serway1 (1)

Atomic Physics 501

P42.25 (a) n = 1: For n = 1, = 0, m = 0, ms = ± 1
2
n
1 m ms
1
00 –1/2
00 +1/2

Yields 2 sets; 2n2 = 2(1)2 = 2

(b) n = 2: For n = 2,
we have

n m ms

2 0 0 ±1/2
2 1 –1 ±1/2
2 1 0 ±1/2
2 1 1 ±1/2

Yields 8 sets; 2n2 = 2(2)2 = 8

Note that the number is twice the number of m values. Also, for each there are (2 + 1)

different m values. Finally, can take on values ranging from 0 to n − 1.

So the general expression is n−1
The series is an arithmetic progression:
number = ∑ 2(2 + 1)
0

2 + 6 + 10 + 14…

the sum of which is number = n [2a + (n − 1) d]
where a = 2, d = 4:
2

number = n [4 + (n − 1) 4] = 2n2

2

(c) n = 3: 2(1) + 2(3) + 2(5) = 2 + 6 + 10 = 18 2n2 = 2(3)2 = 18

(d) n = 4: 2(1) + 2(3) + 2(5) + 2(7) = 32 2n2 = 2(4)2 = 32

(e) n = 5: 32 + 2(9) = 32 + 18 = 50 2n2 = 2(5)2 = 50

P42.26 For a 3d state, n = 3 and = 2

Therefore, L = ( + 1) = 6 = 2.58 × 10−34 J ⋅ s

m can have the values –2, –1, 0, 1, and 2

so Lz can have the values − 2 , − , 0, and 2 .
Using the relation cosθ = Lz
we find the possible values of q
L
145°, 114°, 90.0°, 65.9°, and 35.3°

502 Chapter 42

m 1.67 × 10−27 kg 3.99 × 1017 kg m3
V
3)π 1.00 × 10−15
*P42.27 (a) Density of a proton: ( )ρ= = (4 m 3 =
(b)
( ( ) )Size of model electron: 1 3 ⎞1 3
⎛ 3m ⎞ ⎛ 3 9.11 × 10−31 kg ⎠⎟
r = ⎝⎜ 4π ρ ⎟⎠ = ⎝⎜ 4π 3.99 × 1017 kg m3 = 8.17 × 10−17 m

(c) Moment of inertia: ( )( )I = 2 mr2 = 2 9.11 × 10−31 kg 8.17 × 10−17 m 2
55
= 2.43 × 10−63 kg ⋅ m2

Lz = Iω = 2 = Iv
r

( )( )v = r =
( )2I
Therefore, 6.626 × 10−34 J ⋅ s 8.17 × 10−17 m = 1.77 × 1012 ms
2π 2 × 2.43 × 10−63 kg ⋅ m2

(d) This is 5.91 × 103 times larger than the speed of light. So the spinning-solid-ball model
of an electron with spin angular momentum is absurd.

P42.28 In the N shell, n = 4. For n = 4, can take on values of 0, 1, 2, and 3. For each value of , m
can be − to in integral steps. Thus, the maximum value for m is 3. Since Lz = m , the
maximum value for Lz is Lz = 3 .

P42.29 The 3d subshell has = 2, and n = 3. Also, we have s = 1.
Therefore, we can have n = 3, = 2; m = −2, − 1, 0, 1, 2; s = 1; and ms = −1, 0, 1

leading to the following table:

n 3 3 3 3 3 3 333 333 333
2 2 2 2 2 2 222 222 222

m –2 –2 –2 –1 –1 –1 0 0 0 1 1 1 2 2 2
s 1 1 1 1 1 1 111 111 111
ms –1 0 1 –1 0 1 –1 0 1 –1 0 1 –1 0 1

Atomic Physics 503

Section 42.7 The Exclusion Principle and the Periodic Table

P42.30 (a) 1s2 2s2 2 p4

(b) For the 1s electrons, n = 1, = 0, m = 0, ms = + 1 and −1
2 and 2
or
For the two 2s electrons, n = 2, = 0, m = 0, ms = + 1 −1
2 2

For the four 2p electrons, n = 2; = 1; m = −1, 0, or 1; and ms = + 1 −1
2 2

So one possible set of quantum numbers is

n1 1 22 2 222
00
0 0 00 1 111
–1 1
m0 0 –1 0 1 –1
22
ms 1 – 1 1 –1 1 1
2 2 2 2
22

P42.31 The 4s subshell fills first , for potassium and calcium, before the 3d subshell starts to fill for

scandium through zinc. Thus, we would first suppose that [Ar]3d4 4s2 would have lower energy
than [Ar]3d5 4s1. But the latter has more unpaired spins, six instead of four, and Hund’s rule
suggests that this could give the latter configuration lower energy. In fact it must, for [Ar]3d5 4s1

is the ground state for chromium.

P42.32 Electronic configuration: Sodium to Argon

⎣⎡1s2 2s2 2 p6 ⎦⎤ +3s1 → Na11
+3s2 → Mg12
→ Al13
+3s2 3 p1 → Si14
→ P15
+3s2 3 p2 → S16
→ Cl17
+3s2 3 p3 → Ar18
→ K19
+3s2 3 p4

+3s2 3 p5

+3s2 3 p6

⎣⎡1s2 2s2 2 p6 3s2 3 p6 ⎦⎤ 4s1

P42.33 In the table of electronic configurations in the text, or on a periodic table, we look for the element
whose last electron is in a 3p state and which has three electrons outside a closed shell. Its

electron configuration then ends in 3s2 3p1. The element is aluminum .

504 Chapter 42

P42.34 (a) For electron one and also for electron two, n = 3 and = 1 . The possible states are listed
here in columns giving the other quantum numbers:

electron m 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0

one 1 1 1 1 1 −1 −1 −1 −1 −1 1 1 1 1 1
ms 2 2 2 2 2 2 2 2 2 22 2 2 2 2

electron m 1 0 0 –1 –1 1 0 0 –1 –1 1 1 0 –1 –1

two ms −1 1 −1 1 −1 1 1 −1 1 −1 1 −1 −1 1 −1
2 2 2 2 2 2 2 22 22 2 22 2

electron m 0 0 0 0 0 –1 –1 –1 –1 –1 –1 –1 –1 –1 –1

one ms −1 −1 −1 −1 −1 1 1 1 1 1 −1 −1 −1 −1 −1
2 2 2 2 2 2 2 2 2 2 22222

electron m 1 1 0 –1 –1 1 1 0 0 –1 1 1 0 0 –1

two ms 1 −1 1 1 −1 1 −1 1 −1 −1 1 −1 1 −1 1
22 2 2 22 22 2 22 22 22

There are thirty allowed states, since electron one can have any of three possible values for
m for both spin up and spin down, amounting to six states, and the second electron can
have any of the other five states.

(b) Were it not for the exclusion principle, there would be 36 possible states, six for each
electron independently.

P42.35 (a) n+ 12 3 4 5 6 7
subshell
1s 2s 2p, 3s 3p, 4s 3d, 4p, 5s 4d , 5p, 6s 4f , 5d, 6p, 7s

(b) Z = 15: Filled subshells: 1s, 2s, 2 p, 3s (12 electrons)
Z = 47: Valence subshell: 3 electrons in 3p subshell
Prediction: Valence = −3 or +5
Element is phosphorus, Valence = −3 or +5 (Prediction correct)

Filled subshells: 1s, 2s, 2 p, 3s, 3p, 4s, 3d, 4 p, 5s

(38 electrons)

Outer subshell: 9 electrons in 4d subshell
Prediction:
Element is silver, Valence = −1 Valence is +1
(Prediction fails)
Filled subshells:
Z = 86: 1s, 2s, 2 p, 3s, 3p, 4s, 3d, 4 p, 5s, 4d, 5p, 6s,
4 f , 5d, 6 p (86 electrons)
Prediction Outer subshell is full: inert gas
Element is radon, inert (Prediction correct)

Atomic Physics 505

P42.36 Listing subshells in the order of filling, we have for element 110,
1s2 2s2 2 p6 3s2 3 p6 4s2 3d10 4 p6 5s2 4d10 5 p6 6s2 4 f 14 5d10 6 p6 7s2 5 f 14 6d8
In order of increasing principal quantum number, this is
1s2 2s2 2 p6 3s2 3 p6 3d10 4s2 4 p6 4d10 4 f 14 5s2 5 p6 5d10 5 f 14 6s2 6 p6 6d8 7s2

P42.37 In the ground state of sodium, the outermost electron is in an s state. This state is spherically

symmetric, so it generates no magnetic field by orbital motion, and has the same energy no matter

whether the electron is spin-up or spin-down. The energies of the states 3p ↑ and 3p ↓ above 3s

are hf1 = hc and hf2 = hc .
λ λ2

The energy difference is

2µB B = ⎛ 1 − 1 ⎞
hc ⎝⎜ λ1 λ2 ⎠⎟

so B = hc ⎛ 1 − 1 ⎞
2µB ⎝⎜ λ1 λ2 ⎠⎟

( ( )( ) )=
6.63 × 10−34 J ⋅ s 3 × 108 ms ⎜⎛⎝ 1 − 1 m ⎠⎞⎟
2 9.27 × 10−24 J T 588.995 × 589.592 × 10−9
10 −9 m

B = 18.4 T

Section 42.8 More on Atomic Spectra: Visible and X-ray

P42.38 (a) n = 3, = 0, m = 0

n = 3, = 1, m = −1, 0, 1

For n = 3, = 2, m = −2, − 1, 0, 1, 2

(b) ψ 300 corresponds to E300 = − Z 2E0 = − 22 (13.6) = −6.05 eV .
n2
32

ψ 31−1, ψ 310 , ψ 311 have the same energy since n is the same.

ψ 32−2 , ψ 32−1, ψ 320 , ψ 321, ψ 322 have the same energy since n is the same.

All states are degenerate.

506 Chapter 42

*P42.39 For the 3p state, En = −13.6 eV Z2 becomes −3.0 eV = −13.6 eV Z2 so Zeff = 1.4
eff eff

n2 32

For the 3d state −1.5 eV = −13.6 eV Z2 so Zeff = 1.0
eff

32

When the outermost electron in sodium is promoted from the 3s state into a 3p state, its wave
function still overlaps somewhat with the ten electrons below it. It therefore sees the +11e nuclear
charge not fully screened, and on the average moves in an electric field like that created by a
particle with charge +11e – 9.6e = 1.4e. When this valence electron is lifted farther to a 3p state,
it is essentially entirely outside the cloud of ten electrons below it, and moves in the field of a net
charge +11e – 10e = 1e.

*P42.40 (a) Picture all of the energy of an electron after its acceleration going into producing a single

photon. Then we have E= hc = e∆V and
λ

∆V = hc = 6.626 × 10−34 J ⋅ s 2.998 × 108 m/s = 1240 V ⋅ nm
λ λ 1.602 × 10−19 J/eV λ

(b) The potential difference is inversely proportional to the wavelength.

(c) Yes. It predicts a minimum wavelength of 33.5 pm when the accelerating voltage is 37 keV,
in agreement with the minimum wavelength in the figure.

(d) Yes, but it might be unlikely for a very high-energy electron to stop in a single interaction
to produce a high-energy gamma ray; and it might be difficult to observe the very low-
intensity radio waves produced as bremsstrahlung by low-energy electrons. The potential
difference goes to infinity as the wavelength goes to zero. The potential difference goes to
zero as the wavelength goes to infinity.

P42.41 Following Example 42.5 Eγ = 3 (42 − 1)2 (13.6 eV) = 1.71 × 104 eV = 2.74 × 10−15 J

4

f = 4.14 × 1018 Hz

and λ = 0.072 5 nm

P42.42 E = hc = 1 240 eV ⋅ nm = 1.240 keV ⋅ nm
λ λ λ

For λ1 = 0.018 5 nm, E = 67.11 keV

λ2 = 0.020 9 nm, E = 59.4 keV FIG. P42.42
λ3 = 0.021 5 nm, E = 57.7 keV

The ionization energy for the K shell is 69.5 keV, so the ionization energies for the other shells are:

L shell = 11.8 keV M shell = 10.1 keV N shell = 2.39 keV

Atomic Physics 507

P42.43 The Kβ x-rays are emitted when there is a vacancy in the ( n = 1) K shell and an electron from the
(n = 3) M shell falls down to fill it. Then this electron is shielded by nine electrons originally and

by one in its final state.

hc = − 13.6 ( Z − 9)2 eV + 13.6 (Z − 1)2 eV
λ
32 12

( )( )6.626 × 10−34 J ⋅ s 3.00 × 108 m s = (13.6 eV ) ⎛ − Z2 + 18Z − 81 + Z2 − 2Z + 1⎠⎞⎟
( )( )0.152 × 10−9 m 1.60 × 10−19 J eV ⎜⎝ 9 9 9

⎛ 8Z 2 − 8⎠⎞⎟
⎝⎜ 9
8.17 × 103 eV = (13.6 eV )

so 601 = 8Z 2 − 8
9

and Z = 26 Iron

Section 42.9 Spontaneous and Stimulated Transitions

Section 42.10 Lasers

P42.44 The photon energy is E4 − E3 = (20.66 − 18.70) eV = 1.96 eV = hc
λ

) )( (6.626 × 10−34 J ⋅ s 3.00 × 108 m s
( )λ = 1.96 1.60 × 10−19 J = 633 nm

P42.45 f = E = 0.117 eV ⎛ 1.60 × 10−19 C⎞ ⎛ 1J ⎞ = 2.82 × 1013 s−1
h 6.630 × 10−34 J ⋅ s ⎜⎝ e ⎟⎠ ⎝1 V ⋅ C⎠

λ = c = 3.00 × 108 m s = 10.6 µm , infrared
f 2.82 × 1013 s−1

( ( ) ( ) )I =
P42.46 (a) 3.00 × 10−3 J 2⎤ = 4.24 × 1015 W m2
⎦
1.00 × 10−9 s ⎣⎡π 15.0 × 10−6 m

( ) (( ))(b)
3.00 × 10−3 J 0.600 × 10−9 m 2 1.20 × 10−12 J = 7.50 MeV
30.0 × 10−6 m 2 =

P42.47 E = P ∆t = (1.00 × 106 W)(1.00 × 10−8 s) = 0.010 0 J

( )( )Eγ
= hf = hc = 6.626 × 10−34 3.00 × 108 J = 2.86 × 10−19 J
λ 694.3 × 10−9

N= E = 0.010 0 = 3.49 × 1016 photons
Eγ 2.86 × 10−19

508 Chapter 42

N N e3 g − E3 (kB⋅300 K) −(E3 −E2 ) (kB⋅300 K) − hc λ(kB⋅300 K)

P42.48 (a) = = e = e− E2 (kB⋅300 K)
N N e2 g

where l is the wavelength of light radiated in the 3 → 2 transition.

N = e ( )( ) ( )( )3 − 6.63×10−34 J⋅s 3×108 m s 632.8×10−9 m 1.38×10−23 J K (300 K)
N2
N3 = e−75.9 = 1.07 × 10−33
N2

(b) Nu = e−(Eu −E ) kBT
N

where the subscript u refers to an upper energy state and the subscript to a lower energy state.

Since Eu − E = Ephoton = hc Nu = e−hc λkBT
λ N

Thus, we require 1.02 = e−hc λkBT
or
) ) ))( ( ( (ln(1.02) = −
6.63 × 10−34 J ⋅ s 3 × 108 m s T
632.8 × 10−9 m 1.38 × 10−23 J K

T = − 2.28 × 104 = −1.15 × 106 K

ln (1.02)

A negative-temperature state is not achieved by cooling the system below 0 K, but by
heating it above T = ∞, for as T → ∞ the populations of upper and lower states approach
equality.

(c) Because Eu − E > 0, and in any real equilibrium state T > 0 ,
e−(Eu −E ) kBT < 1 and Nu < N
Thus, a population inversion cannot happen in thermal equilibrium.

P42.49 (a) The light in the cavity is incident perpendicularly on the mirrors, 12
although the diagram shows a large angle of incidence for clarity. SiO2
We ignore the variation of the index of refraction with wavelength.
To minimize reflection at a vacuum wavelength of 632.8 nm, FIG. P42.49
the net phase difference between rays (1) and (2) should be 180°.
There is automatically a 180° shift in one of the two rays upon
reflection, so the extra distance traveled by ray (2) should be
one whole wavelength:

2t = λ
n

t = λ = 632.8 nm = 217 nm
2n
2(1.458)

(b) The total phase difference should be 360°, including contributions of 180° by reflection and
180° by extra distance traveled

2t = λ
2n

t = λ = 543 nm = 93.1 nm
4n
4 (1.458)

Atomic Physics 509

Additional Problems

P42.50 (a) Using the same procedure that was used in the Bohr model of the hydrogen atom, we
apply Newton’s second law to the Earth. We simply replace the Coulomb force by the
gravitational force exerted by the Sun on the Earth and find

G MS M E = ME v2 (1)
r2 r

where v is the orbital speed of the Earth. Next, we apply the postulate that angular
momentum of the Earth is quantized in multiples of :

M E vr = n (n = 1, 2, 3, …)

Solving for v gives (2)
v= n
MEr

Substituting (2) into (1), we find

r = n2 2 (3)

GMS M 2
E

(b) Solving (3) for n gives

n = GMSr ME (4)

Taking M S = 1.99 × 1030 kg , and M E = 5.98 × 1024 kg, r = 1.496 × 1011 m,
G = 6.67 × 10−11 Nm2 kg2, and = 1.055 × 10−34 Js, we find

n = 2.53 × 1074

(c) We can use (3) to determine the radii for the orbits corresponding to the quantum numbers
n and n + 1:

rn = n2 2 and rn+1 = (n + 1)2 2

GM S M 2 GMS M 2
E E

Hence, the separation between these two orbits is

22

∆r = GMS ME2 ⎡⎣(n + 1)2 − n 2 ⎦⎤ = GMS ME2 (2n + 1)

Since n is very large, we can neglect the number 1 in the parentheses and express the
separation as

2

∆r ≈ GMS M 2 (2n) = 1.18 × 10−63 m
E

( )This number is much smaller than the radius of an atomic nucleus ~10−15 m , so the

distance between quantized orbits of the Earth is too small to observe.

510 Chapter 42

(( ) )∆E
P42.51 (a) = eB 1.60 × 10−19 C 6.63 × 10−34 J ⋅ s (5.26 T) ⎛ N ⋅s ⎞ ⎛ kg ⋅ m ⎞ = 9.75 × 10−23 J
me = 2π 9.11 × 10−31 kg
⎝ T ⋅C⋅m⎠ ⎝ N ⋅s2 ⎠

= 609 µeV

( )( )(b) kBT = 1.38 × 10−23 J K 80 × 10−3 K = 1.10 × 10−24 J = 6.9 µeV

(c) f = ∆E = 9.75 × 10−23 J = 1.47 × 1011 Hz
h 6.63 × 10−34 J ⋅ s

λ = c = 3 × 108 m s = 2.04 × 10−3 m
f 1.47 × 1011 Hz

∞ 4 ∞ ⎡ ⎛ 2r ′2 2r ′ ⎞ e −2 r ′ a0 ⎤∞
a03 ⎢− ⎝⎜ a02 a0 1⎠⎟ ⎥
P1s (r′) dr′ = r ′2e−2r′ a0 dr ′ ⎣ ⎦r

r r
P42.52 (a) Probability ∫ ∫= = + +

using integration by parts, we find

= ⎛ 2r2 + 2r + ⎞ e−2r a0
⎜⎝ a02 a0 1⎠⎟

(b) Probability Curve for Hydrogen
1.2

1

Probability 0.8

0.6

0.4

0.2

0
012345

r /a0

FIG. P42.52

(c) The probability of finding the electron inside or outside the sphere of radius r is 1 .
2

∴ ⎛ 2r 2 + 2r + ⎞ e−2r a0 = 1 or z2 + 2z + 2 = ez where z = 2r
⎜⎝ a02 a0 1⎠⎟ 2 a0

One can home in on a solution to this transcendental equation for r on a calculator,

the result being r = 1.34a0 to three digits.

Atomic Physics 511

P42.53 hf = ∆E = 4π 2me ke2e4 ⎛1 − 1⎞ f = 2π 2me ke2e4 ⎛ 2n −1 ⎞
2h2 n2 ⎟⎠ h3 ⎠⎟
⎝⎜ (n − 1)2 ⎜⎝ (n − 1)2 n2

As n approaches infinity, we have f approaching 2π 2meke2e4 2
h3 n3

The classical frequency is f= v = 1 kee2 1
2π r 2π me r 3 2

where r = n2h2
4π mekee2

Using this equation to eliminate r from the expression for f, we find f = 2π 2me ke2e4 2
h3 n3
in agreement with the Bohr result for large n.

P42.54 (a) The energy difference between these two states is equal to the energy that is absorbed.

Thus, E = E2 − E1 = (−13.6 eV) − (−13.6 eV) = 10.2 eV = 1.63 × 10−18 J

4 1

( ( ) )T
(b) E = 3 kBT or = 2E = 2 1.63 × 10−18 J = 7.88 × 104 K
2 3kB 3 1.38 × 10−23 J K

P42.55 (a) The energy of the ground state is: E1 = − hc = − 1 240 eV ⋅ nm
λseries limit 152.0 nm

= −8.16 eV

From the wavelength of the Lyman a line: E2 − E1 = hc = 1 240 nm ⋅ eV = 6.12 eV
λ 202.6 nm

E2 = E1 + 6.12 eV = −2.04 eV

The wavelength of the Lyman b line gives: E3 − E1 = 1 240 nm ⋅ eV = 7.26 eV
so 170.9 nm
Next, using the Lyman g line gives:
E3 = −0.902 eV

E4 − E1 = 1 240 nm ⋅ eV = 7.65 eV
162.1 nm

and E4 = −0.508 eV
From the Lyman d line,
so E5 − E1 = 1 240 nm ⋅ eV = 7.83 eV
158.3 nm

E5 = −0.325 eV

continued on next page

512 Chapter 42

(b) For the Balmer series, hc = Ei − E2 , or λ = 1 240 nm ⋅ eV
For the a line, Ei = E3 and so λ Ei − E2

λa = 1 240 nm ⋅ eV eV)

(−0.902 eV) − (−2.04

= 1 090 nm

Similarly, the wavelengths of the b line, g line, and the short wavelength limit are found to be:
811 nm , 724 nm , and 609 nm .

(c) Computing 60.0% of the wavelengths of the spectral lines shown on the energy-level
diagram gives:

0.600 (202.6 nm) = 122 nm , 0.600 (170.9 nm) = 103 nm ,

0.600 (162.1 nm) = 97.3 nm , 0.600 (158.3 nm) = 95.0 nm ,

and 0.600 (152.0 nm) = 91.2 nm

These are seen to be the wavelengths of the a, b, g, and d lines as well as the short
wavelength limit for the Lyman series in Hydrogen.

(d) The observed wavelengths could be the result of Doppler shift when the source moves
away from the Earth. The required speed of the source is found from

f′= λ = 1− (v c) = 0.600 yielding v = 0.471c
f λ′ 1+ (v c)

*P42.56 (a) The energy emitted by the atom is ∆E = E4 − E2 = −13.6 eV ⎛ 1 − 1⎞ = 2.55 eV . The
⎝ 42 22 ⎠

wavelength of the photon produced is then

( )( )λ = hc = hc =
( )Eγ ∆E
6.626 × 10−34 J ⋅ s 3.00 × 108 m s = 4.87 × 10−7 m = 487 nm

(2.55 eV) 1.60 × 10−19 J eV

(b) Since momentum must be conserved, the photon and the atom go in opposite directions

with equal magnitude momenta. Thus, p = matom v = h or
λ

6.626 × 10−34 J ⋅ s
1.67 × 10−27 kg 4.87 × 10−7 m
( )( )v = h = = 0.814 m s
matom λ

P42.57 The wave function for the 2s state is given by Equation 42.26:

1 ⎛ 1 ⎞ 3 2 ⎡ r ⎤
2π ⎜⎝ a0 ⎠⎟ ⎢ a0 ⎥
ψ2s (r) = ⎣ 2 − ⎦ e−r 2 a0 .

4

(a) Taking r = a0 = 0.529 × 10−10 m
we find
1 ⎛1 ⎞ 3 2
2π ⎝ 0.529 × 10−10 ⎠
( )ψ 2s a0 = [ 2 − 1] e−1 2 = 1.57 × 1014 m−3 2

4 m

( ) ( )(b) ψ 2s a0 2 = 1.57 × 1014 m−3 2 2 = 2.47 × 1028 m−3

(c) Using Equation 42.24 and the results to (b) gives P2s (a0 ) = 4π a02 ψ 2s (a0 ) 2 = 8.69 × 108 m−1 .

Atomic Physics 513

*P42.58 The average squared separation distance is

∗ ∞ 1 e−r a0 r 2 1 e−r a0 4πr 2dr 4π ∞ 4 −2r a0 dr
1s r=0 π a03 π a03 π a03
∫ ∫ ∫r2= ψ r 2ψ 1s dV = = r e
all space 0

∫We use ∞ n e− ax dx = n! from Table B.6.
an+1
x
0

4 4! a02 96 3a02
a03 2 / a0 32
( )r2 = 5 = =

⎛ 3a0 2⎞1 2 ⎛ 9a02 ⎞1 2 3 1 2
⎜⎝ 2 ⎟⎠ ⎝⎜ 4 ⎠⎟ 4
∆r = r2 − r 2 = 3a02 − ⎛ ⎞ = 3a02 − = ⎛ ⎞ a0
⎝ ⎠ ⎝ ⎠

) )( (P42.59 (a) 3.00 × 108 m s 14.0 × 10−12 s = 4.20 mm

(b) E = hc = 2.86 × 10−19 J
λ

N = 3.00 J J = 1.05 × 1019 photons
2.86 × 10−19

(c) V = (4.20 mm) ⎡⎣π (3.00 mm)2 ⎦⎤ = 119 mm3

n = 1.05 × 1019 = 8.82 × 1016 mm−3
119

P42.60 (a) The length of the pulse is ∆L = c∆t .

(b) The energy of each photon is Eγ = hc so N = E = Eλ
λ Eγ hc

(c) V = ∆Lπ d 2 n= N = ⎛ 4 ⎞ ⎛ Eλ ⎞
4 V ⎝⎜ c∆tπ d 2 ⎟⎠ ⎜⎝ hc ⎟⎠

P42.61 The fermions are described by the exclusion principle. Two of them, one spin-up and one
spin-down, will be in the ground energy level, with

dNN = L = 1λ, λ = 2L = h , and p= h K = 1 mv 2 = p2 = h2
2 p 2L 2 2m 8mL2

The third must be in the next higher level, with

dNN = L = λ , λ = L, and p= h K = p2 = h2
2 2 L 2m 2mL2

The total energy is then h2 + h2 + h2 = 3h2
8mL2 8mL2 2mL2 4mL2

514 Chapter 42

( )∆z at 2 µz ⎛ ∆x⎞2
P42.62 = 2 = 1 ⎛ Fz ⎞ t 2 = dBz dz ⎝v⎠ and µz = e
2 ⎜⎝ m0 ⎟⎠ 2m0 2me

( )( )( )( )( )dBz
( )( )( )dz
= 2m0 (∆z)v 2 2me 2(108) 1.66 × 10−27 kg 10−3 m 104 m2 s2 2 × 9.11 × 10−31 kg

∆ x 2e = 1.00 m2 1.60 × 10−19 C 1.05 × 10−34 J ⋅ s

dBz = 0.389 T m
dz

( )ψ 2s (r) 1 −1 2 ⎛ r ⎞ e−r 2 a0
P42.63 We use = 4 2π a03 ⎝⎜ 2 − a0 ⎠⎟

1 ⎛ r2 ⎞ ⎛ r ⎞ 2
8 ⎜⎝ a03 ⎠⎟ ⎜⎝ 2 a0 ⎟⎠
By Equation 42.24, P(r) = 4πr 2ψ 2 = − e−r a0

dP(r ) 1 ⎡ 2r ⎛ r ⎞2 2r 2 ⎛ 1 ⎞ ⎛ r⎞ r2 ⎛ r ⎞ 2 ⎛ 1 ⎞ ⎤
8 ⎢ a03 ⎜⎝ 2 − a0 ⎟⎠ a03 ⎜⎝ a0 ⎟⎠ ⎝⎜ a0 ⎟⎠ a03 ⎜⎝ a0 ⎟⎠ ⎝⎜ a0 ⎟⎠ ⎥
(a) dr = ⎢⎣ − 2 − − 2 − ⎦⎥ e− r a0 = 0

or 1 ⎛ r ⎞ ⎛ 2 − r ⎞ ⎡ ⎛ 2 − r ⎞ − 2r −r ⎛ 2 − r ⎞ ⎤ e−r a0 =0
8 ⎝⎜ a03 ⎠⎟ ⎝⎜ a0 ⎠⎟ ⎢2 ⎜⎝ a0 ⎠⎟ a0 a0 ⎝⎜ a0 ⎟⎠ ⎥
⎣ ⎦

The roots of dP =0 at r = 0, r = 2a0, and r = ∞ are minima with P (r) = 0.
dr

⎛ 6r ⎞ ⎛ r ⎞ 2
⎜⎝ a0 ⎠⎟ ⎝⎜ a0 ⎠⎟
Therefore we require [.....] = 4 − + = 0

with solutions )(r = 3 ± 5 a0

We substitute the last two roots into P (r) to determine the most probable value:

( )When r = 3 − 5 a0 = 0.763 9a0 , P (r) = 0.051 9

a0

( )When r = 3 + 5 a0 = 5.236a0 , P (r) = 0.191

a0

Therefore, the most probable value of r is ( )3 + 5 a0 = 5.236a0

∞ ∞1⎛ r2 ⎞ ⎛ r ⎞2 e−r a0 dr
0 8 ⎝⎜ a03 ⎠⎟ ⎜⎝ 2 − a0 ⎠⎟
P(r )dr

0
∫ ∫(b) =

Let u = r , dr = a0 du,
a0

∞ 1 u2 e−udr = ∞ 1
∫ ∫ ( ) ∫ ( )∞ e − u du

P(r )dr
= 4 − 4u + u2 08 u4 − 4u3 + 4u2
0 08

( )= − 1 ∞
u4 + 4u2 + 8u + 8
e−u = 1
80

This is as required for normalization.

Atomic Physics 515

P42.64 (a) Suppose the atoms move in the +x direction. The absorption of a photon by an atom is a
completely inelastic collision, described by

( )mviˆi + h −ˆi = mv f ˆi so vf − vi = − h
λ mλ

This happens promptly every time an atom has fallen back into the ground state, so it
happens every 10−8 s = ∆t. Then,

6.626 × 10−34 J ⋅ s
kg 500 × 10−9 m
( )( )( )a = v f − vi
∆t
=− h ~− 10−25 10−8 s ~ −106 m s2
mλ∆t

(b) With constant average acceleration,

v 2 = vi2 + 2a∆x ) )( (0 ~ 103 m s 2 + 2 −106 m s2 ∆x
f )(103 m s 2

so ∆x ~ 106 m s2 ~ 1 m

P42.65 With one vacancy in the K shell, excess energy

∆E ≈ −(Z − 1)2 (13.6 eV ) ⎛ 1 − 1⎞ = 5.40 keV
⎝ 22 12 ⎠

We suppose the outermost 4s electron is shielded by 22 electrons inside its orbit:

Eionization ≈ 22 (13.6 eV) = 3.40 eV

42

As evidence that this is of the right order of magnitude, note that the experimental ionization
energy is 6.76 eV.

K = ∆E − Eionization ≈ 5.39 keV

P42.66 E = hc = 1 240 eV ⋅ nm = ∆E
λλ

λ1 = 310 nm, so ∆E1 = 4.00 eV

λ2 = 400 nm, ∆E2 = 3.10 eV

λ3 = 1 378 nm, ∆E3 = 0.900 eV FIG. P42.66

and the ionization energy = 4.10 eV.

The energy level diagram having the fewest levels and consistent with these energies is shown at
the right.

∞ 4r 2 e−2r a0 dr 1 ∞ z2e−z dz 2r
P42.67 =∫ ∫P2.50 a0a03 = 2 5.00 where z ≡ a0

( )P = − 1 e−z ∞ 1 [0] + 1 (25.0 2.00 ) e −5 ⎛ 37 ⎞ (0.006 74)
2 5.00 ⎝ 2⎠
z2 + 2z + 2 = − 2 2 + 10.0 + = = 0.125

516 Chapter 42

P42.68 (a) One molecule’s share of volume

Al: V = mass per molecule = ⎛ 27.0 g mol ⎞ ⎛ 1.00 × 10−6 m3 ⎞
density ⎜⎝ 6.02 × 1023 molecules mol ⎠⎟ ⎜⎝ 2.70 g ⎟⎠

= 1.66 × 10−29 m3

3 V = 2.55 × 10−10 m~10−1 nm

U: V = ⎛ 238 g ⎞ ⎛ 1.00 × 10−6 m3 ⎞ = 2.09 × 10−29 m3
⎝ 6.02 × 1023 molecules⎠ ⎝⎜ 18.9 g ⎠⎟

3 V = 2.76 × 10−10 m ~ 10−1 nm

(b) The outermost electron in any atom sees the nuclear charge screened by all the electrons
below it. If we can visualize a single outermost electron, it moves in the electric field of

net charge +Ze − (Z − 1) e = +e, the charge of a single proton, as felt by the electron in

hydrogen. So the Bohr radius sets the scale for the outside diameter of every atom. An
innermost electron, on the other hand, sees the nuclear charge unscreened, and the scale
size of its (K-shell) orbit is a0 .

Z

P42.69 ∆E = 2µBB = hf

( ) ( )so 2 9.27 × 10−24 J T (0.350 T) = 6.626 × 10−34 J ⋅ s f

and f = 9.79 × 109 Hz

ANSWERS TO EVEN PROBLEMS

P42.2 (a) 1.94 mm (b) λCB = 1/λCA 1
P42.4 − 1/λBA

(a) 56.8 fm (b) 11.3 N away from the nucleus

P42.6 (a) 2.19 Mm/s (b) 13.6 eV (c) −27.2 eV

P42.8 (a) The atoms must be excited to energy level n = 4, to emit six different photon energies in
the downward transitions 4 → 3, 4 → 2, 4 → 1, 3 → 2, 3 → 1, and 2 → 1. The photon energy
absorbed in the 1 → 4 transition is 12.8 eV, making the wavelength 97.4 nm. (b) 1.88 mm,
infrared, Paschen (c) 97.4 nm, ultraviolet, Lyman

P42.10 (a) 3 (b) 520 km s

P42.12 (a) 153 as (b) 8.18 × 109 revolutions (c) Its lifetime in electron years is comparable to the
lifetime of the Sun in Earth years, so you can think of it as a long time.

P42.14 (a) See the solution. (b) 0.179 nm

P42.16 See the solution.

P42.18 4a0
P42.20 797 times

Atomic Physics 517

P42.22 (a) 6 = 2.58 × 10−34 J ⋅ s (b) 12 = 3.65 × 10−34 J ⋅ s

P42.24 6

P42.26 6 ; −2 , − , 0, , 2 ; 145°, 114°, 90.0°, 65.9°, 35.3°

P42.28 3

P42.30 (a) 1s2 2s2 2 p4

(b) n 11 22 22 22
00 00 11 11
m 00 00 11 0 −1
ms 1 −1 1 −1 1 −1 11
22 22 22 22

P42.32 See the solution.

P42.34 (a) See the solution. (b) 36 states instead of 30

P42.36 1s2 2s2 2 p6 3s2 3 p6 3d10 4s2 4 p6 4d10 4 f 14 5s2 5 p6 5d10 5 f 14 6s2 6 p6 6d8 7s2

P42.38 (a) = 0 with m = 0; = 1 with m = 1, 0, or 1; and = 2 with m = −2, –1, 0, 1, 2 (b) –6.05 eV

P42.40 (a) ∆V = 1240 V ⋅ nm/l. (b) The potential difference is inversely proportional to the wavelength.
(c) Yes. It predicts a minimum wavelength of 33.5 pm when the accelerating voltage is 37 keV,
in agreement with the minimum wavelength in the figure. (d) Yes, but it might be unlikely for a
very high-energy electron to stop in a single interaction to produce a high-energy gamma ray; and
it might be difficult to observe the very low-intensity radio waves produced as bremsstrahlung
by low-energy electrons. The potential difference goes to infinity as the wavelength goes to zero.
The potential difference goes to zero as the wavelength goes to infinity.

P42.42 L shell 11.8 keV, M shell 10.1 keV, N shell 2.39 keV. See the solution.

P42.44 See the solution.

P42.46 (a) 4.24 PW m2 (b) 1.20 pJ = 7.50 MeV

P42.48 (a) 1.07 × 10−33 (b) −1.15 × 106 K (c) Negative temperatures do not describe systems in
thermal equilibrium.

P42.50 (a) See the solution. (b) 2.53 × 1074 (c) 1.18 × 10−63 m , unobservably small
P42.52
P42.54 (a) Probability = ⎛ 2r 2 + 2r + ⎞ e −2r a0 (b) See the solution. (c) 1.34a0
⎝⎜ a02 a0 1⎟⎠

(a) 10.2 eV = 1.63 aJ (b) 7.88 × 104 K

P42.56 (a) 487 nm (b) 0.814 m s

P42.58 3/4 a0 = 0.866 a0

518 Chapter 42

P42.60 (a) c∆t (b) Eλ (c) 4Eλ
P42.62 hc ∆tπ d 2hc2

0.389 T/m

P42.64 (a) ~ –106 m/s2 (b) ~1 m

P42.66 Energy levels at 0, –0.100 eV, –1.00 eV, and –4.10 eV

P42.68 (a) diameter ~10−1 nm for both (b) A K-shell electron moves in an orbit with size on the
order of a0 .

Z

43

Molecules and Solids

CHAPTER OUTLINE ANSWERS TO QUESTIONS

43.1 Molecular Bonds Q43.1 Ionic bonds are ones between oppositely charged ions.
43.2 Energy States and Spectra of A simple model of an ionic bond is the electrostatic
attraction of a negatively charged latex balloon to a
Molecules positively charged Mylar balloon.
43.3 Bonding in Solids
43.4 Free-Electron Theory of Metals Covalent bonds are ones in which atoms share
43.5 Band Theory of Solids electrons. Classically, two children playing a short-range
43.6 Electrical Conduction in Metals, game of catch with a ball models a covalent bond. On
a quantum scale, the two atoms are sharing a wave
Insulators, and Semiconductors function, so perhaps a better model would be two
43.7 Semiconductor Devices children using a single hula hoop.
43.8 Superconductivity
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

2

of excitation. Rotational energy for a diatomic molecule is on the order of , where I is
2I

the moment of 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.

*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 ~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.

The ranking is then b > d > c > a.

519

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.

Molecules and Solids 521

Q43.13 A molecule containing two atoms of 2 H, deuterium, has twice the mass of a molecule
Q43.14
containing two atoms of ordinary hydrogen 1H. The atoms have the same electronic structure, so

the molecules have the same interatomic spacing, and the same spring constant. Then the moment

of inertia of the double-deuteron is twice as large and the rotational energies one-half as large

as for ordinary hydrogen. Each vibrational energy level for D2 is 1 times that of H2.
2

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

SOLUTIONS TO PROBLEMS

Section 43.1 Molecular Bonds

1.60 × 10−19 2 8.99 × 109
5.00 × 10−10 2
( ( ) ( ) )P43.1(a)F=4πq2 r2 = N = 0.921 × 10−9 N toward the other ion.
∈0

( ) ( )(b)
U = −q2 r =− 1.60 × 10−19 2 8.99 × 109 J ≈ −2.88 eV
4π ∈0 5.00 × 10−10

P43.2 We are told K + Cl + 0.7 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 .

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

dU = −12Ar−13 + 6Br−7 = 0 yielding ⎡ 2A ⎤1 6
dr ⎢⎣ B ⎦⎥
r0 =

(b) E =U r=ϱ −U r = r0 = 0 − ⎡ 4 A A B2 − B B ⎤ = − ⎡ 1 − 1 ⎤ B2 = B2
⎣⎢ 2A ⎥⎦ ⎣⎢ 4 2 ⎦⎥ A 4A
2

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) ⎤1 6
⎡ 2 0.124 × 10−120 eV ⋅ m12 ⎥ = 7.42 × 10−11 m =
r0 = ⎢ ⎥⎦ 74.2 pm
⎣⎢ 1.488 × 10−60 eV ⋅ m6

( )1.488 × 10−60 eV ⋅ m6 2
( )E = 4 0.124 × 10−120 eV ⋅ m12 = 4.46 eV

522 Chapter 43

P43.4 (a) We add the reactions K + 4.34 eV → K+ + e−
and I + e− → I− + 3.06 eV

to obtain K + I → K+ + I− + (4.34 − 3.06) eV

The activation energy is 1.28 eV .

dU 4 ∈⎡ ⎜⎛s ⎞⎟ 13 + 6 ⎛⎜s ⎟⎞ 7 ⎤
dr ⎝r ⎠ ⎝r ⎠ ⎥
(b) = s ⎢ −12 ⎦
⎣

At r = r0 we have dU = 0. Here ⎛s ⎞ 13 = 1 ⎛s ⎞7
dr ⎜ ⎟ 2 ⎜ ⎟
⎝ r0 ⎠ ⎝ r0 ⎠

s = 2−1 6 s = 2−1 6 (0.305) nm = 0.272 nm = s
r0

Then also

U (r0 ) = 4 ∈ ⎡⎢⎛⎜ 2−1 6 r0 ⎞12 ⎛ 2−1 6 r0 ⎞6 ⎤ Ea = 4 ∈ ⎡1 − 21 ⎤⎦⎥ + Ea = − ∈ + Ea
⎣⎢⎝ r0 ⎟ −⎜ r0 ⎟ ⎥+ ⎢⎣ 4
⎠ ⎠ ⎥⎦
⎝

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

dU 4 ∈⎡ ⎛⎜s ⎟⎞ 13 − 6 ⎛⎜s ⎞⎟ 7 ⎤
dr ⎝r ⎠ ⎝r ⎠ ⎥
(c) F (r ) = − = s ⎢ 12 ⎦
⎣

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

s = ⎛⎜ 42 ⎟⎞1 6
rrupture ⎝156 ⎠

4 (4.65 eV) ⎡ ⎝⎛⎜ 42 ⎞⎠⎟ 13 6 ⎜⎛⎝ 42 ⎠⎞⎟ 7 6 ⎤ 1.6 × 10−19 Nm
⎢12 156 156 ⎥ 10−9 m
Fmax = 0.272 nm ⎣ − 6 ⎦ = −41.0 eV nm = −41.0

= −6.55 nN

Therefore the applied force required to rupture the molecule is +6.55 nN away from the

center.

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

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

⎡1 ⎛ s + 78 s2 − ⎞ 1 ⎛ s + 21 s2 − ⎞⎤
= 4 ∈ ⎢ ⎜1 − 12 r0 r02 ⎟− 2 ⎜1− 6 r0 r02 ⎟⎥ + Ea
⎣ 4 ⎝ ⎠ ⎝ ⎠⎦

= ∈ −12 ∈ s + 78 ∈ s2 −2 ∈ +12 ∈ s − 42 ∈ s2 + Ea +
r0 r02 r0 r02

= − ∈ +Ea + ⎛ s ⎞ 36 ∈ s2 +
0⎜ ⎟+ r02
⎝ r0 ⎠

U (r0 + s) ≈ U (r0 ) + 1 ks2
2

where k= 72 ∈ = 72 (4.65 eV) = 3 599 eV nm2 = 576 N m
r02 (0.305 nm)2

Molecules and Solids 523

P43.5 At the boiling or condensation temperature, ( )kBT ≈ 10−3 eV = 10−3 1.6 × 10−19 J

T ≈ 1.6 ×10−22 J ~10 K
1.38 ×10−23 J 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

( )mr2 ⎛ 0.75 × 10−10 m⎞2
+ mr2 =2 1.67 × 10−27 kg ⎜⎝ 2 ⎠⎟ = 4.70 × 10−48 kg ⋅ m2

2

The rotational energy is Erot = 2I J (J +1) or it is zero for J = 0 and for J = 1 it is

( )h21(2) J ⋅ s 2 ⎛ 1 eV
kg ⋅ m2 ) ⎜⎝1.6 × 10−19
4π 2 2I
= 6.626 × 10−34 J ⎞ = 0.0148 eV
4π 2 (4.70 × 10−48 ⎠⎟

(b) λ= c = ch = (2.998 ×108 m/s)(6.626 ×10−34 J ⋅ s) 1 eV = 1240 eV ⋅ nm = 83.8 µm
f E 0.0148 eV 1.602 ×10−19 0.0148 eV
J

P43.7 ( )µ = m1m2 = 132.9(126.9) 1.66 × 10−27 kg = 1.08 × 10−25 kg
m1 + m2 132.9 + 126.9

( )( )I = µr2 = 1.08 ×10−25 kg 0.127 ×10−9 m 2 = 1.74 ×10−45 kg ⋅ m2

(a) E = 1 Iω 2 = (Iω )2 = J (J + 1) 2

2 2I 2I
J = 0 gives E = 0

2 6.626 × 10−34 J ⋅ s 2
I = 4π 2 1.74 × 10−45 kg ⋅ m2
( ( ) )J = 1 gives E = = 6.41 × 10−24 J = 40.0 µeV

hf = 6.41 × 10−24 J − 0 gives f = 9.66 × 109 Hz

(b) f = E1 = 2 h ∝ r−2 If r is 10% too small, f is 20% too large.

=
h hI 4π 2µr2

*P43.8 ∆Evib = h k = hf so k = 4π 2 f 2µ
2π µ

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

µ = m1m2 = 14.007 u 15.999 u 1.66 × 10−27 kg = 1.24 × 10−26 kg
m1 + m2 14.007 u +15.999 u 1u

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

524 Chapter 43

P43.9 For the HCl molecule in the J = 1 rotational energy level,
we are given r0 = 0.127 5 nm.

2

Erot = 2I J (J + 1)

Taking J = 1 , we have Erot = 2 = 1 Iω2 or ω = 2 2= 2
2 I2 I
I

The moment of inertia of the molecule is given by Equation 43.3. FIG. P43.9

I = µ r02 = ⎛ m1m2 ⎞ r02
⎜⎝ m1 + m2 ⎠⎟

⎡(1.01 u) ( 35.5 u) ⎤ 1.275 ×10−10 m 2 = 2.65 ×10−47 kg ⋅ m2
⎥
u + 35.5 u ⎦
( )( )I
= ⎢ 1.01 r02 = (0.982 u) 1.66 ×10−27 kg u
⎣

Therefore, ω = 2 I = 2 6.626 × 10−34 J ⋅ s = 5.63 × 1012 rad s
2π (2.65 × 10−47 kg ⋅ m2 )

P43.10 2 22

hf = ∆E = [2(2 + 1)] − [1(1 + 1)] = (4)

2I 2I 2I

= 4(h 2π )2 = h = 6.626 × 10−34 J ⋅ s = 1.46 × 10−46 kg ⋅ m2
2π 2 f 2π 2 2.30 × 1011 Hz
2hf
( )I

P43.11 I = m1r12 + m2r22 where m1r1 = m2r2 and r1 + r2 = r
so and
Then r1 = m2 r2 m2 r2 + r2 = r r2 = m1r
m1 Thus, m1 and m1 + m2

Also, r2 = m1r1 . r1 + m1r1 = r r1 = m2 r
m2 m2 m1 + m2

( ) ( ) ( ( ) )I = m1
m22 r 2 2+ m2 m12r2 2 = m1m2r2 m2 + m1 = m1m2r2 = µr2
m1 + m2 m1 + m2 m1 + m2 2 m1 + m2

22.99(35.45)
(22.99 + 35.45)
( )P43.12 (a) = 2.32 × 10−26
µ = 1.66 × 10−27 kg kg

( )( )I = µr2 = 2.32 × 10−26 kg 0.280 × 10−9 m 2 = 1.81 × 10−45 kg и m2

(b) hc = 2 2 1(1 + 1) = 3 2 − 2 =2 2 = 2h2
I 4π 2 I
2(2 + 1) − I
λ 2I 2I I

( ) ( )λ = c4π 2I =
( )2h
3.00 × 108 m s 4π 2 1.81 × 10−45 kg ⋅ m2 = 1.62 cm
2 6.626 × 10−34 J ⋅ s

Molecules and Solids 525

P43.13 The energy of a rotational transition is ∆E = ⎛ 2⎞ where J is the rotational quantum number of
⎝⎜ I ⎠⎟ J

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

( )( )∆E = hc =
λ
6.626 × 10−34 J ⋅ s 3.00 × 108 ms ⎛⎝⎜ 1 eV J ⎠⎟⎞ .
λ 1.60 × 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. 2

E1 = I = 0.002 56 eV

( )2 2

and I = =
E1
1.055 ×10−34 J ⋅ s ⎛ 1 eV J ⎞ = 2.72 ×10−47 kg ⋅ m2
⎝⎜1.60 ×10−19 ⎟⎠
(0.002 56 eV)

For the HCl molecule, the internuclear radius is r= I= 2.72 ×10−47 m = 0.130 nm
µ 1.62 ×10−27

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

1 kA2 = 1 ω = h k so A= h ⎛ 1 ⎞ 1/4
2 2 4π µ 2π ⎝⎜ kµ ⎠⎟

A= 6.626 × 10−34 J⋅s ⎛ 1 ⎞ 1/4 = 12.1 pm
2π ⎜⎝ (320 N/m)(127/128)(1.66 × 10−27 kg) ⎠⎟

(b) For HF, A = 6.626 × 10−34 J⋅s ⎛ 1 ⎞ 1/4 = 9.23 pm
2π ⎜⎝ (970 N/m)(19/20)(1.66 × 10−27 kg) ⎠⎟

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

P43.15 µ = m1m2 = 35 × 1.66 × 10−27 kg = 1.61 × 10−27 kg
m1 + m2 36

k = 1.055 ×10−34 480
µ 1.61 × 10−27
( )∆Evib = = 5.74 ×10−20 J= 0.358 eV

526 Chapter 43

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

(16 u)(16 u)
(16 u) + (16 u)
( )µ = 1.33×10−26
= = 8 u = 8 1.66 ×10−27 kg kg

The moment of inertia is then

( )( )I = µr2 = 1.33×10−26 kg 1.20 ×10−10 m 2

= 1.91×10−46 kg ⋅ m2

( )2 1.055 ×10−34 J ⋅ s 2
( )The rotational energies are Erot = 2I J (J +1) = 2 1.91×10−46 kg ⋅ m2 J (J +1)

Thus ( )Erot = 2.91×10−23 J J (J +1)

And for J = 0, 1, 2, Erot = 0, 3.64 ×10−4 eV, 1.09 ×10−3 eV

⎛⎝⎜v 1⎞ k = ⎛⎝⎜v + 1⎞ 1177 N m
2 ⎟⎠ µ 2 ⎟⎠ 8 1.66 ×10−27 kg
( ) ( )(b)
Evib = + 1.055 ×10−34 J⋅s

( )Evib ⎜⎛v 1 ⎞⎟ ⎛⎜ 1 eV ⎞⎟ = ⎛⎜v + 1 ⎞⎟ (0.196 eV)
= ⎝ + 2⎠ 3.14 × 10−20 J ⎝1.60 × 10−19
J⎠ ⎝ 2⎠

For v = 0, 1, 2, Evib = 0.098 2 eV, 0.295 eV, 0.491 eV .

P43.17 In Benzene, 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, I = Σmr2:

( )( ) ( )( )I = 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 ⋅ m2

The allowed rotational energies are then

(( ) ) ( )Erot
= 2 J (J + 1) = 2 1.055 × 10−34 J ⋅ s 2 J (J + 1) = 2.95 × 10−24 J J (J + 1)
1.89 × 10−45 kg ⋅ m2
2I

= (18.4 × 10−6 eV) J (J + 1)

Erot = (18.4 µeV) J (J +1) where J = 0, 1, 2, 3, . . .

The first five of these allowed energies are: Erot = 0, 36.9 µeV, 111 µeV, 221 µeV, and 369 µeV.

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

2

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

λ = hc = hc = 4π 2 Ic . If m is the reduced mass, then
∆Emin 2I h

( ) ( )I = µr2 = µ 0.127 46 ×10−9 m 2 = 1.624 605 ×10−20 m2 µ (1)

( ) ( )4π 2 1.624 605 ×10−20 m2 µ 2.997 925 ×108 m s
( )Therefore λ =
6.626 075 ×10−34 J ⋅ s = 2.901 830 ×1023 m kg µ

(a) µ35 = (1.007 825u)(34.968 853u) = 0.979 593u = 1.626 653 × 10−27 kg

1.007 825u + 34.968 853u

( )( )From (1): λ35 = 2.901 830 × 1023 m kg 1.626 653 × 10−27 kg = 472 µm

(b) µ37 = (1.007 825u)(36.965 903u) = 0.981 077u = 1.629 118 × 10−27 kg

1.007 825u + 36.965 903u

( )( )From (1): λ37 = 2.901 830 × 1023 m kg 1.629 118 × 10−27 kg = 473 µm

(c) λ37 − λ35 = 472.742 4 µm − 472.027 0 µm = 0.715 µm

P43.19 We find an average spacing between peaks by counting 22 gaps between 7.96 × 1013 Hz and
9.24 × 1013 Hz:

∆f = (9.24 − 7.96)1013 Hz = 0.058 2 × 1013 Hz = 1 ⎛ h2 ⎞
⎜ ⎟
22 h ⎝ 4 π 2 I ⎠

I = h = 6.63 ×10−34 J ⋅ s = 2.9 ×10−47 kg ⋅ m2
4π 2∆f 4π 2 5.82 ×1011 s

528 Chapter 43

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

(a) EvJ = ⎛⎜v + 1 ⎟⎞ hf + 2 J (J + 1)
⎝ 2⎠
2I

∴ E00 = 1 hf , E11 = 3 hf + 2 E02 = 1 hf + 32
2 2 2 I
,
I

( )( )∴ E11 − E00 = hf +
2 = hc = 6.626 075 × 10−34 J ⋅ s 2.997 925 × 108 ms
Iλ 2.211 2 × 10−6 m

2

∴ hf + = 8.983 573 × 10−20 J (1)
I (2)
( )( )E11
− E02 = hf − 22 = hc = 6.626 075 × 10−34 J ⋅ s 2.997 925 × 108 ms
I λ 2.405 4 × 10−6 m

∴hf − 2 I 2

= 8.258 284 × 10−20 J

Subtract (2) from (1): 3 2 = 7.252 89 × 10−21 J
I

( )3 1.054 573 × 10−34 J ⋅ s 2 4.60 × 10−48 kg ⋅ m2

∴ I = 7.252 89 × 10−21 J =

(b) From (1):

( ( )( ) )f
= 8.983 573 × 10−20 J − 1.054 573 × 10−34 J и s 2
6.626 075 × 10−34 J и s 4.600 060 × 10−48 kg и m2 6.626 075 × 10−34 J и s

= 1.32 × 1014 Hz

(c) I = µr2, where m is the reduced mass:

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

So r = I= 4.600 060 × 10−48 kg ⋅ m2 = 0.074 1 nm .
µ 8.367 669 × 10−28 kg

P43.21 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 ∆J = ±1, so it must end with J = 1. To be lower in
energy, it must be (v = 0, J = 1). The emitted photon energy is therefore

( ) ( ) ( ) ( )hfphoton =
E + Evib v=1 rot J =0 − E + Evib v=0 rot J =1 = E − Evib v=1 vib v=0 − E − Erot J=1 rot J =0

hfphoton = hfvib − hfrot

Thus, fphoton = fvib − frot = 6.42 × 1013 Hz − 1.15 × 1011 Hz = 6.41 × 1013 Hz .

= 2 mr2 + 2 mr2 4m 2
55 5
( )The moment of inertia about the molecular axis is .
P43.22 Ix = 2.00 × 10−15 m

( )The moment of inertia about a perpendicular axis is ⎛ R ⎞2 ⎛ R ⎞2 m 2
Iy = m ⎜⎝ 2 ⎟⎠ + m ⎝⎜ 2 ⎠⎟ = 2 2.00 × 10−10 m
.

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

2

E1 = I . The ratio is therefore

2.00 ×10−10 m 2
2.00 ×10−15 m 2
( ) ( ) ( )E1, x =
( ) ( )E1, y
2 Ix = Iy (1 2) m =5 105 2= 6.25 ×109
2 Iy Ix 8
=

(4 5)m

Molecules and Solids 529

Section 43.3 Bonding in Solids

1.60 ×10−19 2
0.281 × 10−9
( ) (( ))P43.23
U = − α kee2 ⎛⎜⎝1 − 1 ⎞ = − (1.747 6) 8.99 ×109 ⎛⎜⎝1 − 1 ⎞ = −1.25 × 10−18 J= −7.84 eV
r0 m ⎟⎠ 8 ⎠⎟

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

The number of atoms is ⎛ 10−4 m ⎞3 ~ 1017
⎜ ⎟
⎝ 0.261 × 10−9 m⎠

This number of salt crystals would have volume ( )10−4 m 3 ⎛ 10−4 m ⎞3 ~ 105 m3
⎜ ⎟
⎝ 0.261 × 10−9 m⎠

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

P43.25 U = − kee2 − kee2 + kee2 + kee2 − kee2 − kee2 + kee2 + kee2 − e –e e –e e –e
r r 2r 2r 3r 3r 4r 4r +–+–+–

= − 2kee2 ⎛ − 1 + 1 − 1 + ⎞ r
r ⎝⎜1 2 3 4 ⎟⎠ 2r
3r
But, ln (1+ x) = 1− x2 + x3 − x4 +
FIG. P43.25
234

so, U = − 2kee2 ln 2, or U = −keα e2 where α = 2 ln 2
r r

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 nm) = 0.444 nm

Distance ac is 2 (0.314 nm) = 0.628 nm

Distance ad is ( )22 + 2 2 (0.314 nm) = 0.769 nm FIG. P43.26

Section 43.4 Free-Electron Theory of Metals

Section 43.5 Band Theory of Solids

h2 ⎛⎝⎜ 3ne ⎠⎟⎞ 2 3
2m 8π
P43.27 The density of conduction electrons n is given by EF =

( ( ) ( ) )or ⎦⎤3 2
8π ⎛ 2mEF ⎞3 2 = 8π ⎡⎣2 9.11×10−31 kg (5.48) 1.60 ×10−19 J
3 ⎜⎝ h2 ⎟⎠ 3 J⋅s 3
ne = 6.626 ×10−34 = 5.80 ×1028 m−3

The number-density of silver atoms is

( )nAg = ⎛1 atom ⎞ ⎛ 1 u ⎞ 1028 m−3
10.6 ×103 kg m3 ⎜⎝ ⎠⎟ ⎜ ⎟ = 5.91 ×
108 u ⎝ 1.66 × 10 −27 kg ⎠

So an average atom contributes 5.80 = 0.981 electron to the conduction band .
5.91

530 Chapter 43

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

P43.29 (a) 6.626 × 10−34 J ⋅ s 2
9.11 × 10−31 kg 1.60 × 10−19
( ( )( ) )EFh2⎜⎛⎝ 3ne ⎞2 3 ⎜⎛⎝ 3 ⎞⎠⎟ 2 3
2m 8π ⎟⎠ 8π
= = ne2 3 becomes

2 J eV

( )EF = 3.65 × 10−19 ne2 3 eV with n measured in electrons m3.

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.

1 mv2 = 7.05 eV
2

( )v =
2(7.05 eV) 1.60 × 10−19 J eV = 1.57 × 106 ms

9.11 × 10−31 kg

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

room temperature is typically kBT = 1 eV.
40

*P43.30 The occupation probability is

1 1
(0.99EF −EF )/kBT + 1
(E−EF )/kBT
e ef (E) =
=
+1

= exp[−0.01(7.05 eV)(1.6 ×10−19 1 J/K)300 = 1 = 0.938
J/eV)/(1.38 ×10−23 K] + 1 e−2.72 +1

*P43.31 (a) Eav = 3 EF = 0.6(7.05 eV) = 4.23 eV
(b) 5

The average energy of a molecule in an ideal gas is 3 kBT so we have
2

T = 2 4.23 eV 1.6 ×10−19 J = 3.27 ×104 K
3 1.38 ×10−23 J/K 1 eV

P43.32 For sodium, M = 23.0 g mol and ρ = 0.971 g cm3.

( )( )(a) cm 3
ne = NAρ = 6.02 × 1023 electrons mol 0.971 g
M 23.0 g mol

ne = 2.54 × 1022 electrons cm3 = 2.54 × 1028 electrons m3

( ( )) ( )(b) ⎤2 3
EF = ⎛ h2 ⎞ ⎛ 3ne ⎞ 2 3 = 6.626 × 10−34 J ⋅ s 2 ⎡ 3 2.54 × 1028 m−3 ⎥
⎜⎝ 2m ⎟⎠ ⎝ 8π ⎠ ⎢ 8π ⎦⎥
2 9.11 × 10−31 kg ⎣⎢

= 5.05 × 10−19 J = 3.15 eV

P43.33 Taking EF = 5.48 eV for sodium at 800 K,

f = ⎡⎣e(E−EF ) kBT + 1⎤⎦−1 = 0.950

e =(E−EF ) kBT 1 − 1 = 0.052 6 E = 5.28 eV
0.950

E − EF = ln (0.052 6) = −2.94

kBT

( )1.38 ×10−23 (800) J

E − EF = −2.94 1.60 ×10−19 J eV = −0.203 eV or

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
P43.35
fraction of electrons excited is

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

∫Eav = 1 ∞ EN (E) dE
ne 0

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

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

we can take N (E) = CE1 2 = 8 2π me3 2 E1 2
h3

EF EF

1 CE 3 2dE = C 2C
ne 0 ne 5ne
∫ ∫Eav = E3 2 dE = EF5 2

0

But from Equation 43.25, C = 3 EF−3 2 , so that Eav = ⎛ 2 ⎞⎛ 3 EF−3 2 ⎞ EF5 2 = 3 EF
ne 2 ⎝⎜ 5 ⎟⎠⎜⎝ 2 ⎠⎟ 5

P43.36 d = 1.00 mm, so ( )V = 1.00 ×10−3 m 3 = 1.00 ×10−9 m3
The density of states is
g (E ) = CE1 2 = 8 2π m3 2 E1 2
or h3

( ( ) )g(E) = 8
2π 9.11 × 10−31 kg 3 2 )((4.00 eV) 1.60 × 10−19 J eV
6.626 × 10−34 J ⋅ s 3

g (E) = 8.50 ×1046 m−3 ⋅ J−1 = 1.36 ×1028 m−3 ⋅ eV−1

So, the total number of electrons is

( ) ( )N = ⎣⎡g (E)⎤⎦(∆E)V = 1.36 ×1028 m−3 ⋅ eV−1 (0.025 0 eV) 1.00 ×10−9 m3

= 3.40 ×1017 electrons

*P43.37 (a) The density of states at energy E is g (E) = CE1 2
(b)
Hence, the required ratio is g (8.50 eV) = C (8.50)1 2 = 1.10
g (7.00 eV) C (7.00)1 2

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

( )N E = CE1 2 +1
e(E−EF ) kBT

( ) ( )N 12 ⎡ e(7.00−7.00) kBT + 1⎤⎥
( ) ( )N 12 ⎢ e(8.50−7.00) kB T +1⎦
Hence, the required ratio is 8.50 eV = 8.50 ⎣
7.00 eV 7.00

At T = 300 K, kBT = 4.14 × 10−21 J = 0.025 9 eV,

N (8.50 eV) = (8.50)1 2 ⎡ 2.00 9 ⎤
N (7.00 eV) (7.00)1 2 ⎣⎢ e(1.50) 0.025 + 1⎥⎦

And N (8.50 eV) = 1.47 ×10−25
N (7.00 eV)

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

532 Chapter 43

P43.38 Consider first the wave function in x. At x = 0 and x = L, ψ = 0.

Therefore, sin kx L = 0 and kx L = π, 2π, 3π, . . .

Similarly, sin ky L = 0 and ky L = π, 2π, 3π, . . .

sin kz L = 0 and kz L = π, 2π, 3π, . . .

ψ = A sin ⎛ nxπ x ⎞ sin ⎛ nyπ y ⎞ ⎛ nz π z ⎞
⎜⎝ L ⎠⎟ ⎜ L ⎟sin ⎜⎝ L ⎟⎠
⎝ ⎠

From ∂2ψ + ∂2ψ + ∂2ψ = 2me (U − E)ψ, we have inside the box, where U = 0,
∂x2 ∂y2 ∂z2
2

⎛ nx2π 2 ny2π 2 nz2π 2 ⎞ 2me 2π 2
⎜⎝ − L2 L2 L2 ⎟⎠ 2me L2
2 ( )E
− − ψ = (−E)ψ = nx2 + ny2 + nz2 nx , ny , nz = 1, 2, 3, . . .

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

Section 43.6 Electrical Conduction in Metals, Insulators, and Semiconductors

P43.39 (a) Eg = 1.14 eV for Si
(b)
( )hf = 1.14 eV = (1.14 eV) 1.60 × 10−19 J eV = 1.82 × 10−19 J so f ≥ 2.75 × 1014 Hz

c = λf; λ = c = 3.00 × 108 m s = 1.09 × 10−6 m = 1.09 µm (in the infrared region)
f 2.75 × 1014 Hz

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

( )( )λ = hc =
E
6.626 ×10−34 J ⋅ s 3.00 ×108 ms = 514 nm
2.42 ×1.60 ×10−19 J

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 ×10−34 J ⋅ s 3.00 ×108 ms ⎛ 1 eV ⎞ = 1.24 eV
1.00 ×10−6 m ⎝⎜1.60 ×10−19 ⎟⎠
J

Thus, the energy gap for the collector material should be Eg ≤ 1.24 eV . Since Si has
an energy gap Eg ≈ 1.14 eV, it can absorb nearly all of the photons in sunlight. Therefore,

Si is an appropriate material for a solar collector.

( )( )P43.42
Eg = hc = 6.626 × 10−34 J ⋅ s 3.00 × 108 m s J≈ 1.91 eV
λ 650 × 10−9 m

Molecules and Solids 533

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

( ( )( ) )λmin
(hf )max = hc = 5.5 eV: = hc = 6.626 × 10−34 J ⋅ s 3.00 × 108 m s
λmin 5.5 eV
(5.5 eV) 1.60 × 10−19 J eV

λmin = 2.26 ×10−7 m = 226 nm

P43.44 In the Bohr model we replace ke by ke and me by m * . Then the radius of the first Bohr orbit,
κ

2

a0 = mekee2 in hydrogen, changes to

a′ = 2κ = ⎛ me ⎠⎞⎟κ 2 = ⎛ me ⎞⎟⎠κ a0 ⎛ me ⎞ (0.052 9 nm) = 2.81 nm
m * kee2 ⎝⎜ m* ⎝⎜ m* =⎜ ⎟11.7
me ke e2 ⎝ 0.220me ⎠

The energy levels are in hydrogen En = − kee2 1 and here
2a0 n2

= − kee2 1 = − kee2 = − ⎛ m *⎞ En
κ 2a′ n2 ⎝⎜ me ⎠⎟ κ2
(me m *
)En′ κ 2 κ a0

For n = 1, E1′ = −0.220 13.6 eV = −0.021 9 eV .
11.72

Section 43.7 Semiconductor Devices

( )P43.45 I = I0 ee(∆V) kBT −1 Thus, ee(∆V ) kBT = 1 + I
I0
and
∆V = kBT ⎛ I ⎞
At T = 300 K, ln ⎜1 + ⎟
e ⎝ I0 ⎠

( )∆V =
1.38 ×10−23 J K (300 K) ⎛ I ⎞ = (25.9 mV) ln ⎛ I ⎞
1.60 ×10−19 ln ⎜1 + I0 ⎟ ⎜1+ I0 ⎟
C ⎠ ⎝ ⎠
⎝

(a) If I = 9.00I0 , ∆V = (25.9 mV) ln (10.0) = 59.5 mV

(b) If I = −0.900I0 , ∆V = (25.9 mV) ln (0.100) = −59.5 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 A)(150 Ω) = 3.75 V. Thus, ε − 0.6 V − 3.8 V = 0 and ε = 4.4 V .

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

T = 300 K.

( )e(∆V ) =
( )kBT
1.60 × 10−19 C (0.100 V) I0 = I −1 = 200 mA = 4.28 mA
1.38 × 10−23 J K (300 K) = 3.86 so ee(∆V ) kBT e3.86 − 1

If ∆V = −100 mV, e(∆V ) = −3.86; and the current will be

kBT

( ) ( )( )I = I0 ee(∆V) kBT −1 = 4.28 mA e−3.86 −1 = −4.19 mA

534 Chapter 43

P43.48 (a) The currents to be plotted are Diode and Wire Currents

( )( )ID = 10−6 A e∆V 0.025 V − 1 , Current (mA) 20
10
IW = 2.42 V − ∆V Diode
745 Ω Wire

The two graphs intersect at
∆V = 0.200 V. The currents

are then 0 0.1 0.2 0.3
0 ∆V(volts)
( )( )ID = 10−6 A e − 10.200 V 0.025 V

= 2.98 mA FIG. P43.48

IW = 2.42 V − 0.200 V = 2.98 mA. They agree to three digits.
745 Ω

∴ ID = IW = 2.98 mA

(b) ∆V = 0.200 V A = 67.1 Ω
ID 2.98 × 10−3

(c) d (∆V ) = ⎡ dI D ⎤−1 = ⎡ 10−6 A e0.200 V 0.025 V ⎤−1 = 8.39 Ω
⎣⎢ ⎦⎥ ⎢⎣ 0.025 V ⎥⎦
dI D d ( ∆V )

Section 43.8 Superconductivity

P43.49 By Faraday’s law (from Chapter 32) ∆ΦB = L ∆I = A ∆B
∆t ∆t ∆t

Thus, ∆I = A (∆B) = π (0.010 0 m)2 (0.020 0 T) = 203 A

L 3.10 ×10−8 H

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 = 1 = ∆I = (155 − 57.8) mA = 43.1 Ω−1
R ∆V (3.61 − 1.356) mV

R = 0.023 2 Ω

(c) Expulsion of magnetic flux and therefore fewer FIG. P43.50
current-carrying paths could explain the
decrease in current.

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 = Nµ0I
or NI = B = (0.540 T) 2.50 ×10−2 m = 10.7 kA
µ0
4π ×10−7 T⋅ m A

FIG. P43.51

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 = k = 4.45 × 1014 rad s , I = µr2 = 1.16 ×10−26 kg 1.20 ×10−10 m 2 = 1.67 ×10−46 kg ⋅ m2
µ

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

( )( )2

J (J + 1) =

2I
ω so J (J + 1) = 2Iω = 2 1.67 × 10−46 4.45 × 1014 = 1 410.
1.055 × 10−34

Thus J = 37 .

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

Then f = 1 k where µ12 = (12 u)(16 u) = 6.86 u and µ14 = (14 u)(16 u) = 7.47 u.
2π µ
12 u + 16 u 14 u + 16 u

Therefore,

1 k =1 k ⎛ µ12 ⎞ µ12 = 6.42 ×1013 Hz 6.86 u = 6.15 ×1013 Hz
2π µ14 2π µ12 ⎜ ⎟ µ14 7.47 u
( )f14= ⎠ = f12
⎝ µ14

The equilibrium distance is the same for both molecules.

I14 = µ14 r 2 = ⎛ µ14 ⎞ µ12 r 2 = ⎛ µ14 ⎞ I12
⎝⎜ µ12 ⎠⎟ ⎝⎜ µ12 ⎠⎟

( )I14 ⎛ 7.47 u ⎞ 1.59 × 10−46 kg ⋅ m2
= ⎝ 6.86 u ⎠ 1.46 × 10−46 kg ⋅ m2 =

The molecule can move to the (v = 1, J = 9) state or to the (v = 1, J = 11) state. The

energy it can absorb is either

∆E = hc = ⎡⎣⎢⎛⎜⎝1+ 1 ⎞ + 9 (9 +1) 2 ⎤ − ⎡⎛ − 1 ⎞ + 10 (10 +1) 2 ⎤
λ 2 ⎟⎠ hf14 ⎥ ⎢⎣⎝⎜0 2 ⎠⎟ hf14 ⎥
2 I14 ⎦ 2 I14 ⎦

or ∆E = hc = ⎣⎢⎡⎝⎛⎜1+ 1 ⎞ +11(11+1) 2 ⎤ − ⎡⎛ + 1 ⎞ + 10 (10 +1) 2 ⎤
λ 2 ⎠⎟ hf14 ⎥ ⎢⎣⎜⎝0 2 ⎟⎠ hf14 ⎥
2 I14 ⎦ 2 I14 ⎦

The wavelengths it can absorb are then

λ = f14 c )I14 or λ = f14 + 11 c

− 10 (2π (2π I14 )

These are:

( ) ( )λ = 3.00 × 108 m s = 4.96 µm
6.15 × 1013 Hz − ⎣⎡10 1.055 × 10−34 J ⋅ s ⎤⎦ ⎣⎡2π 1.59 × 10−46 kg ⋅ m2 ⎤⎦

3.00 × 108 m s
( ) ( )and λ = = 4.79 µm
6.15 × 1013 Hz + ⎣⎡11 1.055 × 10−34 J ⋅ s ⎤⎦ ⎣⎡2π 1.59 × 10−46 kg ⋅ m2 ⎤⎦

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
P43.55
( )E = 2 1.74 ×10−23 J ⎛ 6.02 ×1023 atoms ⎞ =
atom ⎜ ⎟ 5.23 J g
⎝ 4.00 g ⎠

( ( )( ) )so
∆Emax = 4.5 eV = ⎛⎝⎜v + 1 ⎞ ω (4.5 eV) 1.6 × 10−19 J eV ≥ ⎛ v + 1⎞
2 ⎠⎟ ⎝ 2⎠
1.055 × 10−34 J ⋅ s 8.28 × 1014 s−1

8.25 > 7.5 v=7

P43.56 Suppose it is a harmonic-oscillator potential well. Then, 1 hf + 4.48 eV = 3 hf + 3.96 eV is the
22

depth of the well below the dissociation point. We see hf = 0.520 eV, so the depth of the well is

1 hf + 4.48 eV = 1 (0.520 eV) + 4.48 eV = 4.74 eV .

22

P43.57 The total potential energy is given by Equation 43.17: U total = −α kee2 + B .
r rm

The total potential energy has its minimum value U0 at the equilibrium spacing, r = r0 . At this

point, dU = 0,
dr r=r0

or dU = d ⎛ kee2 + B ⎞ =α kee2 − mB =0
dr ⎜−α r rm ⎟ r02 r m+1
dr ⎝ ⎠
r=r0 r=r0 0

Thus, B =α kee2 r m−1
m 0

Substituting this value of B into Utotal, U0 = −α kee2 +α kee2 r m−1 ⎛ 1 ⎞ = −α kee2 ⎜⎝⎛1 − 1 ⎞
r0 m 0 ⎜ r0m ⎟ r0 m ⎟⎠
⎝ ⎠

*P43.58 (a) The total potential energy −α kee2 + B has its minimum value at
r rm

the equilibrium spacing, r = r0 . At this point, F = − dU = 0, or
dr
r = r0

F =− d ⎛ kee2 + B ⎞ = −α kee2 + mB = 0
dr ⎜−α r rm ⎟ r02 r m+1
⎝ ⎠ r=r0
0

Thus, B=α kee2 r m−1 .
m 0

Substituting this value of B into F, F = −α kee2 + m α kee2 r m−1 = −α kee2 ⎡ − ⎛ r0 ⎞ m−1 ⎤ .
r2 r m+1 m 0 r2 ⎢1 ⎝ r ⎥
⎣ ⎠ ⎦

(b) Let r = r0 + x so r0 = r – x Then assuming x is small we have

F = −α kee2 ⎡ − ⎛ r − x ⎞ m−1 ⎤ = −α kee2 ⎡ − ⎛⎝ 1 − x ⎞ m−1 ⎤ ≈ −α kee2 ⎡⎢⎣1− 1+ (m − 1) x ⎤
r2 ⎢1 ⎝ r ⎥ r2 ⎢1 ⎥ r2 r ⎥⎦
⎣ ⎠ ⎦ ⎣ r⎠ ⎦

≈ −α kee2 (m − 1)x
r03

This is of the form of Hooke’s law with spring constant K = kea e2(m – 1)/r03.

(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 = α kee2 (m − 1) = 1.7476 8.99 × 109 N ⋅ m2 (1.6 × 10−19 C)2 (8 − 1) = 127 N/m
r03 C2 (2.81 × 10−10 m)3

The vibration frequency of a sodium ion within the crystal is

f= 1 K= 1 127 N/m kg = 9.18 THz
2π m 2π 23.0 × 1.66 × 10−27

Molecules and Solids 537

P43.59 (a) For equilibrium, dU = 0 : ( )d Ax−3 − Bx−1 = −3Ax−4 + Bx−2 = 0
dx
dx
x → ∞ describes one equilibrium position, but the stable equilibrium position is at

3Ax0−2 = B.

)(x0 =
3A = 3 0.150 eV ⋅ nm3 = 0.350 nm
B 3.68 eV ⋅ nm

(b) The depth of the well is given by U0 =U = A −B = AB3 2 − BB1 2
x = x0 x03 x0 33 2 A3 2 31 2 A1 2

3.68 eV ⋅ nm 3 2
0.150 eV ⋅ nm3 1 2
( )( )U0 = U x=x0= − 2 B3 2 2 =− 2 = −7.02 eV
33 2 A1 33 2

(c) Fx = − dU = 3 Ax −4 − Bx −2
dx

To find the maximum force, we determine finite xm such that dF = 0.
dx
x=xm

⎛ 6A ⎞1 2
⎝⎜ B ⎟⎠
Thus, ⎣⎡−12Ax−5 + 2Bx−3 ⎤⎦x=xm =0 so that xm =

= 3A ⎛ B ⎞2 − ⎛ B⎞ = − B2 = − (3.68 eV ⋅ nm)2
⎝ 6A⎠ ⎝ 6A⎠ 12A
12 0.150 eV ⋅ nm3
Then ( )Fmax B

⎛ 1.60 × 10 −19 J ⎞ ⎛ 1 nm ⎞
⎜ ⎟ ⎜⎝10−9 m ⎠⎟
or Fmax = −7.52 eV nm ⎝ 1 eV ⎠ = −1.20 × 10−9 N= −1.20 nN

P43.60 (a) For equilibrium, dU = 0: ( )d Ax−3 − Bx−1 = −3Ax−4 + Bx−2 = 0
dx
dx

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

3Ax0−2 = B or x0 = 3A
B

(b) The depth of the well is given by U0 =U x = x0 = A − B = AB3 2 − BB1 2 = −2 B3 .
x03 x0 33 2 A3 2 31 2 A1 2 27 A

(c) Fx = − dU = 3Ax−4 − Bx−2
dx

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

dFx = ⎣⎡−12Ax−5 + 2Bx−3 ⎦⎤x=x0 =0 then Fmax = 3A ⎛ B ⎞2 − B ⎛ B ⎞ = − B2
dx ⎜⎝ 6A ⎟⎠ ⎜⎝ 6A ⎟⎠ 12 A
x = xm

538 Chapter 43

P43.61 (a) At equilibrium separation, r = re , dU = −2aB⎡⎣e−a(re−r0) −1⎦⎤e−a(re−r0 ) =0
dr
r=re

We have neutral equilibrium as re → ∞ and stable equilibrium at e−a(re−r0) = 1,

or re = r0

(b) At r = r0 , 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 (r0 ) + dU (r − r0 ) + 1 d 2U (r − r0 )2
dr 2 dr 2
r=r0 r=r0

≈ 0+0+1
2
( ) ( ) ( ) ( ) ( )U
r −2Ba ⎡⎣−ae−2(r−r0 ) − ae−(r−r0 ) e−2(r−r0 ) − 1 ⎤⎦r−r0 r − r0 2 ≈ Ba2 r − r0 2

This is of the form 1 kx 2 = 1 k(r − )r0 2
2 2

for a simple harmonic oscillator with k = 2Ba2

Then the molecule vibrates with frequency f = 1 k = a 2B = a B
2π µ 2π µ π 2µ

(d) The zero-point energy is 1 ω = 1 hf = ha B
2 2 π 8µ

Therefore, to dissociate the molecule in its ground state requires energy B − ha B .
π 8µ

Molecules and Solids 539

P43.62 T =0 T = 0.1TF T = 0.2TF T = 0.5TF

E ( )e⎡⎣(E EF )−1⎦⎤(TF T ) f E ( ) ( )e⎡⎣(E EF )−1⎦⎤(TF T ) f E
e⎣⎡(E EF )−1⎤⎦(TF T ) f E
( )EF e⎡⎣(E EF )−1⎦⎤(TF T ) f E

0 e−∞ 1.00 e−10.0 1.000 e−5.00 0.993 e−2.00 0.881

0.500 e−∞ 1.00 e−5.00 0.993 e−2.50 0.924 e−1.00 0.731

0.600 e−∞ 1.00 e−4.00 0.982 e−2.00 0.881 e−0.800 0.690

0.700 e−∞ 1.00 e−3.00 0.953 e−1.50 0.818 e−0.600 0.646

0.800 e−∞ 1.00 e−2.00 0.881 e−1.00 0.731 e−0.400 0.599

0.900 e−∞ 1.00 e−1.00 0.731 e−0.500 0.622 e−0.200 0.550

1.00 e0 0.500 e0 0.500 e0 0.500 e0 0.500

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

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

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

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

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

FIG. P43.62

ANSWERS TO EVEN PROBLEMS
P43.2 4.3 eV
P43.4 (a) 1.28 eV (b) σ = 0.272 nm, ∈= 4.65 eV (c) 6.55 nN (d) 576 N m
P43.6 (a) 0.0148 eV (b) 83.8 mm
P43.8 12.2 × 10–27 kg; 12.4 × 10–27 kg; They agree, because the small apparent difference can be

attributed to uncertainty in the data.

540 Chapter 43

P43.10 1.46 × 10−46 kg ⋅ m2

P43.12 (a) 1.81 × 10−45 kg ⋅ m2 (b) 1.62 cm

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

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

P43.18 (a) 472 µm (b) 473 µm (c) 0.715 µm

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
P43.62 (a) x0 = 3A (b) −2 B3 (c) − B2
B 27A 12A

See the solution.

44

Nuclear Structure

CHAPTER OUTLINE ANSWERS TO QUESTIONS

44.1 Some Properties of Nuclei Q44.1 Because of electrostatic repulsion between the
44.2 Nuclear Binding Energy positively-charged nucleus and the +2e alpha particle.
44.3 Nuclear Models To drive the a -particle into the nucleus would require
44.4 Radioactivity extremely high kinetic energy.
44.5 The Decay Processes
44.6 Natural Radioactivity *Q44.2 (a) X has a mass number less by 2 than the others.
44.7 Nuclear Reactions (b) The ranking is W = Y = Z > X.
44.8 Nuclear Magnetic Resonance Y has a greater atomic number, because a neutron
(c) in the parent nucleus has turned into a proton.
and Magnetic Resonance X has an atomic number less by two than W,
Imagining so the ranking is Y > W = Z > X.
Y has one fewer neutron compared to the parent
nucleus W, and X has two fewer neutrons than W.
The ranking is W = Z > Y > X.

Q44.3 The nuclear force favors the formation of neutron-proton pairs, so a stable nucleus cannot be too
far away from having equal numbers of protons and neutrons. This effect sets the upper boundary
of the zone of stability on the neutron-proton diagram. All of the protons repel one another
electrically, so a stable nucleus cannot have too many protons. This effect sets the lower boundary
of the zone of stability.

Q44.4 Nuclei with more nucleons than bismuth-209 are unstable because the electrical repulsion forces
among all of the protons is stronger than the nuclear attractive force between nucleons.

Q44.5 Nucleus Y will be more unstable. The nucleus with the higher binding energy requires more
energy to be disassembled into its constituent parts.

Q44.6 Extra neutrons are required to overcome the increasing electrostatic repulsion of the protons.
The neutrons participate in the net attractive effect of the nuclear force, but feel no Coulomb
repulsion.

*Q44.7 (i) Answer (a). The liquid drop model gives a simpler account of a nuclear fission reaction,
including the energy released and the probable fission product nuclei.

(ii) Answer (b). The shell model predicts magnetic moments by necessarily describing the spin
and orbital angular momentum states of the nucleons.

(iii) Answer (b). Again, the shell model wins when it comes to predicting the spectrum of an
excited nucleus, as the quantum model allows only quantized energy states, and thus only specific
transitions.

541

542 Chapter 44

Q44.8 The statement is false. Both patterns show monotonic decrease over time, but with very different
shapes. For radioactive decay, maximum activity occurs at time zero. Cohorts of people now
living will be dying most rapidly perhaps forty years from now. Everyone now living will be
dead within less than two centuries, while the mathematical model of radioactive decay tails off
exponentially forever. A radioactive nucleus never gets old. It has constant probability of decay
however long it has existed.

*Q44.9 (i) Answer (b). Since the samples are of the same radioactive isotope, their half-lives are the same.

(ii) Answer (b). When prepared, sample G has twice the activity (number of radioactive
decays per second) of sample H. After 5 half-lives, the activity of sample G is decreased
by a factor of 25, and after 5 half-lives the activity of sample H is decreased by a factor of 25.
So after 5 half-lives, the ratio of activities is still 2:1.

Q44.10 After one half-life, one half the radioactive atoms have decayed. After the second half-life, one
half of the remaining atoms have decayed. Therefore 1 + 1 = 3 of the original radioactive atoms

24 4
have decayed after two half-lives.

Q44.11 Long-lived progenitors at the top of each of the three natural radioactive series are the sources
of our radium. As an example, thorium-232 with a half-life of 14 Gyr produces radium-228 and
radium-224 at stages in its series of decays, shown in Figure 44.17.

*Q44.12 Answer (d). A free neutron decays into a proton plus an electron and an antineutrino. This
implies that a proton is more stable than a neutron, and in particular the proton has lower mass.
Therefore a proton cannot decay into a neutron plus a positron and a neutrino. This reaction
satisfies every conservation law except for energy conservation.

*Q44.13 The alpha particle and the daughter nucleus carry equal amounts of momentum in opposite
p2

directions. Since kinetic energy can be written as , the small-mass alpha particle has much
2m

more of the decay energy than the recoiling nucleus.

Q44.14 Yes. The daughter nucleus can be left in its ground state or sometimes in one of a set of excited
states. If the energy carried by the alpha particle is mysteriously low, the daughter nucleus can
quickly emit the missing energy in a gamma ray.

*Q44.15 Answer (d). The reaction energy is the amount of energy released as a result of a nuclear reaction.
Equation 44.28 in the text implies that the reaction energy is (initial mass − final mass) c2.
The Q-value is taken as positive for an exothermic reaction.

*Q44.16 The samples would have started with more carbon-14 than we first thought. We would increase
our estimates of their ages.

Q44.17 Iz may have 6 values for I = 5 , namely 5 , 3, 1 ,− 1 ,− 3 , and − 5. Seven Iz values are possible
22 2 222 2
for I = 3.

Nuclear Structure 543

*Q44.18 Answer (b). The frequency increases linearly with the magnetic field strength.

Q44.19 The decay of a radioactive nucleus at one particular moment instead of at another instant cannot
be predicted and has no cause. Natural events are not just like a perfect clockworks. In history,
the idea of a determinate mechanical Universe arose temporarily from an unwarranted wild
extrapolation of Isaac Newton’s account of planetary motion. Before Newton’s time [really you
can blame Pierre Simon de Laplace] and again now, no one thought of natural events as just like
a perfect row of falling dominos. We can and do use the word “cause” more loosely to describe
antecedent enabling events. One gear turning another is intelligible. So is the process of a hot
dog getting toasted over a campfire, even though random molecular motion is at the essence of
that process. In summary, we say that the future is not determinate. All natural events have
causes in the ordinary sense of the word, but not necessarily in the contrived sense of a cause
operating infallibly and predictably in a way that can be calculated. We have better reason now
than ever before to think of the Universe as intelligible. First describing natural events, and
second determining their causes form the basis of science, including physics but also scientific
medicine and scientific bread-baking. The evidence alone of the past hundred years of discoveries
in physics, finding causes of natural events from the photoelectric effect to x-rays and jets
emitted by black holes, suggests that human intelligence is a good tool for figuring out how
things go. Even without organized science, humans have always been searching for the causes of
natural events, with explanations ranging from “the will of the gods” to Schrödinger’s equation.
We depend on the principle that things are intelligible as we make significant strides towards
understanding the Universe. To hope that our search is not futile is the best part of human nature.

SOLUTIONS TO PROBLEMS

Section 44.1 Some Properties of Nuclei

P44.1 An iron nucleus (in hemoglobin) has a few more neutrons than protons, but in a typical water
molecule there are eight neutrons and ten protons.
So protons and neutrons are nearly equally numerous in your body, each contributing mass
(say) 35 kg:

⎛ 1 nucleon ⎞ ~1028 protons
35 kg⎝⎜ 1.67 × 10−27 kg⎠⎟

and ~1028 neutrons

The electron number is precisely equal to the proton number, ~1028 electrons .

544 Chapter 44

P44.2 1 mv2 = q∆V and mv2 = qvB
2 r

2m∆V = qr2B2: 2m∆V = 2(1 000 V)
qB2 1.60 × 10−19 C (0.200 T)2 m
( )r =

( )r = 5.59 × 1011 m kg m

( ) ( )(a) For 12C, m = 12 u and r = 5.59 × 1011 m kg 12 1.66 × 10−27 kg

r = 0.078 9 m = 7.89 cm

For 13C: ( ) ( )r = 5.59 × 1011 m kg 13 1.66 × 10−27 kg

r = 0.082 1 m = 8.21 cm

(b) With r1 = 2m1∆V and r2 = 2 m2 ∆V
qB2 qB2

the ratio gives r1 = m1
r2 m2

r1 = 7.89 cm = 0.961
r2 8.21 cm

and m1 = 12 u = 0.961
so they do agree. m2 13 u

*P44.3 (a) Let V represent the volume of the tank. The number of moles present is

n = PV = 1.013 × 105 NV mol ⋅K = 44.6 V mol/m3
RT 8.314 N ⋅ m m2 273 K

Then the number of molecules is 6.023 × 1023 × 44.6 V /m3.

The volume of one molecule is 2 4 πr3 = 8π ⎛ 10−10 m⎞3 = 1.047 × 10−30 m3.
3 3 ⎝⎜ 2 ⎟⎠

The volume of all the molecules is 2.69 × 1025 V (1.047 × 10−30) = 2.81 × 10−5 V.
So the fraction of the volume occupied by the hydrogen molecules is 2.81 × 10−5 .
An atom is precisely one half of a molecule.

nuclear volume 43πr3 ⎛ 1.20 × 10−15 m ⎞ 3
atomic volume ⎜⎝ 0.5 × 10−10 m ⎠⎟
(b) = 4 π (d / 2)3 = = 1.38 × 10−14
3

In linear dimension, the nucleus is small inside the atom in the way a fat strawberry is small
inside the width of the Grand Canyon. In terms of volume, the nucleus is really small.

Nuclear Structure 545

P44.4 Eα = 7.70 MeV

( () ( ) )(a) 2
4 ke Ze2 2ke Ze2 2 8.99 × 109 (79) 1.60 × 10−19
dmin = mv2 = Eα = = 29.5 × 10−15 m= 29.5 fm
7.70 1.60 × 10−13

(b) The de Broglie wavelength of the α is

6.626 × 10−34
( ) ( )λ = h = h = = 5.18 × 10−15 m = 5.18 fm
mα vα 2mα Eα 2 6.64 × 10−27 7.70 1.60 × 10−13

(c) Since λ is much less than the distance of closest approach , the α may be considered a
particle.

P44.5 (a) The initial kinetic energy of the alpha particle must equal the electrostatic potential energy
at the distance of closest approach.

Ki = Uf = keqQ
rmin

( ( ) ( ) )rmin
= keqQ = 8.99 × 109 N ⋅ m2 C2 (2)(79) 1.60 × 10−19 C 2 = 4.55 × 10−13 m
Ki
(0.500 MeV) 1.60 × 10−13 J MeV

(b) Since Ki = 1 mα vi2 = keqQ ,
2 rmin

( ( ) )( ( ) )vi =
2keqQ = 2 8.99 × 109 N ⋅ m2 C2 (2)(79) 1.60 × 10−19 C 2 6.04 × 106 ms
mα rmin (4.00 u) 1.66 × 10−27 kg u 3.00 × 10−13 m =

P44.6 ( )(a) r = r0 A1 3 = 1.20 × 10−15 m (4)1 3 = 1.90 × 10−15 m
( )(b) r = r0 A1 3 = 1.20 × 10−15 m (238)1 3 = 7.44 × 10−15 m

P44.7 The number of nucleons in a star of two solar masses is

( )A
= 2 1.99 × 1030 kg = 2.38 × 1057 nucleons
1.67 × 10−27 kg nucleon

( )( )Therefore r = r0 A1 3 = 1.20 × 10−15 m 2.38 × 1057 1 3 = 16.0 km

P44.8 V = 4 π r4 = 4 π (0.021 5 m)3 = 4.16 × 10−5 m3

33
We take the nuclear density from Example 44.2

( )( )m = ρV = 2.3 × 1017 kg m3 4.16 × 10−5 m3 = 9.57 × 1012 kg

9.57 × 1012 kg 2

(1.00 m)2
( )( )and
F = G m1m2 = 6.67 × 10−11 N ⋅ m2 kg2
r2

F = 6.11 × 1015 N toward the other ball.

546 Chapter 44

Section 44.2 Nuclear Binding Energy

P44.9 Using atomic masses as given in the table in the text,

(a) For 2 H: −2.014 102 + 1(1.008 665) + 1(1.007 825)
1
2

Eb = (0.001194 u)⎛⎝ 931.5 MeV ⎞ = 1.11 MeV nucleon
A u ⎠

(b) For 4 He: 2(1.008 665) + 2(1.007 825) − 4.002 603
2
4

Eb = 0.007 59 uc2 = 7.07 MeV nucleon
A

(c) For 56 Fe: 30(1.008 665) + 26(1.007 825) − 55.934 942 = 0.528 u
26

Eb = 0.528 = 0.009 44 uc2 = 8.79 MeV nucleon
A 56

(d) For 238 U: 146(1.008 665) + 92(1.007 825) − 238.050 783 = 1.934 2 u
92

Eb = 1.934 2 = 0.008 13 uc2 = 7.57 MeV nucleon
A 238

*P44.10 ∆M = ZmH + Nmn − M Eb = ∆M (931.5)

AA

Nuclei Z N M in u ∆M in u Eb in MeV
A

55 Mn 25 30 54.938 050 0.517 5 8.765
56 Fe 26 30 55.934 942 0.528 46 8.790
59 Co 27 32 58.933 200 0.555 35 8.768
∴ 56 Fe has a greater Eb than its neighbors. This tells us finer detail than is shown in Figure 44.5.

A

P44.11 (a) The neutron-to-proton ratio A − Z is greatest for 139 Cs and is equal to 1.53.
Z 55

(b) 139 La has the largest binding energy per nucleon of 8.378 MeV.

(c) 139Cs with a mass of 138.913 u. We locate the nuclei carefully on Figure 44.4, the
neutron−proton plot of stable nuclei. Cesium appears to be farther from the center of the
zone of stability. Its instability means extra energy and extra mass.

P44.12 Use Equation 44.2.

Then for 23 Na, Eb = 8.11 MeV nucleon
11 A

and for 23 Mg, Eb = 7.90 MeV nucleon
12 A

The binding energy per nucleon is greater for 23 Na by 0.210 MeV . There is less proton
11

repulsion in Na23. It is the more stable nucleus.

Nuclear Structure 547

)(P44.13 ⎦⎤ (931.494 u)
The binding energy of a nucleus is Eb (MeV) = ⎡⎣ZM (H) + Nmn − M A X MeV
Z

For 15 O: Eb = ⎣⎡8(1.007 825 u) + 7(1.008 665 u) − 15.003 065 u⎦⎤(931.494 MeV u) = 111.96 MeV
8

For 15 N: Eb = ⎣⎡7(1.007 825 u) + 8(1.008 665 u) − 15.000 109 u⎦⎤(931.494 MeV u) = 115.49 MeV
7

Therefore, the binding energy of 15 N is larger by 3.54 MeV .
7

P44.14 (a) The radius of the 40Ca nucleus is: ( )R = r0 A1 3 = 1.20 × 10−15 m (40)1 3 = 4.10 × 10−15 m

The energy required to overcome electrostatic repulsion is

( ) ( )U = 3keQ2 = 3 8.99 × 109 ⎤⎦2
( )5R
N ⋅ m2 C2 ⎣⎡20 1.60 × 10−19 C = 1.35 × 10−11 J = 84.1 MeV
5 4.10 × 10−15 m

(b) The binding energy of 40 Ca is
20

Eb = ⎡⎣20(1.007 825 u) + 20(1.008 665 u) − 39.962 591 u⎦⎤(931.5 MeV u) = 342 MeV

(c) The nuclear force is so strong that the binding energy greatly exceeds the minimum energy
needed to overcome electrostatic repulsion.

P44.15 Removal of a neutron from 43 Ca would result in the residual nucleus, 42 Ca. If the required
20 20

separation energy is Sn , the overall process can be described by

( ) ( )mass43 42 + mass(n)
20 Ca + Sn = mass 20 Ca

Sn = (41.958 618 + 1.008 665 − 42.958 767) u = (0.008 516 u)(931.5 MeV u) = 7.93 MeV

Section 44.3 Nuclear Models

P44.16 (a) The first term overstates the importance of volume and the second term subtracts this

overstatement.

(b) For spherical volume (4 3)π R3 = R For cubical volume R3 = R
. 6R2 .
4π R2
3 6

The maximum binding energy or lowest state of energy is achieved by building “nearly”

spherical nuclei.

P44.17 ∆Eb = Ebf − Ebi A = 200, Eb = 7.8 MeV
For A
so
For Ebi = 200(7.8 MeV) = 1 560 MeV
so
A ≈ 100, Eb = 8.7 MeV
∆Eb = Ebf − Ebi : A

Ebf = 2(100)(8.7 MeV) FIG. P44.17

= 1 740 MeV

Eb = 1 740 MeV − 1 560 MeV
≈ 200 MeV

548 Chapter 44

P44.18 (a) “Volume” term: E1 = C1A = (15.7 MeV)(56) = 879 MeV

“Surface” term: E2 = −C2A2 3 = − (17.8 MeV)(56)2 3 = −260 MeV
“Coulomb” term:
E3 = −C3 Z (Z − 1) = − (0.71 MeV) (26)(25) = −121 MeV
(56)1 3
A1 3

“Asymmetry” term: E4 = C4 (A − 2Z )2 = − (23.6 MeV) (56 − 52)2 = −6.74 MeV
Eb = 491 MeV
A 56

(b) E1 = 179%; E2 = −53.0%; E3 = −24.6%; E4 = −1.37%
Eb Eb Eb Eb

Section 44.4 Radioactivity

P44.19 dN = −λ N
dt

( )( )so 1 ⎜⎝⎛ − dN ⎟⎞⎠
λ= N dt = 1.00 × 10−15 6.00 × 1011 = 6.00 × 10−4 s−1

T1 2 = ln 2 = 1.16 × 103 s (= 19.3 min)
λ

( )P44.20 5 mCi)⎝⎜⎛ 1 ⎞⎠⎟
R = R0e−λ t = (6.40 mCi) e−(ln2 8.04 d)(40.2 d) = (6.40 mCi) e− ln 2 = (6.40 25 = 0.200 mCi

P44.21 (a) From R = R0e−λ t ,

λ = 1 ln ⎛ R0 ⎞ = ⎛1 ⎞ ln ⎛ 10.0 ⎞ = 5.58 × 10−2 h−1 = 1.55 × 10−5 s−1
t ⎝ R⎠ ⎝ 4.00 h⎠ ⎝ 8.00 ⎠

T1 2 = ln 2 = 12.4 h
λ

(b) N0 = R0 = 10.0 × 10−3 Ci ⎛ 3.70 × 1010 s⎞ = 2.39 × 1013 atoms
λ 1.55 × 10−5 s ⎜⎝ 1 Ci ⎠⎟

( )(c) R = R0e−λt = (10.0 mCi)exp −5.58 × 10−2 × 30.0 = 1.88 mCi

Nuclear Structure 549

*P44.22 From the law of radioactive decay N = N0 e−lt the decay rate is originally R0 = −dN/dt and
decreases according to R = −dN/dt = +N0 λ e−lt R = R0 e−lt

Then algebra to isolate the decay constant gives R0/R = elt ln(R0/R) = λt λ = 1 ln ⎛⎝⎜ R0 ⎞⎟⎠
t R

Now λ = ln 2 gives ln 2 = 1 ln ⎛ R0 ⎞ T1/2 = (ln 2)t
T1/2 T1/2 t ⎝ R ⎠ ln(R0 /R)

where t = ∆t is the time interval during which the activity decreases from R to R.
0

( )P44.23 The number of nuclei that decay during the interval will be N1 − N2 = N0 e−λt1 − e−λt2 .

First we find λ: λ = ln 2 = 0.693 = 0.010 7 h−1 = 2.97 × 10−6 s−1
T1 2 64.8 h

( )and
N0 = R0 = (40.0 µCi) 3.70 × 104 s−1 µCi = 4.98 × 1011 nuclei
λ
2.97 × 10−6 s−1

( )N1 − N2 = ⎡ e−(0.010 7 )h−1 (10.0 h) e−(0.010 7 )h−1 (12.0 h) ⎤
Substituting these values, 4.98 × 1011 ⎣⎢ − ⎦⎥

Hence, the number of nuclei decaying during the interval is N1 − N2 = 9.47 × 109 nuclei .

P44.24 ( )The number of nuclei that decay during the interval will be N1 − N2 = N0 e−λ t1 − e−λ t2 .

First we find λ: λ = ln 2
T1 2

so e−λ t = e ( )ln 2 −t T1 2 = 2−t T1 2

and N0 = R0 = R0T1 2
λ ln 2

R0T1 2 R T 2 − 20 1 2 −t1 T1 2
( ) ( )Substituting in these values ln 2
N1 − N2 = e−λ t1 − e−λ t2 = −t2 T1 2

ln 2

*P44.25 The number remaining after time T1/2 = ln 2
2 2λ

( )is 1/ 2 ⎜⎝⎛ 1 ⎠⎞⎟ 1/ 2
2
N = N0e−λt = N0 e−λ ln 2/2λ = N0 e− ln 2 = N0 = 0.7071 N0

The number decaying in this first half of the first half-life is N0 − 0.7071 N0 = 0.2929 N0.
The number remaining after time T1/2 is 0.500 N0, so the number decaying in the second half
of the first half-life is 0.7071 N0 − 0.500 N0 = 0.2071 N0.

The ratio required is then 0.2929 N0/0.2071 N0 = 1.41

550 Chapter 44

P44.26 (a) dN2 = rate of change of N2
dt

= rate of production of N2 − rate of decay of N2

= rate of decay of N1 − rate of decay of N2

= λ1N1 − λ2 N2

(b) From the trial solution

= N10λ1
λ1 − λ2
( )( )N2
t e−λ2t − e−λ1t

( )∴ dN2 = N10λ1
dt λ1 − λ2
− λ2 e− λ2t + λ e−λ1t (1)
1

N10 λ1
λ1 − λ2
( )∴dN2
dt
+ λ2N2 = −λ2e−λ2t + λ1e−λ1t + λ2e−λ2t − λ2e−λ1t

( )= N10λ1 e− λ1t

λ1 − λ2
λ1 − λ2

= λ1N1

So dN2 = λ1N1 − λ2 N2 as required.
dt

(c) The functions to be plotted are Decay of 218Po and 214Pb

( )N1 t = 1 000 (e− 0.223 6 )min−1 t 1 200 Po
1 000 Pb
( )N2 t = 1130.8 ⎢⎣⎡e−(0.223 6 )min−1 t − (e− 0.025 9 )min−1 t ⎤ Number of nuclei
⎦⎥ 800 10 20 30
600 time (min)
From the graph: tm ≈ 10.9 min 400
200
40
0
0

FIG. P44.26(c)

( )(d)
From (1), dN2 =0 if λ e−λ2t = λ e−λ1t . ∴ e(λ1−λ2 )t = λ1 . Thus, t = tm = ln λ1 λ2 .
dt 2 1 λ2 λ1 − λ2

With λ1 = 0.223 6 min−1, λ2 = 0.025 9 min−1, this formula gives tm = 10.9 min ,
in agreement with the result of part (c).

P44.27 We have all this information: Nx (0) = 2.50Ny (0)

Nx (3d) = 4.20Ny (3d)

Nx ( )0 e−λx 3d = 4.20 N y ( )0 e−λy 3d = 4.20 ( )Nx 0 e−λy 3d

2.50

e3dλx = 2.5 e3dλy
4.2

3dλx = ln ⎝⎜⎛ 2.5 ⎟⎠⎞ + 3dλy
4.2

3d 0.693 = ln ⎛⎝⎜ 42..25 ⎞⎠⎟ + 3d 0.693 = 0.781
T1 2 x 1.60 d

T1 2 x = 2.66 d