43

CHAPTER OUTLINE

43.1 Molecular Bonds43.2 Energy States and Spectra of Molecules43.3 Bonding in Solids43.4 Free-Electron Theory of Metals43.5 Band Theory of Solids43.6 Electrical Conduction in Metals, Insulators, and Semiconductors43.7 Semiconductor Devices43.8 Superconductivity

Molecules and Solids

ANSWERS TO QUESTIONS

Q43.1 Rotational, vibrational and electronic (as discussed inChapter 42) are the three major forms of excitation. Rotational

energy for a diatomic molecule is on the order of 2

2I, where I is

the moment of inertia of the molecule. A typical value for asmall molecule is on the order of 1 10 3 meV eV= − . Vibrationalenergy is on the order of hf, where f is the vibration frequencyof the molecule. A typical value is on the order of 0.1 eV.Electronic energy depends on the state of an electron in themolecule and is on the order of a few eV. The rotational energycan be zero, but neither the vibrational nor the electronicenergy can be zero.

Q43.2 The Pauli exclusion principle limits the number of electrons inthe valence band of a metal, as no two electrons can occupy thesame state. If the valence band is full, additional electrons mustbe in the conduction band, and the material can be a goodconductor. For further discussion, see Q43.3.

Q43.3 The conductive properties of a material depend on the electron population of the conduction bandof the material. If the conduction band is empty and a full valence band lies below the conductionband by an energy gap of a few eV, then the material will be an insulator. Electrons will be unable tomove easily through the material in response to an applied electric field. If the conduction band ispartly full, states are accessible to electrons accelerated by an electric field, and the material is a goodconductor. If the energy gap between a full valence band and an empty conduction band iscomparable to the thermal energy k TB , the material is a semiconductor.

Q43.4 Thermal excitation increases the vibrational energy of the molecules. It makes the crystal lattice lessorderly. We can expect it to increase the width of both the valence band and the conduction band, todecrease the gap between them.

545

546 Molecules and Solids

Q43.5 First consider electric conduction in a metal. The number of conduction electrons is essentially fixed.They conduct electricity by having drift motion in an applied electric field superposed on theirrandom thermal motion. At higher temperature, the ion cores vibrate more and scatter moreefficiently the conduction electrons flying among them. The mean time between collisions isreduced. The electrons have time to develop only a lower drift speed. The electric current isreduced, so we see the resistivity increasing with temperature.

Now consider an intrinsic semiconductor. At absolute zero its valence band is full and itsconduction band is empty. It is an insulator, with very high resistivity. As the temperature increases,more electrons are promoted to the conduction band, leaving holes in the valence band. Then bothelectrons and holes move in response to an applied electric field. Thus we see the resistivitydecreasing as temperature goes up.

Q43.6 In a metal, there is no energy gap between the valence and conduction bands, or the conductionband is partly full even at absolute zero in temperature. Thus an applied electric field is able to injecta tiny bit of energy into an electron to promote it to a state in which it is moving through the metalas part of an electric current. In an insulator, there is a large energy gap between a full valence bandand an empty conduction band. An applied electric field is unable to give electrons in the valenceband enough energy to jump across the gap into the higher energy conduction band. In asemiconductor, the energy gap between valence and conduction bands is smaller than in aninsulator. At absolute zero the valence band is full and the conduction band is empty, but at roomtemperature thermal energy has promoted some electrons across the gap. Then there are somemobile holes in the valence band as well as some mobile electrons in the conduction band.

Q43.7 Ionic bonds are ones between oppositely charged ions. A simple model of an ionic bond is theelectrostatic attraction of a negatively charged latex balloon to a positively charged Mylar balloon.

Covalent bonds are ones in which atoms share electrons. Classically, two children playing ashort-range game of catch with a ball models a covalent bond. On a quantum scale, the two atoms aresharing a wave function, so perhaps a better model would be two children using a single hula hoop.

Van der Waals bonds are weak electrostatic forces: the dipole-dipole force is analogous to theattraction between the opposite poles of two bar magnets, the dipole—induced dipole force issimilar to a bar magnet attracting an iron nail or paper clip, and the dispersion force is analogous toan alternating-current electromagnet attracting a paper clip.

A hydrogen atom in a molecule is not ionized, but its electron can spend more time elsewherethan it does in the hydrogen atom. The hydrogen atom can be a location of net positive charge, andcan weakly attract a zone of negative charge in another molecule.

Q43.8 Ionically bonded solids are generally poor electric conductors, as they have no free electrons. Whilethey are transparent in the visible spectrum, they absorb infrared radiation. Physically, they formstable, hard crystals with high melting temperatures.

Q43.9 Covalently bonded solids are generally poor conductors, as they form structures in which the atomsshare several electrons in the outer shell, leaving no room for conducting electrons. Depending onthe structure of the solid, they are usually very hard and have high melting points.

Q43.10 Metals are good conductors, as the atoms have many free electrons in the conduction band. Metallicbonds allow the mixing of different metals to form alloys. Metals are opaque to visible light, and canbe highly reflective. A metal can bend under stress instead of fracturing like ionically and covalentlybonded crystals. The physical properties vary greatly depending on the composition.

Chapter 43 547

Q43.11 The energy of the photon is given to the electron. The energy of a photon of visible light is sufficientto promote the electron from the lower-energy valence band to the higher-energy conduction band.This results in the additional electron in the conduction band and an additional hole—the energystate that the electron used to occupy—in the valence band.

Q43.12 Along with arsenic (As), any other element in group V, such as phosphorus (P), antimony (Sb), andbismuth (Bi), would make good donor atoms. Each has 5 valence electrons. Any element in group IIIwould make good acceptor atoms, such as boron (B), aluminum, (Al), gallium (Ga), and indium (In).They all have only 3 valence electrons.

Q43.13 The two assumptions in the free-electron theory are that the conduction electrons are not bound toany particular atom, and that the nuclei of the atoms are fixed in a lattice structure. In this model, itis the “soup” of free electrons that are conducted through metals. The energy band model is morecomprehensive than the free-electron theory. The energy band model includes an account of themore tightly bound electrons as well as the conduction electrons. It can be developed into a theoryof the structure of the crystal and its mechanical and thermal properties.

Q43.14 A molecule containing two atoms of 2 H, deuterium, has twice the mass of a molecule containingtwo atoms of ordinary hydrogen 1 H . The atoms have the same electronic structure, so the moleculeshave the same interatomic spacing, and the same spring constant. Then the moment of inertia of thedouble-deuteron is twice as large and the rotational energies one-half as large as for ordinary

hydrogen. Each vibrational energy level for D2 is12

times that of H 2 .

Q43.15 Rotation of a diatomic molecule involves less energy than vibration. Absorption of microwavephotons, of frequency ~ Hz1011 , excites rotational motion, while absorption of infrared photons, offrequency ~ Hz1013 , excites vibration in typical simple molecules.

Q43.16 Yes. A material can absorb a photon of energy greater than the energy gap, as an electron jumps intoa higher energy state. If the photon does not have enough energy to raise the energy of the electronby the energy gap, then the photon will not be absorbed.

Q43.17 From the rotational spectrum of a molecule, one can easily calculate the moment of inertia of themolecule using Equation 43.7 in the text. Note that with this method, only the spacing betweenadjacent energy levels needs to be measured. From the moment of inertia, the size of the moleculecan be calculated, provided that the structure of the molecule is known.

SOLUTIONS TO PROBLEMS

Section 43.1 Molecular Bonds

P43.1 (a) Fq

r=

∈=

× ×

×= ×

−

−

−2

02

19 2 9

10 29

4

1 60 10 8 99 10

5 00 100 921 10

π

. .

..

e j e je j

N N toward the other ion.

(b) Uq

r=

−∈

= −× ×

×≈ −

−

−

2

0

19 2 9

104

1 60 10 8 99 10

5 00 102 88

π

. .

..

e j e j J eV

548 Molecules and Solids

P43.2 We are told K Cl eV K Cl+ + → ++ −0 7.

and Cl e Cl eV+ → +− − 3 6.

or Cl Cl e eV− −→ + − 3 6. .

By substitution, K Cl eV K Cl+e eV+ + → + −+ −0 7 3 6. .

K eV K +e+ → + −4 3.

or the ionization energy of potassium is 4 3. eV .

P43.3 (a) Minimum energy of the molecule is found from

dUdr

Ar Br= − + =− −12 6 013 7 yielding rA

B0

1 62= LNMOQP .

(b) E U UA

A BBA B

BA

BAr r r= − = − −

LNMM

OQPP= − −LNM

OQP ==∞ = 0

04 2

14

12 42 2

2 2

This is also the equal to the binding energy, the amount of energy given up by the twoatoms as they come together to form a molecule.

(c) r0

120

60

1 6

112 0 124 10

1 488 107 42 10 74 2=

× ⋅

× ⋅

L

NMM

O

QPP = × =

−

−−

.

.. .

eV m

eV m m pm

12

6

e j

E =× ⋅

× ⋅=

−

−

1 488 10

4 0 124 104 46

60 2

120

.

..

eV m

eV m eV

6

12

e je j

*P43.4 (a) We add the reactions K eV K e+ → ++ −4 34.

and I e I eV+ → +− − 3 06.

to obtain K I K I eV+ → + + −+ − 4 34 3 06. .a f .

The activation energy is 1 28. eV .

(b)dUdr r r

=∈

− FHGIKJ + FHG

IKJ

LNMM

OQPP

412 6

13 7

σσ σ

At r r= 0 we have dUdr

= 0 . Here σ σr r0

13

0

712

FHGIKJ =FHGIKJ

σr0

1 62= − σ σ= = =−2 0 305 0 2721 6 . .a f nm nm .

Then also

U rr

rr

rE E E

E U r

a a a

a

0

1 60

0

12 1 60

0

6

0

42 2

414

12

1 28 3 37

b g

b g

= ∈FHG

IKJ −FHG

IKJ

LNMM

OQPP + = ∈ −LNM

OQP + = − ∈+

∈= − = + =∈

− −

. . eV eV = 4.65 eV

continued on next page

Chapter 43 549

(c) F rdUdr r r

a f = − =∈ FHGIKJ − FHG

IKJ

LNMM

OQPP

412 6

13 7

σσ σ

To find the maximum force we calculate dFdr r r

=∈

− FHGIKJ + FHG

IKJ

LNMM

OQPP =

4156 42 02

14 8

σσ σ

when

σrrupture

= FHGIKJ

42156

1 6

Fmax.

.. .

.

.

= FHGIKJ − FHG

IKJ

LNMM

OQPP = − = −

×

= −

−

−

4 4 650 272

1242

1566

42156

41 0 41 01 6 10

6 55

13 6 7 6 19

9

eV nm

eV nm Nm

10 m

nN

a f

Therefore the applied force required to rupture the molecule is +6 55. nN away from the

center.

(d) U r sr s r s

Er

r s r sEa a0

0

12

0

6 1 60

0

12 1 6

0

6

4 42 2

+ = ∈+FHGIKJ −

+FHGIKJ

LNMM

OQPP + = ∈

+

FHG

IKJ −

+

FHGIKJ

LNMM

OQPP +

− −

b g σ σ

= ∈ +FHGIKJ − +

FHGIKJ

LNMM

OQPP +

= ∈ − + −FHG

IKJ − − + −FHG

IKJ

LNMM

OQPP +

=∈− ∈ + ∈ − ∈+ ∈ − ∈ + +

= − ∈+ +FHGIKJ + ∈ +

+ ≅

− −

414

112

1

414

1 12 7812

1 6 21

12 78 2 12 42

0 36

0

12

0

6

0

2

02

0

2

02

0

2

02

0

2

02

0

2

02

0

sr

sr

E

sr

sr

sr

sr

E

sr

sr

sr

sr

E

Esr

sr

U r s U r

a

a

a

a

… …

…

…

b g 021

2b g+ ks

where kr

=∈= = =

72 72 4 65

0 3053 599 576

02 2

.

.

eV

nm eV nm N m2a f

a f .

P43.5 At the boiling or condensation temperature, k TB ≈ = ×− − −10 10 1 6 103 3 19 eV J.e j

T ≈××

−

−1 6 10

1022

23.

~ J

1.38 10 J K K .

550 Molecules and Solids

Section 43.2 Energy States and Spectra of Molecules

P43.6 µ =+

=+

× = ×− −m mm m

1 2

1 2

27 25132 9 126 9132 9 126 9

1 66 10 1 08 10. .. .

. .a f e j kg kg

I r= = × × = × ⋅− − −µ 2 25 9 2 451 08 10 0 127 10 1 74 10. . . kg m kg m2e je j

(a) E II

IJ J

I= = =

+12 2

12

22 2

ωωa f a f

J = 0 gives E = 0

J = 1 gives EI

= =× ⋅

× ⋅= × =

−

−−

2 34 2

2 4524

6 626 10

4 1 74 106 41 10 40 0

.

.. .

J s

kg m J eV

2

e je jπ

µ

hf = × −−6 41 10 024. J to f = ×9 66 109. Hz

(b) fEh hI

hr

r= = = ∝ −12

2 22

4π µIf is 10% too small, is 20% too large.r f

P43.7 For the HCl molecule in the J = 1 rotational energy level, we are givenr0 0 127 5= . nm.

EI

J Jrot = +2

21a f

Taking J = 1 , we have EI

Irot = =2

212ω or ω = =

22

2

2I I.

The moment of inertia of the molecule is given by Equation 43.3.

I rm m

m mr= =

+FHG

IKJµ 0

2 1 2

1 202

FIG. P43.7

I r=+

LNM

OQP = × × = × ⋅− − −1 35

1 350 972 1 66 10 1 275 10 2 62 100

2 27 10 2 47 u u u u

u kg u m kg m2a fa f a fe je j. . . . .

Therefore, ω = =× ⋅

× ⋅= ×

−

−2 21 055 10

105 69 10

34

4712

I.

. J s

2.62 kg m rad s2 .

P43.8 hf EI I I

= = + − + =∆2 2 2

22 2 1

21 1 1

24a f a f a f

Ih

hfh

f= = =

× ⋅

×= × ⋅

−−4 2

2 26 626 10

2 30 101 46 10

2

2

34

1146π

π π

b ge j

.

..

J s

2 Hz kg m

22

Chapter 43 551

P43.9 I m r m r= +1 12

2 22 where m r m r1 1 2 2= and r r r1 2+ = .

Then rm rm1

2 2

1= so

m rm

r r2 2

12+ = and r

m rm m2

1

1 2=

+.

Also, rm rm2

1 1

2= . Thus, r

m rm

r11 1

2+ = and r

m rm m1

2

1 2=

+

I mm r

m m

m m r

m m

m m r m m

m m

m m rm m

r=+

++

=+

+=

+=1

22 2

1 22

2 12 2

1 22

1 22

2 1

1 22

1 22

1 2

2

b g b gb g

b gµ .

P43.10 (a) µ =+

× = ×− −22 99 35 4522 99 35 45

1 66 10 2 32 1027 26. .. .

. .a f

a f e j kg kg

I r= = × × = × ⋅− − −µ 2 26 9 2 452 32 10 0 280 10 1 81 10. . . kg m kg m2e je j

(b)hc

I I I I Ih

Iλ π= + − + = − = =

2 2 2 2 2 2

222 2 1

21 1 1

3 2 24

a f a f

λπ π

= =× × ⋅

× ⋅=

−

−

c Ih

42

3 00 10 4 1 81 10

2 6 626 101 62

2 8 2 45

34

. .

..

m s kg m

J s cm

2e j e je j

P43.11 The energy of a rotational transition is ∆EI

J=FHGIKJ

2

where J is the rotational quantum number of the

higher energy state (see Equation 43.7). We do not know J from the data. However,

∆Ehc

= =× ⋅ ×

×

FHG

IKJ

−

−λ λ

6 626 10 3 00 10 110

34 8

19

. . J s m s eV1.60 J

e je j.

For each observed wavelength,

λ (mm) ∆E (eV)0.120 4 0.010 320.096 4 0.012 880.080 4 0.015 440.069 0 0.018 000.060 4 0.020 56

The ∆E’s consistently increase by 0.002 56 eV. EI1

2

0 002 56= = . eV

and IE

= =× ⋅

×

FHG

IKJ = × ⋅

−

−−

2

1

34 2

1947

1 055 10

0 002 561

102 72 10

.

..

J s

eV eV

1.60 J kg m2e j

b g .

For the HCl molecule, the internuclear radius is rI

= =××

=−

−µ2 72 101 62 10

0 13047

27..

. m nm.

552 Molecules and Solids

P43.12 (a) Minimum amplitude of vibration of HI is

12

12

2kA hf= : Ahfk

= =× ⋅ ×

= × =−

−6 626 10 6 69 10

1 18 10 0 011 834 13

11. .

. . J s s

320 N m m nm

e je j.

(b) For HF, A =× ⋅ ×

= × =−

−6 626 10 8 72 10

9707 72 10 0 007 72

34 1312

. .. .

J s s

N m m nm

e je j.

Since HI has the smaller k, it is more weakly bound.

P43.13 µ =+

= × × = ×− −m mm m

1 2

1 2

27 273536

1 66 10 1 61 10. . kg kg

∆Ek

vib J eV= = ××

= × =−−

−

µ1 055 10

4801 61 10

5 74 10 0 3583427

20..

. .e j

P43.14 (a) The reduced mass of the O2 is µ =+

= = × = ×− −16 1616 16

8 8 1 66 10 1 33 1027 26 u u u u

u kg kga fa fa f a f e j. . .

The moment of inertia is then I r= = × × = × ⋅− − −µ 2 26 10 2 461 33 10 1 20 10 1 91 10. . .kg m kg m2e je j .

The rotational energies are EI

J J J Jrot 2

J s

kg m= + =

× ⋅

× ⋅+

−

−

2 34 2

4621

1 055 10

2 1 91 101a f e j

e ja f.

..

Thus E J Jrot J= × +−2 91 10 123.e j a f .

And for J = 0 1 2, , , Erot eV, eV= × ×− −0 3 64 10 1 09 104 3, . . .

(b) E vk

vvib J s N m

kg= +FHG

IKJ = +FHG

IKJ × ⋅

×−

−

12

12

1 055 101 177

8 1 66 1034

27µ.

.e j e j

E v vvib J eV

1.60 J eV= +FHG

IKJ ×

×

FHG

IKJ = +FHG

IKJ

−−

12

3 14 101

1012

0 1962019. .e j a f

For v = 0 1 2, , , Evib eV, 0.295 eV, 0.491 eV= 0 098 2. .

Chapter 43 553

P43.15 In Benzene, the carbon atoms are each 0.110 nm from the axis and each hydrogen atom is0 110 0 100 0 210. . .+ = nm nma f from the axis. Thus, I mr= ∑ 2 :

I = × × + × × = × ⋅− − − − −6 1 99 10 0 110 10 6 1 67 10 0 210 10 1 89 1026 9 2 27 9 2 45. . . . . kg m kg m kg m2e je j e je j .

The allowed rotational energies are then

EI

J J J J J J J J

E J J J

rot 2

rot

J s

kg m J eV

eV where

= + =× ⋅

× ⋅+ = × + = × +

= + =

−

−− −

…

2 34 2

4524 6

21

1 055 10

2 1 89 101 2 95 10 1 18 4 10 1

18 4 1 0 1 2 3

a f e je j

a f e j a f e j a f

b g a f

.

.. .

. , , , ,µ

The first five of these allowed energies are: Erot eV, 111 eV, 221 eV, and 369 eV= 0 36 9, . µ µ µ µ .

*P43.16 We carry extra digits through the solution because part (c) involves the subtraction of two closenumbers. The longest wavelength corresponds to the smallest energy difference between the

rotational energy levels. It is between J = 0 and J = 1 , namely 2

I

λπ

= = =hcE

hcI

Ich∆ min

2

24. If µ is the reduced mass, then

I r= = × = ×

∴ =× ×

× ⋅= ×

− −

−

−

µ µ µ

λπ µ

µ

2 9 2 20

2 20 8

3423

0 127 46 10 1 624 605 10

4 1 624 605 10 2 997 925 10

6 626 075 102 901 830 10

. .

. .

..

m m

m m s

J s m kg

2

2

e j e je j e j e j

(1)

(a) µ35271 007 825 34 968 853

1 007 825 34 968 8530 979 593 1 626 653 10=

+= = × −. .

. .. .

u u

u uu kg

b gb g

From (1): λ µ3523 272 901 830 10 1 626 653 10 472= × × =−. . m kg kg me je j

(b) µ37271 007 825 36 965 903

1 007 825 36 965 9030 981 077 1 629 118 10=

+= = × −. .

. .. .

u u

u uu kg

b gb g

From (1): λ µ3723 272 901 830 10 1 629 118 10 473= × × =−. . m kg kg me je j

(c) λ λ µ µ µ37 35 472 742 4 472 027 0 0− = − =. . m m .715 m

P43.17 hfh

IJ=

2

24π where the rotational transition is from J −1 to J,

where f = ×6 42 1013. Hz and I = × ⋅−1 46 10 46. kg m2 from Example 43.1.

JIf

h= =

× ⋅ ×

× ⋅=

−

−4 4 1 46 10 6 42 10

6 626 10558

2 2 46 13

34π π . .

.

kg m s

J s

2e je j

554 Molecules and Solids

*P43.18 We find an average spacing between peaks by counting 22 gaps between 7 96 1013. × Hz and9 24 1013. × Hz :

∆

∆

fh

hI

Ih

f

=−

= × =FHGIKJ

= =× ⋅

×= × ⋅

−−

9 24 7 96 100 058 2 10

14

46 63 104 5 82 10

2 9 10

1313

2

2

2

34

2 1147

. ..

..

.

a f Hz22

Hz

J ss

kg m2

π

π π

*P43.19 We carry extra digits through the solution because the given wavelengths are close together.

(a) E v hfI

J JvJ = +FHGIKJ + +

12 2

12

a f

∴ = = + = +

∴ − = + = =× ⋅ ×

×

−

−

E hf E hfI

E hfI

E E hfI

hc

00 11

2

02

2

11 00

2 34 8

6

12

32

12

3

6 626 075 10 2 997 925 10

2 211 2 10

, ,

. .

.λ

J s m s

m

e je j

∴ + = × −hfI

2208 983 573 10. J (1)

E E hfI

hc11 02

2 34 8

62 6 626 075 10 2 997 925 10

2 405 4 10− = − = =

× ⋅ ×

×

−

−λ

. .

.

J s m s

m

e je j

∴ − = × −hfI

28 258 284 10

220. J (2)

Subtract (2) from (1): 3

7 252 89 102

21

I= × −. J

∴ =× ⋅

×= × ⋅

−

−−I

3 1 054 573 10

7 252 89 104 60 10

34 2

2148

.

..

J s

J kg m2e j

(b) From (1):

f =×

× ⋅−

× ⋅

× ⋅ × ⋅

= ×

−

−

−

− −

8 983 573 1010

1 054 573 10

4 600 060 10 6 626 075 10

1 32 10

20

34

34 2

48 34

14

. .

. .

.

J6.626 075 J s

J s

kg m J s

Hz

2

e je je j

(c) I r= µ 2 , where µ is the reduced mass:

µ = = = × −12

12

1 007 825 8 367 669 10 28mH . .u kgb g .

So rI

= =× ⋅

×=

−

−µ4 600 060 10

8 367 669 100

48 2

28

..

kg m kg

.074 1 nm .

Chapter 43 555

P43.20 The emission energies are the same as the absorption energies, but the final state must be belowv J= =1 0,b g . The transition must satisfy ∆J = ±1, so it must end with J = 1 . To be lower in energy, it

must be v J= =0 1,b g . The emitted photon energy is therefore

hf E E E E E E E E

hf hf hfv J v J v v J Jphoton vib rot vib rot vib vib rot rot

photon vib rot

= + − + = − − −

= −= = = = = = = =1 0 0 1 1 0 1 0e j e j e j e j

Thus, f f fphoton vib rot Hz Hz Hz= − = × − × = ×6 42 10 1 15 10 6 41 1013 11 13. . . .

P43.21 The moment of inertia about the molecular axis is I mr mr mx = + = × −25

25

45

2 00 102 2 15 2. me j .

The moment of inertia about a perpendicular axis is I mR

mR m

y =FHGIKJ + FHG

IKJ = × −

2 2 22 00 10

2 210 2

. me j .

The allowed rotational energies are EI

J Jrot =FHGIKJ +

2

21a f , so the energy of the first excited state is

EI1

2

= . The ratio is therefore

E

E

I

I

I

I

m

m

x

y

x

y

y

x

1

1

2

2

10 2

15 25 2 9

1 2 2 00 10

4 5 2 00 10

58

10 6 25 10,

,

.

..= = =

×

×= = ×

−

−

e je j

b g e jb g e j

e j m

m.

Section 43.3 Bonding in Solids

P43.22 Consider a cubical salt crystal of edge length 0.1 mm.

The number of atoms is10

104

9

317

−

−×

FHG

IKJ

m0.261 10 m

~ .

This number of salt crystals would have volume 1010

10104 3 4

9

35−

−

−×

FHG

IKJ m

m0.261 m

m3e j ~ .

If it is cubic, it has edge length 40 m.

P43.23 Uk er m

e= − −FHGIKJ = − ×

×

×−FHGIKJ = − × = −

−

−−α 2

0

919 2

9181

11 747 6 8 99 10

1 60 10

0 281 101

18

1 25 10 7 84. ..

.. .b ge j e j

e j J eV

P43.24 Visualize a K + ion at the center of each shaded cube, a Cl− ion at the centerof each white one.

The distance ab is 2 0 314 0 444. . nm nma f = .

Distance ac is 2 0 314 0 628. . nm nma f = .

Distance ad is 2 2 0 314 0 7692 2+ =e j a f. . nm nm . FIG. P43.24

556 Molecules and Solids

P43.25 Uk e

rk e

rk e

rk e

rk e

rk e

rk e

rk e

rk er

e e e e e e e e

e

= − − + + − − + + −

= − − + − +FHG

IKJ

2 2 2 2 2 2 2 2

22 2 3 3 4 4

21

12

13

14

…

…

But, ln 1 12 3 4

2 3 4

+ = − + − +xx x xa f …

so, Uk ere= −

22

2

ln , or U kere= − =α α2

2 2 where ln .

– ++ – + –

3r2r

r

–e –e –ee e e

FIG. P43.25

Section 43.4 Free-Electron Theory of Metals

Section 43.5 Band Theory of Solids

P43.26 Ehm

nne

eF

J s

kg J eV= FHGIKJ =

× ⋅

× ×

L

NMMM

O

QPPPFHGIKJ

−

− −

2 2 3 34 2

31 19

2 32 3

238

6 626 10

2 9 11 10 1 60 10

38π π

.

. .

e je je j

E neF eV= × −3 65 10 19 2 3.e j with n measured in electrons m3 .

P43.27 The density of conduction electrons n is given by Ehm

neF = FHGIKJ

2 2 3

238π

or nmEhe =FHGIKJ =

× ×

× ⋅= ×

− −

−

−83

2 83

2 9 11 10 5 48 1 60 10

6 626 105 80 102

3 2 31 19 3 2

34 328 3π πF

kg J

J s m

. . .

..

e ja fe je j

.

The number-density of silver atoms is

nAg3 kg m

atom108 u

u1.66 10 kg

m= × FHG

IKJ ×

FHG

IKJ = ×−

−10 6 101 1

5 91 10327

28 3. .e j .

So an average atom contributes 5 805 91

0 981..

.= electron to the conduction band .

P43.28 (a)12

7 052mv = . eV

v =×

×= ×

−

−

2 7 05 1 60 10

9 11 101 57 10

19

316

. .

..

eV J eV

kg m s

a fe j

(b) Larger than m s by ten orders of magnitude.10 4− However, the energy of an electron at

room temperature is typically k TB =140

eV.

Chapter 43 557

P43.29 For sodium, M = 23 0. g mol and ρ = 0 971. g cm3 .

(a) nN

Me = =×

A3 electrons mol g cm

g molρ 6 02 10 0 971

23 0

23. .

.e je j

ne = × = ×2 54 10 2 54 1022 28. . electrons cm electrons m3 3

(b) Ehm

neF

J s

kg

m J eV=

FHGIKJFHGIKJ =

× ⋅

×

×L

NMM

O

QPP = × =

−

−

−−

2 2 3 34 2

31

28 3 2 3

19

238

6 626 10

2 9 11 10

3 2 54 10

85 05 10 3 15

π π

.

.

.. .

e je j

e j

P43.30 The melting point of silver is 1 234 K. Its Fermi energy at 300 K is 5.48 eV. The approximate fractionof electrons excited is

k TEB

F

J K K

eV J eV=

×

×≈

−

−

1 38 10 1 234

5 48 1 60 102%

23

19

.

. .

e jb ga fe j

.

P43.31 Taking EF eV= 5 48. for sodium at 800 K,

f e

e

E Ek T

E E k T

E E k T

B

B

B

= + =

= − =

−= = −

−−

−

F

F

F

b g

b g

b g

1 0 950

10 950

1 0 052 6

0 052 6 2 94

1.

..

ln . .

E E− = −×

×= −

−

−F

J

1.60 10 J eV eV2 94

1 38 10 8000 203

23

19..

.e ja f

or E = 5 28. eV .

P43.32 d = 1 00. mm , so V = × = ×− −1 00 10 1 00 103 3 9. . m m3e j .

The density of states is g E CEm

hEa f = =1 2

3 2

31 28 2π

or g Ea f e je j

a fe j=×

× ⋅×

−

−

−8 2 9 11 10

6 626 104 00 1 60 10

31 3 2

34 319

π .

.. .

kg

J s eV J eV

g Ea f = × ⋅ = × ⋅− − − −8 50 10 1 36 1046 3 1 28 3 1. . m J m eV .

So, the total number of electrons is

N g E E V= = × ⋅ × = ×− − −a f a f e jb ge j∆ 1 36 10 0 025 0 1 00 10 3 40 1028 3 1 9 17. . . . m eV eV m electrons3 .

558 Molecules and Solids

P43.33 En

EN E dEe

av =∞z1

0

a fAt T = 0 , N Ea f = 0 for E E> F;.

Since f Ea f = 1 for E EEF< and f Ea f = 0 for E E> F, we can take N E CEm

hEea f = =1 2

3 2

31 28 2π

En

CE dECn

E dECn

Ee

E

e

E

eav F

5 2F F

= = =z z1 25

3 2

0

3 2

0

.

But from Equation 43.24, Cn

Ee= −3

23 2

F , so that E E E Eav F F5 2

F= FHGIKJFHG

IKJ =−2

532

35

3 2 .

P43.34 Consider first the wave function in x. At x = 0 and x L= , ψ = 0.

Therefore, sin k Lx = 0 and k Lx = π π π, , ,2 3 ….

Similarly, sin k Ly = 0 and k Ly = π π π, , ,2 3 …

sin k Lz = 0 and k Lz = π π π, , ,2 3 …

ψπ π π

= FHGIKJFHGIKJFHGIKJA

n xL

n y

Ln z

Lx y zsin sin sin .

From ∂∂

+∂∂

+∂∂

= −2

2

2

2

2

2 22ψ ψ ψ

ψx y z

mU Ee a f , we have inside the box, where U = 0,

− − −FHG

IKJ

= −n

L

n

Ln

Lm

Ex y z e2 2

2

2 2

2

2 2

2 22π π π

ψ ψa f Em L

n n n n n ne

x y z x y z= + + = …2 2

22 2 2

21 2 3

π e j , , , , , .

Outside the box we require ψ = 0.

The minimum energy state inside the box is n n nx y z= = = 1, with Em Le

=32

2 2

2π

P43.35 (a) The density of states at energy E is g E CEa f = 1 2 .

Hence, the required ratio isgg

C

C

8 507 00

8 50

7 001 10

1 2

1 2

.

..

..

eV eV

a fa f

a fa f= = .

(b) From Eq. 43.22, the number of occupied states having energy E is

N ECE

e E E k TBa f b g=

+−

1 2

1F.

Hence, the required ratio isNN

e

e

k T

k T

B

B

8 507 00

8 50

7 00

1

1

1 2

1 2

7 00 7 00

8 50 7 00

.

..

.

. .

. .

eV eV

a fa f

a fa f

a fa f=

+

+

LNMM

OQPP

−

−.

At T = 300 K , k TB = × =−4 14 10 0 025 921. . J eV ,NN e

8 507 00

8 50

7 00

2 00

1

1 2

1 2 1.50 0 025 9

.

..

.

..

eV eV

a fa f

a fa f a f=

+

LNM

OQP .

AndNN

8 507 00

1 55 10 25..

. eV eV

a fa f = × − .

Comparing this result with that from part (a), we conclude that very few states with E E> F

are occupied.

Chapter 43 559

Section 43.6 Electrical Conduction in Metals, Insulators, and Semiconductors

P43.36 (a) Eg = 1 14. eV for Si

hf = = × = ×− −1 14 1 14 1 60 10 1 82 1019 19. . . . eV eV J eV Ja fe j so f ≥ ×2 75 1014. Hz

(b) c f= λ ; λ µ= =×

×= × =−c

f3 00 102 75 10

1 09 10 1 098

146.

.. .

m s Hz

m m (in the infrared region)

P43.37 Photons of energy greater than 2.42 eV will be absorbed. This means wavelength shorter than

λ = =× ⋅ ×

× ×=

−

−hcE

6 626 10 3 00 10

2 42 1 60 10514

34 8

19

. .

. .

J s m s

J nm

e je j.

All the hydrogen Balmer lines except for the red line at 656 nm will be absorbed.

P43.38 Ehc

g = =× ⋅ ×

×≈

−

−λ

6 626 10 3 00 10

650 101 91

34 8

9

. ..

J s m s

m J eV

e je j

P43.39 If λ ≤ × −1 00 10 6. m, then photons of sunlight have energy

Ehc

≥ =× ⋅ ×

× ×

FHG

IKJ =

−

− −λmax

. .

..

6 626 10 3 00 10

1 00 101

1 2434 8

6 19

J s m s

m eV

1.60 10 J eV

e je j.

Thus, the energy gap for the collector material should be Eg ≤ 1 24. eV . Since Si has an energy gap

Eg ≈ 1 14. eV , it will absorb radiation of this energy and greater. Therefore, Si is acceptable as a

material for a solar collector.

P43.40 If the photon energy is 5.5 eV or higher, the diamond window will absorb. Here,

hfhcb gmaxmin

.= =λ

5 5 eV : λmin .

. .

. .= =

× ⋅ ×

×

−

−

hc5 5

6 626 10 3 00 10

5 5 1 60 10

34 8

19 eV

J s m s

eV J eV

e je ja fe j

λmin .= × =−2 26 10 2267 m nm .

*P43.41 In the Bohr model we replace ke by ke

κ and me by m* . Then the radius of the first Bohr orbit,

am k ee e

0

2

2= in hydrogen, changes to

′ = = FHGIKJ = FHG

IKJ =FHG

IKJ =∗ ∗ ∗a

m k emm m k e

mm

am

me

e

e e

e e

e

2

2

2

2 0 0 22011 7 0 052 9 2 81

κκ κ

.. . . nm nmb g .

The energy levels are in hydrogen Ek e

a nne= −

2

022

1 and here

′ = −′

= − = −FHGIKJ∗

∗

Ek e

a nk e

m m a

mm

En

e e

e e

n2

2

2

022

1

2κ κ κ κe jFor n = 1 , ′ = − = −E1 0 220

13 60 021 9.

..

eV11.7

eV2 .

560 Molecules and Solids

Section 43.7 Semiconductor Devices

P43.42 I I ee V k TB= −0 1∆a fe j. Thus, eI

Ie V k TB∆a f = +1

0

and ∆Vk T

eII

B= +FHGIKJln 1

0.

At T = 300 K , ∆VII

II

=×

×+FHGIKJ = +

FHGIKJ

−

−

1 38 10 300

1 60 101 25 9 1

23

190 0

.

.ln . ln

J K K

C mV

e ja f a f .

(a) If I I= 9 00 0. , ∆V = =25 9 10 0 59 5. ln . . mV mVa f a f .

(b) If I I= −0 900 0. , ∆V = = −25 9 0 100 59 5. ln . . mV mVa f a f .

The basic idea behind a semiconductor device is that a large current or charge can becontrolled by a small control voltage.

P43.43 The voltage across the diode is about 0.6 V. The voltage drop across the resistor is0 025 150 3 75. . A Va fa fΩ = . Thus, ε − − =0 6 3 8 0. . V V and ε = 4 4. V .

P43.44 First, we evaluate I0 in I I ee V k TB= −0 1∆a fe j , given that I = 200 mA when ∆V = 100 mV and T = 300 K .

e Vk TB

∆a f e ja fe ja f=

×

×=

−

−

1 60 10 0 100

1 38 10 3003 86

19

23

. .

..

C V

J K K so I

I

e ee V k TB0 3 861

2001

4 28=−

=−

=∆a f

mA mA. .

If ∆V = −100 mV , e Vk TB

∆a f= −3 86. ; and the current will be

I I e ee V k TB= − = − = −−0

3 861 4 28 1 4 19∆a fe j a fe j. .. mA mA .

*P43.45 (a) The currents to be plotted are

I eDV= −−10 16 0 025 A Ve je j∆ . ,

IV

W =−2 42

745. V

∆

Ω

The two graphs intersect at∆V = 0 200. V . The currentsare then

I eD = −

=

−10 1

2 98

6 0 200 A

mA

V 0.025 Ve je j.

.

0

10

20

0 0.1 0.2 0.3

Diode and Wire Currents

∆V(volts)

DiodeWire

FIG. P43.45

IW =−

=2 42 0 200

2 98. .

. V V745

mAΩ

. They agree to three digits.

∴ = =I ID W 2 98. mA

(b)∆

ΩV

ID=

×=−

0 2002 98 10

67 13.

..

V A

(c)d V

dIdI

d Ve

D

D∆∆

Ωa f

a f=LNMOQP

=LNM

OQP

=− − −1 6

0 2001

108 39

A0.025 V

V 0.025 V. .

Chapter 43 561

Section 43.8 Superconductivity

P43.46 (a) See the figure at right.

(b) For a surface current around the outside of the cylinder as shown,

BN I

=µ 0 or NI

B= =

×

× ⋅=

−

−µ π0

2

7

0 540 2 50 10

4 1010 7

. ..

T m

T m A kA

a fe je j

.

FIG. P43.46

P43.47 By Faraday’s law (Equation 32.1),∆Φ∆

∆∆

∆∆

B

tL

It

ABt

= = .

Thus, ∆∆

IA B

L= =

×=−

a f b g b gπ 0 010 0 0 020 0

3 10 10203

2

8

. .

.

m T

H A .

The direction of the induced current is such as to maintain the B– field through the ring.

P43.48 (a) ∆V IR=

If R = 0 , then ∆V = 0 , even when I ≠ 0 .

(b) The graph shows a direct proportionality.

Slope mA mV

= = =−−

=

=

−1 155 57 83 61 1 356

43 1

0 023 2

1

RIV

R

∆∆

Ω

Ω

.. .

.

.

a fa f

(c) Expulsion of magnetic flux and therefore fewercurrent-carrying paths could explain thedecrease in current.

FIG. P43.48

Additional Problems

P43.49 (a) Since the interatomic potential is the same for both molecules, the spring constant is thesame.

Then fk

=1

2π µ where µ12

12 1612 16

6 86=+

= u u u u

ua fa f

. and µ1414 1614 16

7 47=+

= u u u u

ua fa f

. .

Therefore,

fk k

f1414 12

12

1412

12

14

13 1312

12

6 42 106 86

6 15 10= =FHGIKJ = = × = ×

π µ π µµµ

µµ

..

. Hz u

7.47 u Hze j .

continued on next page

562 Molecules and Solids

(b) The equilibrium distance is the same for both molecules.

I r r I

I

14 142 14

1212

2 14

1212

1446 467 47

1 46 10 1 59 10

= =FHGIKJ =

FHGIKJ

= FHGIKJ × ⋅ = × ⋅− −

µµµ

µµµ

.. .

u6.86 u

kg m kg m2 2e j

(c) The molecule can move to the v J= =1 9,b g state or to the v J= =1 11,b g state. The energy itcan absorb is either

∆Ehc

hfI

hfI

= = +FHGIKJ + +

LNM

OQP− −FHG

IKJ + +

LNM

OQPλ

112

9 9 12

012

10 10 1214

2

1414

2

14a f a f ,

or ∆Ehc

hfI

hfI

= = +FHGIKJ + +

LNM

OQP− +FHG

IKJ + +

LNM

OQPλ

112

11 11 12

012

10 10 1214

2

1414

2

14a f a f .

The wavelengths it can absorb are then

λπ

=−

cf I14 1410 2b g or λ

π=

+c

f I14 1411 2b g .

These are: λπ

µ=×

× − × ⋅ × ⋅=

− −

3 00 10

6 15 10 10 1 055 10 2 1 59 104 96

8

13 34 46

.

. . ..

m s

Hz J s kg m m

2e j e j

and λπ

µ=×

× + × ⋅ × ⋅=

− −

3 00 10

6 15 10 11 1 055 10 2 1 59 104 79

8

13 34 46

.

. . ..

m s

Hz J s kg m m

2e j e j.

P43.50 For the N 2 molecule, k = 2 297 N m, m = × −2 32 10 26. kg , r = × −1 20 10 10. m , µ =m2

ωµ

= = ×k

4 45 1014. rad s, I r= = × × = × ⋅− − −µ 2 26 10 2 461 16 10 1 20 10 1 67 10. . . kg m kg m2e je j .

For a rotational state sufficient to allow a transition to the first exited vibrational state,

2

21

IJ J + =a f ω so J J

I+ = =

× ×

×=

−

−12 2 1 67 10 4 45 10

1 055 101 410

46 14

34a f e je jω . .

..

Thus J = 37 .

P43.51 ∆E vmax .= = +FHGIKJ4 5

12

eV ω so4 5 1 6 10

1 055 10 8 28 10

12

19

34 14 1

. .

. .

eV J eV

J s s

a fe je je j

×

× ⋅ ×≥ +FHG

IKJ

−

− −v

8 25 7 5. .> v = 7

P43.52 With 4 van der Waal bonds per atom pair or 2 electrons per atom, the total energy of the solid is

E = ××F

HGIKJ =

−2 1 74 106 02 10

5 232323

..

. J atom atoms

4.00 g J ge j .

Chapter 43 563

P43.53 The total potential energy is given by Equation 43.17: Uk e

rB

re

mtotal = − +α2

.

The total potential energy has its minimum value U0 at the equilibrium spacing, r r= 0 . At this point,dUdr r r=

=0

0 ,

ordUdr

ddr

k er

Br

k er

mBrr r

em

r r

em

= =+= − +

FHG

IKJ = − =

00

2 2

02

01 0α α .

Thus, Bk em

re m= −α2

01.

Substituting this value of B into Utotal , Uk er

k em

rr

k er m

e e mm

e0

2

0

2

01

0

2

0

11

1= − +

FHGIKJ = − −FHG

IKJ

−α α α .

P43.54 Suppose it is a harmonic-oscillator potential well. Then, 12

4 4832

3 96hf hf+ = +. . eV eV is the depth

of the well below the dissociation point. We see hf = 0 520. eV , so the depth of the well is

12

4 4812

0 520 4 48 4 74hf + = + =. . . . eV eV eV eVa f .

P43.55 (a) For equilibrium, dUdx

= 0 :ddx

Ax Bx Ax Bx− − − −− = − + =3 1 4 23 0e jx→∞ describes one equilibrium position, but the stable equilibrium position is at3 0

2Ax B− = .

xAB0

3 3 0 150

3 680 350= =

⋅

⋅=

.

..

eV nm

eV nm nm

3e j

(b) The depth of the well is given by U UAx

Bx

ABA

BBAx x0

03

0

3 2

3 2 3 2

1 2

1 2 1 20 3 3= = − = −=

U UB

Ax x0

3 2

3 2 1 2

3 2

3 2 1 20

23

2 3 68

3 0 1507 02= = − = −

⋅

⋅= −=

.

..

eV nm

eV nm eV

3

a fe j

.

(c) FdUdx

Ax Bxx = − = −− −3 4 2

To find the maximum force, we determine finite xm such that dFdx x xm=

= 0 .

Thus, − + =− −

=12 2 05 3Ax Bx

x xm

so that xA

Bm = FHGIKJ

6 1 2

.

Then F ABA

BBA

BAmax

.

.= FHG

IKJ − FHG

IKJ = − = −

⋅

⋅3

6 6 123 68

12 0 150

2 2 2 eV nm

eV nm3

a fe j

or Fmax ..

. .= −×F

HGIKJFHG

IKJ = − × = −

−

−−7 52

1 60 10 11 20 10 1 20

19

99 eV nm

J1 eV

nm10 m

N nN .

564 Molecules and Solids

P43.56 (a) For equilibrium, dUdx

= 0 :ddx

Ax Bx Ax Bx− − − −− = − + =3 1 4 23 0e jx→∞ describes one equilibrium position, but the stable equilibrium position is at

3 02Ax B− = or x

AB0

3= .

(b) The depth of the well is given by U UAx

Bx

ABA

BBA

BAx x0

03

0

3 2

3 2 3 2

1 2

1 2 1 2

3

0 3 32

27= = − = − = −= .

(c) FdUdx

Ax Bxx = − = −− −3 4 2

To find the maximum force, we determine finite xm such that

dFdx

Ax Bxx

x x x xm=

− −

== − + =12 2 05 3

0

then F ABA

BBA

BAmax =

FHGIKJ − FHG

IKJ = −3

6 6 12

2 2

.

P43.57 (a) At equilibrium separation, r re= ,dUdr

aB e er r

a r r a r r

e

e e

=

− − − −= − − =2 1 00 0b g b g .

We have neutral equilibrium as re →∞ and stable equilibrium at e a r re− − =0 1b g ,

or r re = 0 .

(b) At r r= 0 , U = 0. As r →∞ , U B→ . The depth of the well is B .

(c) We expand the potential in a Taylor series about the equilibrium point:

U r U rdUdr

r rd Udr

r r

U r Ba ae ae e r r Ba r r

r r r r

r r r r r r

r r

a f b g b g b g

a f a f e j b g b gb g b g b g

≈ + − + −

≈ + + − − − −LNM

OQP − ≈ −

= =

− − − − − −

=

0 0

2

2 02

2 20

2 20

2

0 0

0 0 0

0

12

0 012

2 1

This is of the form12

12

20

2kx k r r= −b gfor a simple harmonic oscillator with k Ba= 2 2 .

Then the molecule vibrates with frequency fk a B a B

= = =1

2 22

2π µ π µ π µ.

(d) The zero-point energy is12

12 8

ωπ µ

= =hfha B

.

Therefore, to dissociate the molecule in its ground state requires energy Bha B

−π µ8

.

Chapter 43 565

P43.58 T = 0 T TF= 0 1. T TF= 0 2. T TF= 0 5.E

EF e E E T TF Fb g b g−1 f Ea f e E E T TF Fb g b g−1 f Ea f e E E T TF Fb g b g−1 f Ea f e E E T TF Fb g b g−1 f Ea f0 e−∞ 1.00 e−10 0. 1.000 e−5 00. 0.993 e−2 00. 0.8810.500 e−∞ 1.00 e−5 00. 0.993 e−2 50. 0.924 e−1.00 0.7310.600 e−∞ 1.00 e−4.00 0.982 e−2 00. 0.881 e−0 800. 0.6900.700 e−∞ 1.00 e−3 00. 0.953 e−1.50 0.818 e−0 600. 0.6460.800 e−∞ 1.00 e−2 00. 0.881 e−1.00 0.731 e−0 400. 0.5990.900 e−∞ 1.00 e−1.00 0.731 e−0 500. 0.622 e−0 200. 0.5501.00 e0 0.500 e0 0.500 e0 0.500 e0 0.5001.10 e+∞ 0.00 e1.00 0.269 e0 500. 0.378 e0 200. 0.4501.20 e+∞ 0.00 e2 00. 0.119 e1.00 0.269 e0 400. 0.4011.30 e+∞ 0.00 e3 00. 0.047 4 e1.50 0.182 e0 600. 0.3541.40 e+∞ 0.00 e4.00 0.018 0 e2 00. 0.119 e0 800. 0.3101.50 e+∞ 0.00 e5 00. 0.006 69 e2 50. 0.075 9 e1.00 0.269

FIG. P43.58

P43.59 (a) There are 6 Cl− ions at distance r r= 0 . Thecontribution of these ions to the electrostatic

potential energy is −6 2

0

k er

e .

There are 12 Na+ ions at distance r r= 2 0 .Their contribution to the electrostatic potential

energy is +12

2

2

0

k ere . Next, there are 8 Cl− ions

at distance r r= 3 0 . These contribute a term of−8

3

2

0

k ere to the electrostatic potential energy.

FIG. P43.59

To three terms, the electrostatic potential energy is:

Uk er

k er

e e= − + −FHG

IKJ = −6

122

83

2 132

0

2

0. or U

k ere= − =α α

2

02 13 with . .

continued on next page

566 Molecules and Solids

(b) The fourth term consists of 6 Na+ at distance r r= 2 0 . Thus, to four terms,

Uk er

k er

e e= − + =2 13 3 0 8662

0

2

0. .a f .

So we see that the electrostatic potential energy is not even attractive to 4 terms, and that theinfinite series does not converge rapidly when groups of atoms corresponding to nearest

neighbors, next-nearest neighbors, etc. are added together.

ANSWERS TO EVEN PROBLEMS

P43.2 4.3 eV P43.30 2%

P43.4 (a) 1.28 eV; (b) σ = 0 272. nm, ∈= 4 65. eV ; P43.32 3 40 1017. × electrons(c) 6.55 nN; (d) 576 N m

P43.34 see the solutionP43.6 (a) 40 0. eVµ , 9.66 GHz;

P43.36 (a) 275 THz; (b) 1 09. mµ(b) If r is 10% too small, f is 20% too large.

P43.8 1 46 10 46. × ⋅− kg m2 P43.38 1.91 eV

P43.40 226 nmP43.10 (a) 1 81 10 45. × ⋅− kg m2; (b) 1.62 cm

P43.42 (a) 59.5 mV; (b) –59.5 mVP43.12 (a) 11.8 pm; (b) 7.72 pm; HI is more looselybound

P43.44 4.19 mA

P43.14 (a) 0, 364 eVµ , 1.09 meV;P43.46 (a) see the solution; (b) 10.7 kA

(b) 98.2 meV, 295 meV, 491 meV

P43.48 see the solutionP43.16 (a) 472 mµ ; (b) 473 mµ ; (c) 0 715. mµ

P43.50 37P43.18 2 9 10 47. × ⋅− kg m2

P43.52 5 23. J gP43.20 only 64.1 THz

P43.54 4.74 eVP43.22 (a) ~1017 ; (b) ~105 m3

P43.56 (a) xAB0

3= ; (b) −2

27

3BA

; (c) −B

A

2

12P43.24 (a) 0.444 nm, 0.628 nm, 0.769 nm

P43.26 see the solutionP43.58 see the solution

P43.28 (a) 1 57. Mm s;(b) larger by 10 orders of magnitude