ISM chapter43 tủ tài liệu training

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

* An asterisk indicates a question or problem new to this edition.

ANSWERS TO OBJECTIVE QUESTIONS
OQ43.1

(a) False. An infinite current would produce an infinite magnetic
field that would penetrate the surface of the superconductor and
destroy the superconducting properties. (b) False. There is no
physical requirement that a superconductor carry a current. (c) True.
(d) True. (e) True. Collisions do not occur between Cooper pairs and
the lattice ions.

OQ43.2

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

OQ43.3

(i)

Answer (c). Think of aluminum foil.

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

1054
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Chapter 43
OQ43.4

(i)

1055

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.
OQ43.5

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.

OQ43.6

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

OQ43.7

The ranking is then b > d > c > a. 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 ~1011
Hz, accompany excitation and de-excitation of rotational motion,
while infrared photons, of frequency ~1013 Hz, accompany changes in
the vibration state of typical simple molecules.

ANSWERS TO CONCEPTUAL QUESTIONS
CQ43.1

A material can absorb a photon of energy greater than the energy
gap, as an electron jumps into a higher energy state; therefore, silicon
can absorb visible light, thus appearing opaque. 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; therefore,
diamond cannot absorb visible light, thus appearing transparent.

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1056

Molecules and Solids

CQ43.2

Rotational, vibrational, and electronic (as discussed in Chapter 42)
are the three major forms of excitation. Rotational energy for a
2
diatomic molecule is on the order of
, where I is the moment of
2I
inertia of the molecule. A typical value for a small molecule is on the
order of 1 meV = 10–3 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.

CQ43.3

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.

CQ43.4

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.

CQ43.5

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.

CQ43.6

(a)

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.

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Chapter 43

CQ43.7

CQ43.8

1057

(b)

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.

(a)

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.

(b)

The energy band model is more comprehensive than the freeelectron 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.

A molecule containing two atoms of D = 2H, deuterium, has twice the
mass of a molecule containing two atoms of ordinary hydrogen 1H;
therefore the deuterium molecule has twice the reduced mass of the
hydrogen molecule. The atoms have the same electronic structure, so
the molecules have the same interatomic spacing, and the same
spring constant. Therefore, each vibrational energy level for D2 is
1 2 times that of H2. The moment of inertia of deuterium is twice
as large and the rotational energies one-half as large as for the
ordinary hydrogen molecule.

CQ43.9

Ionic bonds are ones between oppositely charged ions. One atom
essentially steals an electron from another; for example, in table salt,
NaCl, the chlorine atom takes the outer 3s electron from the sodium
atom, resulting in two ions Cl– and Na+. 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 electric
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 electromagnet
attracting a paper clip.

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1058

Molecules and Solids
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.

CQ43.10

The atoms of crystalline substances form a regular array of ions in a
lattice structure, and the atoms are close enough together to allow
energy bands to form. The atoms of amorphous solids do not form a
regular array, but they are close enough to produce energy bands.

The atoms of gases do not form regular arrays and are too far apart
to form energy bands.

SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 43.1
P43.1

Molecular Bonds

At the boiling or condensation temperature,
E=

3
kBT ≈ 10−3 eV = 10−3 ( 1.6 × 10−19 J )
2

Solving for the temperature T gives,

2 ( 1.6 × 10−22 J )
E
T=
≈
~ 10 K
kB 3 ( 1.38 × 10−23 J K )
P43.2

(a)

The electrostatic force is
F=

q2
4π ∈0 r 2

( 8.99 × 10
=

9

N ⋅ m 2 /C2 ) ( 1.60 × 10−19 C )

( 5.00 × 10

−10

m)

2

2

= 9.21 × 10−10 N = 921 × 10−12 N

or 921 pN toward the other ion.
(b)

The potential energy of the ion pair is

U=

−q 2
4π ∈0 r

⎡ ( 8.99 × 109 N ⋅ m 2 /C2 ) ( 1.60 × 10−19 C )2 ⎤ ⎛
⎞
1 eV
⎥⎜
= −⎢
−10
−19 ⎟
5.00 × 10 m
⎢⎣
⎥⎦ ⎝ 1.60 × 10 J ⎠
= −2.88 eV

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Chapter 43
P43.3

1059

We are told that

K + Cl + 0.70 eV → K + + Cl −
and

Cl + e− → Cl − + 3.6 eV

or

Cl − → Cl + e− − 3.6 eV

By substitution,

K + Cl + 0.7 eV → K + + Cl + e− − 3.6 eV
K + 4.3 eV → K + + e−
or the ionization energy of potassium is 4.3 eV .
P43.4

(a)

Because the ionization energy of K is 4.34 eV, we have the relation

K + 4.34 eV → K + + e−

[1]

and because the electron affinity of I is 3.06 eV, we have the
relation

I + e− → I − + 3.06 eV
I − 3.06 eV → I − − e−

[2]

Adding equations [1] and [2] gives

( K + 4.34 eV ) + ( I − 3.06 eV ) → ( K + + e− ) + ( I − − e)

K + I + ( 4.34 eV − 3.06 eV ) → K + + I −
K + I + 1.28 eV → K + + I −

Therefore, the activation energy is Ea = 1.28 eV .
(b)

We differentiate the given function:
13
7
dU 4 ∈ ⎡
⎛σ ⎞
⎛σ ⎞ ⎤
=
−12
+
6
⎜⎝ ⎟⎠
⎜⎝ ⎟⎠ ⎥
⎢
dr
σ ⎣
r
r ⎦

Setting the expression above equal to 0, at r = r0 we have

dU
=0 →
dr

⎛σ ⎞
⎜⎝ r ⎟⎠
0

13

1⎛σ ⎞
= ⎜ ⎟
2 ⎝ r0 ⎠

7

which gives
6

⎛σ ⎞
−1
→ σ = 2 −1 6 r0 = 2 −1 6 ( 0.305 ) nm
⎜⎝ r ⎟⎠ = 2
0

or

σ = 0.272 nm

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1060

Molecules and Solids
Then also

⎡⎛ 2 −1 6 r ⎞ 12 ⎛ 2 −1 6 r ⎞ 6 ⎤
0
0
U ( r0 ) = 4 ∈ ⎢⎜
⎟⎠ − ⎜⎝ r ⎟⎠ ⎥ + Ea
r
⎝
⎢⎣
⎥⎦
0
0
1 1
= 4 ∈ ⎡⎢ − ⎤⎥ + Ea = − ∈ + Ea
⎣4 2⎦
solving for ∈ gives

∈= Ea − U ( r0 ) = 1.28 eV + 3.37 eV
= 4.65 eV
(c)

The force of attraction between the atoms is
13
7
dU 4 ∈ ⎡ ⎛ σ ⎞
⎛σ ⎞ ⎤
F (r ) = −
=

⎢12 ⎜ ⎟ − 6 ⎜⎝ ⎟⎠ ⎥
dr
σ ⎣ ⎝ r⎠
r ⎦

To find the maximum force we calculate
14
8
dF 4 ∈ ⎡
⎛σ ⎞
⎛σ ⎞ ⎤
=
⎢ −156 ⎜⎝ ⎟⎠ + 42 ⎜⎝ ⎟⎠ ⎥ = 0
dr σ 2 ⎣
r
r ⎦

σ
rbreak

⎛ 42 ⎞
=⎜
⎝ 156 ⎟⎠

16

So at r = rbreak, the force is a maximum:
Fmax

4 ( 4.65 eV ) ⎡ ⎛ 42 ⎞

=
⎢12 ⎜
⎟
0.272 nm ⎣ ⎝ 156 ⎠
=

13 6

⎛ 42 ⎞
− 6⎜
⎝ 156 ⎟⎠

76

⎤
⎥
⎦

−41.0 eV ⎛ 1.60 × 10−19 N ⋅ m ⎞ ⎛ 1 nm ⎞
⎟⎠ ⎜⎝ 10−9 m ⎟⎠ = − 6.55 nN
nm ⎜⎝
1 eV

Therefore the applied force required to break the molecule is
+ 6.55 nN away from the center.
(d) To calculate the force constant, we expand U(r) as suggested in
the problem statement:

⎡⎛ σ ⎞ 12 ⎛ σ ⎞ 6 ⎤
U ( r0 + s ) = 4 ∈ ⎢⎜

⎟ − ⎜⎝ r + s ⎟⎠ ⎥ + Ea
⎢⎣⎝ r0 + s ⎠
⎥⎦
0
⎡⎛ 2 −1 6 r ⎞ 12 ⎛ 2 −1 6 ⎞ 6 ⎤
0
= 4 ∈ ⎢⎜
⎟⎠ − ⎜⎝ r + s ⎟⎠ ⎥ + Ea
r
+
s
⎝
⎢⎣ 0
⎥⎦
0

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Chapter 43

1061

Expanding,
−12
−6
⎡1⎛
s⎞
1⎛
s⎞ ⎤

U ( r0 + s ) = 4 ∈ ⎢ ⎜ 1 + ⎟ − ⎜ 1 + ⎟ ⎥ + Ea
r0 ⎠
2⎝
r0 ⎠ ⎥⎦
⎢⎣ 4 ⎝

⎡1⎛
⎞
s
s2
= 4 ∈ ⎢ ⎜ 1 − 12 + 78 2 −⎟
r0
r0
⎠
⎣4⎝
⎞⎤
1⎛
s
s2
− ⎜ 1 − 6 + 21 2 −⎟ ⎥ + Ea
2⎝
r0
r0
⎠⎦
s
s2
s
s2
= ∈ −12 ∈ + 78 ∈ 2 − 2 ∈ +12 ∈ − 42 ∈ 2 + Ea +
r0

r0
r0
r0
⎛ s⎞
s2
= − ∈ +Ea + 0 ⎜ ⎟ + 36 ∈ 2 +
r
⎝r ⎠
0

or

1
U ( r0 + s ) ≈ U ( r0 ) + ks 2
2

where k =
P43.5

(a)

0

72 ∈ 72 ( 4.65 eV )
2
=
2 = 3 599 eV nm = 576 N m
2
r0
( 0.305 nm )

The minimum energy of the molecule at r = r0 is found from
dU
= −12Ar0−13 + 6Br0−7 = 0
dr

yielding

⎡ 2A ⎤
r0 = ⎢
⎣ B ⎥⎦

16

⎡ 2 ( 0.124 × 10−120 eV ⋅ m12 ) ⎤
=⎢
⎥
−60
6
⎢⎣ 1.488 × 10 eV ⋅ m
⎥⎦

(b)

16

= 7.42 × 10−11 m = 74.2 pm

The energy required to break up the molecule would separate the
atoms from r = r0 to r = ∞:

2
⎡ A
B ⎤
B2
⎡1 1⎤ B
E = U r = ∞ − U r = r0 = 0 − ⎢ 2 2 −
=
−
−
=
⎥
⎢⎣ 4 2 ⎥⎦ A 4A
2A B ⎦
⎣ 4A B

(1.488 × 10
E=
4 ( 0.124 × 10

−60

eV ⋅ m6 )

−120

2

eV ⋅ m12 )

= 4.46 eV

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.
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1062
P43.6

Molecules and Solids
(a)

The minimum energy of the molecule at r = r0 is found from
dU
= −12Ar0−13 + 6Br0−7 = 0
dr

yielding

⎡ 2A ⎤
r0 = ⎢
⎣ B ⎥⎦
(b)

16

The energy required to break up the molecule would separate the
atoms from r = r0 to r = ∞:
2

⎡ A
B ⎤
B2
⎡1 1⎤ B
E = U r = ∞ − U r = r0 = 0 − ⎢ 2 2 −
=
−
−
=
⎥
⎢⎣ 4 2 ⎥⎦ A
2A B ⎦
4A
⎣ 4A B

Section 43.2
P43.7

(a)

Energy States and Spectra of Molecules
Recall from Chapter 42 that the energy of the photon is given by

hf = ΔE =

2
2
2
2 ( 2 + 1)] − [ 1( 1 + 1)] = ( 4 )
[

2I
2I
2I

Then,

4 ( h 2π )
h
6.626 × 10−34 J ⋅ s
I=
=
=
2hf
2π 2 f 2π 2 ( 2.30 × 1011 Hz )
2

= 1.46 × 10−46 kg ⋅ m 2
(b)

The results are the same, suggesting that the bond length
of the molecule does not change measurably between the
two transitions.

P43.8

From Equations 43.4 and 43.3, the reduced mass and moment of inertia
of CsI are
m1m2
( 132.9 u )( 126.9 u ) ⎛ 1.66 × 10−27 kg ⎞
µ=

=
⎟⎠
m1 + m2 132.9 u + 126.9 u ⎜⎝
u
= 1.08 × 10−25 kg

and
I = µ r 2 = ( 1.08 × 10−25 kg ) ( 0.127 × 10−9 m )

2

= 1.74 × 10−45 kg ⋅ m 2
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Chapter 43

1063

The allowed rotational energies (from Equation 43.6) are

Erot

⎡ ( 6.626 × 10−34 J ⋅ s 2π )2 ⎤
2
⎥
=
J ( J + 1) = J ( J + 1) ⎢
−45
2

2I
⎢⎣ 2 ( 1.74 × 10 kg ⋅ m ) ⎥⎦
⎛
⎞
1 eV
= J ( J + 1) ( 3.20 × 10−24 J ) ⋅ ⎜
−19 ⎟
⎝ 1.602 × 10 J ⎠
= J ( J + 1) ( 2.00 × 10−5 eV )

(a)

J = 2 gives

Erot
(b)

2
= 2 ( 3 ) = 1.20 × 10−4 eV = 0.120 meV
2I

The photon that can cause the transition J = 1 → 2 has energy

⎛ 2 ⎞
⎛ 2 ⎞
⎛ 2 ⎞
hf = ΔErot = 2 ( 2 + 1) ⎜ ⎟ − 1( 1 + 1) ⎜ ⎟ = 4 ⎜ ⎟
⎝ 2I ⎠
⎝ 2I ⎠
⎝ 2I ⎠

= 4 ( 3.20 × 10−24 J ) = 1.28 × 10−23 J = 7.99 × 10−2 eV

The frequency of the photon is

f=
*P43.9

ΔErot
1.28 × 10−23 J
=
= 1.93 × 1010 s −1 = 19.3 GHz
h
6.626 × 10−34 J ⋅ s

For the HCl molecule in the J = 2 rotational energy level, we are given
the distance between nuclei, r0 = 0.127 5 nm. From Equation 43.6, the
allowed rotational energies are

Erot =

2
J ( J + 1)
2I

Taking J = 2, we have Erot = 6
or

ω=

 2 3 2 1 2
=
= Iω ,
2I
I
2

6 2

= 6
2
I
I

The moment of inertia of the molecule is given by Equation 43.3:
⎛ mm ⎞
I = µ r02 = ⎜ 1 2 ⎟ r02
⎝ m1 + m2 ⎠

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1064

Molecules and Solids
Substituting numerical values,

⎡ ( 1.008 u )( 35.45 u ) ⎤ 2
I=⎢
r0

⎣ 1.008 u + 35.45 u ⎥⎦

= ( 0.980 u )( 1.66 × 10−27 kg u ) ( 1.275 × 10−10 m )

2

= 2.64 × 10−47 kg ⋅ m 2
Therefore,

ω= 6
P43.10

(a)

6 ( 6.626 × 10−34 J ⋅ s )

=
= 9.77 × 1012 rad s
−47
2
I 2π (2.64 × 10 kg ⋅ m )

From Equation 43.10, the energy separation between the ground
and first excited state is
h
2π

ΔEvib =

k

= hf
µ

so

k = 4π 2 f 2 µ

and the reduced mass is

µ=

k
4π f
2

2

=

1530 N/m
= 1.22 × 10−26 kg
4π (56.3 × 1012 s −1 )2
2

(b) From Equation 43.4, the reduced mass is
m1m2
( 14.007 u )( 15.999 u ) ⎛ 1.66 × 10−27 kg ⎞
µ=
=
⎟⎠

m1 + m2
14.007 u +15.999 u ⎜⎝
1u
= 1.24 × 10−26 kg

P43.11

(c)

They agree because the small apparent difference can be
attributed to uncertainty in the data.

(a)

With r representing the distance of each atom from the center of
mass, the moment of inertia is

mr 2 + mr 2 = 2mr 2
⎛ 1.66 × 10−27 kg ⎞ ⎛ 0.750 × 10−10 m ⎞
= 2 ( 1.008 u ) ⎜
⎟⎠ ⎜⎝
⎟⎠
⎝
u
2

2

= 4.71 × 10−48 kg ⋅ m 2
The allowed rotational energies (from Equation 43.6) are

Erot

2
=
J ( J + 1)
2I

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Chapter 43

1065

The J = 0 state has energy Erot = 0, and the J = 1 state has energy
Erot

2
2
h 2π )
h 2π )
(
(
=
( 1)( 2 ) =

2I

(6.626 × 10

=
( 4.71 × 10

I

−34

−48

J ⋅ s 2π ) ⎛
1 eV
⎜
2
kg ⋅ m ) ⎝ 1.602 × 10−19
2

⎞
J ⎟⎠

= 1.48 × 10−2 eV = 0.014 8 eV

(b)

The energy of the photon that raises the molecule from 0 to
0.014 8 eV is 0.014 8 eV. The photon’s wavelength is

h 1240 eV ⋅ nm
=
= 8.41 × 10 4 nm
E

0.0148 eV
= 84.1 × 103 nm = 84.1 µm

λ=

*P43.12

From Equation 43.10, the energy separation between the ground and
first excited state is

ΔEvib  = 

k ( m1  + m2 )
k
 = 
µ
m1m2

Substituting numerical values,

ΔEvib  =  ( 1.055 × 10−34  J ⋅ s )

( 480 N/m )( 35 + 1)  
( 35 )( 1)( 1.66 × 10−27  kg )

= 5.75 × 10−20  J =  0.359 eV
To excite a transition with this energy difference, the wavelength of
incident photons must be

λ  =

1 240 eV ⋅ nm
hc
= 
 = 3.45 × 103  nm
ΔEvib  
0.359 eV

The incident photons have a wavelength longer than this, which
means they have less energy than 0.359 eV. Therefore, these photons
cannot excite the molecule to the first excited state.
P43.13

The mass to consider is the molecule’s reduced mass. Iodine has atomic
mass 126.90 u and a hydrogen atom is 1.007 9 u, so the reduced mass of
HI is

µ=

m1m2
( 126.90 u )( 1.007 9 u )
=
= 0.999 96 u
m1 + m2 126.90 u + 1.007 9 u

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1066

Molecules and Solids
Now for the energy of the ground state we have
E=

1 2 ⎛
1⎞
1 h
kA = ⎜ 0 + ⎟ hf =
⎝
2
2⎠
2 2π

k
µ

So the amplitude is

k ⎛ h ⎞
=⎜
⎟
µ ⎝ 2π ⎠

1/2

⎛ 6.626 × 10−34 J ⋅ s ⎞
A=⎜
⎟⎠
⎝
2π

1/2

h
A=
2π k
(a)

⎛ 1 ⎞
⎜⎝ k µ ⎟⎠

1/4

For HI we have

⎛
⎞
1
×⎜
−27
⎝ (320 N/m)(0.999 96)(1.66 × 10 kg) ⎟⎠

1/4

= 1.20 × 10−11 m = 12.0 pm
(b)

Flourine has an atomic mass of 18.998 4 u, so, for HF,

µ=

m1m2
( 18.998 4 u )( 1.007 9 u )
=
= 0.957 12 u
m1 + m2 18.998 4 u + 1.007 9 u

and

⎛ 6.626 × 10−34 J ⋅ s ⎞
A=⎜
⎟⎠
⎝
2π

1/2

⎛
⎞
1
×⎜
−27
⎝ (970 N/m)(0.957 12)(1.66 × 10 kg) ⎟⎠

1/4

= 9.22 × 10−12 m = 9.22 pm
P43.14

⎛ 2 ⎞

The energy of a rotational transition is ΔE = ⎜ ⎟ J, where J is the
⎝ I ⎠
rotational quantum number of the higher energy state (see Equation
43.7). We do not know J from the data. However,

hc
λ
(6.626 × 10−34 J ⋅ s )( 2.998 × 108 m/s ) ⎛ 1 eV ⎞
=
⎜⎝ 1.602 × 10−19 J ⎟⎠
λ

ΔE =

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Chapter 43

1067

For each observed wavelength,

λ (mm)

E (eV)

0.120 4

0.010 30

0.096 4

0.012 86

0.080 4

0.015 42

0.069 0

0.017 97

0.060 4

0.020 53

The ΔE′s consistently increase by 0.002 56 eV.

E1 =

2
= 0.002 56 eV
I

and
−34
⎞
 2 ( 1.055 × 10 J ⋅ s ) ⎛
1 eV

I=
=
−19 ⎟
⎜
E1
( 0.002 56 eV ) ⎝ 1.60 × 10 J ⎠
2

= 2.72 × 10−47 kg ⋅ m 2
For the HCl molecule, the internuclear radius is

r=
P43.15

(a)

I
=
µ

2.72 × 10−47
m = 0.130 nm
1.62 × 10−27

The reduced mass of NaCl is

µ=

mNa mCl
( 22.99 u )( 35.45 u ) ⎛ 1.66 × 10−27 kg ⎞

=
⎟⎠
mNa + mCl
22.99 u + 35.45 u ⎜⎝
u

= 2.32 × 10−26 kg

(b)

Its moment of inertia is
I = µ r 2 = ( 2.32 × 10−26 kg ) ( 0.280 × 10−9 m )

2

= 1.82 × 10−45 kg ⋅ m 2

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1068

Molecules and Solids
(c)

The wavelength of the emitted photon is found from:

hc
2
2

3 2  2 2 2
2h2
= ΔE = 2 ( 2 + 1) − 1( 1 + 1) =
−
=
=
λ
2I
2I
I
I
I
4π 2 I
then,
8
2
−45
2
c4π 2 I ( 3.00 × 10 m/s ) 4π ( 1.82 × 10 kg ⋅ m )
λ=
=
2h
2 ( 6.626 × 10−34 J ⋅ s )

= 1.62 cm
P43.16

Masses m1 and m2 have the respective distances r1 and r2 from the
center of mass. Then,
m1r1 = m2r2

and

r1 + r2 = r

m2 r2
m1

So,

r1 =

and thus,

m2 r2
m1r
+ r2 = r → r2 =
m1
m1 + m2

Also,

r2 =

m1r1
m2

thus,

r1 +

m1r1
m2 r
= r → r1 =
m2
m1 + m2

The moment of inertia of the molecule is then

I = m1r12 + m2 r22 = m1
=
P43.17

(a)

( m1 + m2 )2

m1m2 ( m2 + m1 ) r 2

( m1 + m2 )2

m22 r 2

=

+ m2

m12 r 2

( m1 + m2 )2

m1m2 r 2
= µ r2
m1 + m2

The reduced mass of the O2 is

µ=

mO mO
( 16.00 u )( 16.00 u )
=
=8u
mO + mO ( 16.00 u ) + ( 16.00 u )

= 8 ( 1.66 × 10−27 kg ) = 1.33 × 10−26 kg

The moment of inertia is then
I = µ r 2 = ( 1.33 × 10−26 kg ) ( 1.20 × 10−10 m )

2

= 1.91 × 10−46 kg ⋅ m 2

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Chapter 43

1069

The rotational energies are

6.626 × 10−34 J ⋅ s 2π )
(
2
=
J ( J + 1) =
J ( J + 1)
2I
2 ( 1.91 × 10−46 kg ⋅ m 2 )
2

Erot
Thus,

Erot = ( 2.91 × 10−23 J ) J ( J + 1)
and for J = 0, 1, 2,

Erot = 0, 3.63 × 10−4 eV, 1.09 × 10−3 eV
(b)

The vibrational energies are given by

1⎞
k
⎛
Evib = ⎜ v + ⎟ 
⎝
2⎠

µ
1 ⎞ ⎛ 6.626 × 10
⎛
=⎜v+ ⎟⎜
⎝
2⎠ ⎝
2π

−34

J⋅s⎞
1 177 N m
⎟⎠ 8 1.66 × 10−27 kg
(
)

⎛
⎞
1⎞
1 eV
⎛
= ⎜ v + ⎟ ( 3.14 × 10−20 J ) ⎜
−19
⎝
⎝ 1.602 × 10 J ⎟⎠
2⎠
1⎞
⎛
= ⎜ v + ⎟ ( 0.196 eV )
⎝

2⎠
For v = 0, 1, 2, Evib = 0.098 0 eV, 0.294 eV, 0.490 eV .
P43.18

(a)

In benzene, the dashed lines form equilateral triangles, so the
carbon atoms are each 0.110 nm from the axis and each hydrogen
atom is (0.110 + 0.100 nm) = 0.210 nm from the axis. Thus, the
moment of inertia is given by
I = ∑ mr 2 = 6 ( 1.99 × 10−26 kg ) ( 0.110 × 10−9 m )

2

+ 6 ( 1.67 × 10−27 kg ) ( 0.210 × 10−9 m )

2

= 1.89 × 10−45 kg ⋅ m 2

(b)

The allowed rotational energies are then

(1.055 × 10−34 J ⋅ s ) J ( J + 1)
2
=
J ( J + 1) =
2I
2 ( 1.89 × 10−45 kg ⋅ m 2 )

2

Erot

= ( 2.95 × 10−24 J ) J ( J + 1) = ( 18.4 × 10−6 eV ) J ( J + 1)
Erot = 18.4J ( J + 1) , where Erot is in microelectron volts and
J = 0, 1, 2, 3,. . . .

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1070

Molecules and Solids
The first five of these allowed energies are:

Erot = 0, 36.9 µeV, 111 µeV, 221 µeV, and 369 µeV
P43.19

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

ΔEmin

2
=
I

The wavelength is then

hc
hc
4π 2 Ic
= 2 =
ΔEmin  I
h

λ=
If µ =

mH mCl
is the reduced mass, then
mH + mCl
I = µ r 2 = µ ( 0.127 46 × 10−9 m )

2

and therefore,
2
4π 2 ⎡ µ ( 0.127 46 × 10−9 m ) ⎤ ( 2.997 925 × 108 m/s )
⎣
⎦
λ=
6.626 075 × 10−34 J ⋅ s

= ( 2.901 830 × 1023 m/kg ) µ

(a)

[1]

For 35Cl,

µ35 =

(1.007 825u )( 34.968 853u ) ⎛ 1.660540 × 10−27 kg ⎞
1.007 825u + 34.968 853u ⎜⎝

u

⎟⎠

= 1.626 653 × 10−27 kg

From equation [1],

λ35 = ( 2.901 830 × 1023 m/kg ) ( 1.626 653 × 10−27 kg )
= 472 µm

(b)

For 37Cl,

µ37 =

(1.007 825u )( 36.965 903u ) ⎛ 1.660540 × 10−27 kg ⎞
1.007 825u + 36.965 903u ⎜⎝

u

⎟⎠

= 1.629 118 × 10−27 kg

From equation [1],

λ37 = ( 2.901 830 × 1023 m/kg ) ( 1.629 118 × 10−27 kg ) = 473 µm
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 43
(c)

1071

The separation in wavelength is

λ37 − λ35 = 472.742 4 µm − 472.027 0 µm = 0.715 µm
P43.20

We find an average spacing between peaks by counting 22 gaps
(counting the central gap as two) between 7.96 × 1013 Hz and
9.24 × 1013 Hz:

Δf =

( 9.24 − 7.96 ) × 1013 Hz
22

= 0.058 2 × 1013 Hz

1 ⎛ h2 ⎞
= 5.82 × 10 Hz = ⎜ 2 ⎟
h ⎝ 4π I ⎠
11

The moment of inertia is then
I=

P43.21

h
6.626 × 10−34 J ⋅ s
=
= 2.88 × 10−47 kg ⋅ m 2
4π 2 Δf 4π 2 ( 5.82 × 1011 s −1 )

We carry extra digits through the solution because the given
wavelengths are close together.
(a)

The energy levels are given by

1⎞
2
⎛
EvJ = ⎜ v + ⎟ hf + J ( J + 1)
⎝

2⎠
2I
Therefore,

1
3
2
1
3 2
E00 = hf , E11 = hf + , and E02 = hf +
2
2
I
2
I
Then,
 2 hc
=
I
λ1

ΔE1 = E11 − E00 = hf +

(6.626 075 × 10
=

−34

J ⋅ s ) ( 2.997 925 × 108 m/s )

2.211 2 × 10−6 m

ΔE1 = hf +

2
= 8.983 573 × 10−20 J
I

[1]

and
2 2 hc
ΔE2 = E11 − E02 = hf −
=
I
λ2
=

(6.626 075 × 10

−34

J ⋅ s ) ( 2.997 925 × 108 m/s )

2.405 4 × 10−6 m

2 2
ΔE2 = hf −
= 8.258 284 × 10−20 J
I

[2]

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1072

Molecules and Solids
Subtracting equation [2] from [1] gives,

⎛
2 ⎞ ⎛
2 2 ⎞ hc hc
ΔE1 − ΔE2 = ⎜ hf + ⎟ − ⎜ hf −
=
−
⎝
I ⎠ ⎝
I ⎟⎠ λ1 λ2
solving,
⎛ 1 1⎞
3 2
= hc ⎜ − ⎟
I
⎝ λ1 λ 2 ⎠

→

⎛ 1 1⎞

3h
= c⎜ − ⎟
2
4π I
⎝ λ1 λ 2 ⎠

Then,

⎛ 1 1⎞
3h
= c⎜ − ⎟
2
4π I
⎝ λ1 λ 2 ⎠

= ( 2.997 925 × 108 m/s )
⎛
⎞
1
1
×⎜
−
−6
−6
⎝ 2.211 2 × 10 m 2.405 4 × 10 m ⎟⎠
= 1.0946 × 1013 s −1

Solving for the moment of inertia then gives

I=

(b)

3 ( 6.626 075 × 10−34 J ⋅ s )
4π 2 ( 1.094 6 × 1013 s −1 )

= 4.60 × 10−48 kg ⋅ m 2

From equation [1]:

f1 =

ΔE1  2
−
h
2π I

6.626 075 × 10−34 J ⋅ s )
(
8.983 573 × 10−20 J
=
−
6.626 075 × 10−34 J ⋅ s 2π ( 4.600 060 × 10−48 kg ⋅ m 2 )
= 1.32 × 1014 Hz
(c)

The moment of inertia of the molecule is given by I = µr2, where µ
is the reduced mass,

µ=

1
1
mH = ( 1.007 825u ) = 8.367 669 × 10−28 kg
2
2

The equilibrium separation distance is then,

r=

I
=
µ

4.600 060 × 10−48 kg ⋅ m 2
= 0.074 1 nm
8.367 669 × 10−28 kg

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Chapter 43
P43.22

1073

The emission energies are the same as the absorption energies, but the
final state must be below (v = 1, J = 0). The transition must satisfy
ΔJ = ±1, so it must end with J = 1. To be lower in energy, the state must
be (v = 0, J = 1). The emitted photon energy is therefore

(

hf photon = Evib v=1 + Erot

J=0

(

= Evib v=1 − Evib v=0

) − (E
) − (E

vib v=0

+ Erot

J=1

rot J=1

− Erot

J=0

)
)

hf photon = hf vib − hfrot

Thus,

f photon = f vib − frot = 6.42 × 1013 Hz − 1.15 × 1011 Hz
= 6.41 × 1013 Hz = 64.1 THz

P43.23

The moment of inertia about the molecular axis is
Iy =

2
2 2 2 2 4
mr + mr = m ( 2.00 × 10−15 m )
5
5
5

The moment of inertia about a perpendicular axis is
2

2

2
m
⎛ R⎞
⎛ R⎞
I x = m ⎜ ⎟ + m ⎜ ⎟ = ( 2.00 × 10−10 m )
⎝ 2⎠
⎝ 2⎠

2

⎛ 2 ⎞
The allowed rotational energies are Erot = ⎜ ⎟ J ( J + 1) , so the energy
⎝ 2I ⎠

2
. The ratio is therefore
I

of the first excited state is E1 =

E1, y
E1, x

(  I ) = I = (1 2 ) m( 2.00 × 10
=
2

(

=

Section 43.3
P43.24

(a)

2

y

Ix )

x

Iy

−10

m)

2

( 4 5) m ( 2.00 × 10−15 m )

2

2
5
9
105 ) = 6.25 × 10
(
8

Bonding in Solids
Consider a cubical salt crystal of edge length 0.1 mm.
3

⎛

⎞
10−4 m
= 6 × 1016 ~ 1017 .
The number of atoms is ⎜
−9
⎟
⎝ 0.261 × 10 m ⎠
(b)

This number of salt crystals would have volume

(10

−4

m ) 6 × 1016 = 6 × 10 4 ~ 105 m 3
3

If it is cubic, it has edge length 40 m.
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1074
P43.25

Molecules and Solids
The ionic cohesive energy is
U=−

α ke e 2 ⎛

1⎞
⎜⎝ 1 − ⎟⎠
r0
m

(1.60 × 10 ) ⎛ 1 − 1 ⎞
= − ( 1.747 6 ) ( 8.99 × 10 )
( 0.281 × 10 ) ⎜⎝ 8 ⎟⎠
−19 2

9

−9

= −1.25 × 10−18 J = −7.83 eV

P43.26

We assume the ions are all singly ionized. The total potential energy is
obtained by summing over all interactions of our ion with others:
U = ∑ ke
i≠ j

qi q j
rij

⎡ e2 e2 e2
e2
e2
e2

e2
e2
⎤
= – ke ⎢ +
–
–
+
+
–
–
+⎥
r
2r 2r 3r 3r 4r 4r
⎣r
⎦

e2
U = – 2ke ⎡⎣1 –
r

1
2

+

1
3

–

1
4

+ ⎤⎦

But from Appendix B.5,

ln(1 + x) =  x –

x2 x3 x4
+
–
+
2
3
4

Our series follows this pattern with x = 1, so the
potential energy of one ion due to its
interactions with all the others is
U = (–2 ln 2)ke

e2
e2
= −keα
where α = 2 ln 2
r
r

Section 43.4

Free-Electron Theory of Metals

Section 43.5

Band Theory of Solids

P43.27

ANS. FIG. P43.26

Taking EF = 5.48 eV for sodium at 800 K,

f (E) =
Then,

1
e

(E−EF )/kB T

+1

so

e(E−EF )/kBT =

1
−1
f (E)

⎛ 1
⎞
E − EF
= ln ⎜
− 1⎟
kBT
⎝ f (E) ⎠

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Chapter 43

and

1075

⎛ 1
⎞
E = EF + kBT ln ⎜
− 1⎟
⎝ f (E) ⎠

substituting numerical values,
E = 5.48 eV + ( 1.38 × 10−23 J/K )
⎛ 1
⎞
× ( 1.602 × 10−19 eV/J )( 800 K ) ln ⎜
− 1⎟ = 5.28 eV

⎝ 0.950 ⎠

P43.28

(a)

The Fermi energy is proportional to the spatial concentration of free electrons to the two-thirds power.

(b)

From Equation 43.25,

h2 ⎛ 3ne ⎞
EF =
⎜
⎟
2m ⎝ 8π ⎠

23

(6.626 × 10 J ⋅ s )
=
2 ( 9.11 × 10 kg ) ( 1.60 × 10
−34

−31

2

−19

⎛ 3 ⎞
⎜ ⎟
J eV ) ⎝ 8π ⎠

23

ne2 3

becomes

EF = ( 3.65 × 10−19 ) ne2 3
where EF is in electron volts and ne in electrons per cubic meter.
(c)

Copper has the greater concentration of free electrons by a factor
of

ne ( Cu ) 8.46 × 10−19 m−3
=
= 6.04
ne ( K ) 1.40 × 10−19 m−3
(d)

Copper has the greater Fermi energy, 7.05 eV.

(e)

The Fermi energy is larger by a factor of 7.05 eV 2.12 eV = 0.333 .

(f)

This behavior agrees with the proportionality because EF  ne2 3
and 6.042/3 = 3.32.

P43.29

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
−23
kBT ( 1.38 × 10 J K ) ( 1 234 K )
=
≈ 2%
EF
( 5.48 eV ) (1.60 × 10−19 J eV )

P43.30

(a)

Setting the kinetic energy equal to the Fermi energy,
1
mv 2 = 7.05 eV
2

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1076

Molecules and Solids
we solve for the speed of the conduction electron as

2 ( 7.05 eV )( 1.60 × 10−19 J eV )

v=

9.11 × 10−31 kg

= 1.57 × 106 m/s = 1.57 Mm/s

P43.31

(b)

Compared to the drift velocity of 0.1 mm/s = 10–4 mm/s, the
speed is larger by ten orders of magnitude. The energy of an
1
eV.
electron at room temperature is typically kBT =
40

(a)

From Equation 43.26,
Eavg =

(b)

3

EF = 0.6(7.05 eV) = 4.23 eV
5

The average energy of a molecule in an ideal gas is

3
kBT so we
2

have

4.23 eV
1.6 × 10−19 J
T=
= 3.27 × 10 4 K
−23
1 eV
3 1.38 × 10 J/K
2

P43.32

For edge d = 1.00 mm,

V = d 3 = ( 1.00 × 10−3 m ) = 1.00 × 10−9 m 3
3

The density of states is

g ( E ) = CE

12

8 2π me3 2 1 2
=
E
h3

or

g (E ) =

8 2π ( 9.11 × 10−31 kg )

(6.626 × 10

−34

J ⋅ s)

3

32

( 4.00 eV ) (1.60 × 10−19

J eV )

g ( E ) = 8.50 × 10 46 m −3 ⋅ J −1 = 1.36 × 1028 m −3 ⋅ eV −1

So, the total number of electrons is

N = [ g ( E )]( ΔE )V

= ( 1.36 × 1028 m −3 ⋅eV −1 ) ( 0.025 0 eV ) ( 1.00 × 10−9 m 3 )
= 3.40 × 1017 electrons

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 43
P43.33

1077

For sodium, M = 23.0 g/mol and ρ = 0.971 g/cm3. Sodium contributes
one electron per atom to the conduction band.
(a)

The density of conduction electrons is
23
3
N A ρ ( 6.02 × 10 electrons mol ) ( 0.971 g/cm )
ne =
=
M
23.0 g mol

ne = 2.54 × 1022 electrons/cm 3 = 2.54 × 1028 m −3
(b)

From Equation 43.25,

⎛ h2 ⎞ ⎛ 3ne ⎞
EF = ⎜
⎜
⎟
⎝ 2m ⎟⎠ ⎝ 8π ⎠

(6.626 × 10
=
2 ( 9.11 × 10

23

J ⋅ s ) ⎡ 3 ( 2.54 × 1028 m −3 ) ⎤
⎢
⎥
−31
8π
kg ) ⎢⎣
⎥⎦
2

−34

23

= 5.05 × 10−19 J = 3.15 eV
P43.34

From Equation 43.24, the number density of free electrons is
ne =

2
3

=

2
3

8 2π me3/2 3/2
EF
h3
8 2π ( 9.11 × 10−31 kg )

(6.626 × 10

−34

J ⋅ s)

3

3/2

( 5.48eV )

3/2

⎛ 1.602 × 10−19 J ⎞
⎜⎝
⎟⎠
1 eV

3/2

= 5.83 × 1028 m –3

Then, the number density of atoms in the metal is

natoms  = 
 = 

nN A
mN A
ρNA
 = 
 = 
V
MV
M
3
( 4.90 × 10  kg/m3 )(6.02 × 1023  mol –1 )
0.100 kg/mol

 

= 2.95 × 1028  m −3

Then the number of free electrons per atom is

5.83 × 1028  m –3
 = 
 = 1.97
natoms
2.95 × 1028  m –3
ne

Therefore, there are approximately two free electrons per atom for this
metal, not one.

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1078
P43.35

Molecules and Solids
From Table 43.2, the Fermi energy for copper at 300 K is 7.05 eV. From
Equation 43.19, the Fermi-Dirac distribution function, the occupation
probability is
f (E) =
=
=

P43.36

e

+1

=

1

e

(0.99EF −EF )/kBT

+1

1
e

⎡⎣ −0.01(7.05 eV)(1.602×10−19 J/eV) ⎤⎦/⎡⎣(1.381×10−23 J/K)( 300 K )⎤⎦

1
e −2.72 + 1

+1

= 0.939

From Equation 43.19, the Fermi-Dirac distribution function, the
occupation probability is
f (E) =

P43.37

1
(E−EF )/kBT

1

e

(E−EF )/kBT

+1

=

1

e

( β EF −EF )/kBT

+1

=

1

e

( β −1)EF /kB T

+1

Consider first the wave function in x. At x = 0 and x = L, ψ = 0.
Therefore,
sin kxL = 0 and kxL = π, 2π, 3π, …
Similarly, sin kyL = 0 and kyL = π, 2π, 3π, …
and

sin kzL = 0 and kzL = π, 2π, 3π, …

Then,
⎛ nyπ y ⎞
⎛ n π x⎞
⎛ n π z⎞
ψ = A sin ⎜ x ⎟ sin ⎜
sin ⎜ z ⎟
⎟
⎝ L ⎠
⎝ L ⎠
⎝ L ⎠

∂ 2ψ ∂ 2ψ ∂ 2ψ 2me
+
+
= 2 (U − E )ψ , we
From Schrödinger’s Equation,
∂x 2 ∂y 2 ∂z 2

have inside the box, where U = 0,

⎛ nx2π 2 ny2π 2 nz2π 2 ⎞

2me
⎜ − L2 − L2 − L2 ⎟ ψ =  2 ( −E )ψ
⎝
⎠
Therefore,

E=

 2π 2 2
nx + ny2 + nz2
2
2me L

(

)

nx , ny , nz = 1, 2, 3, …

Outside the box we require ψ = 0. The minimum energy state inside
3 2π 2
the box is nx = ny = nz = 1, with E =
.
2me L2

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