sm chapter43

43Molecules and Solids

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

ANSWERS TO QUESTIONS

Q43.1 Ionic bonds are ones between oppositely charged ions. A simple model of an ionic bond is the electrostatic 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 a short-range game of catch with a ball models a covalent bond. On a quantum scale, the two atoms are sharing 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 the attraction between the opposite poles of two bar magnets, the dipole-induced dipole force is similar to a bar magnet attracting an iron nail or paper clip, and the dispersion force is analogous to an alternating-current electro-magnet attracting a paper clip.

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

Q43.2 Rotational, vibrational, and electronic (as discussed in Chapter 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 a small molecule is on the order of1 10 3meV eV.= − Vibrational energy is on the order of hf, where f is the vibration frequency of the molecule. A typical value is on the order of 0.1 eV. Electronic energy depends on the state of an electron in the molecule and is on the order of a few eV. The rotational energy can be zero, but neither the vibrational nor the electronic energy can be zero.

*Q43.3 If you start with a solid sample and raise its temperature, it will typically melt fi rst, then start emitting lots of far infrared light, then emit light with a spectrum peaking in the near infrared, and later have its molecules dissociate into atoms. Rotation of a diatomic molecule involves less energy than vibration. Absorption and emission of microwave photons, of frequency ~ Hz,1011

accompany excitation and de-excitation of rotational motion, while infrared photons, of frequency ~ Hz,1013 accompany changes in the vibration state of typical simple molecules. The ranking is then b > d > c > a.

519

ISMV2_5104_43.indd 519ISMV2_5104_43.indd 519 02/07/2007 14:28:4202/07/2007 14:28:42

520 Chapter 43

Q43.4 From the rotational spectrum of a molecule, one can easily calculate the moment of inertia of the molecule using Equation 43.7 in the text. Note that with this method, only the spacing between adjacent energy levels needs to be measured. From the moment of inertia, the size of the molecule can be calculated, provided that the structure of the molecule is known.

*Q43.5 Answer (b). At higher temperature, molecules are typically in higher rotational energy levels before as well as after infrared absorption.

*Q43.6 (i) Answer (a). An example is NaCl, table salt. (ii) Answer (b). Examples are elemental silicon and carborundum (silicon carbide).

(iii) Answer (c). Think of aluminum foil.

*Q43.7 (i) Answer (b). The density of states is proportional to the energy to the one-half power. (ii) Answer (a). Most states well above the Fermi energy are unoccupied.

Q43.8 In a metal, there is no energy gap between the valence and conduction bands, or the conduction band is partly full even at absolute zero in temperature. Thus an applied electric fi eld is able to inject a tiny bit of energy into an electron to promote it to a state in which it is moving through the metal as part of an electric current. In an insulator, there is a large energy gap between a full valence band and an empty conduction band. An applied electric fi eld is unable to give electrons in the valence band enough energy to jump across the gap into the higher energy conduction band. In a semiconductor, the energy gap between valence and conduction bands is smaller than in an insulator. At absolute zero the valence band is full and the conduction band is empty, but at room temperature thermal energy has promoted some electrons across the gap. Then there are some mobile holes in the valence band as well as some mobile electrons in the conduction band.

*Q43.9 Answer (b). First consider electric conduction in a metal. The number of conduction electrons is essentially fi xed. They conduct electricity by having drift motion in an applied electric fi eld superposed on their random thermal motion. At higher temperature, the ion cores vibrate more and scatter more effi ciently the conduction electrons fl ying among them. The mean time between collisions is reduced. The electrons have time to develop only a lower drift speed. The electric current is reduced, so we see the resistivity increasing with temperature.

Now consider an intrinsic semiconductor. At absolute zero its valence band is full and its conduction 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 both electrons and holes move in response to an applied electric fi eld. Thus we see the resistivity decreasing as temperature goes up.

Q43.10 The energy of the photon is given to the electron. The energy of a photon of visible light is suffi cient to 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 energy state that the electron used to occupy—in the valence band.

Q43.11 Along with arsenic (As), any other element in group V, such as phosphorus (P), antimony (Sb), and bismuth (Bi), would make good donor atoms. Each has 5 valence electrons. Any element in group III would make good acceptor atoms, such as boron (B), aluminum, (Al), gallium (Ga), and indium (In). They all have only 3 valence electrons.

Q43.12 The two assumptions in the free-electron theory are that the conduction electrons are not bound to any particular atom, and that the nuclei of the atoms are fi xed in a lattice structure. In this model, it is the “soup” of free electrons that are conducted through metals. The energy band model is more comprehensive than the free-electron theory. The energy band model includes an account of the more tightly bound electrons as well as the conduction electrons. It can be developed into a theory of the structure of the crystal and its mechanical and thermal properties.


Molecules and Solids 521

Q43.13 A molecule containing two atoms of 2 H, deuterium, has twice the mass of a molecule containing two atoms of ordinary hydrogen 1H. The atoms have the same electronic structure, so the molecules have the same interatomic spacing, and the same spring constant. Then the moment of inertia of the double-deuteron is twice as large and the rotational energies one-half as large

as for ordinary hydrogen. Each vibrational energy level for D2 is 1

2 times that of H2 .

Q43.14 Yes. A material can absorb a photon of energy greater than the energy gap, as an electron jumps into a higher energy state. If the photon does not have enough energy to raise the energy of the electron by the energy gap, then the photon will not be absorbed.

*Q43.15 (i) and (ii) Answer (a) for both. Either kind of doping contributes more mobile charge carriers, either holes or electrons.

*Q43.16 (a) false (b) false (c) true (d) true (e) true

Section 43.1 Molecular Bonds

P43.1 (a) Fq

r=

∈=

×( ) ×( )×

−

−

2

02

19 2 9

14

1 60 10 8 99 10

5 00 10π. .

. 00 290 921 10

( )= × −N N.

toward the other ion.

(b) Uq

r= −

∈= −

×( ) ×( )×

−

−

2

0

19 2 9

4

1 60 10 8 99 10

5 00 10π. .

. 110 2 88J eV≈ − .

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

dU

drAr Br= − + =− −12 6 013 7

yielding r

A

B0

1 62=⎡⎣⎢

⎤⎦⎥

(b) E U UA

A B

B

A Br r r= − = − −⎡

⎣⎢

⎤

⎦⎥ = − −⎡

⎣⎢⎤

= = 00

4 2

1

4

1

22 2 ⎦⎦⎥=

B

A

B

A

2 2

4

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

(c) r0

120

60

2 0 124 10

1 488 10=

× ⋅( )× ⋅

⎡ −

−

.

.

eV m

eV m

12

6

⎣⎣⎢⎢

⎤

⎦⎥⎥

= × =−

1 6

117 42 10 74 2. .m pm

E =× ⋅( )× ⋅(

−

−

1 488 10

4 0 124 10

60 2

120

.

.

eV m

eV m

6

12 )) = 4 46. eV

SOLUTIONS TO PROBLEMS


522 Chapter 43

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. .

The activation energy is 1 28. .eV

(b) dU

dr r r=

∈− ⎛

⎝⎜ ⎞

⎠⎟ + ⎛

⎝⎜ ⎞

⎠⎟

⎡

⎣⎢

⎤

⎦⎥

412 6

13 7

ss s

At r r= 0 we have dU

dr= 0. Here s s

r r0

13

0

71

2

⎛

⎝⎜

⎞

⎠⎟ =

⎛

⎝⎜

⎞

⎠⎟

sr0

1 62= − s s= ( ) = =−2 0 305 0 2721 6 . . nm nm

Then also

U rr

r

r

r0

1 60

0

12 1 60

0

6

42 2( ) = ∈⎛

⎝⎜

⎞

⎠⎟ −

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢

− −

⎢⎢

⎤

⎦⎥⎥+ = ∈ −

⎡⎣⎢

⎤⎦⎥+ = − ∈ +

∈= − ( ) =

E E E

E U r

a a a

a

41

4

1

2

0 11 28 3 37. . eV eV= 4.65 eV+ = ∈

(c) F rdU

dr r r( ) = − =

∈ ⎛⎝⎜ ⎞

⎠⎟ − ⎛

⎝⎜ ⎞

⎠⎟

⎡

⎣⎢

⎤

⎦⎥

412 6

13 7

ss s

To fi nd the maximum force we calculate dF

dr r r=

∈− ⎛

⎝⎜ ⎞

⎠⎟ + ⎛

⎝⎜ ⎞

⎠⎟

⎡

⎣⎢

⎤

⎦⎥ =

4156 42 02

14 8

ss s

when

s

rrupture

= ⎛⎝⎜ ⎞

⎠⎟42

156

1 6

Fmax

.

.= ( ) ⎛

⎝⎜⎞⎠⎟ −4 4 65

0 27212

42

1566

4213 6eV

nm 115641 0 41 0

1 6 107 6 1⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ = − = − × −

. ..

eV nm99

9

6 55

Nm

10 m

nN

−

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

center.

(d) U r sr s r s0

0

12

0

6

4+( ) = ∈+

⎛

⎝⎜

⎞

⎠⎟ −

+⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥

s s⎥⎥+ = ∈

+⎛

⎝⎜

⎞

⎠⎟ −

+⎛

⎝⎜

⎞

⎠⎟

⎡ − −

Er

r s r sa 42 21 6

0

0

12 1 6

0

6

⎣⎣⎢⎢

⎤

⎦⎥⎥+ Ea

= ∈ +⎛

⎝⎜

⎞

⎠⎟ − +

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

− −

41

41

1

21

0

12

0

6s

r

s

r++

= ∈ − + −⎛

⎝⎜

⎞

⎠⎟− − +

E

s

r

s

r

s

r

a

41

41 12 78

1

21 6 2

0

2

02

0

11

12 78

2

02

0

2

02

s

rE

s

r

s

r

a−

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢

⎤

⎦⎥+

= ∈ − ∈ + ∈

−− ∈ + ∈ − ∈ + +

= − ∈ + +⎛

⎝⎜

⎞

⎠⎟+

2 12 42

0

0

2

02

0

s

r

s

rE

Es

r

a

a

336

1

2

2

02

0 02

∈ +

+( ) ( )+≈

s

r

U r s U r ks

where kr

= ∈ = ( )( )

= =72 72 4 65

0 3053 599

02 2

.

.

eV

nm eV nm2 5576 N m



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

T ≈ ××

−

−

1 6 1010

22

23

.~

J

1.38 10 J K K

Section 43.2 Energy States and Spectra of Molecules

*P43.6 (a) With r representing the distance of each atom from the center of mass, the moment of inertia is

mr mr2 2 2710

2 1 67 100 75 10+ = ×( ) ×⎛

⎝⎜⎞⎠⎟

−−

..

kgm

2

22

484 70 10= × ⋅−. kg m2

The rotational energy is EI

J Jrot = +( )2

21 or it is zero for J = 0 and for J = 1 it is

h

I

2

2

34

2 4

1 2

4 2

6 626 10

4 4 70 10

( ) .

( .π π=

× ⋅( )×

−

−

J s2

88 19

1

100 0148

kg m

eV

1.6 J eV

2⋅ ×⎛⎝⎜

⎞⎠⎟ =−)

.

(b) λ = = = × × ⋅−c

f

ch

E

( .2 998 10 108 34 m/s)(6.626 J s)

0.00148 eV

eV

1.602 J

eV nm

0.0148

1

10

124019×

= ⋅− eeV

m= 83 8. µ

P43.7 µ =+

= ( )+

× −m m

m m1 2

1 2

132 9 126 9

132 9 126 91 66 10

. .

. .. 227 251 08 10kg kg( ) = × −.

I r= = ×( ) ×( ) = ×− −µ 2 25 9 21 08 10 0 127 10 1 74 1. . . kg m 00 45− ⋅ kg m2

(a) E II

I

J J

I= = ( ) =

+( )1

2 2

1

22

2 2

ω ω

J = 0 gives E = 0

J = 1 gives EI

= =× ⋅( )× ⋅

−

−

2 34 2

2 45

6 626 10

4 1 74 10

.

.

J s

kg m2π (( ) = × =−6 41 10 40 024. .J eVµ

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

(b) fE

h hI

h

rr= = = ∝ −1

2

2 22

4

π µ

If is 10% too small, is 20% too large.r f

*P43.8 ∆Eh k

hf kibv = = =2

4π µ

so ππ µ2 2f

µπ π

= =×

= × −k

f4

1530

4 56 3 101 22 102 2 2 12 2

N/m

s)( . /. 226 kg

µ =+

=m m

m m1 2

1 2

14 007

14 007

.

.

u 15.999 u

u +15.9999 u

kg

ukg

1 66 10

11 24 10

2726.

.× = ×

−−

The reduced masses agree, because the small apparent difference can be attributed to uncertainty in the data.


524 Chapter 43

P43.9 For the HCl molecule in the J = 1 rotational energy level, we are given r0 0 127 5= . nm.

EI

J Jrot = +( )2

21

Taking J = 1, we have EI

Irot = =221

2ω or ω = =2

22

2

I I

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

I rm m

m mr= =

+⎛⎝⎜

⎞⎠⎟

µ 02 1 2

1 202

I r= ( )( )+

⎡

⎣⎢

⎤

⎦⎥ =

1 01 35 5

1 01 35 500

2. .

. ..

u u

u u9982 1 66 10 1 275 10 227 10 2

u kg u m( ) ×( ) ×( ) =− −. . .665 10 47× ⋅− kg m2

Therefore, ωπ

= = × ⋅× ⋅

−

−22 6 626 10

10

34

47

I

J s

2 (2.65 kg m2

.

)== ×5 63 1012. rad s

P43.10 hf EI I I

= = +( )[ ]− +( )[ ] = ( )∆ 2 2 2

22 2 1

21 1 1

24

Ih

hf

h

f= ( ) = = × ⋅

×

−4 2

2 2

6 626 10

2 30 1

2

2

34ππ π

.

.

J s

2 2 001 46 10

1146

Hzkg m2

( ) = × ⋅−.

P43.11 I m r m r= +1 12

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

Then rm r

m12 2

1

= so m r

mr r2 2

12+ = and r

m r

m m21

1 2

=+

Also, rm r

m21 1

2

= . Thus, rm r

mr1

1 1

2

+ = and rm r

m m12

1 2

=+

I mm r

m m

m m r

m m

m m r m=

+( )+

+( )=

+1

22 2

1 2

22 1

2 2

1 2

21 2

22 mm

m m

m m r

m mr1

1 2

21 2

2

1 2

2( )+( )

=+

= µ

P43.12 (a) µ = ( )+( ) ×( ) =−22 99 35 45

22 99 35 451 66 10 27. .

. .. kg 22 32 10 26. × − kg

I r= = ×( ) ×( ) = ×− −µ 2 26 9 22 32 10 0 280 10 1 81 1. . .kg m 00 45− kg m2

(b) hc

I I I I I

h

λ= +( ) − +( ) = − = = 2 2 2 2 2 2

22 2 1

21 1 1

3 2 2

44 2π I

λ π π= =

×( ) × ⋅( −c I

h

4

2

3 00 10 4 1 81 102 8 2 45. .m s kg m2 ))× ⋅( ) =−2 6 626 10

1 6234.

.J s

cm

FIG. P43.9



P43.13 The energy of a rotational transition is ∆EI

J=⎛⎝⎜

⎞⎠⎟

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= =

× ⋅( ) ×( )−

λ λ6 626 10 3 00 10 1

34 8. .J s m s eV

1..60 J×⎛⎝⎜

⎞⎠⎟−10 19 .

For each observed wavelength,

l (mm) ∆E (eV)

0.120 4 0.010 32

0.096 4 0.012 88

0.080 4 0.015 44

0.069 0 0.018 00

0.060 4 0.020 56

The ∆Es consistently increase by 0.002 56 eV. EI1

2

0 002 56= =. eV

and IE

= =× ⋅( )

( )

−2

1

34 21 055 10

0 002 56

1.

.

J s

eV

eV

1.660 J kg m2

×⎛⎝⎜

⎞⎠⎟ = × ⋅−

−

102 72 10

1947.

For the HCl molecule, the internuclear radius is rI= = ×

×=

−

−µ2 72 10

1 62 100 130

47

27

.

.. m nm

*P43.14 (a) Minimum amplitude of vibration of HI is characterized by

1

2

1

2 4 22kA

h kA

h= = =ωπ µ π

so11

1 4

kµ⎛⎝⎜

⎞⎠⎟

/

A = × ⋅−6 626 10

2

1

320

34.

(

J s

N/m)(127/128)(1.66π ××⎛⎝⎜

⎞⎠⎟

=−1012 127

1 4

kg)pm

/

.

(b) For HF, A = × ⋅×

−6 626 10

2

1

970 1

34.

(

J s

N/m)(19/20)(1.66π 009 2327

1 4

−

⎛⎝⎜

⎞⎠⎟

=kg)

pm/

.

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

P43.15 µ =+

= × × = ×− −m m

m m1 2

1 2

27 2735

361 66 10 1 61 10. .kg kgg

∆Ek

vib = = ×( ) ×= ×−

−µ

1 055 10480

1 61 105 74 134

27.

.. 00 0 35820− = J eV.


526 Chapter 43

P43.16 (a) The reduced mass of the O2 is

µ = ( )( )( )+ ( )

= = × −16 16

16 168 8 1 66 10 27 u u

u u u . kg kg( ) = × −1 33 10 26.

The moment of inertia is then

I r= = ×( ) ×( )= ×

− −

−

µ 2 26 10 21 33 10 1 20 10

1 91 10

. .

.

kg m

446 kg m2⋅

The rotational energies are EI

J Jrot

J s= +( ) =

× ⋅( )×

−

−

2 34 2

21

1 055 10

2 1 91 10

.

. 4461

kg m2⋅( ) +( )J J

Thus E J Jrot J= ×( ) +( )−2 91 10 123.

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

(b) Ek

vib J= +⎛⎝⎜

⎞⎠⎟ = +

⎛⎝⎜

⎞⎠⎟ × ⋅−v v

1

2

1

21 055 10 34

µ. ss

N m

kg( )

×( )−

1177

8 1 66 10 27.

Evib JeV

1.60= +⎛⎝⎜ ⎞

⎠⎟ ×( )

×−

−v1

23 14 10

1

1020

19.J

eV⎛⎝⎜ ⎞

⎠⎟ = +⎛

⎝⎜ ⎞

⎠⎟ ( )v

1

20 196.

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

P43.17 In Benzene, the carbon atoms are each 0.110 nm from the axis and each hydrogen atom is 0 110 0 100 0 210. . .+( ) =nm nm from the axis. Thus, I mr= Σ 2:

I = ×( ) ×( ) + ×− − −6 1 99 10 0 110 10 6 1 67 1026 9 2. . . kg m 227 9 2

45

0 210 10

1 89 10

kg m

kg m2

( ) ×( )= × ⋅

−

−

.

.

The allowed rotational energies are then

E

IJ Jrot

J s= +( ) =

× ⋅( )×

−

−

2 34 2

21

1 055 10

2 1 89 10

.

. 445241 2 95 10 1

18 4

kg mJ2⋅( ) +( ) = ×( ) +( )

=

−J J J J.

. ××( ) +( )−10 16 eV J J

E J J Jrot eV where = ( ) +( ) =18 4 1 0 1 2 3. , , , , . . .µ

The fi rst fi ve of these allowed energies are: Erot eV, 111 eV, 221 eV, and 369= 0 36 9, . µ µ µ µµeV.



P43.18 We carry extra digits through the solution because part (c) involves the subtraction of two close numbers. 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

λ π= = =

hc

E

hc

I

Ic

h∆ min

.2

24 If m is the reduced mass, then

I r= = ×( ) = ×(− −µ µ2 9 2 200 127 46 10 1 624 605 10. . m m2 ))

=×( )−

µ

λπ µ

Therefore m24 1 624 605 10 2 992 20. . 77 925 10

6 626 075 102 901 830 1

8

34

×( )× ⋅

= ×−

m s

J s.. 0023 m kg( )µ

(1)

(a) µ35

1 007 825 34 968 853

1 007 825 34 968= ( )( )

+. .

. .

u u

u 88530 979 593 1 626 653 10 27

uu kg= = × −. .

From (1): λ3523 272 901830 10 1 626 653 10= ×( ) ×( ) =−. .m kg kg 4472 mµ

(b) µ37

1 007 825 36 965 903

1 007 825 36 965= ( )( )

+. .

. .

u u

u 99030 981 077 1 629 118 10 27

uu kg= = × −. .

From (1): λ3723 272 901830 10 1 629 118 10= ×( ) ×( ) =−. .m kg kg 4473 mµ

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

P43.19 We fi nd an average spacing between peaks by counting 22 gaps between 7 96 1013. × Hz and9 24 1013. × Hz:

∆f

h

h=−( ) = × =

9 24 7 96 100 058 2 10

11313

2. ..

Hz

22 Hz

44

4

6 63 10

4 5 82 1

2

2

34

2

π

π π

I

Ih

f

⎛

⎝⎜

⎞

⎠⎟

= = × ⋅×

−

∆.

.

J s

002 9 10

1147

s kg m2= × ⋅−.


528 Chapter 43

P43.20 We carry extra digits through the solution because the given wavelengths are close together.

(a) E hfI

J JJv v= +⎛⎝⎜ ⎞

⎠⎟ + +( )1

2 21

2

∴ = = + = +

∴ −

E hf E hfI

E hfI

E E

00 11

2

02

2

11 0

1

2

3

2

1

2

3, ,

00

2 346 626 075 10 2 997 925= + = =

× ⋅( ) ×−

hfI

hcλ

. .J s 110

2 211 2 10

8

6

m s

m

( )× −.

∴ + = × −hfI

2208 983 573 10. J (1)

E E hfI

hc11 02

2 342 6 626 075 10 2 9

− = − = =× ⋅( )−

λ. .J s 997 925 10

2 405 4 10

8

6

×( )× −

m s

m.

∴ − = × −hf

I

28 258 284 10

220

. J (2)

Subtract (2) from (1): 3

7 252 89 102

21I

= × −. J

∴ =× ⋅( )×

=−

−I3 1 054 573 10

7 252 89 104

34 2

21

.

..

J s

J660 10 48× ⋅− kg m2

(b) From (1):

f =

××

−−

−

8 983 573 10

10

1 054 520

34

. .J

6.626 075 J s

773 10

4 600 060 10 6 626

34 2

48

×( )×( )

−

−

J s

kg m2

. . 0075 10

1 32 10

34

14

×( )= ×

− J s

Hz

.

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

µ = = ( ) = × −1

2

1

21 007 825 8 367 669 10 28mH . .u kg

So rI= = × ⋅

×=

−

−µ4 600 060 10

8 367 669 10

48 2

28

.

.

kg m

kg00.074 1 nm .

P43.21 The emission energies are the same as the absorption energies, but the fi nal state must be belowv = =( )1 0, .J The transition must satisfy ∆J =±1, so it must end with J = 1. To be lower in

energy, it must be v = =( )0 1, .J The emitted photon energy is therefore

hf E E E E

J Jphoton vib rot vib rot= +( ) − +(= = = =v v1 0 0 1 )) = −( ) − −( )= = = =E E E E

hf

J Jvib vib rot rot

pho

v v1 0 1 0

tton vib rot= −hf hf

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

P43.22 The moment of inertia about the molecular axis is I mr mr mx = + = ×( )−2

5

2

5

4

52 00 102 2 15 2

. . m

The moment of inertia about a perpendicular axis is I mR

mR m

y =⎛⎝⎜

⎞⎠⎟ +

⎛⎝⎜

⎞⎠⎟ = ×( )−

2 2 22 00 10

2 210 2

. . m

The allowed rotational energies are EI

J Jrot =⎛⎝⎜

⎞⎠⎟

+( )2

21 , so the energy of the fi rst excited state is

EI1

2

= . The ratio is therefore

E

E

I

I

I

I

mx

y

x

y

y

x

1

1

2

2

101 2 2 00 10,

,

.= ( )( ) = =

( ) × −

mm

m

( )( ) ×( )

= ( ) = ×−

2

15 25 2 9

4 5 2 00 10

5

810 6 25 10

m ..



Section 43.3 Bonding in Solids

P43.23 Uk e

r me= − −

⎛⎝⎜

⎞⎠⎟ = −( ) ×( )α 2

0

911

1 747 6 8 99 101 6

. .. 00 10

0 281 101

1

81 25 10

19 2

9

×( )×( ) −

⎛⎝⎜

⎞⎠⎟ = − ×

−

−−

.. 118 7 84 J eV= − .

P43.24 Consider a cubical salt crystal of edge length 0.1 mm.

The number of atoms is 10

104

9

3

17−

−×⎛

⎝⎜

⎞

⎠⎟

m

0.261 10 m~

This number of salt crystals would have volume 1010

10104 3

4

9

3

5−−

−( ) ×⎛

⎝⎜

⎞

⎠⎟ m

m

0.261 m m3~

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

P43.25

Uk e

r

k e

r

k e

r

k e

r

k e

r

k e

re e e e e e= − − + + − − +

2 2 2 2 2 2

2 2 3 3

kk e

r

k e

r

k e

r

e e

e

2 2

2

4 42

11

2

1

3

1

4

+ −

= − − + − +⎛⎝⎜

⎞⎠⎟

But, ln 1 12 3 4

2 3 4

+( ) = − + − +xx x x

so, Uk e

re= − 2

22

ln , or U ke

re= − =α α2

2 2 where ln

P43.26 Visualize a K+ ion at the center of each shaded cube, a Cl− ion at the center of each white one.

The distance ab is 2 0 314 0 444. . nm nm( ) =

Distance ac is 2 0 314 0 628. . nm nm( ) =

Distance ad is 2 2 0 314 0 76922

+ ( ) ( ) =. . nm nm

Section 43.4 Free-Electron Theory of Metals

Section 43.5 Band Theory of Solids

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

m

neF =

⎛⎝⎜

⎞⎠⎟

2 2 3

2

3

8π

or nmE

he =⎛⎝⎜

⎞⎠⎟ =

×( )−

8

3

2 8

3

2 9 11 10 5 4

2

3 231

π πF kg. . 88 1 60 10

6 626 10

19 3 2

34 3

( ) ×( )⎡⎣ ⎤⎦

× ⋅( )=

−

−

.

.

J

J s55 80 1028 3. × − m

The number-density of silver atoms is

nAg3 kg m

atom

108 u

u

1.66= ×( )⎛⎝⎜

⎞⎠⎟10 6 10

1 13.××

⎛

⎝⎜

⎞

⎠⎟ = ×−

−

10 kg m

2728 35 91 10.

So an average atom contributes 5 80

5 910 981

.

..= electron to the conduction baand .

FIG. P43.25

– ++ – + –

3r2r

r

–e –e –ee e e

FIG. P43.26


530 Chapter 43

*P43.28 (a) The Fermi energy is proportional to the spatial concentration of free electrons to the two-thirds power.

(b) Eh

m

neF

J s= ⎛

⎝⎜⎞⎠⎟ =

× ⋅( )−2 2 3 34 2

2

3

8

6 626 10

2 9 1π.

. 11 10 1 60 10

3

831 19

2 32

×( ) ×( )⎛⎝⎜

⎞⎠⎟− −kg J eV. π

ne33 becomes

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

(c) Copper has the greater concentration of free electrons by a factor of 8.49 × 1028/1.40 × 1028 = 6.06 Copper has the greater Fermi energy by a factor of 7.05 eV/2.12 eV = 3.33. This behavior

agrees with the proportionality because 6.062/3 = 3.33.

P43.29 (a) 1

27 052mv = . eV

v =( ) ×( )

×=

−

−

2 7 05 1 60 10

9 11 101

19

31

. .

.

eV J eV

kg..57 106× m s

(b) Larger than m s by ten orders of magnit10 4− uude. However, the energy of an electron at

room temperature is typically k TB = 1

40eV.

*P43.30 The occupation probability is

f E

e eE E k T E E k TF B F F B( ) ( )/ ( . )/=

+=

+− −1

1

1

10 99

==− × −

1

0 01 7 05exp[ . ( . eV)(1.6 10 J/eV)/(1.3819 ×× +=

+=− −10 J/K)300 K]23 1

1

10 938

2 72e ..

*P43.31 (a) E Ea Fv = =3

50 6 7 05. ( . eV) = 4.23 eV

(b) The average energy of a molecule in an ideal gas is 3

2k TB so we have

T =×

× =−

−2

3

4 23 1 63

. ..

eV

1.38 10 J/K

10 J

1 eV23

19

227×10 K4

P43.32 For sodium, M = 23 0. g mol and ρ = 0 971. .g cm3

(a) nN

Me = =×( ) ( )A

3electrons mol g cmρ 6 02 10 0 971

2

23. .

33 0

2 54 10 2 54 1022 28

.

. .

g mol

electrons cm e3ne = × = × llectrons m3

(b) Eh

m

neF

J s=⎛⎝⎜

⎞⎠⎟⎛⎝

⎞⎠ =

× ⋅( )−2 2 3 34 2

2

3

8

6 626 10

2π.

99 11 10

3 2 54 10

831

28 3 2 3

.

.

×( )×( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

−

−

kg

m

π

55 05 10 3 1519. .× =− J eV

P43.33 Taking EF eV= 5 48. for sodium at 800 K,

f e

e

E E k T

E E k T

B

B

= +⎡⎣

⎤⎦ =

=

−( ) −

−( )

F

F

1 0 950

1

0 950

1

.

.−− =

− = ( ) = −

1 0 052 6

0 052 6 2 94

.

ln . .E E

k TB

F

E E− = −×( )( )×

=−

−F

J

1.60 10 J eV2 94

1 38 10 80023

19.

.−− =0 203 5 28. . eV or eVE



P43.34 The melting point of silver is 1 234 K. Its Fermi energy at 300 K is 5.48 eV. The approximate fraction of electrons excited is

k T

EB

F

J K K

eV=

×( )( )( ) ×

−1 38 10 1 234

5 48 1 60 1

23.

. . 002

19−( ) ≈J eV%

P43.35 En

EN E dEe

av = ( )∞

∫1

0

At T = 0, N E E E( ) = >0 for F

Since f E( ) =1 for E EEF< and f E( ) = 0 for E E> F ,

we can take N E CEm

hEe( ) = =1 2

3 2

31 28 2π

E

nCE dE

C

nE dE

C

nE

e

E

e

E

e

av F5 2

F F

= = =∫ ∫1 2

53 2

0

3 2

0

But from Equation 43.25, C

nE

e

= −3

23 2

F , so that E E E Eav F F5 2

F=⎛⎝⎜

⎞⎠⎟⎛⎝⎜

⎞⎠⎟ =−2

5

3

2

3

53 2

P43.36 d =1 00. mm, so V = ×( ) = ×− −1 00 10 1 00 103 3 9. . m m3

The density of states is g E CEm

hE( ) = =1 2

3 2

31 28 2π

or g E( ) =×( )

× ⋅( )−

−

8 2 9 11 10

6 626 10

31 3 2

34 3

π .

.

kg

J s44 00 1 60 10 19. .eV J eV( ) ×( )−

g E( ) = × ⋅ = × ⋅− − − −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= ( )⎡⎣ ⎤⎦( ) = × ⋅( )− −∆ 1 36 10 0 025 028 3 1. . m eV eeV m

electrons

3( ) ×( )= ×

−1 00 10

3 40 10

9

17

.

.

*P43.37 (a) The density of states at energy E is g E CE( ) = 1 2

Hence, the required ratio is g

g

C

C

8 50

7 00

8 50

7 001 1

1 2

1 2

.

.

.

..

eV

eV

( )( )

= ( )( )

= 00

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

N ECE

e E E k TB( ) =

+−( )

1 2

1F

Hence, the required ratio is N

N

e8 50

7 00

8 50

7 00

1 2

1 2

7 00.

.

.

.

. eV

eV

( )( )

= ( )( )

−77 00

8 50 7 00

1

1

.

. .

( )

−( )++

⎡

⎣⎢

⎤

⎦⎥

k T

k T

B

Be

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

N

N e

8 50

7 00

8 50

7 00

2 001 2

1 2 1

.

.

.

.

. eV

eV

( )( )

= ( )( ) .. .50 0 025 9 1( ) +

⎡⎣⎢

⎤⎦⎥

And N

N

8 50

7 001 47 10 25.

..

eV

eV

( )( )

= × −

This result is vastly smaller than that in part (a). We conclude that very few states well above the Fermi energy are occupied at room temperature.


532 Chapter 43

P43.38 Consider fi rst 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

ψ π π π=⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜

⎞

⎠⎟

⎛⎝⎜

⎞A

n x

L

n y

L

n z

Lx y zsin sin sin

⎠⎠⎟

From ∂∂

∂∂

∂∂

2

2

2

2

2

2 2

2ψ ψ ψ ψx y z

mU Ee+ + = −( )

, we have inside the box, where U = 0,

− − −⎛

⎝⎜⎞⎠⎟

= −( )n

L

n

L

n

L

mEx y z e

2 2

2

2 2

2

2 2

2 2

2π π πψ ψ

Em L

n n n n n ne

x y z x y z= + +( ) =2 2

22 2 2

21 2 3

π, , , , , . . .

Outside the box we require ψ = 0.

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

= 3

2

2 2

2

π

Section 43.6 Electrical Conduction in Metals, Insulators, and Semiconductors

P43.39 (a) Eg = 1 14. eV for Si

hf = = ( ) ×( ) = ×−1 14 1 14 1 60 10 1 82 119. . . .eV eV J eV 00 19− J

so f ≥ ×2 75 1014. Hz

(b) c f= λ ; λ = = ××

= × =−c

f

3 00 10

2 75 101 09 10 1

8

146.

.. .

m s

Hzm 009 mµ (in the infrared region)

P43.40 Photons of energy greater than 2.42 eV will be absorbed. This means wavelength shorter than

λ = =× ⋅( ) ×( )

×

−hc

E

6 626 10 3 00 10

2 42 1

34 8. .

. .

J s m s

660 10514

19×=− J

nm

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

*P43.41 If λ ≤ × −1 00 10 6. m, then photons of sunlight have energy

Ehc≥ =

× ⋅( ) ×( )−

λmax

. .

.

6 626 10 3 00 10

1 00

34 8 J s m s

×× ×⎛⎝⎜

⎞⎠⎟ =− −10

11 24

6 19 m

eV

1.60 10 J eV.

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 can absorb nearly all of the photons in sunlight. Therefore,

Si is an appropriate material for a solar collector.

P43.42 Ehc

g = =× ⋅( ) ×( )

×

−

λ6 626 10 3 00 10

650 10

34 8. .J s m s−− ≈9 1 91

mJ eV.



P43.43 If the photon energy is 5.5 eV or higher, the diamond window will absorb. Here,

hfhc( ) = =

maxmin

.λ

5 5 eV: λmin .

. .= =

× ⋅( ) ×(−hc

5 5

6 626 10 3 00 1034 8

eV

J s m s))( ) ×( )−5 5 1 60 10 19. .eV J eV

λmin .= × =−2 26 10 2267 m nm

P43.44 In the Bohr model we replace ke by ke

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

am k ee e

0

2

2=

in hydrogen, changes to

′ = =⎛⎝⎜

⎞⎠⎟ =

⎛⎝⎜

⎞⎠⎟a

m k e

m

m m k e

m

me

e

e e

e 2

2

2

2

κ κ* * *

κκ am

me

e

0 0 22011 7 0 052 9 2 81=

⎛

⎝⎜

⎞

⎠⎟ ( ) =

.. . . nm nm

The energy levels are in hydrogen Ek e

a nne= −

2

022

1 and here

′ = −′

= − ( ) = −⎛⎝⎜

⎞E

k e

a n

k e

m m a

m

mne e

e e

2

2

2

02

1

2κ κ κ*

*

⎠⎠⎟En

κ 2

For n = 1, ′ = − = −E1 0 22013 6

0 021 9..

. .eV

11.7eV2

Section 43.7 Semiconductor Devices

P43.45 I I ee V k TB= −( )( )0 1∆ Thus, e

I

Ie V k TB∆( ) = +1

0

and ∆Vk T

e

I

IB= +

⎛

⎝⎜

⎞

⎠⎟ln 1

0

At T = 300 K, ∆VI

I=

×( )( )×

+−

−

1 38 10 300

1 60 101

23

19

.

.ln

J K K

C 00 0

25 9 1⎛

⎝⎜

⎞

⎠⎟ = ( ) +

⎛

⎝⎜

⎞

⎠⎟. ln mV

I

I

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

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

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

P43.46 The voltage across the diode is about 0.6 V. The voltage drop across the resistor is

0 025 150 3 75. .A V.( )( ) =Ω Thus, ε − − =0 6 3 8 0. . V V and ε = 4 4. .V

P43.47 First, we evaluate I0 in I I ee V k TB= −( )( )0 1∆ , given that I = 200 mA when ∆V = 100 mV and

T = 300 K.

e V

k TB

∆( )=

×( )( )×

−

−

1 60 10 0 100

1 38 10

19

23

. .

.

C V

JJ K K( )( )=

3003 86.

so II

e ee V k TB0 3 861

200

14 28=

−=

−=( )∆

mAmA. .

If ∆V = −100 mV, e V

k TB

∆( )= −3 86. ; and the current will be

I I e ee V k TB= −( ) = ( ) −( ) = −( ) −0

3 861 4 28 1 4 19∆ . .. mA mA


534 Chapter 43

P43.48 (a) The currents to be plotted are

I eDV= ( ) −( )−10 16 0 025A V∆ . ,

IV

W = −2 42

745

. V ∆Ω

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

I eD = ( ) −( )

=

−10 1

2 98

6 0 200A

mA

V 0.025 V.

.

IW = − =2 42 0 2002 98

. ..

V V

745mA.

Ω They agree to three digits.

∴ = =I ID W 2 98. mA

(b) ∆ ΩV

ID

=×

=−

0 200

2 98 1067 13

.

..

V

A

(c) d V

dI

dI

d Ve

D

D∆∆

( )=

( )⎡⎣⎢

⎤⎦⎥ =− −1 6

0 210 A

0.025 V. 000

1

8 39V 0.025 V⎡⎣⎢

⎤⎦⎥

=−

. Ω

Section 43.8 Superconductivity

P43.49 By Faraday’s law (from Chapter 32) ∆Φ∆

∆∆

∆∆

B

tL

I

tA

B

t= =

Thus, ∆∆

IA B

L= ( ) = ( ) ( )

× −

π 0 010 0 0 020 0

3 10 10

2

8

. .

.

m T

H== 203 A

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

*P43.50 (a) In the defi nition of resistance ∆V = IR, if R is zero then ∆V = 0 for any value of current.

(b) The graph shows a direct proportionality with resistance given by the reciprocal of the slope:

Slope

mA

mV= = = −( )

−( ) =1 155 57 8

3 61 1 356R

I

V

∆∆

.

. .443 1

0 023 2

1.

.

Ω

Ω

−

=R

(c) Expulsion of magnetic fl ux and therefore fewercurrent-carrying paths could explain the decrease in current.

FIG. P43.48

0

10

20

0 0.1 0.2 0.3

Diode and Wire Currents

∆V(volts)

DiodeWire

Cur

rent

(m

A)

FIG. P43.50



P43.51 (a) See the figure at right.

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

BN I

=µ0

or NI

B= =( ) ×( )

×( ) ⋅

−

−

µ π0

2

7

0 540 2 50 10

4 10

. . T m

T m AA kA= 10 7.

Additional Problems

P43.52 For the N2 molecule, k = 2 297 N m , m = × −2 32 10 26. kg, r = × −1 20 10 10. m, µ = m

2

ωµ

= = ×k4 45 1014. ,rad s I r= = ×( ) ×( ) = ×− −µ 2 26 10 2

1 16 10 1 20 10 1 67 1. . . kg m 00 46− ⋅ kg m2

For a rotational state suffi cient to allow a transition to the fi rst exited vibrational state,

2

21

IJ J +( ) = ω so J J

I+( ) = =×( ) ×( )

×

−

12 2 1 67 10 4 45 10

1 055 1

46 14ω

. .

. 001 41034− = .

Thus J = 37 .

*P43.53 (a) Since the interatomic potential is the same for both molecules, the spring constant is the same.

Then fk= 1

2π µ where µ12

12 16

12 166 86= ( )( )

+=u u

u uu. and µ14

14 16

14 167 47= ( )( )

+=u u

u uu..

Therefore,

fk k

f1414 12

12

1412

12

14

1

2

1

26= =

⎛

⎝⎜

⎞

⎠⎟ = =

π µ π µµµ

µµ

...

.42 106 86

6 15 1013 13×( ) = × Hz u

7.47 u Hz

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

I r r I14 142 14

1212

2 14

1212= =

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

µ µµ

µ µµ

II14467 47

1 46 10 1 5= ⎛⎝⎞⎠ × ⋅( ) =−.

. .u

6.86 ukg m2 99 10 46× ⋅− kg m2

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

∆Ehc

hfI

= = +⎛⎝⎜

⎞⎠⎟ + +( )

⎡

⎣⎢

⎤

⎦⎥− −

λ1

1

29 9 1

2014

2

14

11

210 10 1

214

2

14

⎛⎝⎜

⎞⎠⎟ + +( )

⎡

⎣⎢

⎤

⎦⎥hf

I

or ∆Ehc

hfI

= = +⎛⎝⎜

⎞⎠⎟ + +( )

⎡

⎣⎢

⎤

⎦⎥−λ

11

211 11 1

214

2

14

00

1

210 10 1

214

2

14

+⎛⎝⎜

⎞⎠⎟ + +( )

⎡

⎣⎢

⎤

⎦⎥hf

I

The wavelengths it can absorb are then

λπ

=− ( )

c

f I14 1410 2 or λ

π=

+ ( )c

f I14 1411 2

These are:

λ = ×× − × ⋅−

3 00 10

6 15 10 10 1 055 10

8

13 34

.

. .

m s

Hz J s(( )⎡⎣ ⎤⎦ × ⋅( )⎡⎣ ⎤⎦=

−2 1 59 104 96

46πµ

..

kg mm

2

and λ = ×× + × ⋅−

3 00 10

6 15 10 11 1 055 10

8

13 34

.

. .

m s

Hz J s(( )⎡⎣ ⎤⎦ × ⋅( )⎡⎣ ⎤⎦=

−2 1 59 104 79

46πµ

..

kg mm

2

FIG. P43.51


536 Chapter 43

P43.54 With 4 van der Waals bonds per atom pair or 2 electrons per atom, the total energy of the solid is

E = ×( ) ×⎛−2 1 74 106 02 1023

23

..

J atom atoms

4.00 g⎝⎝⎜

⎞

⎠⎟ = 5 23. J g

P43.55 ∆Emax .= = +⎛⎝⎜

⎞⎠⎟4 5

1

2 eV v ω so

4 5 1 6 10

1 055 10 8 2

19

34

. .

. .

eV J eV

J s

( ) ×( )× ⋅( )

−

− 88 10

1

214 1×( ) ≥ +⎛⎝

⎞⎠−s

v

8 25 7 5. .> v = 7

P43.56 Suppose it is a harmonic-oscillator potential well. Then, 1

24 48

3

23 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

1

24 48

1

20 520 4 48 4 74hf + = ( ) + =. . . . .eV eV eV eV

P43.57 The total potential energy is given by Equation 43.17: Uk e

r

B

re

mtotal = − +α2

.

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

point, dU

dr r r=

=0

0,

or dU

dr

d

dr

k e

r

B

r

k e

rr r

em

r r

e

= =

= − +⎛

⎝⎜

⎞

⎠⎟ = −

0 0

2 2

02

α α mmB

rm0

10+ =

Thus, Bk e

mre m= −α

2

01

Substituting this value of B into U total , Uk e

r

k e

mr

r

k e

re e m

me

0

2

0

2

01

0

2

0

11= − +

⎛

⎝⎜

⎞

⎠⎟ = −−α α α −−

⎛⎝⎜

⎞⎠⎟

1

m

*P43.58 (a) The total potential energy − +α k e

r

B

re

m

2

has its minimum value at

the equilibrium spacing, r r= 0 . At this point, FdU

dr r r

= − == 0

0, or

Fd

dr

k e

r

B

r

k e

r

mB

re

m

r r

em

= − − +⎛

⎝⎜

⎞

⎠⎟ = − +

=

α α2 2

02

00

++ =1

0

Thus, Bk e

mre m= −α

2

01.

Substituting this value of B into F, Fk e

r

m

r

k e

mr

k e

r

r

re

me m e= − + = − − ⎛⎝+

−α α α2

2 1

2

01

2

201 ⎞⎞⎠

⎡

⎣⎢

⎤

⎦⎥

−m 1

.

(b) Let r = r0 + x so r

0 = r – x Then assuming x is small we have

Fk e

r

r x

r

k e

r

xem

e= − − −⎛⎝

⎞⎠

⎡

⎣⎢

⎤

⎦⎥ = − − −

−

α α2

2

1 2

21 1 1rr

k e

rm

x

r

me⎛

⎝⎞⎠

⎡

⎣⎢

⎤

⎦⎥ − − + −⎡

⎣⎢⎤⎦⎥

−

≈

≈ −

1 2

2 1 1 1α ( )

αα k e

rm xe

2

03 1( )−

This is of the form of Hooke’s law with spring constant K = keae2(m – 1)/r

03.

(c) Section 38.5 on electron diffraction gives the interatomic spacing in NaCl as (0.562 737 nm)/2. Other problems in this chapter give the same information, or we could calculate it from the statement in the chapter text that the ionic cohesive energy for this crystal is –7.84 eV.

The stiffness constant is then

Kk e

rme= − = × ⋅ × −

α2

03

9

1 1 74768 99 10 1 6 10

( ) .. ( .N m2 119

10

8 1

2 81 10127

C)

C m)N/m

2

2 3

( )

( .

−×

=−

The vibration frequency of a sodium ion within the crystal is

fK

m= =

× ×=−

1

2

1

2

127

23 0 1 66 109 1827π π

N/m

kgT

. .. HHz



P43.59 (a) For equilibrium, dU

dx= 0 : d

dxAx Bx Ax Bx− − − −−( ) = − + =3 1 4 23 0

x →∞ describes one equilibrium position, but the stable equilibrium position is at3 0

2Ax B− = .

xA

B0

3 3 0 150

3 680 350= =

⋅( )⋅

=.

..

eV nm

eV nmnm

3

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

x

B

x

AB

A

BB

Ax x003

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 20

2

3

2 3 68

3 0 1= = − = −

⋅( )=

.

.

eV nm

5507 02

1 2 eV nm

eV3⋅( )

= − .

(c) FdU

dxAx Bxx = − = −− −3 4 2

To fi nd the maximum force, we determine fi nite xm such that dF

dx x xm=

= 0.

Thus, − +⎡⎣ ⎤⎦ =− −=

12 2 05 3Ax Bxx xm

so that xA

Bm =⎛⎝⎜

⎞⎠⎟

61 2

Then F AB

AB

B

A

B

Amax

.= ⎛⎝

⎞⎠ − ⎛

⎝⎞⎠ = − = − ⋅

36 6 12

3 682 2 eV nm(( )⋅( )

2

12 0 150. eV nm3

or Fmax ..= − ×⎛

⎝⎜

⎞

⎠⎟

−

7 521 60 10 119

eV nm J

1 eV

nm

10−−−⎛

⎝⎜

⎞⎠⎟ = − × = −

991 20 10 1 20

m N nN. .

P43.60 (a) For equilibrium, dU

dx= 0:

d

dxAx Bx Ax Bx− − − −−( ) = − + =3 1 4 23 0

x →∞ describes one equilibrium position, but the stable equilibrium position is at

3 02Ax B− = or x

A

B0

3=

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

x

B

x

AB

A

BB

A

Bx x0

03

0

3 2

3 2 3 2

1 2

1 2 1 20 3 32= = − = − = −

=

33

27A.

(c) FdU

dxAx Bxx = − = −− −3 4 2

To fi nd the maximum force, we determine fi nite xm such that

dF

dxAx Bxx

x xx x

m=

− −=

= − +⎡⎣ ⎤⎦ =12 2 05 3

0

then F AB

AB

B

A

B

Amax =⎛⎝⎜

⎞⎠⎟ −

⎛⎝⎜

⎞⎠⎟ = −3

6 6 12

2 2


538 Chapter 43

P43.61 (a) At equilibrium separation, r re= , dU

draB e e

r r

a r r a r r

e

e e

=

− −( ) − −( )= − −⎡⎣

⎤⎦ =2 1 00 0

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

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 rdU

drr r

d U

drr r

r r r r

( ) ≈ ( )+ −( )+ −(= =

0 0

2

2 0

0 0

1

2))

( ) ≈ + + −( ) − −− − − − −

2

20 01

22 0 0U r Ba ae ae er r r r( ) ( ) 22

0

2 20

20

0

1( )r r

r rr r Ba r r−

−−( )⎡⎣ ⎤⎦ −( ) ≈ −( )

This is of the form 1

2

1

22

0

2kx k r r= −( )

for a simple harmonic oscillator with k Ba= 2 2

Then the molecule vibrates with frequency fk a B a B= = =1

2 2

2

2π µ π µ π µ

(d) The zero-point energy is 1

2

1

2 8ω

π µ= =hf

ha B

Therefore, to dissociate the molecule in its ground state requires energy Bha B−π µ8

.



P43.62 T = 0 T TF= 0 1. T TF= 0 2. T TF= 0 5.

E

EF e E E T TF F( )−⎡⎣ ⎤⎦( )1 f E( ) e E E T TF F( )−⎡⎣ ⎤⎦( )1 f E( ) e E E T TF F( )−⎡⎣ ⎤⎦( )1 f E( ) e E E T TF F( )−⎡⎣ ⎤⎦( )1 f E( )0 e−∞ 1.00 e−10 0. 1.000 e−5 00. 0.993 e−2 00. 0.881

0.500 e−∞ 1.00 e−5 00. 0.993 e−2 50. 0.924 e−1 00. 0.731

0.600 e−∞ 1.00 e−4 00. 0.982 e−2 00. 0.881 e−0 800. 0.690

0.700 e−∞ 1.00 e−3 00. 0.953 e−1 50. 0.818 e−0 600. 0.646

0.800 e−∞ 1.00 e−2 00. 0.881 e−1 00. 0.731 e−0 400. 0.599

0.900 e−∞ 1.00 e−1 00. 0.731 e−0 500. 0.622 e−0 200. 0.550

1.00 e0 0.500 e0 0.500 e0 0.500 e0 0.500

1.10 e+∞ 0.00 e1 00. 0.269 e0 500. 0.378 e0 200. 0.450

1.20 e+∞ 0.00 e2 00. 0.119 e1 00. 0.269 e0 400. 0.401

1.30 e+∞ 0.00 e3 00. 0.047 4 e1 50. 0.182 e0 600. 0.354

1.40 e+∞ 0.00 e4 00. 0.018 0 e2 00. 0.119 e0 800. 0.310

1.50 e+∞ 0.00 e5 00. 0.006 69 e2 50. 0.075 9 e1 00. 0.269

FIG. P43.62

ANSWERS TO EVEN PROBLEMS

P43.2 4.3 eV

P43.4 (a) 1.28 eV (b) σ = 0 272. nm, ∈= 4 65. eV (c) 6.55 nN (d) 576 N m

P43.6 (a) 0.0148 eV (b) 83.8 mm

P43.8 12.2 × 10–27 kg; 12.4 × 10–27 kg; They agree, because the small apparent difference can be attributed to uncertainty in the data.


540 Chapter 43

P43.10 1 46 10 46. × ⋅− kg m2

P43.12 (a) 1 81 10 45. × ⋅− kg m2 (b) 1.62 cm

P43.14 (a) 12.1 pm (b) 9.23 pm (c) HI is more loosely bound since it has the smaller k value.

P43.16 (a) 0, 364 eV,µ 1.09 meV (b) 98.2 meV, 295 meV, 491 meV

P43.18 (a) 472 mµ (b) 473 mµ (c) 0 715. mµ

P43.20 (a) 4.60 × 10–48 kg · m2 (b) 1.32 ×1014 Hz (c) 0.074 1 nm

P43.22 6.25 × 109

P43.24 (a) ~1017 (b) ~105 m3

P43.26 (a) 0.444 nm, 0.628 nm, 0.769 nm

P43.28 (a) The Fermi energy is proportional to the spatial concentration of free electrons to the two-thirds power. (c) 6.06; copper by 3.33 times; it agrees with the equation because 6.062/3 = 3.33.

P43.30 0.938

P43.32 (a) 2.54 × 1028 electron/m3 (b) 3.15 eV

P43.34 2%

P43.36 3 40 1017. × electrons

P43.38 See the solution.

P43.40 All of the Balmer lines are absorbed, except for the red line at 656 nm, which is transmitted.

P43.42 1.91 eV

P43.44 −0.021 9 eV, 2.81 nm

P43.46 4.4 V

P43.48 (a) At ∆V = 0.200 V, ID = I

W = 2.98 mA, agreeing to three digits. (b) 67.1 Ω (c) 8.39 Ω

P43.50 (a) In the defi nition of resistance ∆V = IR, if R is zero then ∆V = 0 for any value of current. (b) The graph shows direct proportionality with resistance 0.023 2 Ω. (c) Expulsion of magnetic fl ux and therefore fewer current-carrying paths could explain the decrease in current.

P43.52 37

P43.54 5 23. J g

P43.56 4.74 eV

P43.58 (c) 9.18 THz

P43.60 (a) xA

B0

3= (b) −227

3B

A (c) − B

A

2

12

P43.62 See the solution.


sm chapter43

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