventors shared the Nobel prize in 1956.

I SOURCE: Courtesy of Bell Labs.

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