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43
Molecules and Solids
CHAPTER OUTLINE
43.1 Molecular Bonds
43.2 Energy States and Spectra of
Molecules
43.3 Bonding in S...
520 Chapter 43
Q43.4 From the rotational spectrum of a molecule, one can easily calculate the moment of inertia of
the mol...
Molecules and Solids 521
Q43.13 A molecule containing two atoms of 2
H, deuterium, has twice the mass of a molecule
contai...
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+ → + ...
Molecules and Solids 523
P43.5 At the boiling or condensation temperature, k TB ≈ = ×( )− − −
10 10 1 6 103 3 19
eV J.
T ≈...
524 Chapter 43
P43.9 For the HCl molecule in the J = 1 rotational energy level,
we are given r0 0 127 5= . nm.
E
I
J Jrot ...
Molecules and Solids 525
P43.13 The energy of a rotational transition is ∆E
I
J=
⎛
⎝⎜
⎞
⎠⎟
2
where J is the rotational qua...
526 Chapter 43
P43.16 (a) The reduced mass of the O2 is
µ =
( )( )
( )+( )
= = × −16 16
16 16
8 8 1 66 10 27u u
u u
u . kg...
Molecules and Solids 527
P43.18 We carry extra digits through the solution because part (c) involves the subtraction of tw...
528 Chapter 43
P43.20 We carry extra digits through the solution because the given wavelengths are close together.
(a) E h...
Molecules and Solids 529
Section 43.3 Bonding in Solids
P43.23 U
k e
r m
e
= − −
⎛
⎝
⎜
⎞
⎠
⎟ = −( ) ×( )α 2
0
9
1
1
1 747 ...
530 Chapter 43
*P43.28 (a) The Fermi energy is proportional to the spatial concentration of free electrons to the
two-thir...
Molecules and Solids 531
P43.34 The melting point of silver is 1 234 K. Its Fermi energy at 300 K is 5.48 eV. The approxim...
532 Chapter 43
P43.38 Consider first the wave function in x. At x = 0 and x L= , ψ = 0.
Therefore, sin k Lx = 0 and k Lx = ...
Molecules and Solids 533
P43.43 If the photon energy is 5.5 eV or higher, the diamond window will absorb. Here,
hf
hc
( ) ...
534 Chapter 43
P43.48 (a) The currents to be plotted are
I eD
V
= ( ) −( )−
10 16 0 025
A V∆ .
,
I
V
W =
−2 42
745
. V ∆
Ω...
Molecules and Solids 535
P43.51 (a) See the figure at right.
(b) For a surface current around the outside of the cylinder ...
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...
Molecules and Solids 537
P43.59 (a) For equilibrium,
dU
dx
= 0 :
d
dx
Ax Bx Ax Bx− − − −
−( )= − + =3 1 4 2
3 0
x → ∞ desc...
538 Chapter 43
P43.61 (a) At equilibrium separation, r re= ,
dU
dr
aB e e
r r
a r r a r r
e
e e
=
− −( ) − −( )
= − −⎡
⎣
⎤...
Molecules and Solids 539
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(...
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 p...
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Solucionario serway cap 43

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Solucionario serway cap 43

  1. 1. 43 Molecules and Solids CHAPTER OUTLINE 43.1 Molecular Bonds 43.2 Energy States and Spectra of Molecules 43.3 Bonding in Solids 43.4 Free-Electron Theory of Metals 43.5 Band Theory of Solids 43.6 Electrical Conduction in Metals, Insulators, and Semiconductors 43.7 Semiconductor Devices 43.8 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 of 1 10 3 meV 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 first, 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
  2. 2. 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 field 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 field 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 fixed. They conduct electricity by having drift motion in an applied electric field superposed on their random thermal motion. At higher temperature, the ion cores vibrate more and scatter more efficiently the conduction electrons flying 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 field. 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 sufficient 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 fixed 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.
  3. 3. 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 1 H. 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) F q r = ∈ = ×( ) ×( ) × − − 2 0 2 19 2 9 14 1 60 10 8 99 10 5 00 10π . . . 00 2 9 0 921 10 ( ) = × − N N. toward the other ion. (b) U q 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 dr Ar Br= − + =− − 12 6 013 7 yielding r A B 0 1 6 2 = ⎡ ⎣⎢ ⎤ ⎦⎥ (b) E U U A A B B A Br r r= − = − − ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = − − ⎡ ⎣⎢ ⎤ = =ϱ 0 0 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 11 7 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
  4. 4. 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 = ∈ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 4 12 6 13 7 s s s At r r= 0 we have dU dr = 0. Here s s r r0 13 0 7 1 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ s r0 1 6 2= − s s= ( ) = =− 2 0 305 0 2721 6 . .nm nm Then also U r r r r r 0 1 6 0 0 12 1 6 0 0 6 4 2 2 ( )= ∈ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ − − ⎢⎢ ⎤ ⎦ ⎥ ⎥ + = ∈ − ⎡ ⎣⎢ ⎤ ⎦⎥+ = − ∈ + ∈= − ( )= E E E E U r a a a a 4 1 4 1 2 0 11 28 3 37. .eV eV= 4.65 eV+ = ∈ (c) F r dU dr r r ( ) = − = ∈ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 4 12 6 13 7 s s s To find the maximum force we calculate dF dr r r = ∈ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 4 156 42 02 14 8 s s s when s rrupture = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 42 156 1 6 Fmax . . = ( ) ⎛ ⎝⎜ ⎞ ⎠⎟ − 4 4 65 0 272 12 42 156 6 42 13 6 eV nm 1156 41 0 41 0 1 6 10 7 6 1 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = − = − × − . . . eV nm 99 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 s r s r s 0 0 12 0 6 4+( )= ∈ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ s s ⎥⎥ + = ∈ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ − − E r r s r s a 4 2 21 6 0 0 12 1 6 0 6 ⎣⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ + Ea = ∈ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ − − 4 1 4 1 1 2 1 0 12 0 6 s r s r ++ = ∈ − + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟− − + E s r s r s r a 4 1 4 1 12 78 1 2 1 6 2 0 2 0 2 0 11 12 78 2 0 2 0 2 0 2 s r E s r s r a − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥+ = ∈ − ∈ + ∈ −− ∈ + ∈ − ∈ + + = − ∈ + + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟+ 2 12 42 0 0 2 0 2 0 s r s r E E s r a a 336 1 2 2 0 2 0 0 2 ∈ + +( ) ( )+≈ s r U r s U r ks where k r = ∈ = ( ) ( ) = = 72 72 4 65 0 305 3 599 0 2 2 . . eV nm eV nm2 5576 N m
  5. 5. Molecules and Solids 523 P43.5 At the boiling or condensation temperature, k TB ≈ = ×( )− − − 10 10 1 6 103 3 19 eV J. T ≈ × × − − 1 6 10 10 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 27 10 2 1 67 10 0 75 10 + = ×( ) ×⎛ ⎝⎜ ⎞ ⎠⎟ − − . . kg m 2 22 48 4 70 10= × ⋅− . kg m2 The rotational energy is E I J Jrot = +( ) 2 2 1 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 s 2 88 19 1 10 0 0148 kg m eV 1.6 J eV2 ⋅ × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =− ) . (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 1240 19 × = ⋅ − eeV m= 83 8. µ P43.7 µ = + = ( ) + × −m m m m 1 2 1 2 132 9 126 9 132 9 126 9 1 66 10 . . . . . 227 25 1 08 10kg kg( )= × − . I r= = ×( ) ×( ) = ×− − µ 2 25 9 2 1 08 10 0 127 10 1 74 1. . .kg m 00 45− ⋅kg m2 (a) E I I I J J I = = ( ) = +( )1 2 2 1 2 2 2 2 ω ω J = 0 gives E = 0 J = 1 gives E I = = × ⋅( ) × ⋅ − − 2 34 2 2 45 6 626 10 4 1 74 10 . . J s kg m2 π (( ) = × =− 6 41 10 40 024 . .J eVµ hf = × −− 6 41 10 024 . J gives f = ×9 66 109 . Hz (b) f E h hI h r r= = = ∝ −1 2 2 2 2 4π µ If is 10% too small, is 20% too large.r f *P43.8 ∆E h k hf kibv = = = 2 4 π µ so ππ µ2 2 f µ π π = = × = × −k f4 1530 4 56 3 10 1 22 102 2 2 12 2 N/m s)( . / . 226 kg µ = + = m m m m 1 2 1 2 14 007 14 007 . . u 15.999 u u +15.9999 u kg u kg 1 66 10 1 1 24 10 27 26. . × = × − − The reduced masses agree, because the small apparent difference can be attributed to uncertainty in the data.
  6. 6. 524 Chapter 43 P43.9 For the HCl molecule in the J = 1 rotational energy level, we are given r0 0 127 5= . nm. E I J Jrot = +( ) 2 2 1 Taking J = 1, we have E I Irot = = 2 21 2 ω or ω = = 2 2 2 2 I I The moment of inertia of the molecule is given by Equation 43.3. I r m m m m r= = + ⎛ ⎝⎜ ⎞ ⎠⎟µ 0 2 1 2 1 2 0 2 I r= ( )( ) + ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = 1 01 35 5 1 01 35 5 00 2. . . . . u u u u 9982 1 66 10 1 275 10 227 10 2 u kg u m( ) ×( ) ×( ) =− − . . .665 10 47 × ⋅− kg m2 Therefore, ω π = = × ⋅ × ⋅ − − 2 2 6 626 10 10 34 47 I J s 2 (2.65 kg m2 . ) == ×5 63 1012 . rad s P43.10 hf E I I I = = +( )[ ]− +( )[ ]= ( )∆ 2 2 2 2 2 2 1 2 1 1 1 2 4 I h hf h f = ( ) = = × ⋅ × − 4 2 2 2 6 626 10 2 30 1 2 2 34 π π π . . J s 2 2 00 1 46 1011 46 Hz kg m2 ( ) = × ⋅− . P43.11 I m r m r= +1 1 2 2 2 2 where m r m r1 1 2 2= and r r r1 2+ = Then r m r m 1 2 2 1 = so m r m r r2 2 1 2+ = and r m r m m 2 1 1 2 = + Also, r m r m 2 1 1 2 = . Thus, r m r m r1 1 1 2 + = and r m r m m 1 2 1 2 = + I m m r m m m m r m m m m r m = +( ) + +( ) = + 1 2 2 2 1 2 2 2 1 2 2 1 2 2 1 2 2 2 mm m m m m r m m r1 1 2 2 1 2 2 1 2 2( ) +( ) = + = µ P43.12 (a) µ = ( ) +( ) ×( )=−22 99 35 45 22 99 35 45 1 66 10 27. . . . . kg 22 32 10 26 . × − kg I r= = ×( ) ×( ) = ×− − µ 2 26 9 2 2 32 10 0 280 10 1 81 1. . .kg m 00 45− kg m2 и (b) hc I I I I I h λ = +( )− +( ) = − = = 2 2 2 2 2 2 2 2 2 1 2 1 1 1 3 2 2 44 2 π 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
  7. 7. Molecules and Solids 525 P43.13 The energy of a rotational transition is ∆E I 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, ∆E hc = = × ⋅( ) ×( )− λ λ 6 626 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 ∆EЈs consistently increase by 0.002 56 eV. E I 1 2 0 002 56= = . eV and I E = = × ⋅( ) ( ) −2 1 34 2 1 055 10 0 002 56 1. . J s eV eV 1.660 J kg m2 × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = × ⋅− − 10 2 72 1019 47 . For the HCl molecule, the internuclear radius is r I = = × × = − − µ 2 72 10 1 62 10 0 130 47 27 . . .m nm *P43.14 (a) Minimum amplitude of vibration of HI is characterized by 1 2 1 2 4 2 2 kA h k A h = = =ω π µ π so 11 1 4 kµ ⎛ ⎝⎜ ⎞ ⎠⎟ / A = × ⋅− 6 626 10 2 1 320 34 . ( J s N/m)(127/128)(1.66π ×× ⎛ ⎝⎜ ⎞ ⎠⎟ =− 10 12 127 1 4 kg) pm / . (b) For HF, A = × ⋅ × − 6 626 10 2 1 970 1 34 . ( J s N/m)(19/20)(1.66π 00 9 2327 1 4 − ⎛ ⎝⎜ ⎞ ⎠⎟ = kg) pm / . (c) Since HI has the smaller k, it is more weakly bound. P43.15 µ = + = × × = ×− −m m m m 1 2 1 2 27 2735 36 1 66 10 1 61 10. .kg kgg ∆E k vib = = ×( ) × = ×− − µ 1 055 10 480 1 61 10 5 74 134 27 . . . 00 0 35820− =J eV.
  8. 8. 526 Chapter 43 P43.16 (a) The reduced mass of the O2 is µ = ( )( ) ( )+( ) = = × −16 16 16 16 8 8 1 66 10 27u u u u u . kg kg( )= × − 1 33 10 26 . The moment of inertia is then I r= = ×( ) ×( ) = × − − − µ 2 26 10 2 1 33 10 1 20 10 1 91 10 . . . kg m 446 kg m2 ⋅ The rotational energies are E I J Jrot J s = +( )= × ⋅( ) × − − 2 34 2 2 1 1 055 10 2 1 91 10 . . 446 1 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) E k vib J= + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ × ⋅− v v 1 2 1 2 1 055 10 34 µ . ss N m kg ( ) ×( )− 1177 8 1 66 10 27 . Evib J eV 1.60 = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ×( ) × − − v 1 2 3 14 10 1 10 20 19 . J eV ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ( )v 1 2 0 196. 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 I J Jrot J s = +( ) = × ⋅( ) × − − 2 34 2 2 1 1 055 10 2 1 89 10 . . 445 24 1 2 95 10 1 18 4 kg m J2 ⋅( ) +( ) = ×( ) +( ) = − J J J J. . ××( ) +( )− 10 16 eV J J E J J Jrot eV where= ( ) +( ) =18 4 1 0 1 2 3. , , , , ...µ The first five of these allowed energies are: Erot eV, 111 eV, 221 eV, and 369= 0 36 9, . µ µ µ µµeV.
  9. 9. Molecules and Solids 527 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 2 4 If m is the reduced mass, then I r= = ×( ) = ×(− − µ µ2 9 2 20 0 127 46 10 1 624 605 10. .m m2 )) = ×( )− µ λ π µ Therefore m2 4 1 624 605 10 2 992 20 . . 77 925 10 6 626 075 10 2 901830 1 8 34 ×( ) × ⋅ = ×− m s J s. . 0023 m kg( )µ (1) (a) µ35 1 007 825 34 968853 1 007 825 34 968 = ( )( ) + . . . . u u u 8853 0 979 593 1 626 653 10 27 u u kg= = × − . . From (1): λ35 23 27 2 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 9903 0 981077 1 629118 10 27 u u kg= = × − . . From (1): λ37 23 27 2 901830 10 1 629118 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 find an average spacing between peaks by counting 22 gaps between 7 96 1013 . × Hz and 9 24 1013 . × Hz: ∆f h h = −( ) = × = 9 24 7 96 10 0 058 2 10 1 13 13 2 . . . Hz 22 Hz 44 4 6 63 10 4 5 82 1 2 2 34 2 π π π I I h f ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = = × ⋅ × − ∆ . . J s 00 2 9 1011 47 s kg m2 = × ⋅− .
  10. 10. 528 Chapter 43 P43.20 We carry extra digits through the solution because the given wavelengths are close together. (a) E hf I J JJv v= + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + +( ) 1 2 2 1 2 ∴ = = + = + ∴ − E hf E hf I E hf I E E 00 11 2 02 2 11 0 1 2 3 2 1 2 3 , , 00 2 34 6 626 075 10 2 997 925 = + = = × ⋅( ) ×− hf I hc λ . .J s 110 2 2112 10 8 6 m s m ( ) × − . ∴ + = × − hf I 2 20 8 983 573 10. J (1) E E hf I hc 11 02 2 34 2 6 626 075 10 2 9 − = − = = × ⋅( )− λ . .J s 997 925 10 2 405 4 10 8 6 ×( ) × − m s m. ∴ − = × − hf I 2 8 258 284 10 2 20 . J (2) Subtract (2) from (1): 3 7 252 89 10 2 21 I = × − . J ∴ = × ⋅( ) × = − − I 3 1 054 573 10 7 252 89 10 4 34 2 21 . . . J s J 660 10 48 × ⋅− kg m2 (b) From (1): f = × × − − − 8 983 573 10 10 1 054 520 34 . .J 6.626 075 J sи 773 10 4 600 060 10 6 626 34 2 48 ×( ) ×( ) − − J s kg m2 и и. . 0075 10 1 32 10 34 14 ×( ) = × − J s Hz и . (c) I r= µ 2 , where m is the reduced mass: µ = = ( )= × −1 2 1 2 1 007 825 8 367 669 10 28 mH . .u kg So r I = = × ⋅ × = − − µ 4 600 060 10 8 367 669 10 48 2 28 . . kg m kg 00.0741 nm . P43.21 The emission energies are the same as the absorption energies, but the final state must be below v = =( )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 EJ 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 5 2 00 102 2 15 2 . .m The moment of inertia about a perpendicular axis is I m R m R m y = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ×( )− 2 2 2 2 00 10 2 2 10 2 . .m The allowed rotational energies are E I J Jrot = ⎛ ⎝⎜ ⎞ ⎠⎟ +( ) 2 2 1 , so the energy of the first excited state is E I 1 2 = . The ratio is therefore E E I I I I mx y x y y x 1 1 2 2 10 1 2 2 00 10, , . = ( ) ( ) = = ( ) × − mm m ( ) ( ) ×( ) = ( ) = × − 2 15 2 5 2 9 4 5 2 00 10 5 8 10 6 25 10 m . .
  11. 11. Molecules and Solids 529 Section 43.3 Bonding in Solids P43.23 U k e r m e = − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −( ) ×( )α 2 0 9 1 1 1 747 6 8 99 10 1 6 . . . 00 10 0 281 10 1 1 8 1 25 10 19 2 9 ×( ) ×( ) − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − × − − − . . 118 7 84J eV= − . P43.24 Consider a cubical salt crystal of edge length 0.1 mm. The number of atoms is 10 10 4 9 3 17 − − × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ m 0.261 10 m ~ This number of salt crystals would have volume 10 10 10 104 3 4 9 3 5− − −( ) × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟m m 0.261 m m3 ~ If it is cubic, it has edge length 40 m. P43.25 U k e r k e r k e r k e r k e r k e r e 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 4 2 1 1 2 1 3 1 4 + − = − − + − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ But, ln 1 1 2 3 4 2 3 4 +( )= − + − +x x x x so, U k e r e = − 2 2 2 ln , or U k e r e= − =α α 2 2 2where 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 7692 2 +( ) ( )=. .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 E h m ne F = ⎛ ⎝⎜ ⎞ ⎠⎟ 2 2 3 2 3 8π or n mE h e = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ×( )− 8 3 2 8 3 2 9 11 10 5 4 2 3 2 31 π πF kg. . 88 1 60 10 6 626 10 19 3 2 34 3 ( ) ×( )⎡ ⎣ ⎤ ⎦ × ⋅( ) = − − . . J J s 55 80 1028 3 . × − m The number-density of silver atoms is nAg 3 kg m atom 108 u u 1.66 = ×( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟10 6 10 1 13 . ×× ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ×− − 10 kg m27 28 3 5 91 10. So an average atom contributes 5 80 5 91 0 981 . . .= electron to the conduction baand . FIG. P43.25 – ++ – + – 3r 2r r –e –e –ee e e FIG. P43.26
  12. 12. 530 Chapter 43 *P43.28 (a) The Fermi energy is proportional to the spatial concentration of free electrons to the two-thirds power. (b) E h m ne F 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 3 2 ×( ) ×( ) ⎛ ⎝⎜ ⎞ ⎠⎟− − kg J eV. π ne 33 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 2 7 052 mv = . eV v = ( ) ×( ) × = − − 2 7 05 1 60 10 9 11 10 1 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 40 eV. *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 1 0 9382 72 e . . *P43.31 (a) E Ea Fv = = 3 5 0 6 7 05. ( . eV) = 4.23 eV (b) The average energy of a molecule in an ideal gas is 3 2 k TB so we have T = × × =− − 2 3 4 23 1 6 3 . . . 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) n N M e = = ×( )( )A 3 electrons mol g cmρ 6 02 10 0 971 2 23 . . 33 0 2 54 10 2 54 1022 28 . . . g mol electrons cm e3 ne = × = × llectrons m3 (b) E h m ne F 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 eV 2 94 1 38 10 80023 19 . . −− =0 203 5 28. .eV or eVE
  13. 13. Molecules and Solids 531 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 E B F J K K eV = ×( )( ) ( ) × − 1 38 10 1234 5 48 1 60 1 23 . . . 00 219− ( ) ≈ J eV % P43.35 E n EN E dE e 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 CE m h Ee ( ) = =1 2 3 2 3 1 28 2π E n CE dE C n E dE C n E e E e E e av F 5 2 F F = = =∫ ∫ 1 2 5 3 2 0 3 2 0 But from Equation 43.25, C n E e = −3 2 3 2 F , so that E E E Eav F F 5 2 F= ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =−2 5 3 2 3 5 3 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 CE m h E( ) = =1 2 3 2 3 1 28 2π or g E( ) = ×( ) × ⋅( ) − − 8 2 9 11 10 6 626 10 31 3 2 34 3 π . . kg J s 44 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 00 1 1 1 2 1 2 . . . . . eV eV ( ) ( ) = ( ) ( ) = 00 (b) From Equation 43.22, the number of occupied states having energy E is N E CE 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 B e 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 00 1 2 1 2 1 . . . . .eV eV ( ) ( ) = ( ) ( ) .. .50 0 025 9 1( ) + ⎡ ⎣⎢ ⎤ ⎦⎥ And N N 8 50 7 00 1 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.
  14. 14. 532 Chapter 43 P43.38 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 ψ π π π = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ A n x L n y L n z L x y z sin sin sin ⎠⎠ ⎟ From ∂ ∂ ∂ ∂ ∂ ∂ 2 2 2 2 2 2 2 2ψ ψ ψ ψ x y z m U Ee + + = −( ) , we have inside the box, where U = 0, − − − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −( ) n L n L n L m Ex y z e 2 2 2 2 2 2 2 2 2 2 2π π π ψ ψ E m L n n n n n n e x y z x y z= + +( ) = 2 2 2 2 2 2 2 1 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 E m 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 10 1 09 10 1 8 14 6. . . . m s Hz m 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 10 51419 × =− 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 E hc ≥ = × ⋅( ) ×( )− λmax . . . 6 626 10 3 00 10 1 00 34 8 J s m s ×× × ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ =− − 10 1 1 246 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 E hc g = = × ⋅( ) ×( ) × − λ 6 626 10 3 00 10 650 10 34 8 . .J s m s −− ≈9 1 91 m J eV.
  15. 15. Molecules and Solids 533 P43.43 If the photon energy is 5.5 eV or higher, the diamond window will absorb. Here, hf hc ( ) = =max min . 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 first Bohr orbit, a m 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 a m m e e 0 0 220 11 7 0 052 9 2 81= ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ( )= . . . .nm nm The energy levels are in hydrogen E k e a n n e = − 2 0 2 2 1 and here ′ = − ′ = − ( ) = − ⎛ ⎝⎜ ⎞ E k e a n k e m m a m m n e e e e 2 2 2 02 1 2κ κ κ* * ⎠⎠⎟ En κ 2 For n = 1, ′ = − = −E1 0 220 13 6 0 021 9. . . . eV 11.7 eV2 Section 43.7 Semiconductor Devices P43.45 I I e e V k TB = −( )( ) 0 1 ∆ Thus, e I I e V k TB∆( ) = +1 0 and ∆V k T e I I B = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ln 1 0 At T = 300 K, ∆V I I = ×( )( ) × + − − 1 38 10 300 1 60 10 1 23 19 . . ln J K K C 00 0 25 9 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = ( ) + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟. lnmV 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( )( ) = 300 3 86. so I I e ee V k TB0 3 86 1 200 1 4 28= − = − =( )∆ mA mA. . If ∆V = −100 mV, e V k TB ∆( ) = −3 86. ; and the current will be I I e e e V k TB = −( )= ( ) −( )= −( ) − 0 3 86 1 4 28 1 4 19 ∆ . .. mA mA
  16. 16. 534 Chapter 43 P43.48 (a) The currents to be plotted are I eD V = ( ) −( )− 10 16 0 025 A V∆ . , I V 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 0 200 A mA V 0.025 V. . IW = − = 2 42 0 200 2 98 . . . V V 745 mA. Ω They agree to three digits. ∴ = =I ID W 2 98. mA (b) ∆ Ω V ID = × =− 0 200 2 98 10 67 13 . . . V A (c) d V dI dI d V e 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 t L I t A B t = = Thus, ∆ ∆ I A 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–field through the ring. *P43.50 (a) In the definition 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 flux and therefore fewer current-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) Diode Wire Current(mA) FIG. P43.50
  17. 17. Molecules and Solids 535 P43.51 (a) See the figure at right. (b) For a surface current around the outside of the cylinder as shown, B N 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 ω µ = = × k 4 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 sufficient to allow a transition to the first exited vibrational state, 2 2 1 I J J +( ) = ω so J J I +( ) = = ×( ) ×( ) × − 1 2 2 1 67 10 4 45 10 1 055 1 46 14 ω . . . 00 1 41034− = . Thus J = 37 . *P43.53 (a) Since the interatomic potential is the same for both molecules, the spring constant is the same. Then f k = 1 2π µ where µ12 12 16 12 16 6 86= ( )( ) + = u u u u u. and µ14 14 16 14 16 7 47= ( )( ) + = u u u u u.. Therefore, f k k f14 14 12 12 14 12 12 14 1 2 1 2 6= = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = = π µ π µ µ µ µ µ .. . .42 10 6 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 14 2 14 12 12 2 14 12 12= = ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞ ⎠⎟µ µ µ µ µ µ II14 467 47 1 46 10 1 5= ⎛ ⎝ ⎞ ⎠ × ⋅( )=−. . . u 6.86 u kg 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 ∆E hc hf I = = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + +( ) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥− − λ 1 1 2 9 9 1 2 014 2 14 11 2 10 10 1 2 14 2 14 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + +( ) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥hf I or ∆E hc hf I = = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + +( ) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥− λ 1 1 2 11 11 1 2 14 2 14 00 1 2 10 10 1 2 14 2 14 + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + +( ) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥hf I The wavelengths it can absorb are then λ π = − ( ) 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 10 4 9646 π µ . . kg m m2 and λ = × × + × ⋅− 3 00 10 6 15 10 11 1 055 10 8 13 34 . . . m s Hz J s(( )⎡⎣ ⎤⎦ × ⋅( )⎡⎣ ⎤⎦ =− 2 1 59 10 4 7946 π µ . . kg m m2 FIG. P43.51
  18. 18. 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 10 6 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 2 4 48 3 2 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 1 2 4 48 1 2 0 520 4 48 4 74hf + = ( )+ =. . . . .eV eV eV eV P43.57 The total potential energy is given by Equation 43.17: U k e r B r e mtotal = − +α 2 . 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 e m r r e = = = − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − 0 0 2 2 0 2 α α mmB rm 0 1 0+ = Thus, B k e m re m = − α 2 0 1 Substituting this value of B into Utotal , U k e r k e m r r k e r e e m m e 0 2 0 2 0 1 0 2 0 1 1= − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −− α α α −− ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 m *P43.58 (a) The total potential energy − +α k e r B r e m 2 has its minimum value at the equilibrium spacing, r r= 0 . At this point, F dU dr r r = − = = 0 0, or F d dr k e r B r k e r mB r e m r r e m = − − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − + = α α 2 2 0 2 0 0 ++ =1 0 Thus, B k e m re m = − α 2 0 1 . Substituting this value of B into F, F k e r m r k e m r k e r r r e m e m e = − + = − − ⎛ ⎝+ − α α α 2 2 1 2 0 1 2 2 0 1 ⎞⎞ ⎠ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ −m 1 . (b) Let r = r0 + x so r0 = r – x Then assuming x is small we have F k e r r x r k e r xe m e = − − −⎛ ⎝ ⎞ ⎠ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = − − − − α α 2 2 1 2 2 1 1 1 rr k e r m x r m e⎛ ⎝ ⎞ ⎠ ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ − − + −⎡ ⎣⎢ ⎤ ⎦⎥ − ≈ ≈ − 1 2 2 1 1 1α ( ) αα k e r m xe 2 0 3 1( )− This is of the form of Hooke’s law with spring constant K = ke ae2 (m – 1)/r0 3 . (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 K k e r me = − = × ⋅ × − α 2 0 3 9 1 1 7476 8 99 10 1 6 10 ( ) . . ( .N m2 119 10 8 1 2 81 10 127 C) C m) N/m 2 2 3 ( ) ( . − × =− The vibration frequency of a sodium ion within the crystal is f K m = = × × =− 1 2 1 2 127 23 0 1 66 10 9 1827 π π N/m kg T . . . HHz
  19. 19. Molecules and Solids 537 P43.59 (a) For equilibrium, dU dx = 0 : d dx Ax Bx Ax Bx− − − − −( )= − + =3 1 4 2 3 0 x → ∞ describes one equilibrium position, but the stable equilibrium position is at 3 0 2 Ax B− = . x A B 0 3 3 0 150 3 68 0 350= = ⋅( ) ⋅ = . . . eV nm eV nm nm 3 (b) The depth of the well is given by U U A x B x AB A BB Ax x0 0 3 0 3 2 3 2 3 2 1 2 1 2 1 20 3 3 = = − = −= U U B Ax x0 3 2 3 2 1 2 3 2 3 20 2 3 2 3 68 3 0 1 = = − = − ⋅( ) = . . eV nm 550 7 021 2 eV nm eV 3 ⋅( ) = − . (c) F dU dx Ax Bxx = − = −− − 3 4 2 To find the maximum force, we determine finite xm such that dF dx x xm= = 0. Thus, − +⎡⎣ ⎤⎦ =− − = 12 2 05 3 Ax Bx x xm so that x A B m = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 6 1 2 Then F A B A B B A B A max . = ⎛ ⎝ ⎞ ⎠ − ⎛ ⎝ ⎞ ⎠ = − = − ⋅ 3 6 6 12 3 68 2 2 eV nm(( ) ⋅( ) 2 12 0 150. eV nm3 or Fmax . . = − ×⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − 7 52 1 60 10 119 eV nm J 1 eV nm 10−− −⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = − × = −9 9 1 20 10 1 20 m N nN. . P43.60 (a) For equilibrium, dU dx = 0: d dx Ax Bx Ax Bx− − − − −( )= − + =3 1 4 2 3 0 x → ∞ describes one equilibrium position, but the stable equilibrium position is at 3 0 2 Ax B− = or x A B 0 3 = (b) The depth of the well is given by U U A x B x AB A BB A B x x0 0 3 0 3 2 3 2 3 2 1 2 1 2 1 20 3 3 2= = − = − = −= 33 27A . (c) F dU dx Ax Bxx = − = −− − 3 4 2 To find the maximum force, we determine finite xm such that dF dx Ax Bxx x x x x m= − − = = − +⎡⎣ ⎤⎦ =12 2 05 3 0 then F A B A B B A B A max = ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −3 6 6 12 2 2
  20. 20. 538 Chapter 43 P43.61 (a) At equilibrium separation, r re= , dU dr aB 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 r dU dr r r d U dr r r r r r r ( ) ≈ ( )+ −( )+ −( = = 0 0 2 2 0 0 0 1 2 )) ( ) ≈ + + −( ) − −− − − − − 2 2 0 0 1 2 2 0 0 U r Ba ae ae er r r r( ) ( ) 22 0 2 2 0 20 0 1( )r r r r r r Ba r r− − −( )⎡ ⎣ ⎤ ⎦ −( ) ≈ −( ) This is of the form 1 2 1 2 2 0 2 kx k r r= −( ) for a simple harmonic oscillator with k Ba= 2 2 Then the molecule vibrates with frequency f k a B a B = = = 1 2 2 2 2π µ π µ π µ (d) The zero-point energy is 1 2 1 2 8 ω π µ = =hf ha B Therefore, to dissociate the molecule in its ground state requires energy B ha B − π µ8 .
  21. 21. Molecules and Solids 539 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.
  22. 22. 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 = IW = 2.98 mA, agreeing to three digits. (b) 67.1 Ω (c) 8.39 Ω P43.50 (a) In the definition 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 flux 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) x A B 0 3 = (b) −2 27 3 B A (c) − B A 2 12 P43.62 See the solution.

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