fisica - tipler 38 solids

1228

Solids

38-1 The Structure of Solids

38-2 A Microscopic Picture of Conduction

38-3 The Fermi Electron Gas

38-4 Quantum Theory of Electrical Conduction

38-5 Band Theory of Solids

38-6 Semiconductors

*38-7 Semiconductor Junctions and Devices

38-8 Superconductivity

38-9 The Fermi–Dirac Distribution

The first microscopic model of electric conduction in metals was proposedby Paul K. Drude in 1900 and developed by Hendrik A. Lorentz about1909. This model successfully predicts that the current is proportional to

the potential drop (Ohm’s law) and relates the resistivity r of conductors to themean speed vav and the mean free path† l of the free electrons within the conduc-tor. However, when vav and l are interpreted classically, there is a disagreement between the calculated values and the measured values of the resistivity, and asimilar disagreement between the predicted temperature dependence and theobserved temperature dependence. Thus, the classical theory fails to adequatelydescribe the resistivity of metals. Furthermore, the classical theory says nothingabout the most striking property of solids, namely that some materials are conductors, others are insulators, and still others are semiconductors, which arematerials whose resistivity falls between that of conductors and insulators.

SILICON INGOT SILICON IS A

SEMICONDUCTOR, AND SLICES OF

INGOTS LIKE THIS ARE USED TO

PRODUCE TRANSISTORS AND OTHER

ELECTRONIC DEVICES. TO MAKE A

TRANSISTOR, ATOMS OF ARSENIC

AND GALLIUM ARE INJECTED INTO

THE SILICON.

Do you know how many

atoms of arsenic it takes to

increase the charge-carrier density

by a factor of 5 million? For more

on this topic, see Example 38-7.

?

C H A P T E R

38

† The mean free path is the average distance traveled between collisions.

TiplerMosca Physics.5e ch 38 6/26/03 8:46 PM 228

When vav and l are interpreted using quantum theory, both the magnitudeand the temperature dependence of the resistivity are correctly predicted. In addition, quantum theory allows us to determine if a material will be a conduc-tor, an insulator, or a semiconductor. In this chapter, we use our understanding of quantum mechanics to discuss the structure of solids and solid-state semiconducting devices. Muchof our discussion will be qualitative because, as in atomic physics, the quantum-mechanical calculations are very difficult.

38-1 The Structure of Solids

The three phases of matter we observe—gas, liquid, and solid—result from therelative strengths of the attractive forces between molecules and the thermal energy of the molecules. Molecules in the gas phase have a relatively high thermal kinetic energy, and such molecules have little influence on one anotherexcept during their frequent but brief collisions. At sufficiently low temperatures(depending on the type of molecule), van der Waals forces will cause practicallyevery substance to condense into a liquid and then into a solid. In liquids, themolecules are close enough—and their kinetic energy is low enough—that theycan develop a temporary short-range order. As thermal kinetic energy is furtherreduced, the molecules form solids, which are characterized by a lasting order.

If a liquid is cooled slowly so that the kinetic energy of its molecules is reduced slowly, the molecules (or atoms or ions) may arrange themselves in aregular crystalline array, producing the maximum number of bonds and leadingto a minimum potential energy. However, if the liquid is cooled rapidly so that itsinternal energy is removed before the molecules have a chance to arrange them-selves, the solid formed is often not crystalline but instead resembles a snapshotof the liquid. Such a solid is called an amorphous solid. It displays short-rangeorder but not the long-range order (over many molecular diameters) that is char-acteristic of a crystal. Glass is a typical amorphous solid. A characteristic result ofthe long-range ordering of a crystal is that it has a well-defined melting point,whereas an amorphous solid merely softens as its temperature is increased.Many materials may solidify into either an amorphous state or a crystalline statedepending on how the materials are prepared; others exist only in one form orthe other.

Most common solids are polycrystalline; that is, they consist of many singlecrystals that meet at grain boundaries. The size of a single crystal is typically afraction of a millimeter. However, large single crystals do occur naturally and canbe produced artificially. The most important property of a single crystal is thesymmetry and regularity of its structure. It can be thought of as having a singleunit structure that is repeated throughout the crystal. This smallest unit of a crystal is called the unit cell; its structure depends on the type of bonding—ionic,covalent, metallic, hydrogen, van der Waals—between the atoms, ions, or mole-cules. If more than one kind of atom is present, the structure will also depend onthe relative sizes of the atoms.

Figure 38-1 shows the structure of the ionic crystal sodium chloride (NaCl).The Na and Cl ions are spherically symmetric, and the Cl ion is approxi-mately twice as large as the Na ion. The minimum potential energy for this crys-tal occurs when an ion of either kind has six nearest neighbors of the other kind.This structure is called face-centered-cubic (fcc). Note that the Na and Cl ions insolid NaCl are not paired into NaCl molecules.

The net attractive part of the potential energy of an ion in a crystal can bewritten

38-1Uatt ake2

r

S E C T I O N 3 8 - 1 The Structure of Solids 1229

Cl–

Na+

F I G U R E 3 8 - 1 Face-centered-cubicstructure of the NaCl crystal.

TiplerMosca Physics.5e ch 38 6/26/03 8:46 PM Page 1229

where r is the separation distance between neighboring ions (0.281 nm for theNa and Cl ions in crystalline NaCl), and a, called the Madelung constant,depends on the geometry of the crystal. If only the six nearest neighbors of eachion were important, a would be six. However, in addition to the six neighbors ofthe opposite charge at a distance r, there are twelve ions of the same charge at adistance r, eight ions of opposite charge at a distance r, and so on. TheMadelung constant is thus an infinite sum:

38-2

The result for face-centered-cubic structures is a 1.7476.†

a 6 1222

823

p

2322

1230 C H A P T E R 3 8 Solids

Crystal structure. (a) The hexagonal symmetry of a snowflake arisesfrom a hexagonal symmetry in its lattice of hydrogen atoms and oxygenatoms. (b) NaCl (salt) crystals, magnified approximately thirty times.The crystals are built up from a cubic lattice of sodium and chlorideions. In the absence of impurities, an exact cubic crystal is formed. This(false-color) scanning electron micrograph shows that in practice thebasic cube is often disrupted by dislocations, giving rise to crystals witha wide variety of shapes. The underlying cubic symmetry, though,remains evident. (c) A crystal of quartz (SiO2 , silicon dioxide), the mostabundant and widespread mineral on the earth. If molten quartzsolidifies without crystallizing, glass is formed. (d ) A soldering iron tip,ground down to reveal the copper core within its iron sheath. Visible inthe iron is its underlying microcrystalline structure.

(b) (c) (d )

(a)

† A large number of terms are needed to calculate the Madelung constant accurately because the sum convergesvery slowly.


When Na and Cl ions are very close together, they repel each other becauseof the overlap of their electrons and the exclusion-principle repulsion discussedin Section 37-1. A simple empirical expression for the potential energy associatedwith this repulsion that works fairly well is

where A and n are constants. The total potential energy of an ion is then

38-3

The equilibrium separation r r0 is that at which the force F dU/dr is zero.Differentiating and setting dU/dr 0 at r r0, we obtain

38-4

The total potential energy can thus be written

38-5

At r r0, we have

38-6

If we know the equilibrium separation r0, the value of n can be found approxi-mately from the dissociation energy of the crystal, which is the energy needed tobreak up the crystal into atoms.

Calculate the equilibrium spacing r0 for NaCl from the measured density ofNaCl, which is r 2.16 g/cm3.

P I C T U R E T H E P R O B L E M We consider each ion to occupy a cubic volume of side r0. The mass of 1 mol of NaCl is 58.4 g, which is the sum of the atomicmasses of sodium and chlorine. There are 2NA ions in 1 mol of NaCl, where NA 6.02 1023 is Avogadro’s number.

1. The volume v per mol of NaCl equals the number of ionstimes the volume per ion:

2. Relate r0 to the density r:

3. Solve for r and substitute the known values:30

E X A M P L E 3 8 - 1SEPARATION DISTANCE BETWEEN NA AND CL IN NACL

U(r0) ake2

r0

a1 1nb

U ake2

r0

c r0

r

1n

ar0

rbn d

A ake2rn1

0

n

U ake2

r

Arn

Urep Arn

S E C T I O N 3 8 - 1 The Structure of Solids 1231

so

0.282 nm r0 2.82 108 cm

2.25 1023 cm3

r30

M2NAr

58.4 g

2(6.02 1023)(2.16 g/cm3)

r Mv

M

2NAr30

v 2NAr30


The measured dissociation energy of NaCl is 770 kJ/mol. Using 1 eV

1.602 1019 J, and the fact that 1 mol of NaCl contains NA pairs of ions, we can express the dissociation energy in electron volts per ion pair. The conversionbetween electron volts per ion pair and kilojoules per mole is

The result is

38-7

Thus, 770 kJ/mol 7.98 eV per ion pair. Substituting 7.98 eV for U(r0), 0.282 nmfor r0, and 1.75 for a in Equation 38-6, we can solve for n. The result is n 9.35 9.

Most ionic crystals, such as LiF, KF, KCl, KI, and AgCl, have a face-centered-cubic structure. Some elemental solids that have fcc structure are silver, aluminum, gold, calcium, copper, nickel, and lead.

Figure 38-2 shows the structure of CsCl, which is called the body-centered-cubic(bcc) structure. In this structure, each ion has eight nearest neighbor ions of theopposite charge. The Madelung constant for these crystals is 1.7627. Elementalsolids with bcc structure include barium, cesium, iron, potassium, lithium,molybdenum, and sodium.

1eV

ion pair 96.47

kJmol

1eV

ion pair

6.022 1023 ion pairs

1 mol

1.602 1019 J1 eV


Cs+

Cl–

Cs+

Cl–

F I G U R E 3 8 - 2 Body-centered-cubicstructure of the CsCl crystal.

Figure 38-3 shows another important crystal structure: the hexagonal close-packed (hcp) structure. This structure is obtained by stacking identical spheres,such as bowling balls. In the first layer, each ball touches six others; thus, thename hexagonal. In the next layer, each ball fits into a triangular depression of the first layer. In the third layer, each ball fits into a triangular depression of thesecond layer, so it lies directly over a ball in thefirst layer. Elemental solids with hcp structure in-clude beryllium, cadmium, cerium, magnesium,osmium, and zinc.

In some solids with covalent bonding, thecrystal structure is determined by the directionalnature of the bonds. Figure 38-4 illustrates the di-amond structure of carbon, in which each atom isbonded to four other atoms as a result of hybridization, which is discussed in Section 38-2.This is also the structure of germanium andsilicon.

F I G U R E 3 8 - 3 Hexagonal close-packedcrystal structure.

F I G U R E 3 8 - 4 Diamond crystalstructure. This structure can beconsidered to be a combination of two interpenetrating face-centered-cubic structures.


S E C T I O N 3 8 - 2 A Microscopic Picture of Conduction 1233

(a)

(d ) (e)

(b)

(c)

38-2 A Microscopic Picture of Conduction

We consider a metal as a regular three-dimensional lattice of ions filling somevolume V and containing a large number N of electrons that are free to movethroughout the whole metal. Experimentally, the number of free electrons in ametal is approximately one electron to four electrons per atom. In the absence ofan electric field, the free electrons move about the metal randomly, much the way gas molecules move about in a container. We will often refer to these freeelectrons in a metal as an electron gas.

The current in a conducting wire segment is proportional to the voltage dropacross the segment:

The resistance R is proportional to the length L of the wire segment and inverselyproportional to the cross-sectional area A:

R rLA

I VR

, or (V IR)

Carbon exists in three well-definedcrystalline forms: diamond, graphite, and fullerenes (short for“buckminsterfullerenes”), the third ofwhich was predicted and discovered lessthan two decades ago. The forms differ inhow the carbon atoms are packedtogether in a lattice. A fourth form ofcarbon, in which no well-definedcrystalline form exists, is commoncharcoal. (a) Synthetic diamonds,magnified approximately 75,000 times. Indiamond, each carbon atom is centered ina tetrahedron of four other carbon atoms.The strength of these bonds accounts forthe hardness of a diamond. (b) An atomic-force micrograph of graphite. In graphite,carbon atoms are arranged in sheets, witheach sheet made up of atoms inhexagonal rings. The sheets slide easilyacross one another, a property that allowsgraphite to function as a lubricant. (c) Asingle sheet of carbon rings can be closedon itself if certain rings are allowed to bepentagonal, instead of hexagonal. Acomputer-generated image of the smallestsuch structure, C60 , is shown here. Each ofthe sixty vertices corresponds to a carbonatom; twenty of the faces are hexagonsand twelve of the faces are pentagons.The same geometric pattern isencountered in a soccer ball. (d) Fullerenecrystals, in which C60 molecules are close-packed. The smaller crystals tend to formthin brownish platelets; larger crystalsare usually rod-like in shape. Fullerenesexist in which more than sixty carbonatoms appear. In the crystals shown here,about one-sixth of the molecules are C70.(e) Carbon nanotubes have veryinteresting electrical properties. A singlegraphite sheet is a semimetal, whichmeans that it has properties intermediatebetween those of semiconductors andthose of metals. When a graphite sheet isrolled into a nanotube, not only do thecarbon atoms have to line up around thecircumference of the tube, but thewavefunctions of the electrons must alsomatch up. This boundary-matchingrequirement places restrictions on thesewavefunctions, which affects the motionof the electrons. Depending on exactlyhow the tube is rolled up, the nanotubecan be either a semiconductor or a metal.


where r is the resistivity. Substituting rL/A for R, and EL for V, we can write thecurrent in terms of the electric field strength E and the resistivity. We have

Dividing both sides by the area A gives I/A (1/r)E, or J (1/r)E, where J I/Ais the magnitude of the current density vector . The current density vector is defined as

38-8

DEFINITION—CURRENT DENSITY

where q, n, and d are the charge, the number density, and the drift velocity of the charge carrier. (This follows from Equation 25-3). In vector form, the relationbetween the current density and the electric field is

38-9

This relation is the point form of Ohm’s law. The reciprocal of the resistivity iscalled the conductivity.

According to Ohm’s law, the resistivity is independent of both the currentdensity and the electric field . Combining Equation 38-8 and Equation 38-9gives

38-10

where e has been substituted for q. According to Equation 38-10, the drift veloc-ity d is proportional to .

In the presence of an electric field, a free electron experiences a force e . Ifthis were the only force acting, the electron would have a constant acceleratione /me. However, Equation 38-10 implies a steady-state situation with a constant drift velocity that is proportional to the field . In the microscopicmodel, it is assumed that a free electron is accelerated for a short time and thenmakes a collision with a lattice ion. The velocity of the electron immediately afterthe collision is completely unrelated to the drift velocity. The justification for this assumption is that the magnitude of the drift velocity is extremely small com-pared with the random thermal speeds of the electrons.

For a typical electron, its velocity a time t after its last collision is 0 (e /me)t,where 0 is its velocity immediately after that collision. Since the direction of 0 israndom, it does not contribute to the average velocity of the electrons. Thus, theaverage velocity or drift velocity of the electrons is

38-11

where t is the average time since the last collision. Substituting for d in Equa-tion 38-10, we obtain

nee aeE!

me

tb 1r

E!

v!

v!d

eE!

me

t

v!

v! E

!v!

E!E

!E

!E!

v!

enev!d

1r

E!

E!

J!

1r

E!

v!

J! qnv

!d

J!

I VR

EL

rL/A

1r

EA



so

38-12

The time t, called the collision time, is also the average time between collisions.†

The average distance an electron travels between collisions is vavt, which is calledthe mean free path l :

38-13

where vav is the mean speed of the electrons. (The mean speed is many orders ofmagnitude greater than the drift speed.) In terms of the mean free path and themean speed, the resistivity is

38-14

RESISTIVITY IN TERMS OF VAV AND l

According to Ohm’s law, the resistivity r is independent of the electric field .Since me, ne, and e are constants, the only quantities that could possibly dependon are the mean speed vav and the mean free path l. Let us examine these quantities to see if they can possibly depend on the applied field .

Class ica l In terpretat ion of va v and l

Classically, at T 0 all the free electrons in a conductor should have zero kineticenergy. As the conductor is heated, the lattice ions acquire an average kinetic energy of kT, which is imparted to the electron gas by the collisions between the electrons and the ions. (This is a result of the equipartition theorem studied in Chapters 17 and 18.) The electron gas would then have a Maxwell–Boltzmanndistribution just like a gas of molecules. In equilibrium, the electrons would beexpected to have a mean kinetic energy of kT, which at ordinary temperatures(300 K) is approximately 0.04 eV. At T 300 K, their root-mean-square (rms)speed,‡ which is slightly greater than the mean speed, is

38-15

Note that this is about nine orders of magnitude greater than the typical driftspeed of 3.5 105 m/s, which was calculated in Example 25-1. The very smalldrift speed caused by the electric field therefore has essentially no effect on thevery large mean speed of the electrons, so vav in Equation 38-14 cannot dependon the electric field .

The mean free path is related classically to the size of the lattice ions in theconductor and to the number of ions per unit volume. Consider one electronmoving with speed v through a region of stationary ions that are assumed to behard spheres (Figure 38-5). The size of the electron is assumed to be negligible.

E!

1.17 105 m/s

vav vrms B3kTme

B3(1.38 1023 J/K)(300 K)9.11 1031 kg

32

32

E!E

!E

!

r mevav

nee2l

l vavt

r me

nee2t

S E C T I O N 3 8 - 2 A Microscopic Picture of Conduction 1235

† It is tempting but incorrect to think that if t is the average time between collisions, the average time since its lastcollision is t rather than t. If you find this confusing, you may take comfort in the fact that Drude used the incorrect result t in his original work.

‡ See Equation 17-23.)1

2

(12

Area = pr2

Lattice ion

vt1

Electron

vt2 vt3

Radius = r

F I G U R E 3 8 - 5 Model of an electronmoving through the lattice ions of aconductor. The electron, which isconsidered to be a point particle, collides with an ion if it comes within adistance r of the center of the ion, where r is the radius of the ion. If the electronspeed is v, it collides in time t with all the ions whose centers are in the volumepr 2vt. While this picture is in accord with the classical Drude model forconduction in metals, it is in conflict with the current quantum-mechanicalmodel presented later in this chapter.


The electron will collide with an ion if it comes within a distance r from the centerof the ion, where r is the radius of the ion. In some time t1, the electron moves adistance vt1. If there is an ion whose center is in the cylindrical volume pr2vt1, theelectron will collide with the ion. The electron will then change directions andcollide with another ion in time t2 if the center of the ion is in the volume pr2vt2.Thus, in the total time t t1 t2 . . . , the electron will collide with the numberof ions whose centers are in the volume pr2vt. The number of ions in this volumeis nion pr2vt, where nion is the number of ions per unit volume. The total pathlength divided by the number of collisions is the mean free path:

38-16

where A pr2 is the cross-sectional area of a lattice ion.

Successes and Fai lures of the C lass ica l Model

Neither n ion nor r depends on the electric field , so l also does not depend on .Thus, according to the classical interpretation of vav and l, neither depend on ,so the resistivity r does not depend on , in accordance with Ohm’s law. How-ever, the classical theory gives an incorrect temperature dependence for the resis-tivity. Because l depends only on the radius and the number density of the latticeions, the only quantity in Equation 38-14 that depends on temperature in the classical theory is vav, which is proportional to . But experimentally, r varies linearly with temperature. Furthermore, when r is calculated at T 300 Kusing the Maxwell–Boltzmann distribution for vav and Equation 38-16 for l, thenumerical result is about six times greater than the measured value.

The classical theory of conduction fails because electrons are not classicalparticles. The wave nature of the electrons must be considered. Because of thewave properties of electrons and the exclusion principle (to be discussed in thefollowing section), the energy distribution of the free electrons in a metal is noteven approximately given by the Maxwell–Boltzmann distribution. Further-more, the collision of an electron with a lattice ion is not similar to the collisionof a baseball with a tree. Instead, it involves the scattering of electron waves bythe lattice. To understand the quantum theory of conduction, we need a qualita-tive understanding of the energy distribution of free electrons in a metal. Thiswill also help us understand the origin of contact potentials between two dis-similar metals in contact and the contribution of free electrons to the heat capac-ity of metals.

38-3 The Fermi Electron Gas

We have used the term electron gas to describe the free electrons in a metal.Whereas the molecules in an ordinary gas, such as air, obey the classicalMaxwell–Boltzmann energy distribution, the free electrons in a metal do not. Instead, they obey a quantum energy distribution called the Fermi–Dirac distri-bution. Because the behavior of this electron gas is so different from a gas of mol-ecules, the electron gas is often called a Fermi electron gas. The main features ofa Fermi electron gas can be understood by considering an electron in a metal to be a particle in a box, a problem whose one-dimensional version we studied extensively in Chapter 34. We discuss the main features of a Fermi electron gassemiquantitatively in this section and leave the details of the Fermi–Dirac distri-bution to Section 38-9.

2T

E! E

!E!

E!

l vt

nionpr2vt

1nionpr2

1nionA



Energy Quant izat ion in a Box

In Chapter 34, we found that the wavelength associated with an electron of momentum p is given by the de Broglie relation:

38-17

where h is Planck’s constant. When a particle is confined to a finite region ofspace, such as a box, only certain wavelengths ln given by standing-wave conditions are allowed. For a one-dimensional box of length L, the standing-wave condition is

38-18

This results in the quantization of energy:

or

38-19

The wave function for the nth state is given by

38-20

The quantum number n characterizes the wave function for a particular state andthe energy of that state. In three-dimensional problems, three quantum numbersarise, one associated with each dimension.

The Exc lus ion Pr inc ip le

The distribution of electrons among the possible energy states is dominated bythe exclusion principle, which states that no two electrons in an atom can be inthe same quantum state; that is, they cannot have the same set of values for theirquantum numbers. The exclusion principle applies to all “spin one-half” parti-cles, which include electrons, protons, and neutrons. These particles have a spinquantum number ms which has two possible values, and . The quantumstate of a particle is characterized by the spin quantum number ms plus the quan-tum numbers associated with the spatial part of the wave function. Because thespin quantum numbers have just two possible values, the exclusion principle canbe stated in terms of the spatial states:

EXCLUSION PRINCIPLE IN TERMS OF SPATIAL STATES

When there are more than two electrons in a system, such as an atom or metal,only two can be in the lowest energy state. The third and fourth electrons mustgo into the second-lowest state, and so on.

There can be at most two electrons with the same set of values for theirspatial quantum numbers.

12

12

cn(x) B 2L

sinnp x

L

En n2 h2

8mL2

En p2

n

2m

(h/ln)2

2m

h2

2m

1l2

n

h2

2m

1(2L/n)2

nln

2 L

l hp

S E C T I O N 3 8 - 3 The Fermi Electron Gas 1237


Compare the total energy of the ground state of fiveidentical bosons of mass m in a one-dimensional boxwith that of five identical fermions of mass m in the samebox.

P I C T U R E T H E P R O B L E M The ground state is the lowest possible energy state. The energy levels in a one-dimensional box are given by En n2E1, where E1

h2/(8 mL2). The lowest energy for five bosons occurs whenall the bosons are in the state n 1, as shown in Figure 38-6a. For fermions, the lowest state occurs with two fermi-ons in the state n 1, two fermions in the state n 2, andone fermion in the state n 3, as shown in Figure 38-6b.

1. The energy of five bosons in the state n 1 is:

2. The energy of two fermions in the state n 1, two fermions in the state n 2, and one fermion in the state n 3 is:

3. Compare the total energies:

R E M A R K S We see that the exclusion principle has a large effect on the total energy of a multiple-particle system.

The Fermi Energy

When there are many electrons in a box, at T 0 the electrons will occupy thelowest energy states consistent with the exclusion principle. If we have N elec-trons, we can put two electrons in the lowest energy level, two electrons in thenext lowest energy level, and so on. The N electrons thus fill the lowest N/2 energy levels (Figure 38-7). The energy of the last filled (or half-filled) level at T 0 is called the Fermi energy EF. If the electrons moved in a one-dimensionalbox, the Fermi energy would be given by Equation 38-19, with n N/2:

38-21

FERMI ENERGY AT T 0 IN ONE DIMENSION

In a one-dimensional box, the Fermi energy depends on the number of free electrons per unit length of the box.

E X E R C I S E Suppose there is an ion, and therefore a free electron, every 0.1 nm ina one-dimensional box. Calculate the Fermi energy. Hint: Write Equation 38-21 as

(Answer EF 9.4 eV)

In our model of conduction, the free electrons move in a three-dimensional boxof volume V. The derivation of the Fermi energy in three dimensions is somewhatdifficult, so we will just give the result. In three dimensions, the Fermi energy atT 0 is

EF (hc)2

32mec2 aN

Lb2

(1240 eVnm)2

32(0.511 MeV) aN

Lb2

EF aN2b2

h2

8meL2

h2

32me

aNLb2

E X A M P L E 3 8 - 2BOSON-SYSTEM ENERGY VERSUS FERMION-SYSTEM ENERGY


E 5E1

The five identical fermions have3.8 times the total energy of thefive identical bosons.

2E1 8E1 9E1 19E1

E 2E1 2E2 1E3 2E1 2(2)2E1 1(3)2E1

5

4

3

2

1

E

E

E

E

E

2

1 1

3

4

5 5

4

3

2

1

E

E

E

E

E

2

3

4

5

Bosons Fermions

(a) (b)

EF

F I G U R E 3 8 - 7 At T 0 the electronsfill up the allowed energy states to theFermi energy EF . The levels are so closelyspaced that they can be assumed to be continuous.

F I G U R E 3 8 - 6


38-22a

FERMI ENERGY AT T 0 IN THREE DIMENSIONS

The Fermi energy depends on the number density of electrons N/V. Substitutingnumerical values for the constants gives

38-22b

FERMI ENERGY AT T 0 IN THREE DIMENSIONS

The number density for electrons in copper was calculated in Example 25-1and found to be 84.7/nm3. Calculate the Fermi energy at T 0 for copper.

1. The Fermi energy is given by Equation 38-22:

2. Substitute the given number density for copper:

R E M A R K S Note that the Fermi energy is much greater than kT at ordinary temperatures. For example, at T 300 K, kT is only about 0.026 eV.

E X E R C I S E Use Equation 38-22b to calculate the Fermi energy at T 0 for gold,which has a free-electron number density of 59.0/nm3. (Answer 5.53 eV)

Table 38-1 lists the free-electronnumber densities and Fermi energies atT 0 for several metals.

The average energy of a free electroncan be calculated from the complete energy distribution of the electrons,which is discussed in Section 38-9. At T 0, the average energy turns out to be

38-23

AVERAGE ENERGY OF ELECTRONSIN A FERMI GAS AT T 0

For copper, Eav is approximately 4 eV.This average energy is huge comparedwith typical thermal energies of aboutkT 0.026 eV at a normal temperatureof T 300 K. This result is very dif-ferent from the classical Maxwell–Boltzmann distribution result that at T 0, E 0, and that at some temper-ature T, E is of the order of kT.

Eav 35 EF

7.04 eV

EF (0.365 eVnm2)(84.7/nm3)2/3

EF (0.365 eVnm2) aNVb2/3

E X A M P L E 3 8 - 3THE FERMI ENERGY FOR COPPER

EF (0.365 eVnm2) aNVb2/3

EF h2

8me

a 3NpVb2/3


TABLE 38-1Free-Electron Number Densities† and Fermi Energies at T 0 for Selected Elements

Element N/V, electrons/nm3 EF, eV

Al Aluminum 181 11.7

Ag Silver 58.6 5.50

Au Gold 59.0 5.53

Cu Copper 84.7 7.04

Fe Iron 170 11.2

K Potassium 14.0 2.11

Li Lithium 47.0 4.75

Mg Magnesium 86.0 7.11

Mn Manganese 165 11.0

Na Sodium 26.5 3.24

Sn Tin 148 10.2

Zn Zinc 132 9.46

† Number densities are measured using the Hall effect, discussed in Section 26-4.

TiplerMosca Physics.5e ch 38 6/26/03 8:47 PM 39

The Fermi Fac tor a t T 0

The probability of an energy state being occupied is called the Fermi factor, f (E).At T 0 all the states below EF are filled, whereas all those above this energy areempty, as shown in Figure 38-8. Thus, at T 0 the Fermi factor is simply

38-24

The Fermi Fac tor for T > 0

At temperatures greater than T 0, some electrons will occupy higher energystates because of thermal energy gained during collisions with the lattice. How-ever, an electron cannot move to a higher or lower state unless it is unoccupied.Since the kinetic energy of the lattice ions is of the order of kT, electrons cannotgain much more energy than kT in collisions with the lattice ions. Therefore, onlythose electrons with energies within about kT of the Fermi energy can gain energy as the temperature is increased. At 300 K, kT is only 0.026 eV, so the exclu-sion principle prevents all but a very few electrons near the top of the energy dis-tribution from gaining energy through random collisions with the lattice ions.Figure 38-9 shows the Fermi factor for some temperature T. Since for T 0 thereis no distinct energy that separates filled levels from unfilled levels, the defini-tion of the Fermi energy must be slightly modified. At temperature T, the Fermienergy is defined to be that energy for which the probability of being occupied is . For all but extremely high temperatures, the difference between the Fermienergy at temperature T and the Fermi energy at temperature T 0 is very small.

The Fermi temperature TF is defined by

38-25

For temperatures much lower than the Fermi temperature, the average energy ofthe lattice ions will be much less than the Fermi energy, and the electron energydistribution will not differ greatly from that at T 0.

Find the Fermi temperature for copper.

Use EF 7.04 eV and k 8.62 105 eV/K in Equation 38-25:

R E M A R K S We can see from this example that the Fermi temperature of copperis much greater than any temperature T for which copper remains a solid.

Because an electric field in a conductor accelerates all of theconduction electrons together, the exclusion principle does notprevent the free electrons in filled states from participating inconduction. Figure 38-10 shows the Fermi factor in one dimen-sion versus velocity for an ordinary temperature. The factor is

81,700 KTF EF

k

7.04 eV8.62 105 eV/K

E X A M P L E 3 8 - 4THE FERMI TEMPERATURE FOR COPPER

kTF EF

12

f (E) 0, E EF

f (E) 1, E EF


f (E)

1

0E EF

F I G U R E 3 8 - 8 Fermi factor versusenergy at T 0.

f (E)

1

0E EF

1––2

F I G U R E 3 8 - 9 The Fermi factor forsome temperature T. Some electrons with energies near the Fermi energy are excited, as indicated by the shadedregions. The Fermi energy is that value of E for which f (E ) .1

2

No electric field With electric field

f (E)

–u +u vxF F

F I G U R E 3 8 - 1 0 Fermi factor versus velocity in one dimension withno electric field (solid) and with an electric field in the x direction(dashed). The difference is greatly exaggerated.


approximately 1 for speeds vx in the range uF vx uF, where the Fermi speeduF is related to the Fermi energy by EF mu . Then

38-26

Calculate the Fermi speed for copper.

Use Equation 38-26 with EF 7.04 eV:

The dashed curve in Figure 38-10 shows theFermi factor after the electric field has been act-ing for some time t. Although all of the free elec-trons have their velocities shifted in the directionopposite to the electric field, the net effect isequivalent to shifting only the electrons near theFermi energy.

Contac t Potent ia l

When two different metals are placed in contact,a potential difference Vcontact called the contactpotential develops between them. The contactpotential depends on the work functions of thetwo metals, f1 and f2 (we encountered workfunctions when the photoelectric effect was intro-duced in Chapter 34), and the Fermi energies of the two metals. When the metalsare in contact, the total energy of the system is lowered if electrons near theboundary move from the metal with the higher Fermi energy into the metal withthe lower Fermi energy until the Fermi energies of the two metals are the same,as shown in Figure 38-11. When equilibrium is established, the metal with thelower initial Fermi energy is negatively charged and the other is positivelycharged, so that between them there is a potential difference Vcontact given by

38-27

Table 38-2 lists the work functions for several metals.

Vcontact f1 f2

e

1.57 106 m/suF B 2(7.04 eV)9.11 1031 kg

a1.60 1019 J1 eV

b

E X A M P L E 3 8 - 5THE FERMI SPEED FOR COPPER

uF B2EF

me

2F

12


TABLE 38-2Work Functions for Some Metals

Metal f, eV

Ag Silver 4.7

Au Gold 4.8

Ca Calcium 3.2

Cu Copper 4.1

Metal 1 Metal 2 Metal 1 Metal 2

+++++

–––––

Allowed

Occupied

φ

Electrons at restoutside either metal

1φ 2

φ1 φ2

EF2

EF1EF

–

Touching

(a) (b)

F I G U R E 3 8 - 1 1 (a) Energy levels fortwo different metals with different Fermienergies and work functions. (b) Whenthe metals are in contact, electrons flowfrom the metal that initially has thehigher Fermi energy to the metal thatinitially has the lower Fermi energy untilthe Fermi energies are equal.

Metal f, eV

K Potassium 2.1

Mn Manganese 3.8

Na Sodium 2.3

Ni Nickel 5.2


The threshold wavelength for the photoelectric effect is 271 nm for tungstenand 262 nm for silver. What is the contact potential developed when silver andtungsten are placed in contact?

P I C T U R E T H E P R O B L E M The contact potential is proportional to the differ-ence in the work functions for the two metals. The work function f can be foundfrom the given threshold wavelengths using f hc/lt (Equation 34-4).

1. The contact potential is given by Equation 38-27:

2. The work function is related to the threshold wavelength (Equation 34-4):

3. Substitute lt 271 nm for tungsten:

4. Substitute lt 262 nm for silver:

5. The contact potential is thus:

Heat Capaci ty Due to E lec t rons in a Meta l

The quantum-mechanical modification of the electron distribution in metals allows us to understand why the contribution of the electron gas to the heat capacity of a metal is much less that of the ions. According to the classicalequipartition theorem, the energy of the lattice ions in n moles of a solid is 3nRT,and thus the molar specific heat is c 3R, where R is the universal gas constant(see Section 18-7). In a metal, there is a free electron gas containing a number ofelectrons approximately equal to the number of lattice ions. If these electronsobey the classical equipartition theorem, they should have an energy of nRTand contribute an additional R to the molar specific heat. But measured heat capacities of metals are just slightly greater than those of insulators. We can understand this because at some temperature T, only those electrons with ener-gies near the Fermi energy can be excited by random collisions with the latticeions. The number of these electrons is of the order of (kT/EF )N, where N is the total number of electrons. The energy of these electrons is increased from that at T 0 by an amount that is of the order of kT. So the total increase in thermalenergy is of the order of (kT/EF )N kT. We can thus express the energy of N elec-trons at temperature T as

38-28

where a is some constant that we expect to be of the order of 1 if our reasoning is correct. The calculation of a is quite difficult. The result is a p 2/4. Using this result and writing EF in terms of the Fermi temperature, EF kTF , we obtainthe following for the contribution of the electron gas to the heat capacity at con-stant volume:

E NEav(0) aNkTEF

kT

32

32

0.15 V

Vcontact fAg fW

e 4.73 V 4.58 V

fAg 1240 eVnm

262 nm 4.73 eV

fW hclt

1240 eVnm

271 nm 4.58 eV

f hclt

Vcontact f1 f2

e

E X A M P L E 3 8 - 6CONTACT POTENTIAL BETWEEN SILVER AND TUNGSTEN



where we have written Nk in terms of the gas constant R (Nk nR). The molarspecific heat at constant volume is then

38-29

We can see that because of the large value of TF, the contribution of the electrongas is a small fraction of R at ordinary temperatures. Because TF 81,700 K forcopper, the molar specific heat of the electron gas at T 300 K is

which is in good agreement with experiment.

38-4 Quantum Theory of Electrical Conduction

We can use Equation 38-14 for the resistivity if we use the Fermi speed uF in placeof vav:

38-30

We now have two problems. First, since the Fermi speed uF is approximately independent of temperature, the resistivity given by Equation 38-30 is indepen-dent of temperature unless the mean free path depends on it. The second problem concerns magnitudes. As mentioned earlier, the classical expression forresistivity using vav calculated from the Maxwell–Boltzmann distribution givesvalues that are about 6 times too large at T 300 K. Since the Fermi speed uF

is about 16 times the Maxwell–Boltzmann value of vav, the magnitude of r

predicted by Equation 38-30 will be approximately 100 times greater than the experimentally determined value. The resolution of both of these problems liesin the calculation of the mean free path l.

The Scat ter ing of E lec t ron Waves

In Equation 38-16 for the classical mean free path l 1/(nionA), the quantity A pr2 is the area of the lattice ion as seen by an electron. In the quantum calcu-lation, the mean free path is related to the scattering of electron waves by thecrystal lattice. Detailed calculations show that, for a perfectly ordered crystal, l ; that is, there is no scattering of the electron waves. The scattering of electron waves arises because of imperfections in the crystal lattice, which havenothing to do with the actual cross-sectional area A of the lattice ions. Accordingto the quantum theory of electron scattering, A depends merely on deviations ofthe lattice ions from a perfectly ordered array and not on the size of the ions. Themost common causes of such deviations are thermal vibrations of the lattice ionsor impurities.

We can use l 1/(nionA) for the mean free path if we reinterpret the area A.Figure 38-12 compares the classical picture and the quantum picture of this area.In the quantum picture, the lattice ions are points that have no size but present an

r meuF

nee2l

c œV

p 2

2 a 300 K

81,700b R 0.02 R

c œV

p 2

2 R

TTF

CV dEdT

2aNkkTEF

p 2

2 nR

TTF

S E C T I O N 3 8 - 4 Quantum Theory of Electrical Conduction 1243

A = πr2

r

A = πr20

r0

F I G U R E 3 8 - 1 2 (a) Classical picture ofthe lattice ions as spherical balls of radiusr that present an area p r2 to the electrons.(b) Quantum-mechanical picture of thelattice ions as points that are vibrating in three dimensions. The area presentedto the electrons is p r , where r0 is theamplitude of oscillation of the ions.

20

(a)

(b)


area A pr , where r0 is the amplitude of thermal vibrations. InChapter 14, we saw that the energy of vibration in simple harmonicmotion is proportional to the square of the amplitude, which is pr .Thus, the effective area A is proportional to the energy of vibrationof the lattice ions. From the equipartition theorem,† we know thatthe average energy of vibration is proportional to kT. Thus, A is pro-portional to T, and l is proportional to 1/T. Then the resistivitygiven by Equation 38-14 is proportional to T, in agreement withexperiment.

The effective area A due to thermal vibrations can be calculated,and the results give values for the resistivity that are in agreementwith experiment. At T 300 K, for example, the effective area turnsout to be about 100 times smaller than the actual area of a latticeion. We see, therefore, that the free-electron model of metals gives agood account of electrical conduction if the classical mean speed vav is replacedby the Fermi speed uF and if the collisions between electrons and the lattice ionsare interpreted in terms of the scattering of electron waves, for which only devia-tions from a perfectly ordered lattice are important.

The presence of impurities in a metal also causes deviations from perfect regularity in the crystal lattice. The effects of impurities on resistivity are approx-imately independent of temperature. The resistivity of a metal containing impuri-ties can be written r rt r i , where rt is the resistivity due to the thermal motion of the lattice ions and r i is the resistivity due to impurities. Figure 38-13shows typical resistance curves versus temperature curves for metals with impurities. As the absolute temperature approaches zero, rt approaches zero,and the resistivity approaches the constant r i due to impurities.

38-5 Band Theory of Solids

Resistivities vary enormously between insulators and conductors. For a typi-cal insulator, such as quartz, r 1016 m, whereas for a typical conductor, r 108 m. The reason for this enormous variation is the variation in the number density of free electrons ne. To understand this variation, we consider theeffect of the lattice on the electron energy levels.

We begin by considering the energy levels of the individual atomsas they are brought together. The allowed energy levels in an isolatedatom are often far apart. For example, in hydrogen, the lowest al-lowed energy E1 13.6 eV is 10.2 eV below the next lowest allowedenergy E2 (13.6 eV)/4 3.4 eV.‡ Let us consider two identicalatoms and focus our attention on one particular energy level. Whenthe atoms are far apart, the energy of a particular level is the same foreach atom. As the atoms are brought closer together, the energy levelfor each atom changes because of the influence of the other atom. Asa result, the level splits into two levels of slightly different energiesfor the two-atom system. If we bring three atoms close together, aparticular energy level splits into three separate levels of slightly dif-ferent energies. Figure 38-14 shows the energy splitting of two en-ergy levels for six atoms as a function of the separation of the atoms.

If we have N identical atoms, a particular energy level in the iso-lated atom splits into N different, closely spaced energy levels whenthe atoms are close together. In a macroscopic solid, N is very large—of the order of 1023—so each energy level splits into a very large

20

20


† The equipartition theorem does hold for the lattice ions, which obey the Maxwell–Boltzmann energy distribution.‡ The energy levels in hydrogen are discussed in Chapter 36.

R/R

T, K

290 K

4.0 × 10 –3

3.0 × 10 –3

2.0 × 10 –3

1.0 × 10 –3

2 4 6 8 10 12 14 16 18 20

F I G U R E 3 8 - 1 3 Relative resistanceversus temperature for three samples ofsodium. The three curves have the sametemperature dependence but differentmagnitudes because of differing amountsof impurities in the samples.

Allowed energy bands

Separation of atoms

Energy

Level 2

Level 1

F I G U R E 3 8 - 1 4 Energy splitting of twoenergy levels for six atoms as a functionof the separation of the atoms. Whenthere are many atoms, each level splitsinto a near-continuum of levels called aband.


number of levels called a band. The levels are spaced almost continuously withinthe band. There is a separate band of levels for each particular energy level of theisolated atom. The bands may be widely separated in energy, they may be closetogether, or they may even overlap, depending on the kind of atom and the typeof bonding in the solid.

The lowest energy bands, corresponding to the lowest energy levels of theatoms in the lattice, are filled with electrons that are bound to the individualatoms. The electrons that can take part in conduction occupy the higher energybands. The highest energy band that contains electrons is called the valenceband. The valence band may be completely filled with electrons or only partiallyfilled, depending on the kind of atom and the type of bonding in the solid.

We can now understand why some solids are conductors and why others areinsulators. If the valence band is only partially full, there are many availableempty energy states in the band, and the electrons in the band can easily beraised to a higher energy state by an electric field. Accordingly, this material is a good conductor. If the valence band is full and there is a large energy gap between it and the next available band, a typical applied electric field will be tooweak to excite an electron from the upper energy levels of the filled band acrossthe large gap into the energy levels of the empty band, so the material is an insulator. The lowest band in which there are unoccupied states is called the conduction band. In a conductor, the valence band is only partially filled, so thevalence band is also the conduction band. An energy gap between allowed bandsis called a forbidden energy band.

The band structure for a conductor, such as copper, is shown in Figure 38-15a.The lower bands (not shown) are filled with the inner electrons of the atoms. Thevalence band is only about half full. When an electric field is established in theconductor, the electrons in the conduction band are accelerated, which meansthat their energy is increased. This is consistent with the Pauli exclusion principlebecause there are many empty energy states just above those occupied by elec-trons in this band. These electrons are thus the conduction electrons.

Figure 38-15b shows the band structure for magnesium, which is also a conductor. In this case, the highest occupied band is full, but there is an emptyband above it that overlaps it. The two bands thus form a combined valence–conduction band that is only partially filled.

Figure 38-15c shows the band structure for a typical insulator. At T 0 K, thevalence band is completely full. The next energy band containing empty energystates, the conduction band, is separated from the valence band by a large energygap. At T 0, the conduction band is empty. At ordinary temperatures, a fewelectrons can be excited to states in this band, but most cannot be excited to statesbecause the energy gap is large compared with the energy an electron might obtain by thermal excitation. Very few electrons can be thermally excited to thenearly empty conduction band, even at fairly high temperatures. When an elec-tric field of ordinary magnitude is established in the solid, electrons cannot be

S E C T I O N 3 8 - 5 Band Theory of Solids 1245

F I G U R E 3 8 - 1 5 Four possible bandstructures for a solid. (a) A typicalconductor. The valence band is onlypartially full, so electrons can be easilyexcited to nearby energy states. (b) Aconductor in which the allowed energybands overlap. (c) A typical insulator.There is a forbidden band with a largeenergy gap between the filled valenceband and the conduction band. (d ) A semiconductor. The energy gapbetween the filled valence band and theconduction band is very small, so someelectrons are excited to the conductionband at normal temperatures, leavingholes in the valence band.

Closely spacedenergy levelswithin the bands

SemiconductorInsulatorConductorConductor

OverlapAllowed, empty

Forbidden

Allowed, occupied

(a) (b) (c) (d)


accelerated because there are no empty energy states at nearby energies. We de-scribe this by saying that there are no free electrons. The small conductivity thatis observed is due to the very few electrons that are thermally excited into thenearly empty conduction band. When an electric field applied to an insulator issufficiently strong to cause an electron to be excited across the energy gap to theempty band, dielectric breakdown occurs.

In some materials, the energy gap between the filled valence band and theempty conduction band is very small, as shown in Figure 38-15d. At T 0, thereare no electrons in the conduction band and the material is an insulator. However, at ordinary temperatures, there are an appreciable number of electronsin the conduction band due to thermal excitation. Such a material is called an intrinsic semiconductor. For typical intrinsic semiconductors, such as siliconand germanium, the energy gap is only about 1 eV. In the presence of an electricfield, the electrons in the conduction band can be accelerated because there areempty states nearby. Also, for each electron in the conduction band there is a vacancy, or hole, in the nearly filled valence band. In the presence of an electricfield, electrons in this band can also be excited to a vacant energy level. This con-tributes to the electric current and is most easily described as the motion of a holein the direction of the field and opposite to the motion of the electrons. The holethus acts like a positive charge. To visualize the conduction of holes, think of atwo-lane, one-way road with one lane full of parked cars and the other laneempty. If a car moves out of the filled lane into the empty lane, it can move aheadfreely. As the other cars move up to occupy the space left, the empty space propa-gates backward in the direction opposite the motion of the cars. Both the forwardmotion of the car in the nearly empty lane and the backward propagation of theempty space contribute to a net forward propagation of the cars.

An interesting characteristic of semiconductors is that the resistivity of thematerial decreases as the temperature increases, which is contrary to the case for normal conductors. The reason is that as the temperature increases, thenumber of free electrons increases because there are more electrons in the con-duction band. The number of holes in the valence band also increases, of course.In semiconductors, the effect of the increase in the number of charge carriers,both electrons and holes, exceeds the effect of the increase in resistivity due to the increased scattering of the electrons by the lattice ions due to thermalvibrations. Semiconductors therefore have a negative temperature coefficient ofresistivity.

38-6 Semiconductors

The semiconducting property of intrinsic semiconductors materials makes themuseful as a basis for electronic circuit components whose resistivity can be con-trolled by application of an external voltage or current. Most such solid-state devices, however, such as the semiconductor diode and the transistor, make use of impurity semiconductors, which are created through the controlled additionof certain impurities to intrinsic semiconductors. This process is called doping.Figure 38-16a is a schematic illustration of silicon doped with a small amount ofarsenic so that the arsenic atoms replace a few of the silicon atoms in the crystallattice. The conduction band of pure silicon is virtually empty at ordinary tem-peratures, so pure silicon is a poor conductor of electricity. However, arsenic hasfive valence electrons rather than the four valence electrons of silicon. Four ofthese electrons take part in bonds with the four neighboring silicon atoms, andthe fifth electron is very loosely bound to the atom. This extra electron occupiesan energy level that is just slightly below the conduction band in the solid, and itis easily excited into the conduction band, where it can contribute to electricalconduction.


F I G U R E 3 8 - 1 6 (a) A two-dimensionalschematic illustration of silicon dopedwith arsenic. Because arsenic has fivevalence electrons, there is an extra,weakly bound electron that is easilyexcited to the conduction band, where it can contribute to electrical conduction.(b) Band structure of an n-typesemiconductor, such as silicon dopedwith arsenic. The impurity atoms providefilled energy levels that are just belowthe conduction band. These levels donateelectrons to the conduction band.

Si Si Si Si Si

SiSi

Si

Si

Si

As

AsSi

Si

Si

SiSiSiSiSi

Extra electron Extra electron

Empty conductionband

Impuritydonor levels

Filled valence band

(a)

(b)


The effect on the band structure of a silicon crystal achieved by doping it witharsenic is shown in Figure 38-16b. The levels shown just below the conductionband are due to the extra electrons of the arsenic atoms. These levels are calleddonor levels because they donate electrons to the conduction band without leaving holes in the valence band. Such a semiconductor is called an n-type semiconductor because the major charge carriers are negative electrons. The conductivity of a doped semiconductor can be controlled by controlling theamount of impurity added. The addition of just one part per million can increasethe conductivity by several orders of magnitude.

Another type of impurity semiconductor can be made by replacing a siliconatom with a gallium atom, which has three valence electrons (Figure 38-17a). Thegallium atom accepts electrons from the valence band to complete its four cova-lent bonds, thus creating a hole in the valence band. The effect on the band struc-ture of silicon achieved by doping it with gallium is shown in Figure 38-17b. Theempty levels shown just above the valence band are due to the holes from theionized gallium atoms. These levels are called acceptor levels because theyaccept electrons from the filled valence band when these electrons are thermallyexcited to a higher energy state. This creates holes in the valence band that arefree to propagate in the direction of an electric field. Such a semiconductor iscalled a p-type semiconductor because the charge carriers are positive holes. Thefact that conduction is due to the motion of positive holes can be verified by theHall effect. (The Hall effect is discussed in chapter 26.)

S E C T I O N 3 8 - 6 Semiconductors 1247

Synthetic crystal silicon is producedbeginning with a raw material containingsilicon (for instance, common beachsand), separating out the silicon, andmelting it. From a seed crystal, the moltensilicon grows into a cylindrical crystal,such as the one shown here. The crystals(typically about 1.3 m long) are formedunder highly controlled conditions toensure that they are flawless and thecrystals are then sliced into thousands of thin wafers onto which the layers of an integrated circuit are etched.

Si Si Si Si Si

Si

Si

Si

Ga

Si

Si

Si

Si

Ga

Si

SiSiSiSiSi

Hole

Hole

Empty conductionband

Impurityacceptor levels

Filled valence band

(a) (b)

F I G U R E 3 8 - 1 7 (a) A two-dimensional schematic illustration of silicon doped withgallium. Because gallium has only three valence electrons, there is a hole in one of itsbonds. As electrons move into the hole the hole moves about, contributing to theconduction of electrical current. (b) Band structure of a p-type semiconductor, such assilicon doped with gallium. The impurity atoms provide empty energy levels justabove the filled valence band that accept electrons from the valence band.

The number of free electrons in pure silicon is approximately 1010 electrons/cm3

at ordinary temperatures. If one silicon atom out of every million atoms is replaced by an arsenic atom, how many free electrons per cubic centimeter arethere? (The density of silicon is 2.33 g/cm3 and its molar mass is 28.1 g/mol.)

P I C T U R E T H E P R O B L E M The number of silicon atoms per cubic centimeter,nSi can be found from nSi NAr/M. Then, since each arsenic atom contributesone free electron, the number of electrons contributed by the arsenic atoms is 106 nS.

T r y I t Yo u r s e l fE X A M P L E 3 8 - 7NUMBER DENSITY OF FREE ELECTRONS

IN ARSENIC-DOPED SILICON


Cover the column to the right and try these on your own before looking at the answers.

Steps Answers

1. Calculate the number of silicon atoms per cubic centimeter.

ne 106 nSi 2. Multiply by 106 to obtain the number of arsenic atoms per cubic centimeter, which equals the added number of freeelectrons per cubic centimeter.

R E M A R K S Because silicon has so few free electrons per atom, the number ofconduction electrons is increased by a factor of approximately 5 million by dop-ing silicon with just one arsenic atom per million silicon atoms.

E X E R C I S E How many free electrons are there per silicon atom in pure silicon?(Answer 2 1013)

*38-7 Semiconductor Junctions and Devices

Semiconductor devices such as diodes and transistors make use of n-type semi-conductors and p-type semiconductors joined together, as shown in Figure 38-18.In practice, the two types of semiconductors are often incorporated into a singlesilicon crystal doped with donor impurities on one side and acceptor impuritieson the other side. The region in which the semiconductor changes from a p-typesemiconductor to an n-type semiconductor is called a junction.

When an n-type semiconductor and a p-type semiconductor are placed in contact, the initially unequal concentrations of electrons and holes result in thediffusion of electrons across the junction from the n side to the p side and holesfrom the p side to the n side until equilibrium is established. The result of this diffusion is a net transport of positive charge from the p side to the n side. Unlikethe case when two different metals are in contact, the electrons cannot travel veryfar from the junction region because the semiconductor is not a particularly goodconductor. The diffusion of electrons and holes therefore creates a double layer ofcharge at the junction similar to that on a parallel-plate capacitor. There is, thus, apotential difference V across the junction, which tends to inhibit further diffu-sion. In equilibrium, the n side with its net positive charge will be at a higher potential than the p side with its net negative charge. In the junction region, between the charge layers, there will be very few charge carriers of either type, sothe junction region has a high resistance. Figure 38-19 shows the energy-level diagram for a pn junction. The junction region is also called the depletion regionbecause it has been depleted of charge carriers.

*Diodes

In Figure 38-20, an external potential difference has been applied across a pnjunction by connecting a battery and a resistor to the semiconductor. When thepositive terminal of the battery is connected to the p side of the junction, asshown in Figure 38-20a, the junction is said to be forward biased. Forward

4.99 1016 electrons/cm3

4.99 1022 atoms/cm3

(2.33 g/cm3)(6.02 1023 atoms/mol)

28.1 g/mol

nSi rNA

M


–

––––––

++++++

electronsholes

p side n side

+

F I G U R E 3 8 - 1 8 A pn junction. Becauseof the difference in their concentrationson either side of the pn junction, holesdiffuse from the p side to the n side, andelectrons diffuse from the n side to the pside. As a result, there is a double layer ofcharge at the junction, with the p sidebeing negative and the n side beingpositive.

–––––– +

+++++

Conductionband Conduction

band

Valenceband Valence

band

Energy

EEg

EF

0

n sidep side Junction ordepletion region

∆

F I G U R E 3 8 - 1 9 Electron energy levelsfor a pn junction.


biasing lowers the potential across the junction. The diffusion of electrons andholes is thereby increased as they attempt to reestablish equilibrium, resulting ina current in the circuit.

If the positive terminal of the battery is connected to the n sideof the junction, as shown in Figure 38-20b, the junction is said to bereverse biased. Reverse biasing tends to increase the potential dif-ference across the junction, thereby further inhibiting diffusion.Figure 38-21 shows a plot of current versus voltage for a typicalsemiconductor junction. Essentially, the junction conducts only inone direction. A single-junction semiconductor device is called adiode.† Diodes have many uses. One is to convert alternating cur-rent into direct current, a process called rectification.

Note that the current in Figure 38-21 suddenly increases inmagnitude at extreme values of reverse bias. In such large electricfields, electrons are stripped from their atomic bonds and acceler-ated across the junction. These electrons, in turn, cause others tobreak loose. This effect is called avalanche breakdown. Althoughsuch a breakdown can be disastrous in a circuit where it is not intended, the fact that it occurs at a sharply defined voltage makesit of use in a special voltage reference standard known as a Zenerdiode. Zener diodes are also used to protect devices from exces-sively high voltages.

An interesting effect, one that we discuss only qualitatively, oc-curs if both the n side and the p side of a pn-junction diode are soheavily doped that the donors on the n side provide so many elec-trons that the lower part of the conduction band is practically filled, and the ac-ceptors on the p side accept so many electrons that the upper part of the valenceband is nearly empty. Figure 38-22a shows the energy-level diagram for this situ-ation. Because the depletion region is now so narrow, electrons can easily pene-trate the potential barrier across the junction and tunnel to the other side. Theflow of electrons through the barrier is called a tunneling current, and such aheavily doped diode is called a tunnel diode.

At equilibrium with no bias, there is an equal tunneling current in each direc-tion. When a small bias voltage is applied across the junction, the energy-level diagram is as shown in Figure 38-22b, and the tunneling of electrons from the

S E C T I O N 3 8 - 7 Semiconductors Junctions and Devices 1249

F I G U R E 3 8 - 2 0 A pn-junction diode.(a) Forward-biased pn junction. Theapplied potential difference enhances the diffusion of holes from the p side tothe n side and of electrons from the nside to the p side, resulting in a current I.(b) Reverse-biased pn junction. Theapplied potential difference inhibits the further diffusion of holes andelectrons across the junction, so there isno current.

–

–

–

+

+

+

p side n side

Forward bias

E–

+

I

–––

––

–

–

–

–

++ +

+++

+++

p side n side

Reverse bias

E–

+

No current

† The name diode originates from a vacuum tube device consisting of just two electrodes that also conducts electriccurrent in one direction only.

I, mA

30

20

10

10

20

30

V

Breakdownvoltage

30 20

0.2 0.4 0.6 0.8 1.0 V, V

µ A

Reversebias

Forwardbias

F I G U R E 3 8 - 2 1 Plot of current versus applied voltageacross a pn junction. Note the different scales on bothaxes for the forward and reverse bias conditions.

F I G U R E 3 8 - 2 2 Electron energy levelsfor a heavily doped pn-junction tunneldiode. (a) With no bias voltage, someelectrons tunnel in each direction. (b) With a small bias voltage, thetunneling current is enhanced in onedirection, making a sizable contributionto the net current. (c) With furtherincreases in the bias voltage, thetunneling current decreases dramatically.

Conductionband

EmptyEF

Valenceband

p side n side

FilledEF EF

p side n side

Tunneling

p side n side

EF

EF

(a) (b) (c)

(a) (b)


n side to the p side is increased, whereas the tunneling of electrons in the oppo-site direction is decreased. This tunneling current, in addition to the usual cur-rent due to diffusion, results in a considerable net current. When the bias voltageis increased slightly, the energy-level diagram is as shown in Figure 38-22c, andthe tunneling current is decreased. Although the diffusion current is increased,the net current is decreased. At large bias voltages, the tunneling current is com-pletely negligible, and the total current increases with increasing bias voltagedue to diffusion, as in an ordinary pn-junction diode. Figure 38-23 shows the current curve versus the voltage curve for a tunnel diode. Such diodes are usedin electric circuits because of their very fast response time. When operated nearthe peak in the current curve versus the voltage curve, a small change in biasvoltage results in a large change in the current.

Another use for the pn-junction semiconductor is the solar cell, which is il-lustrated schematically in Figure 38-24. When a photon of energy greater thanthe gap energy (1.1 eV in silicon) strikes the p-type region, it can excite an elec-tron from the valence band into the conduction band, leaving a hole in the va-lence band. This region is already rich in holes. Some of the electrons created bythe photons will recombine with holes, but some will migrate to the junction.From there, they are accelerated into the n-type region by the electric field be-tween the double layer of charge. This creates an excess negative charge in then-type region and an excess positive charge in the p-type region. The result is apotential difference between the two regions, which in practice is approxi-mately 0.6 V. If a load resistance is connected across the two regions, a chargeflows through the resistance. Some of the incident light energy is thus con-verted into electrical energy. The current in the resistor is proportional to thenumber of incident photons, which is in turn proportional to the intensity of theincident light.

There are many other applications of semiconductors with pn junctions. Parti-cle detectors, called surface-barrier detectors, consist of a pn-junction semicon-ductor with a large reverse bias so that there is ordinarily no current. When ahigh-energy particle, such as an electron, passes through the semiconductor, itcreates many electron–hole pairs as it loses energy. The resulting current pulsesignals the passage of the particle. Light-emitting diodes (LEDs) are pn-junctionsemiconductors with a large forward bias that produces a large excess concen-tration of electrons on the p side and holes on the n side of the junction. Underthese conditions, the diode emits light as the electrons and holes recombine. Thisis essentially the reverse of the process that occurs in a solar cell, in which electron–hole pairs are created by the absorption of light. LEDs are commonlyused as warning indicators and as sources of infrared light beams.

*Trans is tors

The transistor, a semiconducting device that is used to produce a desired outputsignal in response to an input signal, was invented in 1948 by William Shockley,John Bardeen, and Walter Brattain and has revolutionized the electronics indus-try and our everyday world. A simple bipolar junction transistor† consists of threedistinct semiconductor regions called the emitter, the base, and the collector. Thebase is a very thin region of one type of semiconductor sandwiched between tworegions of the opposite type. The emitter semiconductor is much more heavilydoped than either the base or the collector. In an npn transistor, the emitter andcollector are n-type semiconductors and the base is a p-type semiconductor; in apnp transistor, the base is an n-type semiconductor and the emitter and collectorare p-type semiconductors.


F I G U R E 3 8 - 2 3 Current versus applied(bias) voltage V for a tunnel diode. ForV VA , an increase in the bias voltage Venhances tunneling. For VA VVB , anincrease in the bias voltage inhibitstunneling. For VVB , the tunneling isnegligible, and the diode behaves like anordinary pn-junction diode.

I

VVA VB

p-type

n-typeRL

Incident light

I

F I G U R E 3 8 - 2 4 A pn-junctionsemiconductor as a solar cell. When lightstrikes the p-type region, electron–holepairs are created, resulting in a currentthrough the load resistance RL.

A light-emitting diode (LED).

† Besides the bipolar junction transistor, there are other categories of transistors, notably, the field-effect transistor.


Figure 38-25 and Figure 38-26 show, respectively, apnp transistor and an npn transistor with the symbolsused to represent each transistor in circuit diagrams. Wesee that either transistor consists of two pn junctions. Wewill discuss the operation of a pnp transistor. The opera-tion of an npn transistor is similar.

In normal operation, the emitter–base junction is forward biased, and thebase–collector junction is reverse biased, as shown in Figure 38-27. The heavilydoped p-type emitter emits holes that flow toward the emitter–base junction.This flow constitutes the emitter current Ie. Because the base is very thin, most ofthese holes flow across the base into the collector. This flow in the collector con-stitutes a current Ic. However, some of the holes recombine in the base producinga positive charge that inhibits the further flow of current. To prevent this, some ofthe holes that do not reach the collector are drawn off the base as a base current Ib

in a wire connected to the base. In Figure 38-27, therefore, Ic is almost but notquite equal to Ie, and Ib is much smaller than either Ic or Ie. It is customary to ex-press Ic as

38-31

where b is called the current gain of the transistor. Transistors can be designed tohave values of b as low as ten or as high as several hundred.

Figure 38-28 shows a simple pnp transistor used as an amplifier. A small, time-varying input voltage vs is connected in series with a bias voltage Veb. The basecurrent is then the sum of a steady current Ib produced by the bias voltage Veb

and a varying current ib due to the signal voltage vs. Because vs may at any

Ic bIb

S E C T I O N 3 8 - 7 Semiconductors Junctions and Devices 1251

Collector

Base

Emitter

n-type

p-type

n-type

Collector

Emitter

Base

(a)

(b)

Collector

Base

Emitter

p-type

n-type

p-type

Collector

Emitter

Base

F I G U R E 3 8 - 2 5 A pnp transistor. (a) The heavily doped emitter emits holes that pass through the thin base tothe collector. (b) Symbol for a pnptransistor in a circuit. The arrow points in the direction of the conventionalcurrent, which is the same as that of the emitted holes.

(a)

(b)

F I G U R E 3 8 - 2 6 An npn transistor. (a) The heavily doped emitter emitselectrons that pass through the thin baseto the collector. (b) Symbol for an npntransistor. The arrow points in thedirection of the conventional current,which is opposite the direction of the emitted electrons.

I

V

c

Ie

Ib

eb

Vec

c

b

e

p-type

n-type

p-type

–

–

+

+

F I G U R E 3 8 - 2 7 A pnp transistor biasedfor normal operation. Holes from theemitter can easily diffuse across the base,which is only tens of nanometers thick.Most of the holes flow to the collector,producing the current Ic .

Veb

Rb

RL

vsVec

c

b

e

p

n

pIb ib+

Ic ic+

Input –

+

–

+

Ib ib

Veb

Rb

RL

vs

Vec

cb

e+

Ic ic+

–

+

–

+

F I G U R E 3 8 - 2 8 (a) A pnp transistorused as an amplifier. A small change ib inthe base current results in a large changeic in the collector current. Thus, a smallsignal in the base circuit results in a largesignal across the load resistor RL in thecollector circuit. (b) The same circuit as in Figure 38-28a with the conventionalsymbol for the transistor.(a) (b)


instant be either positive or negative, the bias voltage Veb must be large enoughto ensure that there is always a forward bias on the emitter–base junction. Thecollector current will consist of two parts: a direct current Ic bIb and an alternating current ic bib. We thus have a current amplifier in which the time-varying output current ic is b times the input current ib. In such an amplifier, thesteady currents Ic and Ib, although essential to the operation of the transistor, areusually not of interest. The input signal voltage vs is related to the base current byOhm’s law:

38-32

where rb is the internal resistance of the transistor between the base and emitter.Similarly, the collector current ic produces a voltage vL across the output or loadresistance RL given by

38-33

Using Equation 38-31 and Equation 38-32, we have

38-34

The output voltage is thus related to the input voltage by

38-35

The ratio of the output voltage to the input voltage is the voltage gain of theamplifier:

38-36

A typical amplifier (e.g., in a tape player) has several transistors, similar to theone shown in Figure 38-28, connected in series so that the output of one transis-tor serves as the input for the next. Thus, the very small voltage fluctuations produced by the motion of the magnetic tape past the pickup heads controls thelarge amounts of power required to drive the loudspeakers. The power deliveredto the speakers is supplied by the dc sources connected to each transistor.

The technology of semiconductors extends well beyond individual transistorsand diodes. Many of the electronic devices we now take for granted, such as laptop computers and the processors that govern the operation of vehicles andappliances, rely on large-scale integration of many transistors and other circuitcomponents on a single chip. Large-scale integration combined with advancedconcepts in semiconductor theory has created remarkable new instruments forscientific research.

38-8 Superconductivity

There are some materials for which the resistivity suddenly drops to zero below acertain temperature Tc , which is called the critical temperature. This amazingphenomenon, called superconductivity, was discovered in 1911 by the Dutchphysicist H. Kamerlingh Onnes, who developed a technique for liquefying helium (boiling point 4.2 K) and put his technique to work exploring the proper-

Voltage gain vL

vs

bRL

Rb rb

vL bvs

Rb rb

RL bRL

Rb rb

vs

ic bib bvs

Rb rb

vL icRL

ib vs

Rb rb



ties of materials at temperatures in this range. Figure 38-29 shows Onnes’s plot of the resistance of mercury versus temperature. The critical temperature formercury is approximately the same as the boiling point of helium, which is 4.2 K.Critical temperatures for other superconducting elements range from less than0.1 K for hafnium and iridium to 9.2 K for niobium. The temperature range forsuperconductors goes much higher for a number of metallic compounds. For example, the superconducting alloy Nb3Ge, discovered in 1973, has a criticaltemperature of 25 K, which was the highest known until 1986, when the discov-eries of J. Georg Bednorz and K. Alexander Müller launched the era of high-temperature superconductors, now defined as materials that exhibit supercon-ductivity at temperatures above 77 K (the temperature at which nitrogen boils).To date (April 2003), the highest temperature at which superconductivity hasbeen demonstrated, using thallium doped HgBa2Ca2Cu3O8delta, is 138 K atatmospheric pressure. At extremely high pressures, some materials exhibitsuperconductivity at temperatures as high as 164 K.

The resistivity of a superconductor is zero. There can be a current in a super-conductor even when there is no emf in the superconducting circuit. Indeed, insuperconducting rings in which there was no electric field, steady currents havebeen observed to persist for years without apparent loss. Despite the cost and inconvenience of refrigeration with expensive liquid helium, many supercon-ducting magnets have been built using superconducting materials, because suchmagnets require no power expenditure to maintain the large current needed toproduce a large magnetic field.

S E C T I O N 3 8 - 8 Superconductivity 1253

10 –5

4.00 4.10 4.20 4.30 4.40

Ω

0.15

0.125

0.10

0.075

0.05

0.025

0.00

R, Ω

T, K

F I G U R E 3 8 - 2 9 Plot by H. KamerlinghOnnes of the resistance of mercury versustemperature, showing the suddendecrease at the critical temperature of T 4.2 K.

The wires for the magnetic field of amagnetic resonance imaging (MRI)machine carry large currents. To keep the wires from overheating, they aremaintained at superconductingtemperatures. To accomplish this, theyare immersed in liquid helium.


The discovery of high-temperature superconductors has revolutionized thestudy of superconductivity because relatively inexpensive liquid nitrogen, whichboils at 77 K, can be used for a coolant. However, many problems, such as brittle-ness and the toxicity of the materials, make these new superconductors difficultto use. The search continues for new materials that will superconduct at evenhigher temperatures.

The BCS Theor y

It had been recognized for some time that superconductivity is due to a collectiveaction of the conducting electrons. In 1957, John Bardeen, Leon Cooper, andRobert Schrieffer published a successful theory of superconductivity now knownby the initials of the inventors as the BCS theory. According to this theory, theelectrons in a superconductor are coupled in pairs at low temperatures. The coupling comes about because of the interaction between electrons and the crys-tal lattice. One electron interacts with the lattice and perturbs it. The perturbedlattice interacts with another electron in such a way that there is an attraction be-tween the two electrons that at low temperatures can exceed the Coulomb repul-sion between them. The electrons form a bound state called a Cooper pair. Theelectrons in a Cooper pair have equal and opposite spins, so they form a systemwith zero spin. Each Cooper pair acts as a single particle with zero spin, in otherwords, as a boson. Bosons do not obey the exclusion principle. Any number ofCooper pairs may be in the same quantum state with the same energy. In theground state of a superconductor (at T 0), all the electrons are in Cooper pairsand all the Cooper pairs are in the same energy state. In the superconductingstate, the Cooper pairs are correlated so that they all act together. An electriccurrent can be produced in a superconductor because all of the electrons in thiscollective state move together. But energy cannot be dissipated by individualcollisions of electron and lattice ions unless the temperature is high enough tobreak the binding of the Cooper pairs. The required energy is called the super-conducting energy gap Eg. In the BCS theory, this energy at absolute zero is re-lated to the critical temperature by

38-37

The energy gap can be determined by measuring the current across a junctionbetween a normal metal and a superconductor as a function of voltage. Considertwo metals separated by a layer of insulating material, such as aluminum oxide,that is only a few nanometers thick. The insulating material between the metalsforms a barrier that prevents most electrons from traversing the junction. However, waves can tunnel through a barrier if the barrier is not too thick, evenif the energy of the wave is less than that of the barrier.

When the materials on either side of the gap are normal nonsuperconductingmetals, the current resulting from the tunneling of electrons through the insulat-ing layer obeys Ohm’s law for low applied voltages (Figure 38-30a). When one ofthe metals is a normal metal and the other is a superconductor, there is no current(at absolute zero) unless the applied voltage V is greater than a critical voltage Vc Eg/(2e), where Eg is the superconductor energy gap. Figure 38-30b showsthe plot of current versus voltage for this situation. The current escalates rapidlywhen the energy 2eV absorbed by a Cooper pair traversing the barrier approaches Eg 2eVc, the minimum energy needed to break up the pair. (Thesmall current visible in Figure 38-30b before the critical voltage is reached is present because at any temperature above absolute zero some of the electrons inthe superconductor are thermally excited above the energy gap and are thereforenot paired.) At voltages slightly above Vc, the current versus voltage curve becomes that for a normal metal. The superconducting energy gap can thus bemeasured by measuring the average voltage for the transition region.

Eg 3.5 kTc


I

I

V

V

Vc

F I G U R E 3 8 - 3 0 Tunneling currentversus voltage for a junction of twometals separated by a thin oxide layer. (a) When both metals are normal metals,the current is proportional to the voltage,as predicted by Ohm’s law. (b) When onemetal is a normal metal and anothermetal is a superconductor, the current is approximately zero until the appliedvoltage V approaches the critical voltageVc Eg /(2e).

(a)

(b)


Calculate the superconducting energy gap for mercury (Tc 4.2 K) predictedby the BCS theory.

1. The BCS prediction for the energy gap is: Eg 3.5kTc

2. Substitute Tc 4.2 K:

Note that the energy gap for a typical superconductor is muchsmaller than the energy gap for a typical semiconductor, which is ofthe order of 1 eV. As the temperature is increased from T 0, some of the Cooper pairs are broken. Then there are fewer pairs availablefor each pair to interact with, and the energy gap is reduced until at T Tc the energy gap is zero (Figure 38-31).

The Josephson Ef fec t

When two superconductors are separated by a thin nonsupercon-ducting barrier (e.g., a layer of aluminum oxide a few nanometersthick), the junction is called a Josephson junction, based on the pre-diction in 1962 by Brian Josephson that Cooper pairs could tunnelacross such a junction from one superconductor to the other with noresistance. The tunneling of Cooper pairs constitutes a current, whichdoes not require a voltage to be applied across the junction. The cur-rent depends on the difference in phase of the wave functions that de-scribe the Cooper pairs. Let f1 be the phase constant for the wavefunction of a Cooper pair in one superconductor. All the Cooper pairsin a superconductor act coherently and have the same phase constant.If f2 is the phase constant for the Cooper pairs in the second super-conductor, the current across the junction is given by

38-38

where Imax is the maximum current, which depends on the thickness of the barrier. This result has been observed experimentally and is known as the dcJosephson effect.

Josephson also predicted that if a dc voltage V were applied across a Josephson junction, there would be a current that alternates with frequency fgiven by

38-39

This result, known as the ac Josephson effect, has been observed experimentally,and careful measurement of the frequency allows a precise determination of theratio e/h. Because frequency can be measured very accurately, the ac Josephsoneffect is also used to establish precise voltage standards. The inverse effect, inwhich the application of an alternating voltage across a Josephson junction results in a dc current, has also been observed.

f 2eh

V

I Imax sin(f2 f1)

1.27 103 eV

3.5(1.38 1023 J/K)(4.2 K)a 1 ev1.6 1019 J

bEg 3.5kTc

E X A M P L E 3 8 - 8SUPERCONDUCTING ENERGY GAP FOR MERCURY

S E C T I O N 3 8 - 8 Superconductivity 1255

BCS curveTinTantalumNiobium

E(T

)/E

(0)

gg

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

T/Tc

F I G U R E 3 8 - 3 1 Ratio of the energy gap attemperature T to that at temperature T 0 as afunction of the relative temperature T/Tc. The solidcurve is that predicted by the BCS theory.


Using e 1.602 1019 C and h 6.626 1034 Js, calculate the frequency ofthe Josephson current if the applied voltage is 1 mV.

Substitute the given values into Equation 38-39 to calculate f:

38-9 The Fermi–Dirac Distribution†

The classical Maxwell–Boltzmann distribution (Equation 17-39) gives the number of molecules with energy E in the range between E and E dE. It is theproduct of g(E) dE where g(E) is the density of states (number of energy states inthe range dE) and the Boltzmann factor eE/(kT), which is the probability of a statebeing occupied. The distribution function for free electrons in a metal is calledthe Fermi–Dirac distribution. The Fermi–Dirac distribution can be written in thesame form as the Maxwell–Boltzmann distribution, with the density of states calculated from quantum theory and the Boltzmann factor replaced by the Fermi factor. Let n(E) dE be the number of electrons with energies between E andE dE. This number is written

38-40

ENERGY DISTRIBUTION FUNCTION

where g(E) dE is the number of states between E and E dE and f (E) is the prob-ability of a state being occupied, which is the Fermi factor. The density of statesin three dimensions is somewhat difficult to calculate, so we just give the result.For electrons in a metal of volume V, the density of states is

38-41

DENSITY OF STATES

As in the classical Maxwell–Boltzmann distribution, the density of states is proportional to E1/2.

At T 0, the Fermi factor is given by Equation 38-24:

The integral of n(E) dE over all energies gives the total number of electrons N. Wecan derive Equation 38-22a for the Fermi energy at T 0 by integrating n(E) dEfrom E 0 to E . We obtain

f (E) 0, E EF

f (E) 1, E EF

g(E) 822p m3/2

e V

h3 E1/2

n(E) dE g(E) dE f (E)

483.5 MHz 4.835 108 Hz

f 2eh

V 2(1.602 1019 C)6.626 1034 Js

(106 V)

E X A M P L E 3 8 - 9FREQUENCY OF JOSEPHSON CURRENT


† This material is somewhat complicated. You may wish to read it in two passes, the first to gain an overview of thetopic, and the second to understand it in depth.


Note that at T 0, n(E) is zero for E EF. Solving for EF gives the Fermi energy atT 0:

38-42

which is Equation 38-22a. In terms of the Fermi energy, the density of states(Equation 38-41) is

38-43

DENSITY OF STATES IN TERMS OF EF

which is obtained by solving Equation 38-42 for me , and then substituting for me

in Equation 38-41. The average energy at T 0 is calculated from

where N g(E) dE is the total number of electrons. Performing the integration,we obtain Equation 38-23:

38-44

AVERAGE ENERGY AT T 0

At T 0, the Fermi factor is more complicated. It can be shown to be

38-45

FERMI FACTOR

We can see from this equation that for E greater than EF, e(EEF )/(kt) becomes verylarge as T approaches zero, so at T 0, the Fermi factor is zero for E EF. On theother hand, for E less than EF, e(EEF )/(kt) approaches 0 as T approaches zero, so atT 0, f (E) 1 for E EF. Thus, the Fermi factor given by Equation 38-45 holdsfor all temperatures. Note also that for any nonzero value of T, f (E) at E EF.

The complete Fermi–Dirac distribution function is thus

38-46

FERMI–DIRAC DISTRIBUTION

n(E) dE g(E)f (E) dE 822p m3/2

e V

h3 E1/2 1e(EEF)/(kT) 1

dE

12

f (E) 1

e(EEF)/(kT) 1

Eav 35EF

EF

0

Eav

EF

0

Eg(E) dE

EF

0

g(E) dE

1N

EF

0

Eg(E) dE

g(E) 822pm3/2

e V

h3 E1/2 32

NE3/2F E1/2

EF h2

8me

a 3NpVb2/3

1622p m3/2

e V

3h3 E3/2F

N q

0

n(E) dE EF

0

n(E) dE q

EF

n(E) dE 822p m3/2

e V

h3 EF

0

E1/2 dE 0

S E C T I O N 3 8 - 9 The Fermi-Dirac Distribution 1257


We can see that for those few electrons with energies much greater than the Fermienergy, the Fermi factor approaches 1/e(EEF )/(kt) e(EF E)/(kt) eEF /(kT)eE/(kT),which is proportional to eE/(kT). Thus, the high-energy tail of the Fermi–Dirac energy distribution decreases as eE/(kT), just like the classical Maxwell–Boltzmannenergy distribution. The reason is that in this high-energy region, there are manyunoccupied energy states and few electrons, so the Pauli exclusion principle isnot important, and the distribution approaches the classical distribution. This result has practical importance because it applies to the conduction electrons in semiconductors.

At what energy is the Fermi factor equal to 0.1 for copper at T 300 K?

P I C T U R E T H E P R O B L E M We set f (E) 0.1 in Equation 38-45, using T 300 Kand EF 7.04 eV from Table 38-1 and solve for E.

1. Solve Equation 38-45 for e(EEF )/(kt):

2. Take the logarithm of both sides:

3. Solve for E. For EF, use the value for EF at T 0 K listed in Table 38-1:

R E M A R K S The Fermi factor drops from about 1 to 0.1 at just 0.06 eV above theFermi energy of approximately 7 eV.

Find the probability that an energy state in copper 0.1 eV above the Fermi energy is occupied at T 300 K.

P I C T U R E T H E P R O B L E M The probability is the Fermi factor given in Equa-tion 38-45, with EF 7.04 eV, and E 7.14 eV.

1. The probability of a state being occupied equals theFermi factor:

2. Calculate the exponent in the Fermi factor (exponentsare always dimensionless):

3. Use this result to calculate the Fermi factor:

R E M A R K S The probability of an electron having an energy 0.1 eVabove the Fermi energy at 300 K is only about 2 percent.

WEB

MASTER the CON

CEPT

E X A M P L E 3 8 - 1 1PROBABILITY OF A HIGHER ENERGY STATE BEING OCCUPIED

E X A M P L E 3 8 - 1 0FERMI FACTOR FOR COPPER AT 300 K


so

7.10 eV

7.04 eV (ln 9)(8.62 105 eV/K)(300 K)

E EF (ln 9)kT

E EF

kT ln 9

e(EEF)/(kt) 1

f (E) 1

10.1

1 9

f (E) 1

e(EEF)/(kt) 1

0.0204 1

48 1

f (E) 1

e(EEF)/(kT) 1

1e3.87 1

E EF

kT

7.14 eV 7.04 eV(8.62 105 eV/K)(300 K)

3.87

P f (E) 1

e(EEF)/(kT) 1


Find the probability that an energy state in copper 0.1 eV below the Fermi energy is occupied at T 300 K.

P I C T U R E T H E P R O B L E M The probability is the Fermi factor given in Equa-tion 38-45, with EF 7.04 eV, and E 6.94 eV.

Cover the column to the right and try these on your own before looking at the answers.

Steps Answers

1. Write the Fermi factor.

2. Calculate the exponent in the Fermi factor.

3. Use your result from step 2 to calculate the Fermi factor.

R E M A R K S The probability of an electron having an energy of 0.1 eV below theFermi energy at 300 K is approximately 98 percent.

E X E R C I S E What is the probability of an energy state 0.1 eV below the Fermi energy being unoccupied at 300 K? (Answer 1 0.98 0.02 or 2 percent. This isthe probability of there being a hole at this energy.)

0.979 1

0.021 1

f (E) 1

e(EEF)/(kT) 1

1e3.87 1

E EF

kT

6.94 eV 7.04 eV(8.62 105 eV/K)(300 K)

3.87

f (E) 1

e(EEF)/(kT) 1

T r y I t Yo u r s e l fE X A M P L E 3 8 - 1 2PROBABILITY OF A LOWER ENERGY STATE BEING OCCUPIED

Summary 1259

S U M M A R Y

Topic Relevant Equations and Remarks

1. The Structure of Solids Solids are often found in crystalline form in which a small structure, which is calledthe unit cell, is repeated over and over. A crystal may have a face-centered-cubic,body-centered-cubic, hexagonal close-packed, or other structure depending on thetype of bonding between the atoms, ions, or molecules in the crystal and on the relative sizes of the atoms.

Potential energy U a 38-3

where r is the separation distance between neighboring ions and a is the Madelungconstant, which depends on the geometry of the crystal and is of the order of 1.8, andn is approximately 9.

2. A Microscopic Picture of Conduction

Resistivity r 38-14

where vav is the average speed of the electrons and l is their mean free path betweencollisions with the lattice ions.

mevav

nee2l

Arn

ke2

r



Mean free path l 38-16

where nion is the number of lattice ions per unit volume, r is their effective radius, andA is their effective cross-sectional area.

3. Classical Interpretation of vav and l vav is determined from the Maxwell–Boltzmann distribution, and r is the actual radiusof a lattice ion.

4. Quantum Interpretation of vav and l vav is determined from the Fermi–Dirac distribution and is approximately constant independent of temperature. The mean free path is determined from the scattering ofelectron waves, which occurs only because of deviations from a perfectly ordered array. The radius r is the amplitude of vibration of the lattice ion, which is propor-tional to , so A is proportional to T.

5. The Fermi Electron Gas

Fermi energy EF at T 0 EF is the energy of the last filled (or half-filled) energy state.

EF at T 0 EF is the energy at which the probability of being occupied is .

Approximate magnitude of EF EF is between 5 eV and 10 eV for most metals.

EF 2/3

(0.365 eVnm2)2/3

38-22a, 38-22b

Average energy at T 0 Eav EF 38-23

Fermi factor at T 0 The Fermi factor f (E) is the probability of a state being occupied.

f (E) 1, E EF

f (E) 0, E EF 38-24

Fermi temperature TF 38-25

Fermi speed uF 38-26

Contact potential When two different metals are placed in contact, electrons flow from the metal withthe higher Fermi energy to the metal with the lower Fermi energy until the Fermi energies of the two metals are equal. In equilibrium, there is a potential difference between the metals that is equal to the difference in the work function of the two metals divided by the electronic charge e:

Vcontact 38-27

c R 38-29

6. Band Theory of Solids When many atoms are brought together to form a solid, the individual energy levelsare split into bands of allowed energies. The splitting depends on the type of bondingand the lattice separation. The highest energy band that contains electrons is calledthe valence band. In a conductor, the valence band is only partially full, so there aremany available empty energy states for excited electrons. In an insulator, the valenceband is completely full and there is a large energy gap between it and the next

TTF

p 2

2VSpecific heat due to conduction

electrons

f1 f2

e

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aNVba 3N

pVbh2

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Dependence of EF on the numberdensity of free electrons

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allowed band, the conduction band. In a semiconductor, the energy gap between the filled valence band and the empty conduction band is small; so, at ordinary temperatures, an appreciable number of electrons are thermally excited into the conduction band.

7. Semiconductors The conductivity of a semiconductor can be greatly increased by doping. In an n-typesemiconductor, the doping adds electrons at energies just below that of the conduc-tion band. In a p-type semiconductor, holes are added at energies just above that ofthe valence band.

8. *Semiconductor Junctions and Devices

*Junctions Semiconductor devices such as diodes and transistors make use of n-type semicon-ductors and p-type semiconductors. The two types of semiconductors are typically asingle silicon crystal doped with donor impurities on one side and acceptor impuri-ties on the other side. The region in which the semiconductor changes from a p-typesemiconductor to an n-type semiconductor is called a junction. Junctions are used indiodes, solar cells, surface barrier detectors, LEDs, and transistors.

*Diodes A diode is a single-junction device that carries current in one direction only.

*Zener diodes A Zener diode is a diode with a very high reverse bias. It breaks down suddenly at adistinct voltage and can therefore be used as a voltage reference standard.

*Tunnel diodes A tunnel diode is a diode that is heavily doped so that electrons tunnel through thedepletion barrier. At normal operation, a small change in bias voltage results in alarge change in current.

*Transistors A transistor consists of a very thin semiconductor of one type sandwiched betweentwo semiconductors of the opposite type. Transistors are used in amplifiers because asmall variation in the base current results in a large variation in the collector current.

9. Superconductivity In a superconductor, the resistance drops suddenly to zero below a critical temperatureTc. Superconductors with critical temperatures as high as 138 K have been discovered.

The BCS theory Superconductivity is described by a theory of quantum mechanics called the BCS theory in which the free electrons form Cooper pairs. The energy needed to break upa Cooper pair is called the superconducting energy gap Eg. When all the electrons are paired, individual electrons cannot be scattered by a lattice ion, so the resistance is zero.

Tunneling When a normal conductor is separated from a superconductor by a thin layer of oxide, electrons can tunnel through the energy barrier if the applied voltage acrossthe layer is Eg/(2e), where Eg is the energy needed to break up a Cooper pair. The energy gap Eg can be determined by a measurement of the tunneling current versusthe applied voltage.

Josephson junction A system of two superconductors separated by a thin layer of nonconducting mater-ial is called a Josephson junction.

dc Josephson effect A dc current is observed to tunnel through a Josephson junction even in the absenceof voltage across the junction.

ac Josephson effect When a dc voltage V is applied across a Josephson junction, an ac current is observedwith a frequency

f V 38-39

Measurement of the frequency of this current allows a precise determination of theratio e/h.

2eh



10. The Fermi–Dirac Distribution The number of electrons with energies between E and E dE is given by

n(E) dE g(E) dE f (E) 38-40

where g(E) is the density of states and f (E) is the Fermi factor.

Density of states g(E) E1/2 38-41

Fermi factor at temperature f (E) 38-451

e (EEF)/(kT ) 1

822p m3/2e V

h3

Conceptual Problems

1 • In the classical model of conduction, the electronloses energy on average in a collision because it loses the driftvelocity it had picked up since the last collision. Where doesthis energy appear?

2 • When the temperature of pure copper is low-ered from 300 K to 4 K, its resistivity drops by a much greaterfactor than that of brass when it is cooled the same way.Why?

3 • Thomas refuses to believe that a potential differ-ence can be created simply by bringing two different metalsinto contact with each other. John talks him into making asmall wager, and is about to cash in. (a) Which two metalsfrom Table 38-2 would demonstrate his point most effec-tively? (b) What is the value of that contact potential?

4 • (a) In Problem 3, which choices of different metalswould make the least impressive demonstration? (b) What isthe value of that contact potential?

5 • A metal is a good conductor because the valenceenergy band for electrons is (a) completely full. (b) full, butthere is only a small gap to a higher empty band. (c) partlyfull. (d) empty. (e) None of these is correct.

6 • Insulators are poor conductors of electricity because(a) there is a small energy gap between the valence band andthe next higher band where electrons can exist. (b) there is alarge energy gap between the full valence band and the next

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higher band where electrons can exist. (c) the valence bandhas a few vacancies for electrons. (d) the valence band is onlypartly full. (e) None of these is correct.

7 • True or false:

(a) Solids that are good electrical conductors are usually goodheat conductors.

(b) The classical free-electron theory adequately explains theheat capacity of metals.

(c) At T 0, the Fermi factor is either 1 or 0.(d) The Fermi energy is the average energy of an electron in a

solid.(e) The contact potential between two metals is proportional

to the difference in the work functions of the two metals.( f )At T 0, an intrinsic semiconductor is an insulator.( g) Semiconductors conduct current in one direction only.

8 • How does the change in the resistivity of copper compare with that of silicon when the temperature increases?

9 • Which of the following elements are most likely to act as acceptor impurities in germanium? (a) bromine (b) gallium (c) silicon (d) phosphorus (e) magnesium

10 • Which of the following elements are most likely to serve as donor impurities in germanium? (a) bromine (b) gallium (c) silicon (d) phosphorus (e) magnesium

11 • An electron hole is created when a photon is absorbed by a semiconductor. How does this enable the semi-conductor to conduct electricity?

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P R O B L E M S

• Single-concept, single-step, relatively easy

•• Intermediate-level, may require synthesis of concepts

••• Challenging

Solution is in the Student Solutions Manual

Problems available on iSOLVE online homework service

These “Checkpoint” online homework service problems ask students additional questions about their confidence level, and how they arrived at their answer.

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In a few problems, you are given more

data than you actually need; in a few

other problems, you are required to

supply data from your general

knowledge, outside sources, or

informed estimates.


12 • Examine the positions of phosphorus, P; boron, B;thallium, Tl; and antimony, Sb in Table 36-1. (a) Which ofthese elements can be used to dope silicon to create an n-typesemiconductor? (b) Which of these elements can be used todope silicon to create a p-type semiconductor?

13 • When light strikes the p-type semiconductor in a pn junction solar cell, (a) only free electrons are created. (b) only positive holes are created. (c) both electrons and holesare created. (d) positive protons are created. (e) None of theseis correct.

Estimation and Approximation

14 • The ratio of the resistivity of the most resistive(least conductive) material to that of the least resistive mater-ial (excluding superconductors) is approximately 1024. Youcan develop a feeling for how remarkable this range is by con-sidering what the ratio is of the largest to smallest values ofother material properties. Choose any three properties of mat-ter, and using tables in this book or some other resource, cal-culate the ratio of the largest instance of the property to thesmallest instance of that property (other than zero) and rankthese in decreasing order. Can you find any other propertythat shows a range as large as that of electrical resistivity?

15 • A device is said to be "ohmic" if a graph of currentversus applied voltage is a straight line, and the resistance Rof the device is the slope of this line. A pn junction is an exam-ple of a nonohmic device, as may be seen from Figure 38-21.For nonohmic devices, it is sometimes convenient to definethe differential resistance as the reciprocal of the slope of the I versus V curve. Using the curve in Figure 38-21, estimate the differential resistance of the pn junction at applied volt-ages of –20 V, +0.2 V, +0.4 V, +0.6 V, and +0.8 V.

The Structure of Solids

16 • Calculate the distance r0 between the K

and the Cl ions in KCl, assuming that each ion occupies acubic volume of side r0. The molar mass of KCl is 74.55 g/moland its density is 1.984 g/cm3.

17 • The distance between the Li and Cl ionsin LiCl is 0.257 nm. Use this and the molar mass of LiCl (42.4 g/mol) to compute the density of LiCl.

18 • Find the value of n in Equation 38-6that gives the measured dissociation energy of 741 kJ/mol forLiCl, which has the same structure as NaCl and for which r0 0.257 nm.

19 •• (a) Use Equation 38-6 and calculate U(r0) for calcium oxide, CaO, where r0 0.208 nm. Assume n 8. (b) If n 10, what is the fractional change in U(r0)?

A Microscopic Picture of Conduction

20 • A measure of the density of the free-electron gas ina metal is the distance rs, which is defined as the radius of thesphere whose volume equals the volume per conduction elec-tron. (a) Show that rs [3/(4pn)]1/3, where n is the free-electronnumber density. (b) Calculate rs for copper in nanometers.

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21 • (a) Given a mean free path l 0.4 nmand a mean speed vav 1.17 105 m/s for the current flow incopper at a temperature of 300 K, calculate the classical valuefor the resistivity r of copper. (b) The classical model suggeststhat the mean free path is temperature independent and thatvav depends on temperature. From this model, what would r be at 100 K?

22 •• Silicon has an atomic weight of 28.09 and adensity of 2.41 103 kg/m3. Each atom of silicon has two va-lence electrons and the Fermi energy of the material is 4.88 eV.(a) Given that the electron mean free path at room tempera-ture is l 27.0 nm, estimate the resistivity. (b) The acceptedvalue for the resistivity of silicon is 640 m (at room temper-ature). How does this accepted value compare to the valuecalculated in part (a)?

The Fermi Electron Gas

23 • Calculate the number density of free electrons in (a) Ag (r 10.5 g/cm3) and (b) Au (r 19.3 g/cm3), assum-ing one free electron per atom, and compare your results withthe values listed in Table 38-1.

24 • The density of aluminum is 2.7 g/cm3. How manyfree electrons are present per aluminum atom?

25 • The density of tin is 7.3 g/cm3. How many freeelectrons are present per tin atom?

26 • Calculate the Fermi temperature for(a) Al, (b) K, and (c) Sn.

27 • What is the speed of a conduction electronwhose energy is equal to the Fermi energy EF for (a) Na, (b) Au, and (c) Sn?

28 • Calculate the Fermi energy for (a) Al, (b) K,and (c) Sn using the number densities given in Table 38-1.

29 • Find the average energy of the conductionelectrons at T 0 in (a) copper and (b) lithium.

30 • Calculate (a) the Fermi temperature and(b) the Fermi energy at T 0 for iron.

31 •• (a) Assuming that gold contributes one freeelectron per atom to the metal, calculate the electron densityin gold knowing that its atomic weight is 196.97 and its massdensity is 19.3 103 kg/m3. (b) If the Fermi speed for gold is 1.39 106 m/s, what is the Fermi energy in electron volts? (c) By what factor is the Fermi energy higher than the kTenergy at room temperature? (d) Explain the difference be-tween the Fermi energy and the kT energy.

32 •• The pressure of an ideal gas is related to theaverage energy of the gas particles by PV NEav, where N isthe number of particles and Eav is the average energy. Use thisto calculate the pressure of the Fermi electron gas in copper innewtons per square meter, and compare your result with atmospheric pressure, which is about 105 N/m2. (Note: Theunits are most easily handled by using the conversion factors1 N/m2 1 J/m3 and 1 eV 1.6 1019 J.)

33 •• The bulk modulus B of a material can be defined by

B V0P0V

23

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Problems 1263


(a) Use the ideal-gas relation PV NEav and Equation 38-22and Equation 38-23 to show that

P CV5/3

where C is a constant independent of V. (b) Show that the bulkmodulus of the Fermi electron gas is therefore

B P

(c) Compute the bulk modulus in newtons per square meterfor the Fermi electron gas in copper and compare your resultwith the measured value of 140 109 N/m2.

34 • Calculate the contact potential between (a) Ag andCu, (b) Ag and Ni, and (c) Ca and Cu.

Heat Capacity Due to Electrons in a Metal

35 •• Gold has a Fermi energy of 5.53 eV. Deter-mine the molar specific heat at constant volume and roomtemperature for gold.

Quantum Theory of Electrical Conduction

36 • The resistivities of Na, Au, and Sn at T 273 K are4.2 mcm, 2.04 mcm, and 10.6 mcm, respectively. Usethese values and the Fermi speeds calculated in Problem 27 tofind the mean free paths l for the conduction electrons inthese elements.

37 •• The resistivity of pure copper is increased approximately 1 108 m by the addition of 1 percent (by number of atoms) of an impurity throughout the metal.The mean free path depends on both the impurity and the oscillations of the lattice ions according to the equation 1/l 1/lt 1/li. (a) Estimate li from the data given in Table 38-1. (b) If r is the effective radius of an impurity latticeion seen by an electron, the scattering cross section is pr2. Estimate this area, using the fact that r is related to li by Equation 38-16.

Band Theory of Solids

38 • When light of wavelength 380.0 nm falls on a semi-conductor, electrons are promoted from the valence band tothe conduction band. Calculate the energy gap, in electronvolts, for this semiconductor.

39 • You are an electron sitting at the topof the valence band in a silicon atom, longing to jump acrossthe 1.14-eV energy gap that separates you from the bottom of the conduction band and all of the adventures that it may contain. What you need, of course, is a photon. What is themaximum photon wavelength that will get you across the gap?

40 • Repeat Problem 39 for germanium, for which theenergy gap is 0.74 eV.

41 • Repeat Problem 39 for diamond, for which the energy gap is 7.0 eV.

42 •• A photon of wavelength 3.35 mm hasjust enough energy to raise an electron from the valence band

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2NEF

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to the conduction band in a lead sulfide crystal. (a) Find theenergy gap between these bands in lead sulfide. (b) Find thetemperature T for which kT equals this energy gap.

Semiconductors

43 • The donor energy levels in an n-type semiconduc-tor are 0.01 eV below the conduction band. Find the tempera-ture for which kT 0.01 eV.

44 •• When a thin slab of semiconducting material is illu-minated with monochromatic electromagnetic radiation,most of the radiation is transmitted through the slab if thewavelength is greater than 1.85 mm. For wavelengths lessthan 1.85 mm, most of the incident radiation is absorbed. De-termine the energy gap of this semiconductor.

45 •• The relative binding of the extra electron in the ar-senic atom that replaces an atom in silicon or germanium canbe understood from a calculation of the first Bohr orbit of thiselectron in these materials. Four of arsenic’s outer electronsform covalent bonds, so the fifth electron sees a singly chargedcenter of attraction. This model is a modified hydrogen atom.In the Bohr model of the hydrogen atom, the electron moves infree space at a radius a0 given by Equation 36-12, which is

a0

When an electron moves in a crystal, we can approximate theeffect of the other atoms by replacing e0 with ke0 and me withan effective mass for the electron. For silicon, k is 12 and theeffective mass is approximately 0.2me. For germanium, k is 16and the effective mass is approximately 0.1me. Estimate theBohr radii for the outer electron as it orbits the impurity arsenic atom in silicon and germanium.

46 •• The ground-state energy of the hydrogenatom is given by

E1

Modify this equation in the spirit of Problem 45 by replacinge0 by ke0 and me by an effective mass for the electron to esti-mate the binding energy of the extra electron of an impurityarsenic atom in (a) silicon and (b) germanium.

47 •• A doped n-type silicon sample with 1016 electronsper cubic centimeter in the conduction band has a resistivityof 5 103 m at 300 K. Find the mean free path of the electrons. Use the effective mass of 0.2me for the mass of theelectrons. (See Problem 45.) Compare this mean free path withthat of conduction electrons in copper at 300 K.

48 •• The measured Hall coefficient of a doped siliconsample is 0.04 Vm/AT at room temperature. If all the doping impurities have contributed to the total charge carriers of the sample, find (a) the type of impurity (donor or acceptor) used to dope the sample and (b) the concentration of these impurities.

*Semiconductor Junctions and Devices

49 •• Simple theory for the current versus the bias volt-age across a pn junction yields the equation I I0(eeVb/kT 1).

e2me

8e20 h2

mk2e4

2U2

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4pe0U2

mee2



Sketch I versus Vb for both positive and negative values of Vbusing this equation.

50 • The plate current in an npn transistor circuit is 25.0 mA. If 88 percent of the electrons emitted reach the col-lector, what is the base current?

51 •• In Figure 38-27 for the pnp-transistoramplifier, suppose Rb 2 k and RL 10 k. Suppose fur-ther that a 10-mA ac base current generates a 0.5-mA ac collec-tor current. What is the voltage gain of the amplifier?

52 •• Germanium can be used to measure theenergy of incident particles. Consider a 660-keV gamma rayemitted from 137Cs. (a) Given that the band gap in germaniumis 0.72 eV, how many electron–hole pairs can be generated asthis gamma ray travels through germanium? (b) The numberof pairs N in Part (a) will have statistical fluctuations given by . What then is the energy resolution of this detectorin this photon energy region?

53 •• Make a sketch showing the valence and conductionband edges and Fermi energy of a pn-junction diode when biased (a) in the forward direction and (b) in the reverse direction.

54 •• A “good” silicon diode has the current–voltagecharacteristic given in Problem 49. Let kT 0.025 eV (roomtemperature) and the saturation current I0 1 nA. (a) Showthat for small reverse-bias voltages, the resistance is 25 M.(Hint: Do a Taylor expansion of the exponential function or useyour calculator and enter small values for Vb.) (b) Find the dc re-sistance for a reverse bias of 0.5 V. (c) Find the dc resistance fora 0.5-V forward bias. What is the current in this case? (d) Cal-culate the ac resistance dV/dI for a 0.5-V forward bias.

55 •• A slab of silicon of thickness t 1.0 mm and widthw 1.0 cm is placed in a magnetic field B 0.4 T. The slab isin the xy plane, and the magnetic field points in the positive z direction. When a current of 0.2 A flows through the samplein the positive x direction, a voltage difference of 5 mV devel-ops across the width of the sample with the electric field inthe sample pointing in the positive y direction. Determine the semiconductor type (n or p) and the concentration ofcharge carriers.

The BCS Theory

56 • (a) Use Equation 38-37 to calculate the supercon-ducting energy gap for tin, and compare your result with themeasured value of 6 104 eV. (b) Use the measured value tocalculate the wavelength of a photon having sufficient energyto break up Cooper pairs in tin (Tc 3.72 K) at T 0.

57 • Repeat Problem 56 for lead (Tc 7.19 K),which has a measured energy gap of 2.73 103 eV.

The Fermi–Dirac Distribution

58 •• The number of electrons in the conduction band ofan insulator or intrinsic semiconductor is governed chiefly bythe Fermi factor. Since the valence band in these materials isnearly filled and the conduction band is nearly empty, theFermi energy EF is generally midway between the top of thevalence band and the bottom of the conduction band, that is,

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at Eg/2, where Eg is the band gap between the two bands and the energy is measured from the top of the valence band.(a) In silicon, Eg 1.0 eV. Show that in this case the Fermi factor for electrons at the bottom of the conduction band isgiven by exp(Eg/2kT) and evaluate this factor. Discuss thesignificance of this result if there are 1022 valence electrons percubic centimeter and the probability of finding an electron inthe conduction band is given by the Fermi factor. (b) Repeatthe calculation in Part (a) for an insulator with a band gap of 6.0 eV.

59 •• Approximately how many energy states, with ener-gies between 2.00 eV and 2.20 eV, are available to electrons ina cube of silver measuring 1.00 mm on a side?

60 •• (a) Use Equation 38-22a to calculate the Fermienergy for silver. (b) Determine the average energy of a freeelectron and (c) find the Fermi speed for silver.

61 •• Show that at E EF, the Fermi factor is F 0.5.

62 •• What is the difference between the ener-gies at which the Fermi factor is 0.9 and 0.1 at 300 K in (a) cop-per, (b) potassium, and (c) aluminum.

63 •• What is the probability that a con-duction electron in silver will have a kinetic energy of 4.9 eVat T 300 K?

64 •• Show that g(E) (3N/2)EF3/2E1/2 (Equation 38-43)

follows from Equation 38-41 for g(E), and from Equation 38-22afor EF.

65 •• Carry out the integration Eav (1/N) Eg(e) dE toshow that the average energy at T 0 is EF.

66 •• The density of the electron states in a metal can bewritten g(E) AE1/2, where A is a constant and E is measuredfrom the bottom of the conduction band. (a) Show that the total number of states is AE . (b) Approximately what fraction of the conduction electrons are within kT of the Fermienergy? (c) Evaluate this fraction for copper at T 300 K.

67 •• What is the probability that a conduc-tion electron in silver will have a kinetic energy of 5.49 eV at T 300 K?

68 •• Use Equation 38-41 for the density ofstates to estimate the fraction of the conduction electrons incopper that can absorb energy from collisions with the vibrat-ing lattice ions at (a) 77 K and (b) 300 K.

69 •• In an intrinsic semiconductor, the Fermi energy isabout midway between the top of the valence band and thebottom of the conduction band. In germanium, the forbid-den energy band has a width of 0.7 eV. Show that at roomtemperature the distribution function of electrons in the con-duction band is given by the Maxwell–Boltzmann distribu-tion function.

70 ••• (a) Show that for E 0, the Fermi factor may be written as f (E) 1/(CeE/kT 1). (b) Show that if C eE/(kT), f(E) AeE/(kT) 1; in other words, show thatthe Fermi factor is a constant times the classical Boltzmannfactor if A 1. (c) Use n(E) dE N and Equation 38-41 todetermine the constant A. (d) Using the result obtained in Part (c), show that the classical approximation is applicablewhen the electron concentration is very small and/or the

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Problems 1265


temperature is very high. (e) Most semiconductors have impurities added in a process called doping, which increasesthe free electron concentration so that it is about 1017/cm3 atroom temperature. Show that for these systems, the classicaldistribution function is applicable.

71 ••• Show that the condition for the applicability of the classical distribution function for an electron gas (A 1in Problem 70) is equivalent to the requirement that the aver-age separation between electrons is much greater than theirde Broglie wavelength.

72 ••• The root-mean-square (rms) value of a variable isobtained by calculating the average value of the square of thatvariable and then taking the square root of the result. Use thisprocedure to determine the rms energy of a Fermi distribu-tion. Express your result in terms of EF and compare it to theaverage energy. Why do Eav and Erms differ?

General Problems

73 • The density of potassium is 0.851 g/cm3. Howmany free electrons are there per potassium atom?

74 • Calculate the number density of free electrons for(a) Mg (r 1.74 g/cm3) and (b) Zn (r 7.1 g/cm3), assuming

two free electrons per atom, and compare your results withthe values listed in Table 38-1.

75 •• Estimate the fraction of free electrons incopper that are in excited states above the Fermi energy at (a) 300 K (about room temperature) and (b) at 1000 K.

76 •• Determine the energy that has 10 percent free electron occupancy probability for manganese at T 1300 K.

77 •• The semiconducting compound CdSe is widelyused for light-emitting diodes (LEDs). The energy gap inCdSe is 1.8 eV. What is the frequency of the light emitted by aCdSe LED?

78 ••• A 2-cm2 wafer of pure silicon is irradiatedwith light having a wavelength of 775 nm. The intensity of thelight beam is 4.0 W/m2 and every photon that strikes the sam-ple is absorbed and creates an electron–hole pair. (a) Howmany electron–hole pairs are produced in one second? (b) Ifthe number of electron–hole pairs in the sample is 6.25 1011

in the steady state, at what rate do the electron–hole pairs recombine? (c) If every recombination event results in the radiation of one photon, at what rate is energy radiated by the sample?

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