M42 YOUN7066 13 ISM c42 tủ tài liệu training

MOLECULES AND CONDENSED MATTER

42.1.

42

IDENTIFY: The minimum energy the photon must have is the energy of the covalent bond.
hc
SET UP: The energy of the photon is E = . Visible light has wavelengths between 400 nm and 700 nm.

λ

42.2.

42.3.

EXECUTE: The photon must have energy 4.48 eV. Solving for the wavelength gives
hc 1.24 × 10−6 eV ⋅ m
= 277 nm.
λ= =
E
4.48 eV
EVALUATE: This wavelength is shorter than the wavelengths of visible light so lies in the ultraviolet.
1 q1q2
IDENTIFY and SET UP: U =
. The binding energy of the molecule is equal to U plus the
4πε 0 r
ionization energy of K minus the electron affinity of Br.
1 e2
EXECUTE: (a) U = −
= −5.0 eV.

4πε 0 r
(b) −5.0 eV + (4.3 eV − 3.5 eV) = −4.2 eV.
EVALUATE: We expect the magnitude of the binding energy to be somewhat less than this estimate. At
this separation the two ions don’t behave exactly like point charges and U is smaller in magnitude than our
estimate. The experimental value for the binding energy is −4.0 eV, which is smaller in magnitude than
our estimate.
3
IDENTIFY: Set kT equal to the specified bond energy E.
2
SET UP: k = 1.38 × 10−23 J/K.
EXECUTE: (a) E =
(b) T =

3
2 E 2(7.9 × 10−4 eV)(1.60 × 10−19 J/eV)
kT ⇒ T =
=
= 6.1 K.
2
3k
3(1.38 × 10−23 J/K)

2(4.48 eV)(1.60 × 10−19 J/eV)

= 34,600 K.
3(1.38 × 10−23 J/K)
EVALUATE: (c) The thermal energy associated with room temperature (300 K) is much greater than the
bond energy of He 2 (calculated in part (a)), so the typical collision at room temperature will be more than

enough to break up He2 . However, the thermal energy at 300 K is much less than the bond energy of H 2 ,

42.4.

so we would expect it to remain intact at room temperature.
IDENTIFY: If the photon has too little energy, it cannot alter atomic energy levels.
hc
SET UP: ΔE = . Atomic energy levels are separated by a few eV. Vibrational levels are separated by a

λ

few tenths of an eV. Rotational levels are separated by a few thousandths of an eV or less.
hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
EXECUTE: (a) ΔE =
=
= 4.00 × 10−4 eV. This is a typical
λ
3.10 × 10−3 m
transition energy for a rotational transition.
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42-1


42-2

Chapter 42

(b) ΔE =

42.5.

hc

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)

= 5.99 eV. This is a typical transition energy for a
207 × 10−9 m
transition between atomic energy levels.
EVALUATE: As the transition energy increases, the photon requires a shorter and shorter wavelength to
cause transitions.
IDENTIFY: The energy of the photon is equal to the energy difference between the l = 1 and l = 2 states.
This energy determines its wavelength.
mH mH
1
SET UP: The reduced mass of the molecule is mr =
= mH , its moment of inertia is I = mr r02 ,
mH + mH 2

λ

=

the photon energy is ΔE =

hc

λ

, and the energy of the state l is El = l (l + 1)

2

2I

.

2
1
EXECUTE: I = mr r02 = (1.67 × 10−27 kg)(0.074 × 10−9 m) 2 = 4.57 × 10−48 kg ⋅ m 2. Using El = l (l + 1) ,
2I
2

the energy levels are E2 = 6
E1` = 2

2

2I

2

2I

=6

(1.055 × 10−34 J ⋅ s)2
2(4.57 × 10

−48

2

kg ⋅ m )

= 6(1.218 × 10−21 J) = 7.307 × 10−21 J and

= 2(1.218 × 10−21 J) = 2.436 × 10−21 J. ΔE = E2 − E1 = 4.87 × 10−21 J. Using ΔE =

hc

λ

gives

hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 4.08 × 10−5 m = 40.8 μ m.
ΔE
4.871× 10−21 J
EVALUATE: This wavelength is much longer than that of visible light.
IDENTIFY: The energy decrease of the molecule or atom is equal to the energy of the emitted photon.
From this energy, we can calculate the wavelength of the photon.
hc
SET UP: ΔE = .

λ=

42.6.

λ

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
=
= 4.96 µm.
ΔE
0.250 eV
EVALUATE: This radiation is in the infrared.
hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
(b) λ =
=
= 146 nm.
ΔE
8.50 eV
EVALUATE: This radiation is in the ultraviolet.
hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
(c) λ =
=
= 388 µm.
ΔE
3.20 × 10−3 eV
EVALUATE: This radiation is in the microwave region.
IDENTIFY: The energy given to the photon comes from a transition between rotational states.
ћ2
SET UP: The rotational energy of a molecule is E = l (l + 1)
and the energy of the photon is E = hc /λ .
2I
EXECUTE: Use the energy formula, the energy difference between the l = 3 and l = 1 rotational levels of
EXECUTE: (a) λ =

42.7.

the molecule is ΔE =
I=

5ћ 2
ћ2
[3(3 + 1) − 1(1 + 1)] =
. Since ΔE = hc /λ , we get hc /λ = 5ћ 2 /I . Solving for I gives
2I
I

5ћλ 5(1.055 × 10−34 J ⋅ s)(1.780 nm)
=
= 4.981 × 10−52 kg ⋅ m 2 .
2π c
2π (3.00 × 108 m/s)

Using I = mr r02 , we can solve for r0:
r0 =

I (mN + mH )
(4.981×10−52 kg ⋅ m2 )(2.33 ×10−26 kg + 1.67 ×10−27 kg)
=
r0 = 5.65 × 10−13 m
−26
−27
mNmH
(2.33 ×10 kg)(1.67 ×10 kg)

EVALUATE: This separation is much smaller than the diameter of a typical atom and is not very realistic.

But we are treating a hypothetical NH molecule.

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Molecules and Condensed Matter
42.8.

42-3

IDENTIFY: The transition energy E and the frequency f of the absorbed photon are related by E = hf .
EXECUTE: The energy of the emitted photon is 1.01× 10−5 eV, and so its frequency and wavelength are
E (1.01 × 10−5 eV)(1.60 × 10−19 J/eV)
f = =
= 2.44 GHz = 2440 MHz and
h
(6.63 × 10−34 J ⋅ s)
c (3.00 × 108 m/s)
=
= 0.123 m.
f
(2.44 × 109 Hz)
EVALUATE: This frequency corresponds to that given for a microwave oven.
IDENTIFY: Apply Eq. (42.5).
SET UP: Let 1 refer to C and 2 to O. m1 = 1.993 × 10−26 kg, m2 = 2.656 × 10−26 kg, r0 = 0.1128 nm.

λ=

42.9.

⎛ m2 ⎞
⎛ m1 ⎞
EXECUTE: (a) r1 = ⎜
⎟ r0 = 0.0644 nm (carbon); r2 = ⎜
⎟ r0 = 0.0484 nm (oxygen)
⎝ m1 + m2 ⎠
⎝ m1 + m2 ⎠
(b) I = m1r12 + m2 r22 = 1.45 × 10−46 kg ⋅ m 2 ; yes, this agrees with Example 42.2.
EVALUATE: I = m1r12 + m2 r22 and I = mr r02 give equivalent results.
42.10.

IDENTIFY: I = m1r12 + m2 r22 . Since the two atoms are identical, the center of mass is midway between them.
SET UP: Each atom has a mass m and is at a distance L/2 from the center of mass.
EXECUTE: The moment of inertia is 2( m) = ( L/2)2 = mL2 /2 = 2.21 × 10−44 kg ⋅ m 2.
EVALUATE: r0 = L and mr = m /2, so I = mr r02 gives the same result.

42.11.

IDENTIFY and SET UP: Set K = E1 from Example 42.2. Use K = 12 I ω 2 to solve for ω and v = rω to

solve for v.
EXECUTE: (a) From Example 42.2, E1 = 0.479 meV = 7.674 × 10−23 J and I = 1.449 × 10−46 kg ⋅ m 2

K = 12 I ω 2 and K = E gives ω = 2 E1 /I = 1.03 × 1012 rad/s
(b) v1 = r1ω1 = (0.0644 × 10−9 m)(1.03 × 1012 rad/s) = 66.3 m/s (carbon)
v2 = r2ω2 = (0.0484 × 10−9 m)(1.03 × 1012 rad/s) = 49.8 m/s (oxygen)

(c) T = 2π /ω = 6.10 × 10−12 s
EVALUATE: Even for fast rotation rates, v

42.12.

c.

IDENTIFY: For a n → n − 1 vibrational transition, ΔE =
SET UP: mr =

k′
hc
. ΔE is related to λ of the photon by ΔE = .
mr
λ

mNa mCl
.
mNa + mCl
2

⎛ 2π c ⎞
k ′/mr , and solving for k ′, k ′ = ⎜
m = 205 N/m.
⎝ λ ⎟⎠ r
λ
EVALUATE: The value of k ′ we calculated for NaCl is comparable to that of a fairly stiff lab spring.
EXECUTE: ΔE =

42.13.

hc

=

IDENTIFY and SET UP: The energy of a rotational level with quantum number l is El = l (l + 1)ћ 2 /2 I

(Eq. (42.3)). I = mr r 2, with the reduced mass mr given by Eq. (42.4). Calculate I and ΔE and then use
ΔE = hc /λ to find λ.
EXECUTE: (a) mr =

m1m2
mLi mH
(1.17 × 10−26 kg)(1.67 × 10−27 kg)
=
=
= 1.461 × 10−27 kg
m1 + m2 mLi + mH 1.17 × 10−26 kg + 1.67 × 10−27 kg

I = mr r 2 = (1.461 × 10−27 kg)(0.159 × 10−9 m)2 = 3.694 × 10−47 kg ⋅ m 2
⎛ ћ2 ⎞
⎛ ћ2 ⎞
l = 3 : E = 3(4) ⎜ ⎟ = 6 ⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝ 2I ⎠
⎝ I ⎠

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42-4

Chapter 42

⎛ ћ2 ⎞
⎛ ћ2 ⎞
l = 4 : E = 4(5) ⎜ ⎟ = 10 ⎜ ⎟
⎜ 2I ⎟
⎜ I ⎟
⎝ ⎠
⎝ ⎠
⎛ ћ2 ⎞
⎛ (1.055 × 10−34 J ⋅ s) 2 ⎞
ΔE = E4 − E3 = 4 ⎜ ⎟ = 4 ⎜
= 1.20 × 10−21 J = 7.49 × 10−3 eV
⎜ I ⎟
⎜ 3.694 × 10−47 kg ⋅ m 2 ⎟⎟
⎝ ⎠
⎝
⎠
hc (4.136 × 10−15 eV)(2.998 × 108 m/s)
=
= 166 μ m
ΔE
7.49 × 10−3 eV
EVALUATE: LiH has a smaller reduced mass than CO and λ is somewhat smaller here than the
λ calculated for CO in Example 42.2
IDENTIFY: The vibrational energy of the molecule is related to its force constant and reduced mass, while
the rotational energy depends on its moment of inertia, which in turn depends on the reduced mass.
1⎞
1⎞

k′
⎛
⎛
SET UP: The vibrational energy is En = ⎜ n + ⎟ ћω = ⎜ n + ⎟ ћ
and the rotational energy is
2⎠
2 ⎠ mr
⎝
⎝
(b) ΔE = hc /λ so λ =

42.14.

El = l (l + 1)

ћ2
.
2I

EXECUTE: For a vibrational transition, we have ΔEv = ћ

for a rotational transition is ΔER =
I = mr r0 2 , we have mr r02 =
mr =

2ћ 2

=

k′

, so we first need to find mr . The energy
mr

ћ2
2ћ 2
[2(2 + 1) − 1(1 + 1)] =
. Solving for I and using the fact that
2I
I

2ћ 2
, which gives
ΔER

2(1.055 × 10−34 J ⋅ s)(6.583 × 10−16 eV ⋅ s)

(0.8860 × 10−9 m)2 (8.841 × 10−4 eV)
r02ΔER
Now look at the vibrational transition to find the force constant.

= 2.0014 × 10 –28 kg

k′
m ( ΔE )2 (2.0014 × 10−28 kg)(0.2560 eV) 2
⇒ k′ = r 2 v =
= 30.27 N/m
mr
ћ
(6.583 × 10−16 eV ⋅ s) 2
EVALUATE: This would be a rather weak spring in the laboratory.

IDENTIFY and SET UP: The energy of a rotational level is given in Eq. (42.3). The transition energy ΔE
and the frequency f of the photon are related by ΔE = hf .
ΔE v = ћ

42.15.

2
l (l + 1) 2
l (l − 1) 2
l 2
, El −1 =
⇒ ΔE =
(l 2 + l − l 2 + l ) =
2I
2I
2I
I
ΔE ΔE
l
.
=
=
(b) f =
2π
2π I
h
EVALUATE: ΔE and f increase with l because the separation between adjacent energy levels increases with l.
IDENTIFY: Find ΔE for the transition and compute λ from ΔE = hc /λ.

EXECUTE: (a) El =

42.16.

ћ2
ћ2
, with
= 0.2395 × 10−3 eV. Δ E = 0.2690 eV is the
2I
2I
spacing between vibrational levels. Thus En = ( n + 12 ) ћω , with ћω = 0.2690 eV. By Eq. (42.9),
SET UP: From Example 42.2, El = l (l + 1)

ћ2
.
2I
EXECUTE: (a) n = 0 → n = 1 and l = 1 → l = 2
E = En + El = ( n + 12 ) ћω + l (l + 1)

⎛ ћ2 ⎞
For n = 0, l = 1, Ei = 12 ћω + 2 ⎜ ⎟ .
⎜ 2I ⎟
⎝ ⎠

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Molecules and Condensed Matter

42-5

⎛ ћ2 ⎞
For n = 1, l = 2, E f = 32 ћω + 6 ⎜ ⎟ .
⎝ 2I ⎠
⎛ ћ2 ⎞
ΔE = E f − Ei = ћω + 4 ⎜ ⎟ = 0.2690 eV + 4(0.2395 × 10−3 eV) = 0.2700 eV
⎝ 2I ⎠
hc (4.136 × 10−15 eV ⋅ s)(2.998 × 108 m/s)
=
= 4.592 × 10−6 m = 4.592 μm
λ
0.2700 eV
ΔE
(b) n = 0 → n = 1 and l = 2 → l = 1
hc

= ΔE so λ =

⎛ ћ2 ⎞
For n = 0, l = 2, Ei = 12 ћω + 6 ⎜ ⎟ .
⎜ 2I ⎟
⎝ ⎠
⎛
ћ2 ⎞
For n = 1, l = 1, E f = 32 ћω + 2 ⎜ ⎟ .
⎜ 2I ⎟
⎝ ⎠
⎛ ћ2 ⎞
ΔE = E f − Ei = ћω − 4 ⎜ ⎟ = 0.2690 eV − 4(0.2395 × 10−3 eV) = 0.2680 eV
⎜ 2I ⎟

⎝ ⎠
hc (4.136 × 10−15 eV ⋅ s)(2.998 × 108 m/s)
=
= 4.627 × 10−6 m = 4.627 μ m
0.2680 eV
ΔE
(c) n = 0 → n = 1 and l = 3 → l = 2

λ=

⎛ ћ2 ⎞
For n = 0, l = 3, Ei = 12 ћω + 12 ⎜ ⎟ .
⎜ 2I ⎟
⎝ ⎠
⎛ ћ2 ⎞
For n = 1, l = 2, E f = 32 ћω + 6 ⎜ ⎟ .
⎜ 2I ⎟
⎝ ⎠
⎛ ћ2 ⎞
ΔE = E f − Ei = ћω − 6 ⎜ ⎟ = 0.2690 eV − 6(0.2395 × 10−3 eV) = 0.2676 eV
⎜ 2I ⎟
⎝ ⎠

hc (4.136 × 10−15 eV ⋅ s)(2.998 × 108 m/s)
=
= 4.634 × 10−6 m = 4.634 μ m
0.2676 eV
ΔE
EVALUATE: All three transitions are for n = 0 → n = 1. The spacing between vibrational levels is larger
than the spacing between rotational levels, so the difference in λ for the various rotational transitions is

small. When the transition is to a larger l, Δ E > ћω and when the transition is to a smaller l, Δ E < ћω.

λ=

42.17.

IDENTIFY and SET UP: Find the volume occupied by each atom. The density is the average mass of
Na and Cl divided by this volume.
EXECUTE: Each atom occupies a cube with side length 0.282 nm. Therefore, the volume occupied by
each atom is V = (0.282 × 10−9 m)3 = 2.24 × 10−29 m3. In NaCl there are equal numbers of Na and Cl

atoms, so the average mass of the atoms in the crystal is
m = 12 ( mNa + mCl ) = 12 (3.82 × 10−26 kg + 5.89 × 10−26 kg) = 4.855 × 10−26 kg

The density then is ρ =

m 4.855 × 10−26 kg
=
= 2.17 × 103 kg/m3 .
V 2.24 × 10−29 m3

EVALUATE: The density of water is 1.00 × 103 kg/m3, so our result is reasonable.
42.18.

IDENTIFY and SET UP: For an average spacing a, the density is ρ = m/a3, where m is the average of the
ionic masses.
m (6.49 × 10−26 kg + 1.33 × 10−25 kg)/2
= 3.60 × 10−29 m3 , and
EXECUTE: a 3 = =
3

3
ρ
(2.75 × 10 kg/m )
a = 3.30 × 10−10 m = 0.330 nm.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


42-6

Chapter 42

42.19.

EVALUATE: (b) Exercise 42.17 says that the average spacing for NaCl is 0.282 nm. The larger (higher
atomic number) atoms have the larger spacing.
IDENTIFY: The energy gap is the energy of the maximum-wavelength photon.
SET UP: The energy difference is equal to the energy of the photon, so ΔE = hc /λ .
EXECUTE: (a) Using the photon wavelength to find the energy difference gives
ΔE = hc /λ = (4.136 × 10 –15 eV ⋅ s)(3.00 × 108 m/s)/(1.11 × 10 –6 m) = 1.12 eV
(b) A wavelength of 1.11 µm = 1110 nm is in the infrared, shorter than that of visible light.

42.20.

42.21.

EVALUATE: Since visible photons have more than enough energy to excite electrons from the valence to
the conduction band, visible light will be absorbed, which makes silicon opaque.
hc

IDENTIFY and SET UP: ΔE = , where ΔE is the band gap.

λ

hc
EXECUTE: (a) λ =
= 2.27 × 10−7 m = 227 nm, in the ultraviolet.
ΔE
EVALUATE: (b) Visible light lacks enough energy to excite the electrons into the conduction band, so
visible light passes through the diamond unabsorbed.
(c) Impurities can lower the gap energy making it easier for the material to absorb shorter wavelength
visible light. This allows longer wavelength visible light to pass through, giving the diamond color.
IDENTIFY and SET UP: The energy ΔE deposited when a photon with wavelength λ is absorbed is
hc
ΔE = .

λ

EXECUTE: ΔE =

hc

λ

=

(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
9.31 × 10−13 m

= 2.14 × 10−13 J = 1.34 × 106 eV. So the number

of electrons that can be excited to the conduction band is n =

42.22.

1.34 × 106 eV
= 1.20 × 106 electrons.
1.12 eV

EVALUATE: A photon of wavelength
hc (4.13 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
=
= 1.11 × 10−6 m = 1110 nm can excite one electron. This
λ=
ΔE
1.12 eV
photon is in the infrared.
2
IDENTIFY: Set 32 kT = 12 mvrms
.
SET UP: k = 1.38 × 10−23 J/K. m = 9.11 × 10−31 kg.
EXECUTE: vrms = 3kT/m = 1.17 × 105 m/s, as found in Example 42.8.

42.23.

EVALUATE: Temperature plays a very small role in determining the properties of electrons in metals. Instead,
the average energies and corresponding speeds are determined almost exclusively by the exclusion principle.
IDENTIFY: g ( E ) is given by Eq. (42.10).
SET UP: m = 9.11 × 10−31 kg, the mass of an electron.
EXECUTE:

g (E) =

(2m)3/2V
2π 2

3

E1/2 =

(2(9.11 × 10−31 kg))3/2 (1.0 × 10−6 m3 )(5.0 eV)1/2 (1.60 × 10−19 J/eV)1/2
2π 2 (1.054 × 10−34 J ⋅ s)3

.

g ( E ) = (9.5 × 1040 states/J)(1.60 × 10−19 J/eV) = 1.5 × 1022 states/eV.

42.24.

EVALUATE: For a metal the density of states expressed as states/eV is very large.
IDENTIFY and SET UP: Combine Eqs. (42.11) and (42.12) to eliminate nrs .
EXECUTE: Eq. (42.12) may be solved for nrs = (2mE )1/2 ( L π ), and substituting this into Eq. (42.11),

42.25.

using L3 = V , gives Eq. (42.13).
EVALUATE: n is the total number of states with energy of E or less.
(a) IDENTIFY and SET UP: The electron contribution to the molar heat capacity at constant volume of a
⎛ π 2 KT ⎞
metal is CV = ⎜

⎟⎟ R.
⎜
⎝ 2 EF ⎠

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Molecules and Condensed Matter

EXECUTE: CV =

42-7

π 2 (1.381 × 10−23 J/K)(300 K)

R = 0.0233R.
2(5.48 eV)(1.602 × 10−19 J/eV)
(b) EVALUATE: The electron contribution found in part (a) is 0.0233R = 0.194 J/mol ⋅ K. This is
0.194/25.3 = 7.67 × 10−3 = 0.767% of the total CV .
(c) Only a small fraction of CV is due to the electrons. Most of CV is due to the vibrational motion of
42.26.

the ions.
IDENTIFY: Eq. (42.21) relates Eav and EF0 , the Fermi energy at absolute zero. The speed v is related to
Eav by

1 mv 2
2

= Eav .

SET UP: k = 1.38 × 10−23 J/K.
3
EXECUTE: (a) Eav = EF = 1.94 eV.
5
(b) v = 2 Eav /m =
(c)

2(1.94 eV)(1.60 × 10−19 J/eV)
9.11 × 10−31 kg

= 8.25 × 105 m/s.

EF (3.23 eV)(1.60 × 10−19 J/eV)
=
= 3.74 × 104 K.
k
(1.38 × 10−23 J/K)

EVALUATE: The Fermi energy of sodium is less than that of copper. Therefore, the values of Eav and v
42.27.

we have calculated for sodium are less than those calculated for copper in Example 42.7.
IDENTIFY: The probability is given by the Fermi-Dirac distribution.
1
.
SET UP: The Fermi-Dirac distribution is f ( E ) = ( E − E )/kT
F
e

+1
EXECUTE: We calculate the value of f ( E ), where E = 8.520 eV, EF = 8.500 eV,

k = 1.38 × 10 –23 J/K = 8.625 × 10 –5 eV/K, and T = 20°C = 293 K. The result is f (E ) = 0.312 = 31.2%.

42.28.

EVALUATE: Since the energy is close to the Fermi energy, the probability is quite high that the state is
occupied by an electron.
IDENTIFY and SET UP: Follow the procedure of Example 42.9. Evaluate f ( E ) in Eq. (42.16) for

E − EF = Eg /2, where Eg is the band gap.
EXECUTE: (a) The probabilities are 1.78 × 10−7, 2.37 × 10−6 , and 1.51 × 10−5.
(b) The Fermi distribution, Eq. (42.16), has the property that f ( EF − E ) = 1 − f ( E ) (see Problem (42.50)),

42.29.

and so the probability that a state at the top of the valence band is occupied is the same as the probability
that a state of the bottom of the conduction band is filled (this result depends on having the Fermi energy
in the middle of the gap). Therefore, the probabilities at each T are the same as in part (a).
EVALUATE: The probabilities increase with temperature.
1
. Solve for E − EF .
IDENTIFY: Use Eq. (42.16), f ( E ) = ( E − E )/kT
F
e
+1
1
SET UP: e( E − EF )/kT =
−1

f (E)
The problem states that f ( E ) = 4.4 × 10−4 for E at the bottom of the conduction band.
EXECUTE: e( E − EF )/ kT =

1
4.4 × 10−4

− 1 = 2.272 × 103.

E − EF = kT ln(2.272 × 103 ) = (1.3807 × 10 −23 J/T)(300 K)ln(2.272 × 103 ) = 3.201 × 10−20 J = 0.20 eV

EF = E − 0.20 eV; the Fermi level is 0.20 eV below the bottom of the conduction band.
EVALUATE: The energy gap between the Fermi level and bottom of the conduction band is large
compared to kT at T = 300 K and as a result f ( E ) is small.

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42-8

Chapter 42

42.30.

IDENTIFY: The wavelength of the photon to be detected depends on its energy.
hc
SET UP: ΔE = .

λ

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
=
= 1.9 µm.
ΔE
0.67 eV
⎛ 0.67 eV ⎞
(b) λ = (1.9 µm) ⎜
= 1.1 µm.
⎝ 1.14 eV ⎟⎠
EVALUATE: Both of these photons are in the infrared.
IDENTIFY: Knowing the saturation current of a p-n junction at a given temperature, we want to find the
current at that temperature for various voltages.
SET UP: I = IS (eeV /kT − 1).
EXECUTE: (a) λ =

42.31.

eV (1.602 × 10−19 C)(1.00 × 10−3 V)
=
= 0.0400.
kT
(1.381 × 10−23 J/K)(290 K)

EXECUTE: (a) (i) For V = 1.00 mV,

I = (0.500 mA)(e0.0400 − 1) = 0.0204 mA.
eV
= −0.0400. I = (0.500 mA)(e−0.0400 − 1) = −0.0196 mA.
kT

eV
(iii) For V = 100 mV,
= 4.00. I = (0.500 mA)(e 4.00 − 1) = 26.8 mA.
kT
eV
(iv) For V = −100 mV,
= −4.00. I = (0.500 mA)(e −4.00 − 1) = −0.491 mA.
kT
EXECUTE: (b) For small V, between ±1.00 mV, R = V/I is approximately constant and the diode obeys
Ohm’s law to a good approximation. For larger V the deviation from Ohm’s law is substantial.
IDENTIFY: The current depends on the voltage across the diode and its temperature, so the resistance also
depends on these quantities.
SET UP: The current is I = IS (eeV /kT – 1) and the resistance is R = V/I .
(ii) For V = −1.00 mV,

42.32.

EXECUTE: (a) The resistance is R =

V
V
=
. The exponent is
I I s (eeV / kT − 1)

eV
e(0.0850 V)
85.0 mV
=
= 3.3635, giving R =

= 4.06 Ω.
kT (8.625 × 10−5 eV/K)(293 K)
(0.750 mA)(e3.3635 − 1)

(b) In this case, the exponent is

which gives R =

42.33.

eV
e(−0.050 V)
=
= −1.979
kT (8.625 × 10−5 eV/K)(293 K)

−50.0 mV

= 77.4 Ω
(0.750 mA)(e−1.979 − 1)
EVALUATE: Reversing the voltage can make a considerable change in the resistance of a diode.
IDENTIFY and SET UP: The voltage-current relation is given by Eq. (42.22): I = I s (eeV /kT − 1). Use the
current for V = +15.0 mV to solve for the constant I s .
EXECUTE: (a) Find I s : V = +15.0 × 10−3 V gives I = 9.25 × 10−3 A

eV (1.602 × 10−19 C)(15.0 × 10−3 V)
=
= 0.5800
kT
(1.381 × 10−23 J/K)(300 K)

Is =

I
eeV /kT − 1

=

9.25 × 10−3 A
e0.5800 − 1

= 1.177 × 10−2 = 11.77 mA

Then can calculate I for V = 10.0 mV:

eV (1.602 × 10−19 C)(10.0 × 10−3 V)
=
= 0.3867
kT
(1.381 × 10−23 J/K)(300 K)

I = I s (eeV /kT − 1) = (11.77 mA)(e0.3867 − 1) = 5.56 mA

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Molecules and Condensed Matter

42-9

eV
eV
has the same magnitude as in part (a) but not V is negative so
is negative.
kT
kT
eV
V = −15.0 mV :
= −0.5800 and I = I s (eeV /kT − 1) = (11.77 mA)(e−0.5800 − 1) = −5.18 mA
kT
eV
V = −10.0 mV :
= −0.3867 and I = I s (eeV /kT − 1) = (11.77 mA)(e −0.3867 − 1) = −3.77 mA
kT
EVALUATE: There is a directional asymmetry in the current, with a forward-bias voltage producing more
current than a reverse-bias voltage of the same magnitude, but the voltage is small enough for the
asymmetry not be pronounced.
IDENTIFY: Apply Eq. (42.22).
SET UP: IS = 3.60 mA. ln e x = x
(b)

42.34.

EXECUTE: (a) Solving Eq. (42.22) for the voltage as a function of current,

V=

42.35.

⎞ kT ⎛ 40.0 mA ⎞

kT ⎛ I
ln ⎜ + 1⎟ =
ln ⎜
+ 1⎟ = 0.0645 V.
e
⎝ 3.60 mA ⎠
⎝ IS ⎠ e

(b) From part (a), the quantity eeV /kT = 12.11, so far a reverse-bias voltage of the same magnitude,
⎛ 1
⎞
I = IS (e − eV kT − 1) = IS ⎜
− 1⎟ = −3.30 mA.
⎝ 12.11 ⎠
EVALUATE: The reverse bias current for a given magnitude of voltage is much less than the forward bias
current.
IDENTIFY: During the transition, the molecule emits a photon of light having energy equal to the energy
difference between the two vibrational states of the molecule.
1⎞
1⎞
k′
⎛
⎛
.
SET UP: The vibrational energy is En = ⎜ n + ⎟ ћω = ⎜ n + ⎟ ћ
2⎠
2 ⎠ mr
⎝
⎝
EXECUTE: (a) The energy difference between two adjacent energy states is ΔE = ћ

k′
, and this is the
mr

energy of the photon, so ΔE = hc /λ . Equating these two expressions for ΔE and solving for k′, we have
2

2

mH mO ⎛ ΔE ⎞
ΔE hc/λ 2π c
⎛ ΔE ⎞
k ′ = mr ⎜
=
=
with the appropriate numbers gives us
=
⎜
⎟ , and using
⎝ ћ ⎟⎠
mH + mO ⎝ ћ ⎠
λ
ћ
ћ
2

(1.67 × 10−27 kg)(2.656 × 10−26 kg) ⎡ 2π (3.00 × 108 m/s) ⎤
k′ =
⎢

⎥ = 977 N/m
1.67 × 10−27 kg + 2.656 × 10−26 kg ⎣⎢ 2.39 × 10−6 m ⎦⎥

ω
1
(b) f =
=
2π 2π

k′
1
=
mr 2π

mH mO
mH + mO
. Substituting the appropriate numbers gives us
k′

(1.67 × 10−27 kg)(2.656 × 10−26 kg)
1 1.67 × 10−27 kg + 2.656 × 10−26 kg
= 1.25 × 1014 Hz
f =
2π
977 N/m
EVALUATE: The frequency is close to, but not quite in, the visible range.
42.36.

IDENTIFY and SET UP: El = l (l + 1)
2

2

2

2I

. ΔE for the molecule is related to λ for the photon by ΔE =
2

hc

λ

.

2 2
hλ
=
= 7.14 × 10−48 kg ⋅ m 2 .
I
I
ΔE 2π 2c
I
EVALUATE: The I we calculated is approximately a factor of 20 times smaller than I calculated for the CO
molecule in Example 42.2.

EXECUTE: E2 = 3

and E1 =

, so ΔE =

2

. I=

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42-10
42.37.

Chapter 42
IDENTIFY and SET UP: Eq. (21.14) gives the electric dipole moment as p = qd , where the dipole consists
of charges ± q separated by distance d.
EXECUTE: (a) Point charges +e and −e separated by distance d, so
p = ed = (1.602 × 10−19 C)(0.24 × 10−9 m) = 3.8 × 10−29 C ⋅ m

(b) p = qd so q =
(c)

q
1.3 × 10−19 C
=
= 0.81
e 1.602 × 10−19 C

(d) q =

42.38.

p 3.0 × 10−29 C ⋅ m
=
= 1.3 × 10−19 C
d
0.24 × 10−9 m

p 1.5 × 10−30 C ⋅ m
=
= 9.37 × 10−21 C
d
0.16 × 10−9 m

q 9.37 × 10−21 C
=
= 0.058
e 1.602 × 10−19 C
EVALUATE: The fractional ionic character for the bond in HI is much less than the fractional ionic
character for the bond in NaCl. The bond in HI is mostly covalent and not very ionic.
IDENTIFY: The electric potential energy U, the binding energy EB , the electron affinity EA , and the
ionization energy EI , where EB , EA and EI are positive and U is negative, are related by
EB = −U + EA − EI .
SET UP: For two point charges q1 and q2 separated by a distance r, the electric potential energy is given

by U =

q1q2
.

4πε 0 r
1

e2
= 2.8 × 10−10 m.
4πε 0 U
EVALUATE: We have neglected the kinetic energy of the ions in the molecule. Also, it is an
approximation to treat the two ions as point charges.
(a) IDENTIFY: E (Na) + E (Cl) = E (Na + ) + E (Cl− ) + U (r ). Solving for U ( r ) gives
EXECUTE: The electrical potential energy is U = −5.13 eV, and r = −

42.39.

1

U (r ) = −[ E (Na + ) − E (Na)] + [ E (Cl) − E (Cl− )].
SET UP: [ E (Na + ) − E (Na)] is the ionization energy of Na, the energy required to remove one electron,

and is equal to 5.1 eV. [ E (Cl) − E (Cl− )] is the electron affinity of Cl, the magnitude of the decrease in
energy when an electron is attached to a neutral Cl atom, and is equal to 3.6 eV.
1 e2
EXECUTE: U = −5.1 eV + 3.6 eV = −1.5 eV = −2.4 × 10−19 J, and −
= −2.4 × 10−19 J
4π ⑀ 0 r
⎛ 1 ⎞
e2
(1.602 × 10−19 C)2
= (8.988 × 109 N ⋅ m 2 /C2 )
r =⎜
⎟

−
19
⎝ 4π ⑀ 0 ⎠ 2.4 × 10
J
2.4 × 10−19 J
r = 9.6 × 10−10 m = 0.96 nm
(b) ionization energy of K = 4.3 eV; electron affinity of Br = 3.5 eV
Thus U = −4.3 eV + 3.5 eV = −0.8 eV = −1.28 × 10−19 J, and −

e2
= −1.28 × 10−19 J
4π ⑀0 r
1

⎛ 1 ⎞
e2
(1.602 × 10−19 C) 2
= (8.988 × 109 N ⋅ m 2 /C2 )
r =⎜
⎟
−
19
⎝ 4π ⑀ 0 ⎠ 1.28 × 10
J
1.28 × 10−19 J
r = 1.8 × 10−9 m = 1.8 nm
EVALUATE: K has a smaller ionization energy than Na and the electron affinities of Cl and Br are very
similar, so it takes less energy to make K + + Br − from K + Br than to make Na + + Cl− from Na + Cl.
Thus, the stabilization distance is larger for KBr than for NaCl.

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Molecules and Condensed Matter
42.40.

42-11

IDENTIFY: The rotational energy levels are given by Eq. (42.3). The photon wavelength λ is related to
hc
the transition energy of the atom by ΔE = .

λ

SET UP: For emission, Δl = −1. For such a transition, from state l to state l − 1,

ΔEl = [l (l + 1) − (l − 1)l ]
Δ = ΔEl − ΔEl −1 =

2

2I

=

2

l
I

. The difference in transition energies for adjacent lines in the spectrum is

2

.

I

EXECUTE: The transition energies corresponding to the observed wavelengths are 3.29 × 10−21 J,
2.87 × 10−21 J, 2.47 × 10 −21 J, 2.06 × 10−21 J and 1.65 × 10−21 J. The average spacing of these energies is

0.410 × 10−21 J. Then,
2

EVALUATE: With

42.41.

2

I

= 0.410 × 10−21 J, from which I = 2.71 × 10−47 kg ⋅ m 2 .

= 0.410 × 10−21 J and ΔEl =

l

2

, we find that these wavelengths correspond to
I
I
transitions from levels 8, 7, 6, 5 and 4 to the respective next lower levels.
(a) IDENTIFY: The rotational energies of a molecule depend on its moment of inertia, which in turn
depends on the separation between the atoms in the molecule.
SET UP: Problem 42.40 gives I = 2.71 × 10−47 kg ⋅ m 2 . I = mr r 2 . Calculate mr and solve for r.
EXECUTE: mr =

r=

mH mCl
(1.67 × 10−27 kg)(5.81 × 10−26 kg)
=
= 1.623 × 10−27 kg
mH + mCl 1.67 × 10−27 kg + 5.81 × 10−26 kg

I
2.71 × 10−47 kg ⋅ m 2
=
= 1.29 × 10−10 m = 0.129 nm
mr
1.623 × 10−27 kg

EVALUATE: This is a typical atomic separation for a diatomic molecule; see Example 42.2 for the
corresponding distance for CO.
(b) IDENTIFY: Each transition is from the level l to the level l − 1. The rotational energies are given by
Eq. (42.3). The transition energy is related to the photon wavelength by ΔE = hc /λ .

⎛ ћ2 ⎞ ⎛ ћ2 ⎞
SET UP: El = l (l + 1)ћ 2 /2 I , so ΔE = El − El −1 = [l (l + 1) − l (l − 1)]⎜ ⎟ = l ⎜ ⎟ .
⎜ ⎟ ⎜ ⎟
⎝ 2I ⎠ ⎝ I ⎠
⎛ ћ 2 ⎞ ћc
EXECUTE: l ⎜ ⎟ =
⎜ I ⎟ λ
⎝ ⎠
l=

2π cI 2π (2.998 × 108 m/s)(2.71 × 10−47 kg ⋅ m 2 ) 4.843 × 10−4 m
=
=
λ
ћλ
(1.055 × 10−34 J ⋅ s)λ

For λ = 60.4 μ m, l =
For λ = 69.0 μ m, l =
For λ = 80.4 μ m, l =
For λ = 96.4 μ m, l =

4.843 × 10−4 m
60.4 × 10−6 m
4.843 × 10−4 m
69.0 × 10−6 m
4.843 × 10−4 m
80.4 × 10−6 m

4.843 × 10−4 m

96.4 × 10−6 m

For λ = 120.4 μ m, l =

= 8.
= 7.
= 6.

= 5.

4.843 × 10−4 m

= 4.
120.4 × 10−6 m
EVALUATE: In each case l is an integer, as it must be.

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42-12

Chapter 42
(c) IDENTIFY and SET UP: Longest λ implies smallest ΔE , and this is for the transition from l = 1 to
l = 0.

⎛ ћ2 ⎞
(1.055 × 10−34 J ⋅ s) 2
= 4.099 × 10−22 J
EXECUTE: ΔE = l ⎜ ⎟ = (1)

2
−47
⎜ I ⎟
×
⋅
2.71
10
kg
m
⎝ ⎠
hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 4.85 × 10−4 m = 485 μ m.
ΔE
4.099 × 10−22 J
EVALUATE: This is longer than any wavelengths in part (b).
(d) IDENTIFY: What changes is mr , the reduced mass of the molecule.

λ=

⎛ ћ2 ⎞
2π cI
hc
SET UP: The transition energy is ΔE = l ⎜ ⎟ and ΔE = , so λ =
(part (b)). I = mr r 2 , so λ is
⎜ I ⎟
λ
lћ
⎝ ⎠
λ (HCl)

λ (DCl)
m (DCl)
=
so λ (DCl) = λ (HCl) r
directly proportional to mr .
mr (HCl) mr (DCl)
mr (HCl)
EXECUTE: The mass of a deuterium atom is approximately twice the mass of a hydrogen atom, so
mD = 3.34 × 10−27 kg.
mr (DCl) =

m D m Cl
(3.34 × 10−27 kg)(5.81 × 10−27 kg)
=
= 3.158 × 10−27 kg
m D + m Cl 3.34 × 10−27 kg + 5.81 × 10−26 kg

⎛ 3.158 × 10−27 kg ⎞
= (1.946)λ (HCl)
⎜ 1.623 × 10−27 kg ⎟⎟
⎝
⎠
l = 8 → l = 7; λ = (60.4 μ m)(1.946) = 118 μ m
l = 7 → l = 6; λ = (69.0 μ m)(1.946) = 134 μ m
l = 6 → l = 5; λ = (80.4 μ m)(1.946) = 156 μ m
l = 5 → l = 4; λ = (96.4 μ m)(1.946) = 188 μ m
l = 4 → l = 3; λ = (120.4 μ m)(1.946) = 234 μ m

λ (DCl) = λ (HCl) ⎜

42.42.

EVALUATE: The moment of inertia increases when H is replaced by D, so the transition energies decrease
and the wavelengths increase. The larger the rotational inertia the smaller the rotational energy for a given l
(Eq. 42.3).
l 2
IDENTIFY: Problem 42.15b shows that for the l → l − 1 transition, ΔE =
. I = mr r02 .
I
SET UP: mr =

(3.82 × 10−26 kg)(3.15 × 10−26 kg)

EXECUTE: I =

r0 =

3.82 × 10

−26

kg + 3.15 × 10

−26

kg

= 1.726 × 10−26 kg.

2

l
hl λ
=
= 6.43 × 10−46 kg ⋅ m 2 and from Eq. (42.6) the separation is
ΔE 4π 2 c

I
= 0.193 nm.
mr

EVALUATE: Section 42.1 says r0 = 0.24 nm for NaCl. Our result for NaF is smaller than this. This makes
42.43.

sense, since F is a smaller atom than Cl.
2
L2
l (l + 1)
. Eg = 0 (l = 0), and there is an additional multiplicative factor of 2l + 1
IDENTIFY: Eex =
=
2I
2I
because for each l state there are really (2l + 1) ml -states with the same energy.
SET UP: From Example 42.3, I = 1.449 × 10−46 kg ⋅ m 2.
EXECUTE: (a)

nl
= (2l + 1)e−
n0

2

l (l +1)/(2 IkT )

.

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Molecules and Condensed Matter

(b) (i) El =1 =

2

(1)(1 + 1)

= 7.67 × 10−23 J.

2(1.449 × 10−46 kg ⋅ m 2 )
n
(2l + 1) = 3, so l =1 = (3)e−0.0185 = 2.95.
n0

(ii)

42-13

El =1
7.67 × 10−23 J
=
= 0.0185.
kT
(1.38 × 10−23 J/K)(300 K)

2
El = 2
(2) (2 + 1)
=
= 0.0556. (2l + 1) = 5, so
−
46
kT
2(1.449 × 10
kg ⋅ m 2 )(1.38 × 10−23 J/K)(300 K)

nl =1
= (5)(e−0.0556 ) = 4.73.
n0
2
El =10
(10) (10 + 1)
=
= 1.02.
−46
kT
2(1.449 × 10 kg ⋅ m 2 )(1.38 × 10−23 J/K)(300 K)
n

(2l + 1) = 21, so l =10 = (21)(e−1.02 ) = 7.57.
n0

(iii)

(iv)

2
El = 20
(20)(20 + 1)
=
= 3.89. (2l + 1) = 41, so
46
−
kT
2(1.449 × 10 kg ⋅ m 2 ) (1.38 × 10−23 J/K) (300 K)

nl = 20
= (41)e−3.89 = 0.838.
n0
(v)

2
El = 50
(50)(50 + 1)
=
= 23.6. (2l + 1) = 101, so
46
−
kT

2(1.449 × 10 kg ⋅ m 2 )(1.38 × 10−23 J/K)(300 K)

nl =50
= (101)e−23.6 = 5.69 × 10−9.
n0

42.44.

EVALUATE: (c) There is a competing effect between the (2l + 1) term and the decaying exponential. The
2l + 1 term dominates for small l, while the exponential term dominates for large l.
IDENTIFY: The rotational energy levels are given by Eq. (42.3). The transition energy ΔE for the
hc
molecule and λ for the photon are related by ΔE = .

λ

SET UP: From Example 42.2, I CO = 1.449 × 10
EXECUTE: (a) El =1 =

2

−46

kg ⋅ m 2 .

l (l + 1) (1.054 × 10−34 J ⋅ s) 2 (1)(1 + 1)
=
= 7.67 × 10−23 J. El = 0 = 0.
−46
2

2I
2(1.449 × 10 kg ⋅ m )

ΔE = 7.67 × 10−23 J = 4.79 × 10−4 eV. λ =

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
=
ΔE
(7.67 × 10−23 J)

2.59 × 10−3 m = 2.59 mm.
EVALUATE: (b) Let’s compare the value of kT when T = 20 K to that of ΔE for the l = 1 → l = 0
rotational transition: kT = (1.38 × 10−23 J/K)(20 K) = 2.76 × 10−22 J.

kT
= 3.60. Therefore, although T is quite small, there is still plenty
ΔE
of energy to excite CO molecules into the first rotational level. This allows astronomers to detect the
2.59 mm wavelength radiation from such molecular clouds.
IDENTIFY and SET UP: El = l (l + 1)ћ 2 /2 I , so El and the transition energy ΔE depend on I. Different
ΔE = 7.67 × 10−23 J (from part (a)). So

42.45.

isotopic molecules have different I.
EXECUTE: (a) Calculate I for Na 35Cl:

mr =

mNa mCl
(3.8176 × 10−26 kg)(5.8068 × 10−26 kg)
=
= 2.303 × 10−26 kg
mNa + mCl 3.8176 × 10−26 kg + 5.8068 × 10−26 kg

I = mr r 2 = (2.303 × 10−26 kg)(0.2361 × 10−9 m) 2 = 1.284 × 10−45 kg ⋅ m 2

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42-14

Chapter 42

l = 2 → l = 1 transition
⎛ ћ 2 ⎞ 2ћ 2 2(1.055 × 10−34 J ⋅ s) 2
ΔE = E2 − E1 = (6 − 2) ⎜ ⎟ =
=
= 1.734 × 10−23 J
⎜ 2I ⎟
I
1.284 × 10−45 kg ⋅ m 2
⎝ ⎠
hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 1.146 × 10−2 m = 1.146 cm
ΔE
λ

1.734 × 10−23 J
l = 1 → l = 0 transition
ΔE =

hc

so λ =

⎛ ћ2 ⎞ ћ2 1
ΔE = E1 − E0 = (2 − 0) ⎜ ⎟ =
= (1.734 × 10−23 J) = 8.67 × 10−24 J
⎜ 2I ⎟ I
2
⎝ ⎠

λ=

hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 2.291 cm
ΔE
8.67 × 10−24 J

(b) Calculate I for Na 37Cl: mr =

mNa mCl
(3.8176 × 10−26 kg)(6.1384 × 10−26 kg)
=
= 2.354 × 10−26 kg
mNa + mCl 3.8176 × 10−26 kg + 6.1384 × 10−26 kg

I = mr r 2 = (2.354 × 10−26 kg)(0.2361 × 10−9 m) 2 = 1.312 × 10−45 kg ⋅ m 2
l = 2 → l = 1 transition
ΔE =

2ћ 2 2(1.055 × 10−34 J ⋅ s) 2
=
= 1.697 × 10−23 J
I
1.312 × 10−45 kg ⋅ m 2

hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 1.171 × 10−2 m = 1.171 cm
ΔE
1.697 × 10−23 J
l = 1 → l = 0 transition

λ=

ΔE =

ћ2 1
= (1.697 × 10−23 J) = 8.485 × 10−24 J
I
2

hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
=
= 2.341 cm

ΔE
8.485 × 10−24 J
The differences in the wavelengths for the two isotopes are:
l = 2 → l = 1 transition: 1.171 cm − 1.146 cm = 0.025 cm
l = 1 → l = 0 transition: 2.341 cm − 2.291 cm = 0.050 cm

λ=

EVALUATE: Replacing
but measurable.
42.46.

IDENTIFY: ΔE = hf =
SET UP: mr =

35

Cl by

37

Cl increases I, decreases ΔE and increases λ. The effect on λ is small

k′
.
mr

mO mH
= 1.57 × 10−27 kg
mO + mH

EXECUTE: The vibration frequency is f =

ΔE
= 1.12 × 1014 Hz. The force constant is
h

k ′ = (2π f ) 2 mr = 777 N/m.
42.47.

EVALUATE: This would be a fairly stiff spring in an ordinary physics lab.
1⎞
k′
⎛
. The zero-point energy is
IDENTIFY: The vibrational energy levels are given by En = ⎜ n + ⎟
⎝
2⎠
mr

E0 =

1
2

2k ′
.
mH

SET UP: For H 2 , mr =

mH
.
2

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Molecules and Condensed Matter

EXECUTE: E0 =

42-15

1
2(576 N/m)
(1.054 × 10−34 J ⋅ s)
= 4.38 × 10−20 J = 0.274 eV.
2
1.67 × 10−27 kg

EVALUATE: This is much less than the magnitude of the H 2 bond energy.
42.48.

IDENTIFY: The frequency is proportional to the reciprocal of the square root of the reduced mass. The
hc
transition energy ΔE and the wavelength of the light emitted are related by ΔE = .

λ

SET UP:

f 0 = 1.24 × 1014 Hz.

EXECUTE: (a) In terms of the atomic masses, the frequency of the isotope with the deuterium atom is

⎛ m m /(m + mF ) ⎞
f = f0 ⎜ F H H
⎝ m m /(m + m ) ⎠⎟
F D

D

1/2

F

⎛ 1 + (mF /mD ) ⎞
= f0 ⎜
⎝ 1 + (m /m ) ⎠⎟
F

(b) For the molecule, ΔE = hf . hf =

42.49.

hc

λ

H

, so λ =

1/ 2

. Using f 0 and the given masses, f = 8.99 × 1013 Hz.
c 3.00 × 108 m/s
=
= 3.34 × 10−6 m = 3340 nm. This
f 8.99 × 1013 Hz

wavelength is in the infrared.
EVALUATE: The vibrational frequency of the molecule equals the frequency of the light that is emitted.
IDENTIFY and SET UP: Use Eq. (42.6) to calculate I. The energy levels are given by Eq. (42.9). The
transition energy ΔE is related to the photon wavelength by ΔE = hc /λ .
EXECUTE: (a) mr =

mH mI
(1.67 × 10−27 kg)(2.11 × 10−25 kg)
=
= 1.657 × 10−27 kg
mH + mI 1.67 × 10−27 kg + 2.11 × 10−25 kg

I = mr r 2 = (1.657 × 10−27 kg)(0.160 × 10−9 m)2 = 4.24 × 10−47 kg ⋅ m 2
⎛ ћ2 ⎞
k′
(b) The energy levels are Enl = l (l + 1) ⎜ ⎟ + n + 12 ћ
(Eq. (42.9))

⎜ 2I ⎟
mr
⎝ ⎠

(

)

⎛ ћ2 ⎞
k′
= ω = 2π f so Enl = l (l + 1) ⎜ ⎟ + (n + 12 )hf
⎜ 2I ⎟
m
⎝ ⎠

(i) transition n = 1 → n = 0, l = 1 → l = 0

⎛ ћ2 ⎞
ћ2
Δ E = (2 − 0) ⎜ ⎟ + 1 + 12 − 12 hf =
+ hf
⎜ 2I ⎟
I
⎝ ⎠
hc
hc
c
hc
ΔE =
so λ =

= 2
=
λ
ΔE (ћ /I ) + hf (ћ /2π I ) + f

(

ћ
2π I

=

)

1.055 × 10−34 J ⋅ s
2π (4.24 × 10−47 kg ⋅ m 2 )

= 3.960 × 1011 Hz

c
2.998 + 108 m/s
=
= 4.30 μ m
( ћ /2π I ) + f 3.960 × 1011 Hz + 6.93 × 1013 Hz
(ii) transition n = 1 → n = 0, l = 2 → l = 1

λ=

⎛ ћ2 ⎞
2ћ 2

ΔE = (6 − 2) ⎜ ⎟ + hf =
+ hf
⎜ 2I ⎟
I
⎝ ⎠

λ=

c
2(ћ /2π I ) + f

=

2.998 × 108 m/s

2(3.960 × 1011 Hz) + 6.93 × 1013 Hz
(iii) transition n = 2 → n = 1, l = 2 → l = 3

= 4.28 μ m

⎛ ћ2 ⎞
3ћ 2
ΔE = (6 − 12) ⎜ ⎟ + hf = −
+ hf
⎜ 2I ⎟
I
⎝ ⎠

λ=

c
2.998 × 108 m/s
=
= 4.40 μ m
−3(ћ /2π I ) + f −3(3.960 × 1011 Hz) + 6.93 × 1013 Hz

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42-16

42.50.

42.51.

Chapter 42
EVALUATE: The vibrational energy change for the n = 1 → n = 0 transition is the same as for the
n = 2 → n = 1 transition. The rotational energies are much smaller than the vibrational energies, so the
wavelengths for all three transitions don’t differ much.
1
.
IDENTIFY and SET UP: P ( E ) = f ( E ) = ( E − E )/kT
F
+1
e
EXECUTE: The sum of the probabilities is
1
1
1

e−ΔE /kT
f ( EF − ΔE ) + f ( EF + ΔE ) = −ΔE kT
+ ΔE kT
= −ΔE kT
+
= 1. Therefore,
+1 e
+1 e
+ 1 1 + e−ΔE kT
e
f ( EF − ΔE ) = 1 − f ( EF + ΔE ).
EVALUATE: This result is true for all T, even though P is strongly dependent on temperature.
IDENTIFY: EF0 is given by Eq. (42.20). Since potassium is a metal and E does not change much with T

for metals, we approximate EF by EF0 , so EF =

32/3 π 4/3 2 n 2/3
.
2m

SET UP: The number of atoms per m3 is ρ /m. If each atom contributes one free electron, the electron

concentration is n =
EXECUTE: EF =

ρ
m

=

851 kg/m3
6.49 × 10−26 kg

= 1.31 × 1028 electrons/m3.

32/3 π 4/3 (1.054 × 10−34 J ⋅ s)2 (1.31 × 1028 /m3 )2/3
2(9.11 × 10

−31

kg)

= 3.24 × 10−19 J = 2.03 eV.

EVALUATE: The EF we calculated for potassium is about a factor of three smaller than the EF for
42.52.

copper that was calculated in Example 42.7.
IDENTIFY: The only difference between the two isotopes is their mass, which will affect their reduced
mass and hence their moment of inertia.
ћ2
SET UP: The rotational energy states are given by E = l (l + 1)
and the reduced mass is given by
2I
m1 = m1m2 /(m2 + m2 ).
EXECUTE: (a) If we call m the mass of the H-atom, the mass of the deuterium atom is 2m and the reduced
masses of the molecules are
H 2 (hydrogen): mr (H) = mm/( m + m) = m/2

D 2 (deuterium): mr (D) = (2m)(2m)/(2m + 2m) = m

Using I = mr r02 , the moments of inertia are I H = mr0 2 /2 and I D = mr02 . The ratio of the rotational energies
is then

EH l (l + 1)( ћ 2 /2 I H ) I D mr02
=
=
=
= 2.
ED l (l + 1)( ћ 2 /2 I D ) I H m r 2
0
2

(b) The ratio of the vibrational energies is

42.53.

1⎞
k′
⎛
⎜⎝ n + ⎟⎠ ћ
2
mr (H)

EH
mr (D)
m
=
=
=
= 2.

ED ⎛
mr (H)
m/2
1⎞
k′
⎜⎝ n + ⎟⎠ ћ
2
mr (D)

EVALUATE: The electrical force is the same for both molecules since both H and D have the same charge,
so it is reasonable that the force constant would be the same for both of them.
IDENTIFY and SET UP: Use the description of the bcc lattice in Fig.42.11c in the textbook to calculate the
number of atoms per unit cell and then the number of atoms per unit volume.
EXECUTE: (a) Each unit cell has one atom at its center and 8 atoms at its corners that are each shared by
8 other unit cells. So there are 1 + 8/8 = 2 atoms per unit cell.
n
2
=
= 4.66 × 10−8 atoms/m3
V (0.35 × 10−9 m)3

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Molecules and Condensed Matter

(b) EF0 =

32/3 π 4/3ћ 2 ⎛ N ⎞

⎝⎜ V ⎠⎟
2m

42-17

2/3

In this equation N/V is the number of free electrons per m3. But the problem says to assume one free
electron per atom, so this is the same as n/V calculated in part (a).
m = 9.109 × 10−31 kg (the electron mass), so EF0 = 7.563 × 10−19 J = 4.7 eV
42.54.

EVALUATE: Our result for metallic lithium is similar to that calculated for copper in Example 42.7.
d
IDENTIFY and SET UP: At r where U tot is a minimum,
U tot = 0.
dr
8 A4πε 0
d
α e2 1
1
EXECUTE: (a)
− 8 A 9 . Setting this equal to zero when r = r0 gives r07 =
and
U tot =
dr
4πε 0 r 2
α e2
r

so U tot =

α e2 ⎛ 1 r07 ⎞
7α e2
= −1.26 × 10−18 J = −7.85 eV.
⎜ − + 8 ⎟ . At r = r0 , U tot = −
4πε 0 ⎝ r 8r ⎠
32πε 0 r0

(b) To remove a Na + Cl− ion pair from the crystal requires 7.85 eV. When neutral Na and Cl atoms are

formed from the Na + and Cl− atoms there is a net release of energy −5.14 eV + 3.61 eV = −1.53 eV, so
the net energy required to remove a neutral Na Cl pair from the crystal is 7.85 eV − 1.53 eV = 6.32 eV.
42.55.

EVALUATE: Our calculation is in good agreement with the experimental value.
dE
(a) IDENTIFY and SET UP: p = − tot . Relate Etot to EF0 and evaluate the derivative.
dV
3N
3 ⎛ 32/3 π 4/3 2 ⎞ 5/3 −2/3
EXECUTE: Etot = NEav =
EF0 = ⎜
⎟N V
5
5⎝
2m
⎠

⎛ 32/3 π 4/3ћ 2 ⎞ ⎛ N ⎞ 5/3

dEtot 3 ⎛ 32/3 π 4/3ћ 2 ⎞ 5/3 ⎛ 2 −5/3 ⎞
= ⎜
⎟⎠ so p = ⎜
⎟ N ⎝⎜ − V
⎟ ⎜ ⎟ , as was to be shown.
dV
5⎝
2m
3
5m
⎠
⎝
⎠⎝V ⎠
(b) N /V = 8.45 × 1028 m −3
⎛ 32/3 π 4/3 (1.055 × 10−34 J ⋅ s) 2 ⎞
−3 5/3
28
10
5
p=⎜
⎟ (8.45 × 10 m ) = 3.81 × 10 Pa = 3.76 × 10 atm.
5(9.109 × 10−31 kg)
⎝
⎠

42.56.

(c) EVALUATE: Normal atmospheric pressure is about 105 Pa, so these pressures are extremely large.
The electrons are held in the metal by the attractive force exerted on them by the copper ions.
5/3

32/3 π 4/3 2 ⎛ N ⎞
(a) IDENTIFY and SET UP: From Problem 42.53, p =
⎜⎝ ⎟⎠ . Use this expression to calculate
5m
V
dp /dV .

EXECUTE: (a) B = −V
(b)

⎡ 5 32/3 π 4/3
dp
= −V ⎢ ⋅
5m
dV
⎢⎣ 3

N
5 32/3 π 4/3
= 8.45 × 1028 m−3. B = ⋅
V
3
5m

2

2

⎛ N⎞
⋅⎜ ⎟

⎝V ⎠

2/3

⎛ −N ⎞ ⎤ 5
⎜⎝ 2 ⎟⎠ ⎥ = 3 p.
V ⎥⎦

(8.45 × 1028 m −3 )5/3 = 6.33 × 1010 Pa.

EVALUATE: (c) The fraction of B due to the free electrons is
42.57.

6.33 × 1010 Pa
1.4 × 1011 Pa

= 0.45. The copper ions

themselves make up the remaining fraction.
IDENTIFY and SET UP: Follow the steps specified in the problem.
2/3
32/3 π 4/3 2 ⎛ N ⎞
1
mc 2 .
EXECUTE: (a) EF0 =
⎜⎝ ⎟⎠ . Let EF0 =
2m
V
100

2m 2 c 2
⎛ N⎞ ⎡
⎜⎝ ⎟⎠ = ⎢
2/3 4/3
V
⎢⎣ (100)3 π

⎤
2⎥
⎦⎥

3/2

=

23/2 m3c3
1003/23π 2

3

=

23/2 m3c3
3000π 2

3

= 1.67 × 1033 m −3 .

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42-18

Chapter 42

(b)

8.45 × 1028 m −3
1.67 × 1033 m −3

= 5.06 × 10−5. Since the real concentration of electrons in copper is less than one part in

10−4 of the concentration where relativistic effects are important, it is safe to ignore relativistic effects for
most applications.
6(2 × 1030 kg)
(c) The number of electrons is N e =
= 6.03 × 1056. The concentration is
1.99 × 10−26 kg
Ne
6.03 × 1056
=
= 6.66 × 1035 m −3.
4 π (6.00 × 106 m)3
V
3
EVALUATE: (d) Comparing this to the result from part (a)
42.58.

6.66 × 1035 m −3
1.67 × 1033 m −3

≅ 400 so relativistic effects

will be very important.
IDENTIFY: The current through the diode is related to the voltage across it.
SET UP: The current through the diode is given by I = IS (eeV/kT – 1).
EXECUTE: (a) The current through the resistor is (35.0 V)/(125 Ω) = 0.280 A = 280 mA, which is also the
current through the diode. This current is given by
I = IS (eeV/kT – 1), giving 280 mA = 0.625 mA(eeV/kT – 1) and 1 + (280/0.625) = 449 = eeV/kT . Solving for V
kT ln 449 (1.38 × 10−23 J/K)(293 K)ln 449
=
= 0.154 V
e
1.60 × 10−19 C
(b) R = V/I = (0.154 V)/(0.280 A) = 0.551 Ω

at T = 293 K gives V =

42.59.

EVALUATE: At a different voltage, the diode would have different resistance.
IDENTIFY and SET UP: For a pair of point charges q1 and q2 separated by a distance r12 the electric

potential energy is U =

1 q1q2
. Sum over all pairs of charges.
4πε 0 r12

EXECUTE: (a)
qi q j
q 2 ⎛ −1 1
q2 ⎛ 2 2
1
1
1
1 1⎞
1
1 ⎞
U=
∑
=
−
+ − ⎟=
−
⎜⎝ + −
⎜ − −
⎟.
4πε 0 i < j rij
4πε 0 d r r + d r − d r d ⎠ 4πε 0 ⎝ r d r + d r − d ⎠
⎛
⎞
1
1
1⎜ 1
1 ⎟ 1⎛ d d2
d d 2 ⎞ 2 2d 2
+

= ⎜
+
≈ ⎜1 − + 2 + … + 1 + + 2 ⎟ ≈ + 3
⎟
r + d r − d r ⎜ 1 + d 1 − d ⎟ r ⎜⎝
r r
r r ⎟⎠ r
r
r
r⎠
⎝
2 p2
−2q 2 ⎛ 1 d 2 ⎞ −2 p 2
.
⇒U =
−
⎜ + 3⎟ =
3
4πε 0 ⎝ d r ⎠ 4πε 0 r
4πε 0 d 3

But

(b) U =

1

∑

qi q j

4πε 0 i < j rij

=

q 2 ⎛ −1 1
q 2 ⎛ −2 2 2 2d 2 ⎞
1
1
1 1⎞
+
− − ⎟=
− + + 3 ⎟=
⎜⎝ − +
⎜
4πε 0 d r r + d r − d r d ⎠ 4πε 0 ⎝ d r r
r ⎠

−2 p 2
2 p2
−2q 2 ⎛ 1 d 2 ⎞
+
.
⎜ − 3 ⎟ ⇒U =
4πε 0 ⎝ d r ⎠
4πε 0 d 3 4πε 0 r 3

If we ignore the potential energy involved in forming each individual molecule, which just involves a
different choice for the zero of potential energy, then the answers are:
−2 p 2

(a) U =
. The interaction is attractive.
4πε 0 r 3
(b) U =

+2 p 2

4πε 0 r 3

. The interaction is repulsive.

EVALUATE: In each case the interactions between the two dipoles involve two interactions between like
charges and two between unlike charges. But in part (a) the two closest charges are unlike and in (b) the
two closest charges are alike.
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Molecules and Condensed Matter
42.60.

42-19

IDENTIFY: Follow the procedure specified in the hint.
SET UP: According to Eq. (42.8), the vibrational level spacing is ΔE = ω = k ′/m .
⎛ 1 e2 ⎞
1 e2
1 e2
EXECUTE: (a) Following the hint, k ′dr = − d ⎜
=

dr
k
. The energy
=
and
′
⎟
2πε 0 r03
⎝ 4πε 0 r 2 ⎠ r = r 2πε 0 r03
0

level spacing therefore is ω =

2k ′/m =

1

e

2

πε 0 mr03

= 1.23 × 10−19 J = 0.77 eV, where (m/2) has been

used for the reduced mass.
(b) The reduced mass is doubled, and the energy is reduced by a factor of 2 to 0.54 eV.
EVALUATE: The vibrational level spacing is inversely proportional to the square root of the reduced mass
of the molecule. The force constant depends on the bond between the two atoms.

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