Chapter 38 Quantum Mechanics
G
2m U0 E 2 4 1.67 1027 kg 32.36 MeV 4 MeV 1.60 1013 J MeV
1.055 1034 Js
2.33 1015 m1
1 1 1
fT fe2Gl 8.625 1020 collisions s e 2 2.331015 m1 18.91015 m
T e2Gl
; Lifetime
2.06 1017 s 1yr s 7 109 yr
3.156 107
46. We find the lifetime of the particle from Eq. 38-2.
t
1.055 1034 Js GeV 2.6 1025 s.
E
2.5GeV 1.60 1010 J
47. We use the radius as the uncertainty in position for the neutron. We find the uncertainty in the
momentum from Eq. 38-1. If we assume that the lowest value for the momentum is the least
uncertainty, we estimate the lowest possible kinetic energy (non-relativistic) as
2 1.055 1034 kgm s 2
x 1.67 1027 kg 1.2 1015 m
E p 2 2 2 2.314 1012 J 1MeV J
2m 1.60 1013
2m
14.46 MeV 14 MeV
48. The energy levels for a particle in an infinite potential well are given by Eq. 38-13. The wave
functions are given by Eq. 38-14. with A 2 .
l
6.63 1034 Js 2
2.5 1015 m 2 1.60 1013 J MeV 32.90 MeV 33MeV
(a)E1 h2
8ml2 8 1.67 1027 kg
E2 22 E1 4 32.90 MeV 130 MeV ; E3 32 E1 9 32.90 MeV 300 MeV 2 sig. fig.
E4 42 E1 1632.90 MeV 530 MeV
(b) n 2 sin n x
l l
2
1015 1015
1
2.5 m sin 2.5 m x 2.8 107 m1/ 2 sin 1.3 1015 m1 x
2 2
1015 1015
2
2.5 m sin 2.5 m x 2.8 107 m1/ 2 sin 2.5 1015 m1 x
2 3
1015 1015
3
2.5 m sin 2.5 m x 2.8 107 m1/ 2 sin 3.8 1015 m1 x
2 4
1015 1015
4
2.5 m sin 2.5 m x 2.8 107 m1/ 2 sin 5.0 1015 m1 x
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(c) E E2 E1 4E1 E1 3E1 332.90 MeV 98.7 MeV 99 MeV
E hc
hc 6.631034 Js 3.00 108 m s 1.3 1014 m 13fm
E
98.7 MeV 1.60 1013 J MeV
This is in the gamma-ray region of the EM spectrum, as seen in Fig. 31-12.
49. We find the wavelength of the protons from their kinetic energy, and then use the two-slit
interference formulas from Chapter 34, with a small angle approximation. If the protons were
accelerated by a 650-volt potential difference, then they will have 650 eV of kinetic energy.
h h ; d sin m, m 1,2, ... ; y l tan
p 2m0 K
sin tan m y y m l , m 1, 2, ...
d l d
y
l d hl 8.0 104 m 6.631034 Js 18m
d 2m0 K 2 1.67 1027 kg 650eV 1.60 1019 J eV
2.5 108 m
50. We assume that the particles are not relativistic. Conservation of energy is used to find the speed of
each particle. That speed then can be used to find the momentum and finally the de Broglie
wavelength. We let the magnitude of the accelerating potential difference be V.
U Kinitial eV 1 mv 2 v 2eV ; h h h h x
2 m p mv 2eV 2meV
final m
m
xp h p 2 h
2 x
2 h h
pproton xproton xelectron 2melectron eV mproton 1.67 1027 kg 43
pelectron 2 h xproton h m electron 9.11 1031 kg
xelectron 2mproton eV
51. We use Eq. 37-10, Bohr’s quantum condition.
mvrn n h mvr1 mv p p
2 r1
xp x r1
p r1
The uncertainty in position is comparable to the Bohr radius.
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502
Chapter 38 Quantum Mechanics
52. (a) See the diagram.
(b) We use the solution x AeBx2 in the
Schrödinger equation.
Ae Bx2 ; d 2 ABxeBx2 U (x )
dx
d 2
dx2
2 ABeBx2 2 ABx 2 Bxe Bx2
2 2Bx2 1 ABeBx2 x
2 d 2 2 ABe Bx2 Ae Bx2 EAeBx2
2m dx2 U 2m 2 2Bx2 1 1 Cx2
2
2B E 1 C 22B2 x2 0
m 2 m
This is a solution if 2B E and 1 C 22B2 . Solve these two equations for E in terms of C,
m 2 m
and let C m.
1 C 22 B 2 B mC ; E 2B 2 mC 1 C m 1
2 m 2 m m 2 2 2
B mC m C m m
2 2 2
53. We assume the alpha particle is in the ground state. The energy is given by Eq. 38-13.
6.63 1034 Js 2
1.5 1014 m 2 1.60 1013 J
E1h2 0.2285MeV
8ml2 84 1.67 1027 kg MeV
0.23MeV
The speed can be found from the kinetic energy. The alpha is non-relativistic.
E1
h2 1 mv 2 v h 6.63 1034 Js 3.3106 m s
8ml2 2 2ml 1.67 1027 kg 1.5 1014 m
24
54. From energy conservation, the speed of a ball after falling a height H, or the speed needed to rise to a
height H, is v 2gH . We say that the starting height is H0, and so the speed just before the ball
hits the ground before the first bounce is v0 2gH0 . After that bounce, the ball rebounded to
H1 0.65H0, and so the speed right after the first bounce, and right before the second bounce, is
v1 2gH1 2g 0.65 H0 . Repeated application of this idea gives the maximum height after n
bounces as Hn 0.65n H0, and the maximum speed after n bounces as vn 2g 0.65n H0 . The
uncertainty principle will come into play in the problem when the maximum speed after a bounce is
of the same order as the uncertainty in the speed. We take the maximum height as the uncertainty in
the position.
mvy py ; ypy mv y mvn
y Hn
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m 2g 0.65n H0 H0 2m2 g 0.65n H0 2 H 2
0
0.65n 0.652n
1.055 1034 Js 2
ln 3.0 106 kg 2 9.80 m s2
2 3ln 0.65 2.0 m3
2 ln 2 3
2m2 gH 0
0.653n n 105
2m2 gH 3 3ln 0.65
0
After about 105 bounces, the uncertainty principle will be important to consider.
55. We model the electrons as being restricted from leaving the surface of the sodium by an energy
barrier, similar to Figure 38-15a. The difference between the barrier’s height and the energy of the
electrons is the work function, and so U0 E W0 2.28eV. But quantum mechanically, some
electrons will “tunnel” through that barrier without ever being given the work function energy, and
thus get outside the barrier, as shown in Figure 38-15b. This is the tunneling current as indicated in
Figure 38-18. The distance from the sodium surface to the tip of the microscope is the width of the
barrier, l. We calculate the transmission probability as a function of barrier width by Eqs. 38-17a
and 38-17b. The barrier is then increased to l l, which will lower the transmission probability.
2Gl 2l
2m U0 E 2 0.02 109 m 2 9.111031kg 2.28eV 1.60 1019 J eV
1.055 1034 Js
0.3091
T0 e2Gl ; T e2Gll ; T e 2G l l e2Gl e0.3091 0.734
T0 e2Gl
The tunneling current is caused by electrons that tunnel through the barrier. Since current is directly
proportional to the number of electrons making it through the barrier, any change in the transmission
probability is reflected as a proportional change in current. So we see that the change in the
transmission probability, which will be reflected as a change in current, is a decrease of 27%. Note
that this change is only a fraction of the size of an atom.
56. The time independent Schrödinger equation with U = 0 is 2 d 2 E .
2m dx 2
2 2mE 2
d 2 2 d2 2 2k 2 E
dx2 2m dx2 2m 2m 2m
2 Aeikx k 2 Aeikx
2m
We see that the function solves the Schrödinger equation.
57. The wave functions for the particle in the infinite well are n 2 sin n x , as derived in Section
l l
38-8. A table of integrals was consulted to find x2 sin2 ax dx.
l 2 n 2 2 l n
l l l 0 l
x2 x2n 2 dx x2 sin x dx x 2 sin 2 x dx
0
Note that x2 x3 x2 1 x cos 2ax
sin2 ax dx 6 4a 8a 3 sin 2ax 4a2 .
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504
Chapter 38 Quantum Mechanics
x cos 2n x l
l
x22 l x2 sin2 n x dx 2 x3 x2 1 sin 2n x
l 0 l 4 n l
l 6 8 n 3 4 n 2 0
l l l
l2 1 1 n 2
3 2
See the adjacent graph. The 0.34
spreadsheet used for this 0.33
problem can be found on the
Media Manager, with filename 0.32
“PSE4_ISM_CH38.XLS,” on 0.31
tab “Problem 38.57.” 0.30
0.29
0.28
0 2 4 6 8 10 12 14 16 18 20
n
58. (a) To check that the wave function is normalized, we calculate x 2 dx.
x2 x2 b2 x2
2 dx 1 x b2 2 xe b2 2 e b2
b b b2 b2 0
x e dx dx 2 0 1 1
0
We see that the function is normalized.
(b) The most probable position is that for which x 2 is maximized. That point can be found by
solving d x2 0 for x. Since we are only considering x 0, we need not use the absolute
dx
value signs in the function.
d x e x2
b2
2 b2
d x 1 x2 x 2x x2 2x2
b2 e b2 b2 b2 e b2 0 1 b2
dx
dx
x b 1.0 nm 0.71nm
2 2
This value for x maximizes the function, because the function must be positive, and the function
is 0 at x 0 and x . Thus this single local extreme point must be a maximum.
(c) To find the probability, we integrate the probability density function between the given limits.
0.50 nm x 2 dx 0.50 nm x e x2 dx 1 e x2 0.50 nm 1 e0.25 1 0.11
0 b2 b2 2 b2 2 2
P
0
0
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
59. (a) We assume the pencil is a uniform rod, and that it makes an angle of with the
vertical. If the bottom point is fixed, then the torque due to its weight about the
bottom point will cause an angular acceleration. See the diagram. l
I mg 1 l sin 1 ml2 d 2
2 3 dt 2
From the equation, if the pencil is exactly upright, so that 0, then the mg
angular acceleration will be exactly 0 and the pencil will remain stationary.
But according to the uncertainty principle (as expressed at the bottom of page
1023), the angle cannot be known with 0 uncertainty. Let the z axis be coming out of the page.
Lz 0
Lz
Thus the pencil cannot have exactly 0, and so there will be a torque and hence rotation.
(b) For the initial part of the motion, the angle will be very small, and so the differential equation
can be expressed as 3g d 2 . The solutions to this differential equation are of the form
2l dt 2
Aekt Bekt , where k
3g 3 9.80 m s2 9.037s1 Since the angle will be
2l
2 0.18 m
increasing in time, we ignore the second term, which decreases in time. Thus Aekt , with
t 0 0 A. The angular velocity of the pencil is approximated as d kAekt , and
dt
the initial angular velocity is 0 kA. We take the initial position and the initial angular
velocity as their smallest possible values, which are their uncertainties – the magnitude of a
quantity must be at least as big as its uncertainties. Apply the uncertainty principle in angular
form.
Lz I 1 ml2 kA A 1 ml2kA2
3 3
A
3 3 1.055 1034 Js 3.930 1016 rad 0
ml2k
7.0 103 kg 0.18 m2 9.037s1
0 kA 9.037s1 3.930 1016 rad 3.552 1015 rad s
With this initial position and initial angular velocity, we can then do a numeric integration to
find the time when the angle is 2 rad. For a step size of 0.01 s, the time of fall is about
4.07 s. For a step size of 0.001 s, the time of fall is about 3.99 s. This is only about a 2%
change in the final result, so the time is pretty stable around 4 s. Even changing the starting
angle to a value 100 times bigger than that above (so 0 3.930 1014 rad ) still gives a time of
fall of 3.48 s. So within a factor of 2, we estimate the time of fall as 4 seconds.
Note that if the solution of the approximate differential equation is used, Aekt , we get the
following time of fall.
tmax 1 ln max 1 ln 2 3.98s.
k A 9.037s1 3.930 1016
The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH38.XLS,” on tab “Problem 38.59.”
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506
Chapter 38 Quantum Mechanics
60. The ground state wave function for the particle in the infinite well is 2 sin x . Let
l l
xcenter 1 l, so the region of interest extends from xmin 1 l 1 x to xmax 1 l 1 x . We are to
2 2 2 2 2
find the largest value of x so that the approximate probability of x 1 l 2 x 2x (from
2 l
xmax 2 2
l l
Example 38-7) is no more than 10% different than sin x dx , the exact probability.
xmin
We calculate the value of the integral using numeric integration (as described in Section 2-9), first
finding the number of steps needed between xmin and xmax that gives a stable value. Then we
compare the integral to the approximation. (Note that the integral could be evaluated exactly.)
To aid in the evaluation of the integral, we make the substitution that u x l. Then the integral
becomes as follows.
umax 1 1 x
2 l
2sin2 u du
1 x
umin 1 l
2
In doing the numeric integrations, we found that for any value of x up to l, breaking the numeric
integration up into 50 steps gave the same answer to 3 significant digits as breaking it up into 100
steps. So we did all numeric integrations with 50 steps. We then numerically calculated the integral
for values of x l , starting at 0.01, and increasing by steps of 0.01, until we found a 10% difference
between the approximation and the numeric integration. This happens at x l 0.34, and so the
approximation is good within 10% up to x 0.34l 0.340.10 nm 0.034 nm . This is much
broader than we might have guessed initially, indicating that the wave function is varying rather
slowly over the central region of the potential well. The spreadsheet used for this problem can be
found on the Media Manager, with filename“PSE4_ISM_CH38.XLS,” on tab “Problem 38.60.”
61. (a) See the graph.
(b) From the graph and the 1.0
spreadsheet, we find these 0.8
0.6
results. Probability 0.4 Transmission
T 10% at E U0 0.146 Reflection
T 20% at E U0 0.294 0.2
T 50% at E U0 0.787 0.0 2 4 6 8 10
T 80% at E U0 1.56 0
E /U 0
The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH38.XLS,” on tab “Problem 38.61.”
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507
CHAPTER 39: Quantum Mechanics of Atoms
Responses to Questions
1. The Bohr model placed electrons in definite circular orbits described by a single quantum number
(n). The Bohr model could not explain the spectra of atoms more complex than hydrogen and could
not explain fine structure in the spectra. The quantum-mechanical model uses the concept of electron
“probability clouds,” with the probability of finding the electron at a given position determined by
the wave function. The quantum model uses four quantum numbers to describe the electron (n, l, ml,
ms) and can explain the spectra of more complex atoms and fine structure.
2. The quantity 2 is maximum at r = 0 because of its dependence on the factor er .r0 In the ground
state, the electron is expected to be found near the nucleus. The radial probability density
4 r2 2 gives the probability of finding the electron in a thin spherical shell located at r. Since r = 0
is at the center of the nucleus, the radial probability density is zero here.
3. The quantum-mechanical model predicts that the electron spends more time near the nucleus. In the
Bohr model, the electron in the ground state is in a fixed orbit of definite radius. The electron cannot
come any closer to the nucleus than that distance. In the quantum-mechanical model, the electron is
most often found at the Bohr radius, but it can also be found closer to the nucleus (and farther away).
4. As the number of electrons goes up, the number of protons in the nucleus increases, which increases
the attraction of the electrons to the center of the atom. Even though the outer electrons are partially
screened from the increased nuclear charge by the inner electrons, they are all pulled closer to the
more positive nucleus. Also, more states are available in the upper shells to accommodate many
more electrons at approximately the same radius.
5. Because the nuclei of hydrogen and helium are different, the energy levels of the atoms are different.
The presence of the second electron in helium will also affect its energy levels. If the energy levels
are different, then the energy difference between the levels will be different and the spectra will be
different.
6. The two levels have different orbital quantum numbers. The orbital quantum number for the upper
level is l = 2. This results in five different possible values of ml (–2, –1, 0, 1, and 2) so the energy
level is split into five separate levels in the presence of a magnetic field. The lower level shown has
an orbital quantum number of l = 1, so only three different values of ml (–1, 0 and 1) are possible,
and therefore the energy level is split into only three separate levels.
7. In the time-independent Schrödinger equation, the wave function and the potential depend on the
three spatial variables. The three quantum numbers result from application of boundary conditions to
the wave function.
8. The Zeeman effect is the splitting of an energy level in the presence of a magnetic field. In the
reference frame of the electron, the nucleus orbits the electron. The “internal” Zeeman effect, as seen
in sodium, is caused by the magnetic field produced by the “orbiting” nucleus.
9. (a) and (c) are allowed for atoms in an excited state. (b) is not allowed. Only six electrons are
allowed in the 2p state.
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Chapter 39 Quantum Mechanics of Atoms
10. The complete electron configuration for a uranium atom is as follows.
1s22s22p63s23p63d104s24p64d104f145s25p65d105f36s26p66d17s2
11. (a) Group II; (b) Group VIII; (c) Group I; (d) Group VII.
12. The periodicity of the periodic table depends on the number and arrangements of the electrons in the
atom. It therefore depends on all the factors which determine this arrangement. One of these, the
Pauli exclusion principle, states that no two electrons can occupy the same quantum state. The
number of electrons that can be in any principle state depends on how many different substates are
available, which is determined by the number of possible orbital quantum numbers for each principle
state, the number of possible magnetic quantum numbers for each orbital quantum number, and
finally, the number of spin orientations for each electron. Therefore, quantization of angular
momentum, direction of angular momentum, and spin all play a role in the periodicity of the periodic
table. (See Table 39-1 for a summary of the quantum numbers.)
13. If there were no electron spin, then, according to the Pauli exclusion principle, s-subshells would be
filled with one electron, p-subshells with three electrons, and d-subshells with five electrons. The
first 20 elements of the periodic table would look like the following:
H1 Na 11 Mg 12 Al 13 Si 14 P 15 Li 3 Be 4 B5
1s1 3d1 3d2 3d3 3d4 3d5 2p1 2p2 2p3
He 2 Ca 20 N7 O8 F9
2s1 4d1 3p1 3p2 3p3
C6 S 16 Cl 17 Ar 18
3s1 4p1 4p2 4p3
Ne 10
4s1
K 19
5s1
14. Neon is a noble gas and does not react readily with other elements. Neon has its outermost subshell
completely filled, and so the electron distribution is spherically symmetric, making it harder to
remove an electron. Sodium is in the first column of the periodic table and is an alkali metal. Sodium
has a single outer s electron, which is outside the inner closed shells and shielded from the nuclear
charge by the inner electrons, making it easier to remove. Therefore, neon has a higher ionization
energy than sodium, even though they differ in number of protons by only one.
15. Chlorine and iodine are in the same column of the periodic table. They are each one electron away
from having a complete outermost shell and will react readily with atoms having only one electron in
the outermost shell. Interactions with other atoms depend largely on the outermost electrons;
therefore these two elements will have similar properties because their outermost electrons are in
similar configurations.
16. Potassium and sodium are in the same column of the periodic table. They each have only one
electron in the outermost shell, and their inner shells are completely filled. Sodium has only one
electron in the n = 3 shell, or principle energy level, and potassium has only one electron in the n = 4
level. Interactions with other atoms depend largely on the outermost electrons, and therefore these
two elements will have similar properties because their outermost electrons are in similar
configurations.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
17. Rare earth elements have similar chemical properties because the electrons in the filled 6s or 7s-
suborbitals serve as the valence electrons for all these elements. They all have partially filled inner f-
suborbitals, which are very close together in energy. The different numbers of electrons in the f-
suborbitals have very little effect on the chemical properties of these elements.
18. When we use the Bohr theory to calculate the X-ray line wavelengths, we estimate the nuclear
charge seen by the transitioning electron as Z – 1, assuming that the second electron in the ground
state is partially shielding the nuclear charge. This is only an estimate, so we do not expect the
calculated wavelengths to agree exactly with the measured values.
19. In helium and other complex atoms, electrons interact with other electrons in addition to their
interactions with the nucleus. The Bohr theory only works well for atoms that have a single outer
electron in an s state. X-ray emissions generally involve transitions to the 1s or 2s states. In these
cases the Bohr theory can be modified to correct for screening from a second electron by using the
factor Z – 1 for the nuclear charge and can yield good estimates of the transition energies.
Transitions involving outer electrons in more complex atoms will be affected by additional complex
screening effects and cannot be adequately described by the Bohr theory.
20. The continuous portion of the X-ray spectrum is due to the “bremsstrahlung” radiation. An incoming
electron gives up energy in the collision and emits light. Electrons can give up all or part of their
kinetic energy. The maximum amount of energy an electron can give up is its total amount of kinetic
energy. In the photon description of light, the maximum electron kinetic energy will correspond to
the energy of the shortest wavelength (highest energy) photons that can be produced in the
collisions. The result is the existence of “cut-off” wavelength in the X-ray spectrum. An increase in
the number of electrons will not change the cut-off wavelength. According to wave theory, an
increase in the number of electrons could result in the production of shorter-wavelength photons,
which is not observed experimentally.
21. To figure out which lines in an X-ray spectrum correspond to which transitions, you would use the
Bohr model to estimate the energies of the transitions between levels and match these to the energies
of the observed lines. The energies of transitions to the n = 1 level (K) will be the greatest, followed
by the transitions to the n = 2 level (L). Within a level, the α line will have the lowest energy
(because it corresponds to a transition between adjacent levels), followed by the β line, and so on.
22. The characteristic X-ray spectra occur when inner electrons are knocked out of their shells. X-rays
are the high energy photons emitted when other electrons fall to replace the knocked-out electrons.
Because the shells involved are close to the nucleus, Z will have a direct influence on the energies.
The visible spectral lines due to transitions between upper levels have energies less influenced by Z
because the inner electrons shield the outer electrons from the nuclear charge.
23. The difference in energy between adjacent energy levels in an atom decreases with increasing n.
Therefore, transitions of electrons between inner energy levels will produce higher energy (shorter
wavelength) photons than transitions between outer energy levels.
24. The electron has a negative charge.
25. Consider a silver atom in its ground state for which the entire magnetic moment is due to the spin of
only one of its electrons. In a uniform magnetic field, the dipole will experience a torque that would
tend to align it with the field. In a non-uniform field, each pole of the dipole will experience a force
of different magnitude. Consequently, the dipole will experience a net force that varies with the
spatial orientation of the dipole. The Stern-Gerlach experiment provided the first evidence of space
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510
Chapter 39 Quantum Mechanics of Atoms
quantization, since it clearly indicated that there are two opposite spin orientations for the outermost
electron in the silver atom.
26. Spontaneous emission occurs randomly when an electron in an excited state falls to a lower energy
level and emits a photon. Stimulated emission also results when an electron falls to a lower energy
level, but it occurs when a photon of the same energy as the transition stimulates the electron to fall
sooner than it would have naturally.
27. No. The intensity of a spherical wave, which spreads out in all directions, follows the inverse-square
law. A laser produces light that is very nearly a plane wave; its intensity is nearly constant with
distance.
28. Laser light is monochromatic, coherent, and in a narrow beam that spreads very little if at all.
Ordinary light is usually made up of many different wavelengths, incoherent, and spreads out in all
directions. Both types of light can be created when electrons fall to lower energy levels and emit
photons.
29. Since laser light is a plane wave, its intensity remains approximately constant with distance. The
light produced by a street lamp spreads out with an intensity that decreases as 1/r. Thus, at a
sufficient distance, the laser light will be more intense than the light from a street lamp.
Solutions to Problems
1. The value of l can range from 0 to n 1. Thus for n = 7, l 0, 1, 2, 3, 4, 5, 6 .
2. The value of ml can range from l to l. Thus for l 3, ml 3, 2, 1, 0, 1, 2, 3 .
The possible values of ms are 1 , 1 .
2 2
3. The value of l ranges from 0 to n 1. Thus for n 3, l 0,1,2. For each l the value of ml can
range from l to l, or 2l 1 values. For each ml there are 2 values of ms. Thus the number of
states for each l is 22l 1. The number of states is N 20 1 22 1 24 1
18states . We start with l 0, and list the quantum numbers in the order n, l,ml ,ms .
(3, 0, 0, 1 ), (3, 0, 0, 1 ), (3, 1, –1, 1 ), (3, 1, –1, 1 ), (3, 1, 0, 1 ), (3, 1, 0, 1 ),
2 2 2 2 2 2
(3, 1, 1, 1 ), (3, 1, 1, 1 ), (3, 2, – 2, 1 ), (3, 2, – 2, 1 ), (3, 2, –1, 1 ), (3, 2, –1, 1 ),
2 2 2 2 2 2
(3, 2, 0, 1 ), (3, 2, 0, 1 ), (3, 2, 1, 1 ), (3, 2, 1, 1 ), (3, 2, 2, 1 ), (3, 2, 2, 1 )
2 2 2 2 2 2
4. The value of ml can range from l to l, so we have 4 .
The value of l can range from 0 to n 1. Thus we have n 1minimum5 .
There are two values of ms : ms 1 , 1 .
2 2
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5. The value of l can range from 0 to n 1. Thus for l 5, we have n 6 .
For each l the value of ml can range from l to l : ml 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5 .
There are two values of ms : ms 1 , 1 .
2 2
6. The magnitude of the angular momentum depends only on l.
L l l 1 12 12 1.055 1034 Js 3.65 1034 Js
7. (a) The principal quantum number is n 7 .
(b) The energy of the state is
E7 13.6 eV 13.6 eV 0.278eV .
n2 72
(c) The “g” subshell has l 4 . The magnitude of the angular momentum
depends on l only:
L l l 1 20 20 1.055 1034 Js 4.72 1034 Js
(d) For each l the value of ml can range from l to l : ml 4, 3, 2, 1, 0, 1, 2, 3, 4 .
8. (a) For each l the value of ml can range from l to l, or 2l 1 values. For each of these there
are two values of ms. Thus the total number of states in a subshell is N 22l 1 .
(b) For 0, 1, 2, 3, 4, 5, and 6, N 2, 6, 10, 14, 18, 22, and 26 , respectively.
9. For a given n, 0 l n 1. Since for each l the number of possible states is 22l 1, the number
of possible states for a given n is as follows.
n1 n 1 n n 1
l0 2
22l
l0
n 1 2 4 2n 2n2
1 4 l
l0
10. Photon emission means a jump to a lower state, so for the final state, n = 1, 2, 3, or 4. For a d
subshell, l 2, and because l 1, the new value of l must be 1 or 3.
(a) l 1 corresponds to a p subshell, and l 3 corresponds to an f subshell. Keeping in mind that
0 l n 1, we find the following possible destination states: 2 p,3p,4 p,4 f .
(b) In a hydrogen atom, l has no appreciable effect on energy, and so for energy purposes there are
four possible destination states, corresponding to n = 2, 3, and 4. Thus there are three different
photon wavelengths corresponding to three possible changes in energy.
11. We use Eq. 39-3 to find l and Eq. 39-4 to find ml.
6.84 1034 Js 2
1.055 1034 Js 2 42
L L2
l l 1 l l 1 2 l6
Lz ml ml Lz 2.111034 Js 2
1.055 1034 Js
Since l 6, we must have n 7 .
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Chapter 39 Quantum Mechanics of Atoms
12. To show that the ground-state wave function is normalized, we integrate 100 2 over all space. Use
substitution of variables and an integral from Appendix B-5.
100 2 dV 1 e 2r 4 r 2 dr ; let x 2r r 1 r0 x , dr 1 r0dx
0 r03 r0 r0 2 2
all
space
Note that if r 0, x 0 and if r , x .
1 2r 4
0 r03 r03
2 dV e ex
all r0 r2dr r02
100 4 0 1 x2 1 r0dx 1 ex x2dx 1 2! 1
4 2 2 2
space 0
And so we see that the ground-state wave function is normalized.
13. The ground state wave function is 100 1 er / r0 .
r03
(a) 100 r1.5r0 1 e1.5
r03
(b) 1
2 r03 e3
100 r1.5r0
(c) 1 4 e3
Pr 4 r2 100 2 r 1.5r0 4 r02 r03 e2 r0
14. The state n 2, l 0 must have ml 0 and so the wave function is 200 1 2 r e r .
32 r0 2 r0
r03
(a) 1 2 4r0 e 4 r0 1 e2
200 r4r0 32 r03 r0 2 r0 8 r03
(b) 1 4r0 2 e2 4 r0 1
2 32 r03 2 r0 2 r0 8 r03 e4
200
r 4 r0
(c) 1 8 e4
Pr 4 r2 200 2 4 4r0 2 8 r03 e4 r0
r 4 r0
15. The factor is found from the ratio of the radial probability densities for 100. Use Eq. 39-7.
4 r2 e 2r r02 e 2 r0
r02 r0 rr0 4 r02 r0 rr0
Pr r r0 2r 2 2 2 r0 e2 e2 1.85
Pr r 2r0 r0 r0 4e4 4
4 r2 e 2r0 e
r02 r2r0 4 r02
r 2 r0
16. (a) To find the probability, integrate the radial probability distribution for the ground state. We
follow Example 39-4, and use the last integral in Appendix B-4.
P r0 4 r2 e 2r dr ; let x 2 r
0 r03 r0 r0
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
2
P 1 x 2e x dx 1 e x x2 2x 2 02 1 5e2 0.32 32%
2 2
0
(b) We follow the same process here.
P 2 r0 4 r2 e 2r dr ; let x 2 r
r0 r03 r0 r0
4 x2 4 5e2 13e4
P 1 x 2e x dx 1 e x 2x 2 0.44 44%
2 2 2
2
17. To find the probability for the electron to be within a sphere of radius r, we must integrate the radial
probability density for the ground state from 0 to r. The density is given in Eq. 39-7.
P rsphere 4 r2 e 2r dr ; let x 2 r ; let 2 rsphere x
0 r03 r0 r0 r0
x x
P
1 x2e xdx 1 e x x2 2x 2 0 1 e x x2 2x 2 2 1 ex 1 x2 x 1
2 2 2 2
0
We solve this equation numerically for values of x that give P = 0.50, 0.90, and 0.99.
(a) The equation for P = 0.50 is solved by x = 2.674, and so rsphere 1 2.674 r0 1.3r0 .
2
(b) The equation for P = 0.90 is solved by x = 5.322, and so rsphere 1 5.322 r0 2.7 r0 .
2
(c) The equation for P = 0.99 is solved by x = 8.406, and so rsphere 1 8.406 r0 4.2 r0 .
2
The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH39.XLS,” on tab “Problem 39.17.”
18. (a) To find the probability for the electron to be within a sphere of radius r, we must integrate the
radial probability density for the ground state from 0 to r. The density is given in Eq. 39-7.
Since r r0, we approximate e2r r0 1.
1.11015 m 3
0.529 1010 m
P1.1fm 4 r2 e 2r dr 1.1fm 4 r2 dr 4 r3 4 1.2 1014
0 r03 r0 0 r03 3 r03 3
3
(b) The Bohr radius, r0, is inversely proportional to the mass of the particle. So now the Bohr
radius is smaller by a factor of 207.
1.11015 m 3
0.529 1010 m 207
P1.1fm 4 r2 e 2r dr 1.1fm 4 r2 dr 4 r3 4 1.1 107
0 r03 r0 0 r03 3 r03 3
3
19. We follow the directions as given in the problem. We use the first integral listed in Appendix B-5.
2 4 r2dr 1 2 r 4 2dr r3 2 r ; let 2r
0 r03 r0 r03 r0 r0
r 100 e 4
0 0
r r r e dr x
x 3e x dx
r1r0 1 r0 3! 3 r0
4 4 2
0
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Chapter 39 Quantum Mechanics of Atoms
20. To show that 200 is normalized, we integrate 200 2 over all space. Use substitution of variables and
an integral from Appendix B-5.
200 2 dV 1 2 r e r 4 r2dr ; let x r r r0x , dr r0dx
0 32 r03 r0 r0 r0
all
space
Note that if r 0, x 0 and if r , x .
200 2 dV 1 r 2 e r r2dr 4 2 e xr02 x2r0dx
0 32 r0 r0 32 r03 0
all 2 4 2x
r03
space
1 2 ex x2dx 1 1 1 1
80 80 2 0 2 0 8 0
2x 4x2 4x3 x4 e x dx x 2e x dx x 3e x dx x 4e x dx
1 2! 1 3! 1 4! 1 3 3 1
2 2 8
And so we see that 200 is normalized.
21. We follow the directions as given in the problem. The three wave functions are given in Eq. 39-9.
We explicitly show the expressions involving the complex conjugate.
Pr 4 r2 1 2 1 2 1 2
3
210 3 211 3 211
4 r2 1 z2 e r 1 x iy x iy e r 1 x iy x iy e r
3 32 r05 r0 3 r0 3 r0
64 r05 64 r05
4 r2 1 z2 e r 1 x2 y2 e r 1 x2 y2 e r r2 z2 x2 y2 e r
3 32 r05 r0 3 64 r05 r0 64 r05 r0 24r05 r0
3
r4 e r
24r05 r0
22. From Problem 21, we have that Pr r4 e r for the 2p state. We find the most probable distance
24r05 r0
by setting dPr 0 and solving for r. This is very similar to Example 39-3.
dr
Pr r4 e r dPr 4r3 e r 1 r4 e r r 3e r
24r05 r0 ; dr 24r05 r0 r0 24r05 r0 4r0 r r0 0 r 4r0
r06 24
23. The probability is found by integrating the radial probability density over the range of radii given.
P 1.01r0 4 r2 e 2r dr ; let x 2 r
0.99 r0 r03 r0 r0
2.02 2.02
x2e xdx e x x2 e1.98 e2.02
P 1.98
1 1 2x 2 4.9402 5.0602 0.0108 1.1%
2 2
1.98
Because the range of radii is small and the radial probability density is relatively constant over that
range (see Figure 39-7), we can approximate the probability as follows.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
1.01r0 r2 2r r02 e 2 r0 0.08e2 0.0108 1.1%
0.99 r0 r03 r0 r03 r0
P 4 4
e dr P r r0 r 0.02r0
24. The probability is found by integrating the radial probability density over the range of radii given.
The radial probability density is given after Eq. 39-8.
P 5.00 r0 r2 2 r 2 e r dr ; let x r
4.00r0 8r03 r0 r0 r0
5.00 5.00
2 exdx
P
1 x2 2x 1 x4 4x3 4x2 exdx
8 8
4.00 4.00
There are some difficult integrals to evaluate. We use integration by parts.
x2exdx : u x2; dv ex ; du 2xdx; v ex
x2exdx x2ex 2xexdx x2ex 2 ex x 1 ex x2 2x 2
x3exdx : u x3; dv ex ; du 3x2dx; v ex
x3exdx x3ex 3 x2exdx ex x3 3x2 6x 6
x4exdx : u x4; dv ex ; du 4x3dx; v ex
x4exdx x4ex 4 x3exdx ex x4 4x3 12x2 24x 24
We substitute the integrals above into the expression for the probability. We are not showing the
algebra.
P 5.00
1 e x x4 4x2 8x 8 1 773e5 360e4 0.173 17.3%
8 4.00 8
25. The wave function is given in Eq. 39-5a. Note that r x2 y2 z2 1/2 . We will need the
derivative relationship derived in the first line below.
r x 1 e r r 1 1 e r x
r r0 x r x r0 r0
x
1 x2 y2 z2 2x ; 100 r03 ; r03 r
2
2 1 1 e r x 1 1 e r 1 1 1 e r x r
x2 x r0 r03 r0 r r0 r03 r0 r r r0 r0
r03 r x
1 1 r 1 x 2 1 1 1 1 x 2 1 1
rr0 r03 rr0 r2 rr0 rr0 r2
e r0
Similarly, we would have r y ; 2 1 1 y 2 1 1 ; r z ; and
y r y 2 rr0 rr0 z r
r2
2 1 1 z 2 1 1 . Substitute into the time-independent Schrödinger equation.
z2 rr0 rr0 r2
E 2 2 2 2 1 e2
2m x2 x2 x2 40 r
2 1 1 x 2 1 1 1 y2 1 1 1 z2 1 1 e2
2m rr0 rr0 r2 rr0 r2 rr0 r2 4
0r
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516
Chapter 39 Quantum Mechanics of Atoms
2 1 3 1 1 e2
2m rr0 x2 y2 z2 rr0 r2
4 0r
2 1 3 r2 1 1 e2 2 1 2 r2 e2
rr0 rr0 r2 rr0 rr0
2m 4 0r 2m 4 0r
2 2 e2
2mr02
mrr0 4 0 r
Since the factor in square brackets must be a constant, the terms with the r dependence must cancel.
2 e2 0 r0 4 0 2 0h2
mrr0 4 0 r me2 me2
Note from Equation 37-11 that this expression for r0 is the same as the Bohr radius. Since those two
terms cancel, we are left with the following.
E 2 E 2 2 me4
2mr02 2mr02
2m 0h2 2 8h2 2
me2 0
26. (a) The probability is found by integrating the radial probability density over the range of radii
given. The radial probability density is given after Eq. 39-8.
P r0 r2 2 r 2 e r dr ; let x r
0 8r03 r0 r0 r0
11
P 2
1 x2 2x e x dx 1 x4 4x3 4x2 exdx
8 8
00
The following integrals are derived in the solution to Problem 24.
x2exdx ex x2 2x 2 ; x3exdx ex x3 3x2 6x 6
x4exdx ex x4 4x3 12x2 24x 24
1
P
1 x4 4x3 4x2 exdx
8
0
1
1 ex x4 4x3 12x2 24x 24 4 x3 3x2 6x 6 4 x2 2x 2
8 0
0
1 e x x4 4x2 8x 8 1 8 21e1 0.0343 3.4%
8 1 8
(b) From Problem 21, we have that the radial probability density for this state is Pr r4 r . We
24r05 r0
e
proceed as in part (a).
P r0 r4 e r dr ; let x r
0 24r05 r0 r0
1 x4 10
P 1 x4e xdx 1 e x 4x3 12x2 24x 24 1 24 65e1 3.66 103
24 24 24
0
0.37%
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
27. (a) The radial probability distribution is given by Eq. 39-6. Use the wave function given.
Pr 4 r2 300 2 4 r2 1 1 2r 2r2 2 e 2r 4r2 1 2r 2r2 2 e 2r
27 r03 3r0 27r02 3r0 27r03 3r0 27r02 3r0
(b) See the graph.
(c) The most probable distance is the radius 2.0
for which the radial probability 1.5
distribution has a global maximum. We
find that location by setting dPr 0 and Pr (nm-1) 1.0
dr
solving for r. We see from the graph that 0.5
the global maximum is approximately at 0.0
0 5 10 15 20
r 13r0. 25
r/r0
dPr 8r 1 2r 2r2 2 e 2r 8r2 1 2r 2r2 2 4r 2r
dr 27r03 3r0 27r02 3r0 27r03 3r0 27r02 27r02 3r0
e
3r0
4r2 1 2r 2r2 2 2 e 2r
27r03 3r0 27r02 3r0 3r0
8r 1 2r 2r2 1 5r 12r2 2r3 e 2r 0
27r03 3r0 27r02 3r0 27r02 81r03 3r0
The above system has 6 non-infinite solutions. One solution is r = 0, which leads to Pr 0,
which is not a maximum for the radial distribution. The second-order polynomial,
1 2r 2r2 , is a factor of the radial probability distribution, and so its zeros also give
3r0 27r02
locations where Pr 0. So the maxima must be found from the roots of the third-order
polynomial. A spreadsheet was used to find the roots of 1 5 x 12 x2 2 x3 0. Those roots
3 27 81
are x 0.74, 4.19, and 13.07. So the most probable distance is r 13.1r0 .
The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH39.XLS,” on tab “Problem 39.27.”
28. For oxygen, Z = 8. We start with the n 1 shell, and list the quantum numbers in the order
n, l,ml ,ms .
1, 0, 0, 1 , 1, 0, 0, 1 , 2, 0, 0, 1 , 2, 0, 0, 1 ,
2 2 2 2
2,1, 1, 1 , 2,1, 1, 1 , 2,1, 0, 1 , 2,1, 0, 1
2 2 2 2
Note that, without additional information, there are two other possibilities that could substitute for
any of the last four electrons.
29. (a) For carbon, Z = 6. We start with the n 1 shell, and list the quantum numbers in the order
n, l,ml ,ms .
1, 0, 0, 1 , 1, 0, 0, 1 , 2, 0, 0, 1 , 2, 0, 0, 1 , 2,1, 1, 1 , 2,1, 1, 1
2 2 2 2 2 2
Note that, without additional information, there are other possibilities for the last two electrons.
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518
Chapter 39 Quantum Mechanics of Atoms
(b) For aluminum, Z = 13. We start with the n 1 shell, and list the quantum numbers in the order
n, l,ml ,ms .
1, 0, 0, 1 , 1, 0, 0, 1 , 2, 0, 0, 1 , 2, 0, 0, 1 , 2,1, 1, 1 , 2,1, 1, 1 ,
2 2 2 2 2 2
2,1, 0, 1 , 2,1, 0, 1 , 2,1,1, 1 , 2,1,1, 1 , 3, 0, 0, 1 , 3, 0, 0, 1 , 3,1, 1, 1
2 2 2 2 2 2 2
Note that, without additional information, there are other possibilities for the last electron.
30. The number of electrons in the subshell is determined by the value of l. For each l the value of ml
can range from l to l, which is 2l 1 values. For each ml value there are two values of ms.
Thus the total number of states for a given l is N 22l 1.
N 22l 1 2 24 1 18 electrons
31. Since the electron is in its lowest energy state, we must have the lowest possible value of n. Since
ml 2, the smallest possible value of l is l 2 , and the smallest possible value of n is n 3 .
32. Limiting the number of electron shells to six would mean that the periodic table stops with radon
(Rn), since the next element, francium (Fr), begins filling the seventh shell. Including all elements
up through radon means 86 elements.
33. (a) Nickel has Z = 28.
1s2 2s2 2 p6 3s2 3 p6 3d 8 4s2
(b) Silver has Z = 47.
1s2 2s2 2 p6 3s2 3 p6 3d 10 4s2 4 p6 4d 10 5s1
(c) Uranium has Z = 92.
1s2 2s2 2 p6 3s2 3 p6 3d 10 4s2 4 p6 4d 10 4 f 14 5s2 5 p6 5d 10 6s2 6 p6 5 f 36d 17s2
34. The third electron in lithium is in the 2s subshell, which is outside the more tightly bound filled 1s
shell. This makes it appear as if there is a “nucleus” with a net charge of 1e. Thus we use the
energy of the hydrogen atom.
E2 13.6 eV 13.6 eV 3.4eV
n2 22
We predict the binding energy to be 3.4 eV . Our assumption of complete shielding of the nucleus
by the 2s electrons is probably not correct. The partial shielding means the net charge of the
“nucleus” is higher than +1e, and so it holds the outer electron more tightly, requiring more energy to
remove it.
35. We use Eq. 37-13, which says that the radius of a Bohr orbit is inversely proportional to the atomic
number. We also use Eq. 37-14b, which says that the energy of Bohr orbit is proportional to the
square of the atomic number. The energy to remove the electron is the opposite of the total energy.
rnn2 1 5.75 1013 m
Z 0.529 1010 m 92 0.529 1010 m
En 13.6 eV Z 2 13.6 eV 922 1.15 105eV
n 2 12
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
36. The energy levels of the infinite square well are given in Eq. 38-13. Each energy level can have a
maximum of two electrons, since the only quantum numbers are n and ms. Thus the lowest energy
level will have two electrons in the n = 1 state, two electrons in the n = 2 state, and 1 electron in the
n = 3 state.
E 2E1 2E2 E3 212 2 22 1 32 h2 2 19 h2 2
8ml 8ml
37. In a filled subshell, there are an even number of electrons. All of the possible quantum number
combinations for electrons in that subshell represent an electron that is present. Thus for every m
value, both values of ms are filled, representing a spin “up” state and a spin “down” state. The total
angular momentum of that pair is zero, and since all of the electrons are paired, the total angular
momentum is zero.
38. The shortest wavelength X-ray has the most energy, which is the maximum kinetic energy of the
electron in the tube:
E hc
6.63 1034 Js 3.00 108 m/s 4.6 104 eV 46 keV
1.60 1019 J/eV 0.027 109 m
Thus the operating voltage of the tube is 46 kV .
39. The shortest wavelength X-ray has the most energy, which is the maximum kinetic energy of the
electron in the tube.
hc 6.63 1034 Js 3.00 108 m/s 3.825 1011m 0.0383nm
E 1.60 1019 J/eV 32.5 103 eV
The longest wavelength of the continuous spectrum would be at the limit of the X-ray region of the
electromagnetic spectrum, generally on the order of 1 nm.
40. The energy of the photon with the shortest wavelength must equal the maximum kinetic energy of an
electron. We assume V is in volts.
E hf0 hc eV
0
0
hc 6.63 1034 Js 3.00 108 m s 109 nm m 1243 nm 1240 nm
eV V V
1.60 1019 C V V
41. With the shielding provided by the remaining n 1 electron, we use the energies of the hydrogen
atom with Z replaced by Z 1. The energy of the photon is found, and then the wavelength.
hf E 13.6eV26 12 1 1 6.40 103 eV.
22 12
hc 6.63 1034 Js 3.00 108 m/s 1.94 1010 m 0.194 nm
E 1.60 1019 J/eV 6.40 103 eV
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520
Chapter 39 Quantum Mechanics of Atoms
42. We follow the procedure of Example 39-6, of using the Bohr formula, Eq. 37-15, with Z replaced by
Z – 1.
1 e4m 12 1 1 12 1 1 109 m1
8 02 h 3c n2 n2 12 22
Z 1.097 107 m1 27 5.562
1 1.798 1010 m
5.562 109 m1
43. The wavelength of the K line is calculated for molybdenum in Example 39-6. We use that same
procedure. Note that the wavelength is inversely proportional to Z 12 .
ZFe 1 2
Zunknown 1 2
unknown Zunknown 194 pm
Fe 26 1 229 pm 1 24
The unknown material has Z = 24, and so is chromium.
44. We assume that there is “shielding” provided by the 1s electron that is already at that level. Thus the
effective charge “seen” by the transitioning electron is 42 – 1 = 41. We use Eqs. 37-9 and 37-14b.
hf E 13.6 eV Z 12 1 1
n2 n2
hc hc 6.63 1034 Js 3.00 108 m/s
E
13.6 eV Z 12 1 1 13.6 eV 412 1 1 1.60 1019 J eV
n2 n2 12 32
6.12 1011m 0.0612 nm
We do not expect perfect agreement because there is some partial shielding provided by the n 2
shell, which was ignored when we replaced Z by Z 1. That would make the effective atomic
number a little smaller, which would lead to a larger wavelength. The amount of shielding could be
estimated by using the actual wavelength and solving for the effective atomic number.
45. Momentum and energy will be conserved in any inertial reference frame. Consider the frame of
reference that is moving with the same velocity as the electron’s initial velocity. In that frame of
reference, the initial momentum of the electron is 0, and its initial total energy is mc2. Let the
emitted photon have frequency f, and let the direction of motion of that photon be considered the
positive direction. The momentum of the photon is then p h hf , and so the momentum of the
c
electron must be pe hf . The final energy of the photon is E hf p c, and the final energy
c
of the electron is, from Eq. 36-13, Eelectron pe2c2 m2c4 . We write the conservation conditions,
final
and then solve for the frequency of the emitted photon.
Momentum: 0 hf pe pe hf
c c
Energy: mc2 hf pe2c2 m2c4
mc2 hf hf 2 c 2 m2c2 h2 f 2 m2c4
c
m2c4 2mc2hf h2 f 2 h2 f 2 m2c4 2mc2hf 0 f 0
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
Since the photon must have f = 0, no photon can be emitted and still satisfy the conservation laws.
Another way to consider this situation is that if an electron at rest emits a photon, the energy of the
electron must decrease for energy to be conserved. But the energy of a stationary electron cannot
decrease, unless its mass were to change. Then it would no longer be an electron.
So we conclude that a third object (with mass) must be present in order for both energy and
momentum to be conserved.
46. The Bohr magneton is given by Eq. 39-12.
μB
e 1.602 1019 C 1.054 1034 Js 9.27 1024 J T
2m 2 9.109 1031kg
47. We use Eq. 39-14 for the magnetic moment, since the question concerns spin angular momentum.
The energy difference is the difference in the potential energies of the two spin states.
U
μzB spin down gμ B Bms 2.0023 9.27 1024 J T 2.5T 1 1 2.9 104 eV
spin up 1.60 1019 J eV 2 2
48. (a) The energy difference is the difference in the potential energies of the two spin states. Use Eq.
39-14 for the magnetic moment.
U
μzB spin down gμ B Bms 2.0023 9.27 1024 J T 1.0 T 1 1
spin up 1.60 1019 J eV 2 2
1.160 104 eV 1.2 104 eV
(b) Calculate the wavelength associated with this energy change.
U E h c
hc 6.63 1034 Js 3.00 108 m/s 1.072 102 m 1.1cm
U 1.160 104 eV 1.60 1019 J eV
(c) The answer would be no different for hydrogen. The splitting for both atoms is due to an s-state
electron: 1s for hydrogen, 5s for silver. See the discussion on page 1058 concerning the Stern-
Gerlach experiment.
49. (a) Refer to Figure 39-14 and the equation following it. A constant magnetic field gradient will
produce a constant force on the silver atoms. Atoms with the valence electron in one of the spin
states will experience an upward force, and atoms with the valence electron in the opposite spin
state will experience a downward force. That constant force will produce a constant
acceleration, leading to the deflection from the original direction of the atoms as they leave the
oven. We assume the initial direction of the atoms is the x direction, and the magnetic field
gradient is in the z direction. If undeflected, that atoms would hit the screen at z = 0.
z 1 at 2 1 F x 2 1 μz dBz x 2 1 gμBms dBz x 2
2 2 mAg v 2 dz v 2 dz v
mAg
mAg
One beam is deflected up, and the other down. There separation is the difference in the two
deflections due to the two spin states.
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522
Chapter 39 Quantum Mechanics of Atoms
z dBz
gμ B 1 1 dz x 2
2 2
zms zms 1
1 1 2 mAg v
2 2
m 0.050 m 2
1 2.0023 9.27 1024 J T 1800T 3.833 104 m 0.38 mm
2 107.87 u 1.66 1027 kg u 780 m s
(b) The separation is seen in the above equation to be proportional to the g-factor. So to find the
new deflection, divide the answer to part (a) by the original g-factor.
z g 1 3.833104 m 0.19 mm
2.0023
50. For the 5g state, l 4 and s 1 . Thus the possible values of j are j l s 4 1 7 , 9 .
2 2 2 2
Let j 7 . Then we have the following.
2
mj 7 , 5 , 3 , 1 , 1 , 3 , 5 , 7 J 7 7 1 3 7
2 2 2 2 2 2 2 2 2 2 2
Jz mj 7 , 5 , 3 , 1 , 1 , 3 , 5 , 7
2 2 2 2 2 2 2 2
Let j 9 . Then we have the following.
2
mj 9 , 7 , 5 , 3 , 1 , 1 , 3 , 5 , ,7 9 J 9 9 1 3 11
2 2 2 2 2 2 2 2 2 2 2
22
Jz mj 9 , 7 , 5 , 3 , 1 , 1 , 3 , 5 , 7 , 9
2 2 2 2 2 2 2 2 2 2
51. (a) For the 4p state, l 1. Since s 1 , the possible values for j are j ls 3 and
2 2
j ls 1 .
2
(b) For the 4f state, l 3. Since s 1 , the possible values for j are j ls 7 and
2 2
j ls 5 .
2
(c) For the 3d state, l 2. Since s 1 , the possible values for j are j ls 5 and
2 2
j ls 3 .
2
(d) The values of J are found from Eq. 39-15.
4p: J j j 1 15 and 3
2 2
4f : J j j 1 63 and 35
2 2
3d : J j j 1 35 and 15
2 2
52. (a) Gallium has Z = 31. We list the quantum numbers in the order n, l,ml ,ms .
1, 0, 0, 1 , 1, 0, 0, 1 , 2, 0, 0, 1 , 2, 0, 0, 1 , 2,1, 1, 1 , 2,1, 1, 1 ,
2 2 2 2 2 2
2,1, 0, 1 , 2,1, 0, 1 , 2,1,1, 1 , 2,1,1, 1 , 3, 0, 0, 1 , 3, 0, 0, 1 ,
2 2 2 2 2 2
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
3,1, 1, 1 , 3,1, 1, 1 , 3,1, 0, 1 , 3,1, 0, 1 , 3,1,1, 1 , 3,1,1, 1 ,
2 2 2 2 2 2
3, 2, 2, 1 , 3, 2, 2, 1 , 3, 2, 1, 1 , 3, 2, 1, 1 , 3, 2, 0, 1 , 3, 2, 0, 1 ,
2 2 2 2 2 2
3, 2,1, 1 , 3, 2,1, 1 , 3, 2, 2, 1 , 3, 2, 2, 1 , 4, 0, 0, 1 , 4, 0, 0, 1 ,
2 2 2 2 2 2
4,1, 0, 1
2
The last electron listed could have other quantum numbers for ml and ms.
(b) The 1s, 2s, 2p, 3s, 3p, 3d, and 4s subshells are filled.
(c) For a 4p state, l 1. Since s 1 , the possible values for j are j ls 3 and
2 2
j ls 1 .
2
(d) The 4p electron is the only electron not in a filled subshell. The angular momentum of a filled
subshell is zero, so the total angular momentum of the atom is the angular momentum of the 4p
electron.
(e) When the beam passes through the magnetic field gradient, the deflecting force will be
proportional to mj . If j 1 , the values of mj are 1 , and there will be two lines. If j 3 , the
2 2 2
values of mj are 1 , 3 , and there will be four lines. The number of lines indicates the value
2 2
of j.
53. (a) The additional term for the spin-orbit interaction is given in the text as Uspin μB n μzBn .
orbit
The separation of the energy levels due to the two different electron spins is twice this.
Uspin μ z Bn spin down gμ B Bn ms
orbit spin up
Uspin
orbit
gμ Bms
Bn
5 105eV 1.60 1019 J eV 0.431T 0.4 T
2.0023 9.27 1024 J T 1 1
2
2
(b) If we consider the nucleus to be a loop of current with radius r, then the magnetic field due to
the nucleus at the center of the loop (the location of the electron) is given in Example 28-12 as
Bn 0 I . Model the current as the charge of the nucleus moving in a circle, with a period as
2r
given by circular motion.
I q e e v ev e me r v
t T 2 r 2 r 2 r mer
Note that classically, merv Le, the angular momentum of the electron, and so
merv l l 1 with l 2. Thus we have the following:
Bn 0 I 0 e l l 1 e 0 l l 1 μB0 l l 1
2r 2r 2me
2 mer2 2 r3 2 r3
From Figure 39-9 (b), we see that the most probable radius for the n 2, l 1 state is
approximately r 4r0. We can now calculate the approximate magnetic field.
Bn
μ B0 l l 1 9.27 1024 J T 4 107 Tm A 6
2 4r0 3 2 4 0.529 1010 m 3 0.479T 0.5T
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524
Chapter 39 Quantum Mechanics of Atoms
The two values are about 1 0.479T 0.431T 100 11% different, and so are consistent.
2
0.479T 0.431T
54. The energy of a pulse is the power of the pulse times the duration in time.
E P t 0.63W 23103s 0.01449J 0.014 J
The number of photons in a pulse is the energy of a pulse, divided by the energy of a photon as given
in Eq. 37-3.
N
E E 0.01449 J 640 10-9 m 4.7 1016 photons
hf hc
6.63 1034 Js 3.00 108 m/s
55. The angular half-width of the beam can be found in r
Section 35-4, and is given by 1/ 2 1.22 , where d D
d
is the diameter of the diffracting circle. The angular
width of the beam is twice this. The linear diameter
of the beam is then the angular width times the distance from the source of the light to the
observation point, D r. See the diagram.
(a)
D r r 2.44 380 103 m 2.44 694 109 m 180 m
d 3.6 103 m
(b)
D r r 2.44 384 106 m 2.44 694 109 m 1.8 105 m
d 3.6 103 m
56. Intensity equals power per unit area. The area of the light from the laser is assumed to be in a
circular area, while the area intercepted by the light from a light bulb is the surface area of a sphere.
P P 0.50 103 W m2 m2
S r2 1.5 103 m
(a) I 2 70.74 W 71W
(b) I P P 15 W 0.2984 W m2 0.30 W m2
S 4 r2
4 2.0 m2
The laser beam is more intense by a factor of 70.74 W m2 237 240 .
0.2984 W m2
57. Transition from the E3 state to the E2 state releases photons with energy 1.96 eV, as shown in
Figure 39-21. The wavelength is determined from the energy.
hc 6.63 1034 Js 3.00 108 m/s 6.34 107 m 634 nm
E
1.60 1019 J/eV 1.96eV
58. We use Eq. 39-16b.
N 2
E2 E0 2.2 eV 1.601019 J eV
kT 1.381023 J K 300 K
85.0 37
N e e e 1.2 100
E2 E0 1.8eV 1.601019 J eV
kT
N 1
1.381023 J K 300 K 69.6 31
N e e e 6.1100
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
59. The relative numbers of atoms in two energy states at a given temperature is given by Eq. 39-16b.
From Figure 39-20, the energy difference between the two states is 2.2 eV.
N2
E2 E0 T E2 E0 2.2 eV 1.60 1019 J eV 3.7 104 K
kT k ln N2
1.38 1023 J K ln 0.5
N e0 N0
60. Consider Eq. 39-16b, with En En. To have a population inversion means that Nn Nn.
Nn En En Nn En En En En
kT N n kT kT
N e 1n ln 0 0
Since En En, to satisfy the last condition we must have T < 0, a negative temperature.
This negative “temperature” is not a contradiction. The Boltzmann distribution assumes that a
system is in thermal equilibrium, and the inverted system is not in thermal equilibrium. The
inversion cannot be maintained without adding energy to the system. If left to itself, the excited
states will decay and the inversion will not be maintained.
61. (a) Boron has Z 4, so the outermost electron has n 2. We use the Bohr result with an
effective Z. We might naively expect to get Zeff 1, indicating that the other three electrons
shield the outer electron from the nucleus, or Zeff 2, indicating that only the inner two
electrons accomplish the shielding.
E2 13.6 eV Z eff 2 8.26 eV 13.6 eV Z eff 2 Zeff 1.56
n2 22
This indicates that the second electron in the n 2 shell does partially shield the electron that is
to be removed.
(b) We find the average radius from the expression below.
r 22
n2r1 0.529 1010 m 1.36 1010 m
Z eff
1.56
62. An h subshell has l 5. For a given l value, ml ranges from l to l, taking on 2l 1 different
values. For each ml there are 2 values of ms. Thus the number of states for a given l value is
22l 1. Thus there are 22l 1 211 22 possible electron states.
63. (a) Z = 25 is manganese.
1s2 2s2 2 p6 3s2 3 p6 3d 5 4s2
(b) Z = 34 is selenium.
1s2 2s2 2 p6 3s2 3 p6 3d 10 4s2 4 p4
(c) Z = 39 is yttrium.
1s2 2s2 2 p6 3s2 3 p6 3d 10 4s2 4 p6 4d 15s2
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526
Chapter 39 Quantum Mechanics of Atoms
64. The value of l can range from 0 to n 1. Thus for n 6, we have 0 l 5. The magnitude of L
is given by Eq. 39-3, L l l 1.
Lmin 0 ; Lmax 30
65. (a) We treat the Earth as a particle in rotation about the Sun. The angular momentum of a particle
is given in Example 11-7 as L mvr, where r is the orbit radius. We equate this to the
quantum mechanical expression in Eq. 39-3. We anticipate that the quantum number will be
1
very large, and so approximate l l 12 as l.
M 2 r2 1
T
l 1 2
L
M EarthvrSun l l
Earth
l 2
M Earth 2 r2 5.98 1024 kg 2 1.496 1011m
T 1.055 1034 Js 3.156 107s 2.5255 1074 2.5 1074
(b) There are 2l 1 values of ml for a value of l, so the number of orientations is as follows.
N 2l 1 2 2.5255 1074 1 5.0511074 5.1 1074 .
66. Eq. 37-15 gives the Bohr-theory result for the wavelength of a spectral line. For the Mosley plot, the
wavelengths are for the K line, which has n 2 and n 1. We assume that the shielding of the
other n = 1 electron present reduces the effective atomic number to Z – 1. We use the value of the
Rydberg constant from Section 37-11.
1 Z 2e4m 1 1 1 e4m Z 2 1 1 1 3e4m Z 2
n2 n2 1 4 32 02 h 3c
8 02h3c 8 2 h3c
0
1 3e4m 1 3e4m 1/ 2
32 02h 3c 32 02 h 3c
Z 12 aZ b , a , b1
3e4m 1/ 2 1/ 2 1/ 2
32 02 h 3c
a
3 R 3 1.0974 107 m1 2868.9 m1/ 2
4 4
67. This is very similar to Example 39-3. We find the radial
probability distribution for the n = 2, l = 0 wave function, and
find the position at which that distribution has a maximum. We
see from Figure 39-8 that there will be two local maxima in the
probability distribution function, and the global maximum is at
approximately 5r0. The wave function is given in Eq. 39-8.
200 1 2 r e r
32 r03 r0 2 r0
Pr 4 r2 200 2 r2 2 r 2 e r
8r03 r0 r0
dPr 2r 2 r 2 r r2 2 2 r 1 e r r2 2 r 2 1 r
dr 8r03 r0 r0 8r03 r0 r0 r0 8r03 r0 r0 r0
e e
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
r
re r0
2r0 r
8r06 r2 6r0r 4r02 0 r 0, r 2r0, r2 6r0r 4r02 0
r 6r0 36r02 16r02
2
r2 6r0r 4r02 0 3 5 r 0.764r0, 5.24r0
So there are four extrema: r 0, 0.76r0, 2r0, 5.2r0. From Figure 39-8 we see that the most probable
distance is r 5.2r0 .
68. The “location” of the beam is uncertain in the transverse direction by an amount equal to the aperture
opening, D. This gives a value for the uncertainty in the transverse momentum. The momentum
must be at least as big as its uncertainty, and so we obtain a value for the transverse momentum.
pyy p y py py
y D D
px
The momentum in the forward direction is related to the D
wavelength of the light by px h . See the diagram to
relate the momentum to the angle.
py D ; “spread” 2
px h 2 D D D
69. The magnitude of the angular momentum is given byEq. 39-3, and Lz is
given by Eq. 39-4. The cosine of the angle between L and the z axis is
found from L and Lz.
L l l 1 ; Lz ml ; cos1 Lz cos1 ml
L
l l 1
(a) For l 1, ml 1, 0, 1.
1,1 cos1 1 45 ; 1,0 cos1 0 90 ; y
2 2
1,1 cos1 1 135 x
2
(b) For l 2, ml 2, 1, 0, 1, 2.
2,2 cos1 2 35.3 ; 2,1 cos1 1 65.9 ; 2,0 cos1 0 90
6 6 2
2,1 cos1 1 114.1 ; 2,2 cos1 2 144.7
6 6
(c) For l 3, ml 3, 2, 1, 0, 1, 2, 3.
3,3 cos1 3 30 ; 3,2 cos1 2 54.7 ; 3,1 cos1 1 73.2
12 12 12
3,0 cos1 0 90 ; 3,1 cos1 1 106.8 ; 3,2 cos1 2 125.3 ;
12 12 12
3,3 cos1 3 150
12
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528
Chapter 39 Quantum Mechanics of Atoms
(d) We see from the previous parts that the smallest angle occurs for ml l.
100,100 cos1 100 5.71
100 101
106 0.0573
106 106 1
106,106 cos1
This is consistent with the correspondence principle, which would say that the angle between L
and the z axis could be any value classically, which is represented by letting l (which also
means n ).
70. (a) Since Lz 0, , and so is unknown. We can say nothing about the
value of .
(b) Since is completely unknown, we have no knowledge of the component of the angular
momentum perpendicular to the z-axis. Thus Lx and Ly are unknown.
(c) The square of the total angular momentum is given by L L2x L2y L2z. Use this with the
quantization conditions for L and Lz given in Eqs. 39-3 and 39-4.
L L2x L2y L2z l l 1 2 L2x L2y ml22 L2x L2y l l 1 ml2 2
L2x L2y l l 1 ml2 1 / 2
71. (a) The mean value can be found as described in Problem 19. We use the first definite integral
given in Appendix B-5, with n = 1 and a = 1.
1 1 2 1 1 e2 r r2dr 4 r e2 r r
r 0 r 0 r r03 r0 r02 0 r0 r0 r0
100 4 r2dr 4 dr ; let x 2
1 xe x dx 1 1 1
r0 0 r0 r0
1 e2 U e2 1 2 4 r2dr 1 e2
4 0 r 40 0 r 40 r0
U 100
(b) For the ground state of hydrogen, Eq. 37-14a gives the energy, and Eq. 37-11 gives the Bohr
radius. Substitute those expressions into E U K.
E e4m ; U 1 e2
40 r0
8 2 h 2
0
e4m 1 e2 e2 e4m e2 e 4m h 2 0
802h2 r0 4 0 r0 4 0 r0 me2
K E U
4 8 2 h 2 8 02 h 2 r0
0
0
e2 e2 e2 1 U
4 0 r0 8 0 r0 8 0 r0 2
72. In the Bohr model, LBohr n h 2 . In quantum mechanics, LQM l l 1 . For n 2, l 0
2
or l 1, so that LQM 00 1
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
73. (a) The 4 p 3 p transition is forbidden, because l 0 1.
(b) The 3p 1s transition is allowed, because l 1.
(c) The 4d 3d transition is forbidden, because l 0 1.
(d) The 4d 3s transition is forbidden, because l 2 1.
(e) The 4s 2 p transition is allowed, because l 1.
74. The binding energy is given by the opposite of Eq. 37-14b.
En 13.6 eV Z 2 13.6 eV 6.72 103eV
n 2 452
The radius is given by Eq. 37-13.
rnn2 1.07 107 m
Z 0.529 1010 m 452 0.529 1010 m
The effective cross-sectional area is as follows.
2 3.60 1014 m2
r2 1.07 107 m
75. The wavelengths of emitted lines from one-electron atoms are given by Eq. 37-15. We can simplify
the equation by using the Rydberg constant, so 1 R 1 1 . The Lyman series with
n2 n2
hydrogen has Z = 1 and n Every fourth line from the unknown element match the wavelengths
of the first three Lyman lines. This gives three equations in three unknowns.
1 R 1 1 Z R 1 1 3 Z 2 1 1
Lyman 12 22 n2 m2 4 n2 m2
21
1 R 1 1 Z R 1 1 8 Z 1 1
Lyman 12 32 n2 9 n2
m 42 m 42
31
1 R 1 1 Z R 1 1 15 Z 1 1
Lyman 12 42 n2 16 n2
m 82 m 82
41
Subtract the first equation from each of the other two equations.
8 3 Z 1 1 Z 2 1 1 5 Z 1 1
9 4 n2 n2 m2 36 m2
m 42 m 42
15 3 Z 1 1 Z 2 1 1 3 Z 1 1
16 4 n2 n2 m2 16 m2
m 82 m 82
Divide the resulting equations to eliminate Z, solve for m, and then substitute to find Z.
5 Z 1 1 m 42 m2
36 m2
3 m 42 20 m2 m 42 m 2m 82
16 27 m 82 m2 2m 43
Z 1 1
m2
m 82 m2 m 82
40m 43 27m 2m 82 13m3 514m2 608m 896 0
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530
Chapter 39 Quantum Mechanics of Atoms
This is a cubic equation, which can be solved by numerical techniques. We first drew a graph, and
saw that two of the 0’s of the function were negative, and one was near m = 8. The only acceptable
results are for m > 0, and substitution verifies that m = 8 solves the equation. We use that result to
find Z.
3 Z 1 1 Z 3 1 1 1 3 1 1 1 4
16 m2 16 m2 16 82
m 82 m 82 8 82
Thus the element is beryllium. The spreadsheet used for this problem can be found on the Media
Manager, with filename “PSE4_ISM_CH39.XLS,” on tab “Problem 39.75.”
76. (a) The additional energy due to the presence of a magnetic field is derived in Section 39-7, as
U μBmlBz. We use this to calculate the energy spacing between adjacent ml values.
U μBml Bz
U μBBzml
9.27 1024 J T 1.6T 9.27 105eV 9.3105eV
1.60 1019 J eV
(b) As seen in Figure 39-4, the n 3, l 2 level will split n = 3, ¬ = 2 +2B
+B
into 5 levels, and the n 2, l 1 level will split into 3
levels. With no restrictions, there would be 15 0
different transitions possible. All transitions would
have the same n 1. Thus there are only three –B
unique wavelengths possible: the one corresponding to –2B
a transition with ml 1 (a slightly larger energy
change than in the B = 0 case), the one corresponding n = 2, ¬ = 1 +B
to a transition with ml 0 (the same energy change 0
as in the B = 0 case), and the one corresponding to a
m² ml = –1 0 +1 –B
transition with ml 1 (a slightly smaller energy
change than in the B = 0 case). See the diagram,
showing 9 possible transitions grouped into 3 actual energy changes. The value along the right
side is the change in energy level due to the magnetic field interaction.
(c) Eq. 37-15 gives the wavelength for hydrogen, considering only a change in principal quantum
number. The energies for those transitions is on the order of eV. The energy change due to the
magnetic field interaction is much smaller than that, so we can use an approximation, knowing
E from part (a). We obtain E from Eq. 37-14b.
n32 R 1 1 1 1.0974 107 m1 1 1 1 6.56096 107 m 656.10 nm
22 32 4
9
h c hc E E ; En32 13.6 eV 1 1 1.889 eV
E E2 E 22 32
E 656.10 nm 9.27 105eV 0.032 nm
E 1.889 eV
ml 1 n32 656.10 nm 0.032nm 656.07 nm
ml 1 n32 0 656.10 nm
ml 1 n32 656.10 nm 0.032nm 656.13nm
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
77. (a) We use Eq. 39-16b, with the “note” as explained in the problem, multiplying the initial
expression times 8 .
2
13.6 eV 13.6 eV
4 1
N 2 1.601019 J eV
E2 E1 1.381023 J K 300 K
kT
8 394.2
e e 4e2 28
N1
Many calculators will not directly evaluate e394.2 , so we do the following.
x e394.2 ; log x 394.2 log e 394.20.4343 171.2 ;
x 10171.2 100.210171 0.63110171
N2
N1
4e394.2 4 0.631 10171 2.52 10171 3 10171
There are 18 states with n = 3, so we multiply by 18 .
2
13.6 eV 13.6 eV
9 1
N 3 1.601019 J eV
E3 E1 1.381023 J K 300 K
kT
18 467.2 202 202
e e 9e 1.1310 1102 218
N1
(b) We repeat the evaluations for the higher temperature.
13.6 eV 13.6 eV
4 1
N 2 1.601019 J eV
E2 E1 1.381023 J K 6000 K
kT
8 19.71 8 8
e e 4e 1.10 10 1102 28
N1
13.6 eV 13.6 eV
9 1
N 3 1.601019 J eV
E3 E1 1.381023 J K 6000 K
kT
18 23.36 10 10
e e 9e 6.44 10 6 102 218
N1
(c) Since the fraction of atoms in each excited state is very small, we assume that N1 is the number
of hydrogen atoms given. 1.0 g of H atoms contains 6.02 1023 atoms.
N2 N1 1.10 108 6.02 1023 1.10 108 6.62 1015 7 1015
N3 N1 6.44 1010 6.02 1023 6.44 1010 3.88 1014 4 1014
(d) We assume the lifetime of an excited state atom is 108s. Each atom would emit one photon as
its electron goes to the ground state. The number of photons emitted per second can be
estimated by the number of atoms, divided by the lifetime.
n3 N3 4 1014 4 1022 photons s ; n2 N2 7 1015 7 1023 photons s
108 s 108 s
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532
CHAPTER 40: Molecules and Solids
Responses to Questions
1. (a) Covalent; (b) ionic; (c) metallic.
2. A neutral calcium atom has 20 electrons. Its outermost electrons are in the 4s2 state. The inner 18
electrons form a spherically symmetric distribution and partially shield the outer two electrons from
the nuclear charge. A neutral chlorine atom has 17 electrons; it lacks just one electron to have its
outer shell filled. A CaCl2 molecule could be formed when the outer two electrons of the calcium
atom are “shared” with two chlorine atoms. These electrons will be attracted by both the Ca and the
Cl nuclei and will spend part of their time between the Ca and Cl nuclei. The nuclei will be attracted
to this negatively charged area, forming a covalent bond. As is the case with other asymmetric
covalent bonds, this bond will have a partial ionic character as well. The two electrons will partly
orbit the Ca nucleus and partly orbit each of the two Cl nuclei. Since each Cl nucleus will now have
an extra electron part of the time, it will have a net negative charge. The Ca nucleus will “lose” two
electrons for part of the time, giving it a net positive charge.
3. No, neither the H2 nor the O2 molecule has a permanent dipole moment. The outer electrons are
shared equally between the two atoms in each molecule, so there are no polar ends that are more
positively or negatively charged. The H2O molecule does have a permanent dipole moment. The
electrons associated with the hydrogen atoms are pulled toward the oxygen atom, leaving each
hydrogen with a small net positive charge and the oxygen with a small net negative charge. Because
of the shape of the H2O molecule (see Figure 40-6), one end of the molecule will be positive and the
other end will be negative, resulting in a permanent dipole moment.
4. The molecule H3 has three electrons. According to the Pauli exclusion principle, no two of these
electrons can be in the same quantum state. Two of them will be 1s2 electrons and will form a
“closed shell” and a spherically symmetric distribution, and the third one will be outside this
distribution and unpaired. This third electron will be partially shielded from the nucleus and will thus
be easily “lost,” resulting in an unstable molecule. The ion H3+ only has two electrons. These 1s2
electrons will form a closed shell and a spherically symmetric distribution, resulting in a stable
configuration.
5. The energy of a molecule can be divided into four categories: translational kinetic energy, rotational
kinetic energy, vibrational kinetic energy, and electrostatic potential energy.
6. Yes. The electron will spend most of its time between the two nuclei. Both positive nuclei will be
attracted to this negative charge, forming a bond.
7. The carbon atom (Z = 6) usually forms four bonds because carbon requires four electrons to form a
closed 2p shell, and each hydrogen-like atom contributes one electron.
8. The last valence electron of a sodium atom is shielded from most of the sodium nuclear charge and
experiences a net nuclear charge of +1e. The outer shell of a chlorine atom is the 3p shell, which
contains five electrons. Due to shielding effects, the 3p electrons of chlorine experience a net nuclear
charge of +5e. In NaCl, the last valence electron of sodium is strongly bound to a chlorine nucleus.
This strong ionic bonding produces a large energy gap between the valence band and the conduction
band in NaCl, characteristic of a good insulator.
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
9. The conduction electrons are not strongly bound to particular nuclei, so a metal can be viewed as a
collection of positive ions and a negative electron “gas.” (The positive ions are just the metal atoms
without their outermost electrons, since these “free” electrons make up the gas.) The electrostatic
attraction between the freely-roaming electrons and the positive ions keeps the electrons from
leaving the metal.
10. As temperature increases, the thermal motion of ions in a metal lattice will increase. More electrons
will collide with the ions, increasing the resistivity of the metal. When the temperature of a
semiconductor increases, more electrons are able to move into the conduction band, making more
charge carriers available and therefore decreasing the resistivity. (Note: the thermal motion increases
in semiconductors as well, but the increase in the number of charge carriers is a larger effect.)
11. When the top branch of the input circuit is at the high voltage (current is flowing in this direction for
half the cycle), then the bottom branch of the output is at the high voltage. The current follows the
path through the bridge in the diagram on the left. When the bottom branch of the input circuit is at
the high voltage (current is flowing in this direction during the other half of the cycle), then the
bottom branch of the output is still at high voltage. The current follows the path through the bridge
in the diagram on the right.
12. In an ideal gas, it is possible for all of the gas particles to have the same energy. (The velocity
distribution of the particles in an ideal gas usually follows the Maxwell velocity distribution.) As the
temperature of the gas increases, the kinetic energy of the gas increases. In a Fermi electron gas,
only two electrons can have the same energy (Pauli exclusion principle). The electrons fill up the
energy states up to the Fermi level. As the temperature of the Fermi gas increases, only the electrons
in the top few levels can move to higher energy levels. The result is that the energy of the Fermi gas
is not strongly temperature dependent.
13. For an ideal pn junction diode connected in reverse bias, the holes and electrons that would normally
be near the junction are pulled apart by the reverse voltage, preventing current flow across the
junction. The resistance is essentially infinite. A real diode does allow a small amount of reverse
current to flow if the voltage is high enough, so the resistance in this case is very high but not
infinite. A pn junction diode connected in forward bias has a low resistance (the holes and electrons
are close together at the junction) and current flows easily.
14. The general shape of Figure 40-28 is the same for most metals. The scale of the graph (especially the
x-axis scale and the Fermi energy) is peculiar to copper and will change from metal to metal.
15. The base current (between the base and the emitter) controls the collector current (between the
collector and the emitter). If there is no base current, then no collector current flows. Thus,
controlling the relatively small base current allows the transistor to act as a switch, turning the larger
collector current on and off.
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534
Chapter 40 Molecules and Solids
16. The main difference between n-type and p-type semiconductors is the sign on the charge carriers. In
an n-type semiconductor the charge carriers are negative electrons. In a p-type semiconductor, the
charge carriers are positive holes.
17. A transistor can be used as an amplifier because small changes in the base current can make much
larger changes in the collector current. (See Figure 40-43.) For a pnp transistor, both the collector
and the base voltages are negative, and holes move from the emitter to the collector. The diagram for
a pnp amplifier looks just like Figure 40-43, with the polarity of eB and eC reversed, IB and IC
flowing in opposite directions, and the emitter arrow pointing toward the base.
18. In Figure 40-43, the base–collector junction is reverse-biased and the base–emitter junction is
forward-biased.
19. The energy comes from the power supplied by the collector/emitter voltage source. The input signal
to the base just regulates how much current, and therefore power, can be drawn from the collector’s
voltage source.
20. The phosphorus atoms will be donor atoms. Phosphorus has five valence electrons. It will form four
covalent bonds with the silicon atoms around it, and will have one “extra” electron which is weakly
bound to the atom and can be easily excited up to the conduction band. This process results in extra
electrons in the conduction band. Silicon doped with phosphorus is therefore an n-type
semiconductor.
21. No. Ohmic devices (those that obey Ohm’s law) have a constant resistance and therefore a linear
relationship between voltage and current. The voltage-current relationship for diodes is not linear.
The resistance of a diode operated in reverse-bias is very large. The same diode operated in a
forward-bias mode has a much smaller resistance. Since a transistor can be thought of as made up of
diodes, it is also non-ohmic.
22. No. Single diodes can be used to rectify signals, but cannot amplify signals. The diode will allow
the signal to pass, if forward-biased, or not allow the signal to pass, if reverse-biased. Combinations
of diodes with additional power sources, as in a transistor, are able to amplify a signal.
23. Reversing the collector voltage would reverse the roles of the collector and emitter of the transistor.
Unless the base-emitter voltage is also reversed, the transistor cannot act as an amplifier.
Solutions to Problems
Note: A factor that appears in the analysis of electron energies is
e2
4 0
9.00 109 Nm2 C2 1.60 1019 C 2 2.30 1028 Jm.
1. We calculate the binding energy as the opposite of the electrostatic potential energy. We use Eq. 23-
10 for the potential energy.
Binding energy U 1 Q1Q2 e2 1 2.30 1028 Jm
40 r 40 0.28 109 m 0.28 109 m
8.214 1019 J 8.2 1019 J
8.214 1019 J 1eV J 5.134eV 5.1eV
1.60 1019
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535
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
2. From Problem 1, the “point electron” binding energy is 5.134 eV. With the repulsion of the electron
clouds included, the actual binding energy is 4.43 eV. Use these values to calculate the contribution
of the electron clouds.
5.134 eV 4.43eV 0.704 eV 0.70eV
3. We calculate the binding energy as the difference between the energy 0.66e
of two isolated hydrogen atoms and the energy of the bonded
combination of particles. We estimate the energy of the bonded p p
combination as the negative potential energy of the two electron- d
proton combinations, plus the positive potential energy of the proton-
proton combination. We approximate the electrons as a single object with a charge of 0.33 of the
normal charge of two electrons, since the electrons only spend that fraction of time between the
nuclei. A simple picture illustrating our bonded model is shown.
Uisolated 2 13.6eV 27.2 eV
U bonded Up-p 2Ue-p e2 1 2 e 0.66 e 1 e2 1 1 2.64
4 0 d 4 0 d
4 0 1 d
2
1.64
2.30 1028 Jm 31.9eV
0.074 109 m 1.60 1019 J eV
Ubinding Uisolated Ubonded 27.2 eV 31.9 eV 4.7 eV
This is reasonably close to the actual value of 4.5 eV quoted in the text.
4. We follow the procedure outlined in the statement of the problem.
HN: 74 pm 145pm 110 pm
1 dH2 dN2 1
2 2
CN: 154 pm 145pm
1 dC2 dN2 1 150 pm
2 2
NO: 145 pm 121pm
1 dN2 dO2 1 133pm
2 2
5. According to the problem statement, 5.39 eV of energy is required to make an Li ion from neutral
Li, and 3.41 eV of energy is released when an F atom becomes an F ion. That means that a net
energy input of 5.39 eV – 3.41 eV = 1.98 eV is needed to form the ions. We calculate the negative
potential energy of the attraction between the two ions.
U
1 e2 2.30 1028 J m 9.21eV
40 r 0.156 109 m 1.60 1019 J eV
The binding energy should therefore be 9.21 eV – 1.98 eV = 7.23 eV. But the actual binding energy
is only 5.95 eV. Thus the energy associated with the repulsion of the electron clouds is 7.23 eV –
5.95 eV = 1.28 eV.
6. We convert the units from kcal/mole to eV/molecule.
1 kcal 4186 J 1eV J 6.022 1mole 4.339 102 eV
mole 1kcal 1.602 1019 1023 molecules molecule
Now convert 4.43 eV per molecule into kcal per mole.
4.43 eV 4.339 1kcal mol 102 kcal mol
molecule 102 eV molecule
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536
Chapter 40 Molecules and Solids
7. (a) The neutral He atom has two electrons in the ground state, n 1, 0, m 0. Thus the two
electrons have opposite spins, ms 1 . If we try to form a covalent bond, we see that an
2
electron from one of the atoms will have the same quantum numbers as one of the electrons on
the other atom. From the exclusion principle, this is not allowed, so the electrons cannot be
shared.
(b) We consider the He molecular ion to be formed from a neutral He atom and an He ion. It
2
will have three electrons. If the electron on the ion has a certain spin value, it will have the
opposite spin as one of the electrons on the neutral atom. Thus those two electrons can be in the
same spatial region, and so a bond can be formed.
The units of 2 are Js 2 J2 J2 J2
I kgm s2 J
kgm2 Nm
8. m J.
9. The reduced mass is given in Section 40-4 as m1m2 . We calculate in atomic mass units.
m1 m2
(a) KCl: m1m2 39.10 u35.45u 18.59 u
m1 m2 39.10u 35.45u
(b) O2: m1m2 16.00 u16.00u 8.00 u
m1 m2 16.00 u 16.00 u
(c) HCl: m1m2 1.008 u 35.45 u 0.9801u
m1 m2 1.008u 35.45u
10. (a) The moment of inertia of O2 about its CM is given by mO CM mO
I 2mO r 2 mOr 2 .
2 2
r
1.055 1034 Js 2
2 2 1.789 104 eV
2I mOr2 16 1.66 1027 kg 0.121109 m 2 1.60 1019 J eV
1.79 104 eV
(b) From Figure 40-17, we see that the energy involved in the l = 2 to l = 1 transition is 22 .
I
E 22 4 2
I 2I 4 1.789 104 eV 7.156 104 eV 7.16 10-4 eV
E h c
hc 6.63 1034 Js 3.00 108 m s 1.74 103 m
E 7.156 104 eV 1.60 1019 J eV
11. Use the rotational energy and the moment of inertia of N2 about its mN CM mN
CM to find the bond length.
I 2mN r 2 mN r 2 ; Erot 2 2
2 2 2I mN r 2 r
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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
r 1.055 1034 Js 1.10 1010 m
ErotmN
2.48 104 eV 1.60 1019 J eV 14.01u 1.66 1027 kg u
12. The longest wavelength emitted will be due to the smallest energy change. From Figure 40-17, the
smallest rotational energy change is E 2 I . We find the rotational inertia from Eq. 40-4.
E 2 hc
I
hc I 4 2c m1m2 r 2
2 h m1 m2
4 2 3.00 108 m s
6.631034 Js
6.941u1.008u 1.66 1027 kg u 0.16 109 m 2
6.941u 1.008u
6.7 104 m
13. The energies involved in the transitions are given in Figure 40-17. We find the rotational inertial
from Eq. 40-4. The basic amount of rotational energy is 2 I .
2 2 2 22 2 1.055 1034 Js 2
mHr2 1.008 u 1.66 1027 kg u
I
m1m2 r2 1 mH r 2 0.074 109 m 2
m1 m2 2
2.429 1021J
(a) For l = 1 to l = 0:
E 2 2.429 1021 J 1 1.5 102 eV
I 1019
1.60 J eV
hc 6.63 1034 Js 3.00 108 m s 8.2 105 m
E 2.429 1021J
(b) For l = 2 to l = 1:
E 2 1
2 I 2 2.429 1021J 3.0 102 eV
1.60 1019 J eV
hc 6.63 1034 Js 3.00 108 m s 4.1105 m
E 2 2.429 1021J
(c) For l = 3 to l = 2:
E 2 1
3 I 3 2.429 1021J 4.6 102 eV
1.60 1019 J eV
hc 6.63 1034 Js 3.00 108 m s 2.7 105 m
E 3 2.429 1021J
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538
Chapter 40 Molecules and Solids
14. The energy change for transitions between combined rotational and vibrational states is given above
Eq. 40-8a.
E Evib Erot hf Erot
If E hf is to be in the spectrum, then Erot 0. But the selection rules state that l 1 for a
transition. It is not possible to have l 0 for a transition. The only way to have Erot 0 is for
l 0, which is forbidden. Thus E hf is not possible. Here is a mathematical statement as
well.
Erot E l l El l l l l 1 2 l l 1 2 l 2 l l 1 2
2I 2I 2I
For Erot 0, mathematically we must have l 0, which is forbidden.
15. (a) The reduced mass is defined in Eq. 40-4.
mCmO 12.01u 16.00 u 6.86 u
mC mO 12.01u 16.00 u
(b) We find the effective spring constant from Eq. 40-5.
f 1 k
2
k 4 f 2 4 2 6.42 1013 Hz 2 6.86 u 1.66 1027 kg u 1850 N m
The spring constant for H2 is estimated in Example 40-6 as 550 N/m.
kCO 850 N m 3.4
kH2 550 N m
16. The effective spring constant can be found from Eq. 40-5, using the vibrational frequency and the
reduced mass.
f 1 k k 4 f 2 4 2 f 2 m1m2
2 m1 m2
2 6.941u79.904 u
6.941u 79.904 u
4 2
1.7 1013 Hz 1.66 1027 kg u 120 N m
17. We first find the energies of the transitions represented by the wavelengths.
E1
hc 6.63 1034 Js 3.00 108 m/s 5.38 105eV
1.60 1019 J/eV 23.1103 m
E2
hc 6.63 1034 Js 3.00 108 m/s 10.72 105eV
1.60 1019 J/eV 11.6 103 m
E3
hc 6.63 1034 Js 3.00 108 m/s 16.12 105eV
1.60 1019 J/eV 7.71103 m
Since E2 10.72 2 and E3 16.12 3, from the energy levels indicated in Figure 40-17, and
E1 5.38 E1 5.38
from the selection rule that l 1, we see that these three transitions must represent the l 1 to
l 0 transition, the l 2 to l 1 transition, and the l 3 to l 2 transition. Thus E1 2 I .
We use that relationship along with Eq. 40-4 to find the bond length.
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539
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
E1 2 2
I r2
r 1.055 1034 Js
E1 22.990u35.453u
1.66 1027 kg u 5.38 105eV 1.60 1019 J eV
22.990 u 35.453 u
2.36 1010 m
18. (a) The curve for U 1 k x r0 2 4.5eV is shown in Figure 40–18 as a dotted line. Measuring
2
on the graph with a ruler gives the distance from the origin to the 0.1 nm mark as 38 mm. The
measured distance from the origin to the largest x-intercept of the parabola is 45 mm. Taking a
ratio gives the distance from the origin to the largest x-intercept as 0.118 nm. We fit a parabolic
curve to data.
U 1 k x r0 2 4.5eV ; U xr0 1 k 02 4.5eV 4.5eV check
2 2
U x0.118nm 2 4.5eV 0
1 k 0.118nm r0
2
k 2 4.5 eV 4649 eV 1.60 1019 J 109 nm 2 743.8 N m
0.118nm 0.074 nm2 nm2 eV 1m
740 N m
(b) The frequency of vibration is given by Eq. 14-7a, using the reduced mass. Use this relationship
to find the wavelength.
f 1 k c
2
2 c
2 3.00 108 m s 0.51.00794 u 1.66 1027 kg u 2.0 106 m
k
743.8 N m
19. Consider the system in equilibrium, to find the center of l
l2
mass. See the first diagram. The dashed line represents the l2
location of the center of mass. m1 m2
l l1 l2 ; m1l1 m2l2 m2 l l1 m2
x2
l1 m2 l ; l2 m1 l l1
m1 m2 m1 m2 l1
Now let the spring be stretched to the left and right, m1
x1
but let the center of mass be unmoved.
x x1 x2 ; m1 l1 x1 m2 l2 x2
m1l1 m1x1 m2l2 m2 x2 m1x1 m2 x2
This is the second relationship requested in the problem. Now use the differential relationships.
m1 d 2 x1 kx ; m2 d 2 x2 kx d 2 x1 k x ; d 2 x2 k x
dt 2 dt 2 dt 2 m1 dt 2 m2
d 2 x1 d 2 x2 kx 1 1 d 2 x1 x2 kx m1 m2 k x d2x kx
dt 2 dt 2 m1 m2 m1m2 dt 2
dt 2
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540
Chapter 40 Molecules and Solids
This last equation is the differential equation for simple harmonic motion, as in Eq. 14-3, with m
replaced by . The frequency is given by Eq. 14-7a, f 1 k , which is the same as Eq. 40-5.
2
20. The ionic cohesive energy is given right after Eq. 40-9, and derived in the solution to Problem 25.
The Madelung constant is 1.75.
U0
e2 1 1 1.75 2.30 1028 Jm 1 1 7.9 eV
4 r0 m 0.28 109 m 1.60 1019 J eV 8
21. Because each ion occupies a cell of side s, a molecule occupies two cells. Use the value of the
density to solve for the desired distance.
mass mNaCl
volume 2s3
1/ 3
s 1/ 3
mNaCl 58.44 u 1.661 1027 kg u 2.826 1010 m
2
2 2.165 g 1kg 106 cm3
cm3 1000 g 1m3
22. Because each ion occupies a cell of side s, a molecule occupies two cells. Use the value of the
density to solve for the desired distance.
mass mKCl
volume 2s3
1/ 3
s 1/ 3
mKCl 39.10 u 35.45u 1.661 1027 kg u 3.15 1010 m
2
2 1.99 g 1kg 106 cm3
cm3 1000 g 1m3
23. According to Section 40-5, the NaCl crystal is face-centered cubic. It is Na + d – Cl
illustrated in Figure 40-24. We consider four of the labeled ions from Cl – + Na
Figure 40-24. See the adjacent diagram. The distance from an Na ion
to a Cl ion is labeled as d, and the distance from an Na ion to the nearest
neighbor Na ion is called D.
D d 2 0.24 nm 2 0.34 nm
24. See the diagram. Select a charge in the middle of – + – + – +
the chain. There will be two charges of opposite r
sign a distance r away, two charges of the same sign
a distance 2r away, etc. Calculate the potential energy of the chosen charge.
U 1 2e2 1 2e2 1 2e2 1 2e2
4 0 r 4 0 2r 4 0 3r 4 0 4r
2e2 1 1 1 1
4 0 r 2 3 4
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541
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
From Appendix A, ln 1 x x 1 x2 1 x2 1 x4 . Evaluate this expansion at x = 1.
2 3 4
ln 1 1 1 1 1 1 ln 2 U 2e2 1 1 1 1 2e2 ln 2
2 3 4 4 0 r 2 3 4 4 0 r
From Section 40-5, the potential energy is given as U e2 . Equate the two expressions for the
4 0 r
potential energy to evaluate the Madelung constant.
U e2 2e2 ln 2 2ln 2
4 0 r 4 0 r
25. (a) Start with Eq. 40-9 and find the equilibrium distance, which minimizes the potential energy.
Call that equilibrium distance r0.
U e2 B ; dU e2 m B e2 m B 0
4 0 r rm dr 4 r2 r m 1 4 0 r02 r m 1
r r0 0 r r0
0
B e2 r m 1
4 0
0
m
e2 r m 1
0
U0 U r r0 e2 B e2 4 0 m e2 1 1
4 r0 r0m 4 r0 r0m 4 0 r0 m
0 0
(b) For NaI, we evaluate U0 with m = 10, 1.75, and r0 0.33nm.
e2 1 1 1.75 2.30 1028 Jm 1 6.861eV
4 0 r0 m 0.33109 m 1.60 1019 J
U0 eV 1
10
6.9 eV
(c) For MgO, we evaluate U0 with m = 10, 1.75, and r0 0.21nm.
e2 1 1 1.75 2.30 1028 Jm 1 10.78 eV
4 0 r0 m 0.21109 m 1.60 1019 J
U0 eV 1
10
11eV
(d) Calculate the % difference using m = 8 instead of m = 10.
U0 U0 e2 1 1 e2 1 1 1 1 1 1 1 1
m8 m10 4 r0 8 4 r0 10 8 10
U0 10 8
m10 0 0 0.0278
e2 1 1 1 1 1 1
4 r0 10 10 10
0
2.8%
26. We follow Example 40-9. The density of occupied states (number of states per unit volume in an
infinitesimal energy range) is given by Eq. 40-15. Because we are using a small energy range, we
estimate the calculation with a difference expression. We let N represent the number of states, and V
represent the volume under consideration.
N g E V E 8 2 m3/ 2 E1/ 2V E
h3
E 1 EF 0.985EF 0.9925EF ; E EF 0.985EF 0.015EF 0.0822eV
2
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542
Chapter 40 Molecules and Solids
8 2 9.111031kg 3/ 2
N 6.63 1034 Js 3
0.99255.48eV 1.0 106 m3 0.0822eV 1.60 1019 J eV 3/ 2
1.31021states
27. We follow Example 40-9. The density of occupied states (number of states per unit volume in an
infinitesimal energy range) is given by Eq. 40-15. Because we are using a small energy range, we
estimate the calculation with a difference expression. We let N represent the number of states, and V
represent the volume under consideration.
N g E V E 8 2 m3/ 2 E1/ 2V E
h3
8 2 9.111031kg 3/ 2
6.63 1034 Js 3
7.025eV 1.0 106 m3 0.05eV 1.60 1019 J eV 3/ 2
9.0 1020states
28. The density of molecules in an ideal gas can be found from the ideal gas law.
N P 1.013105 Pa 2.576 1025 m3
V gas kT
1.38 1023 J K 285K
PV NkT
We assume that each copper atom contributes one free electron, and use the density of copper as
given in Table 13-1.
N 1e 6.02 1023Cu atoms 8.9 103 kg 8.4311028 m3
V e's Cu atom 63.546 103 kg m3
N
V gas
N 2.576 1025 m3 3.1 104
8.4311028 m3
V e's
29. We use Eq. 40-14 for the occupancy probability, and solve for the energy. The Fermi energy is
7.0 eV.
(a) Evaluate for T = 295 K.
fE 1
eEEF kT 1
E
kT ln f 1 1 EF 1.38 1023 J K 295 K ln 1 1 7.0 eV
1.60 1019 J eV 0.850
E
6.96eV
(b) Evaluate for T = 750 K.
fE 1
eEEF kT 1
E
kT ln f 1 1 EF 1.38 1023 J K 750 K ln 1 1 7.0 eV
1.60 1019 J eV 0.850
E
6.89eV
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543
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
30. We use Eq. 40-14 for the occupancy probability, and solve for the energy. The Fermi energy is
7.0 eV.
(a) Evaluate for T = 295 K.
f E 1
eEEF kT 1
E
kT ln f 1 1 EF 1.38 1023 J K 295 K ln 1 1 7.0 eV
1.60 1019 J eV 0.150
E
7.04eV
(b) Evaluate for T = 950 K.
f E 1
eEEF kT 1
E
kT ln f 1 1 EF 1.38 1023 J K 950 K ln 1 1 7.0 eV
1.60 1019 J eV 0.150
E
7.14eV
31. The occupancy probability is given by Eq. 40-14. The Fermi level for copper is 7.0 eV.
1 1 1 0.0159
eEEF / kT 1 e1.015EF EF / kT
f
1 e0.0157.0eV 1.601019 J eV 1.381023 J K 295 K
1
1.6%
32. We follow Example 40-10.
(a) Because each zinc atom contributes two free electrons, the density of free electrons is twice the
density of atoms.
N m3 6.02 1023 atoms mol 2 atom 1.307 1029 m3
mol
V
7100 kg 65.409 103 kg free electrons
1.3 1029 m3
(b) The Fermi energy is given by Eq. 40-12.
6.63 1034 Js 2
8 9.111031kg
EFh2 3 N 2/3 3 2 / 3 1
8m V
1.307 1029 m3
1.60 1019 J eV
9.414 eV 9.4 eV
(c) The Fermi speed is the speed of electrons with the Fermi energy.
vF
2EF 2 9.414eV 1.60 1019 J eV 1.8 106 m s
m
9.111031 kg
33. We follow Example 40-10. We need the number of conduction electrons per unit volume of sodium.
N m3 6.02 1023 atoms mol 1free atom 2.540 1028 m3
22.99 103 kg mol
V
970 kg electrons
The Fermi energy is given by Eq. 40-12.
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544
Chapter 40 Molecules and Solids
6.63 1034 Js 2
8 9.111031kg
EFh23N 2 / 3 3 2/3 1
8m V 2.540 1028 m3 3.159 eV
1.60 1019 J eV
3.2eV
The Fermi speed is the speed of electrons with the Fermi energy.
vF
2EF 23.159eV 1.60 1019 J eV 1.1106 m s
m
9.111031 kg
34. (a) Find the density of free electrons from Eq. 40-12.
EF h2 3 N 2/3
8m V
N 8 3/2
V
8mEF 3/ 2 9.111031kg 11.63eV 1.60 1019 J eV
3 h2 3
6.63 1034 Js 2
1.7945 1029 m3 1.79 1029 m3
(b) Let n represent the valence number, so there are n free electrons per atom.
N m3 6.02 1023 atoms mol atom
27.0 103 kg mol
V
270 kg n free electrons
27.0 103 kg mol 1
6.02 1023 atoms mol kg
n
270 m3 1.7945 1029 m3 2.981 3
This agrees nicely with aluminum’s position in the periodic table, and its electron configuration
of 1s22s22p63s23p1. The level 3 electrons are the valence electrons.
35. We calculate the given expression, with T = 0, so that the maximum energy is EF. The value of f(E)
at T = 0 is given below Eq. 40-14.
EF EF EF 8 2 m3/ 2 EF
h3
En0 E dE Eg E f E dE E E1/ 2 1 dE E 3/ 2dE
0 0 0
EF EF EF
n0 E dE g E f E dE E1/ 2dE
0 0 0
E 0
EF 8 2 m3/ 2 E1/ 2 1 dE
h3
0
E2 5 / 2 3 EF
5
5F
E2 3/ 2
3F
36. We first find the density of neutrons, and then use Eq. 40-12.
N 1
V
1 neutron kg 2.5 1.99 1030 kg 4 12, 000 m 3 4.103 1044 m3
1.675 1027 3
6.63 1034 Js 2
1.675 1027 kg
EFh23N 2 / 3 3 2/3 1
8m V 4.103 1044 m3
8 1.60 1013 J MeV
109.8 MeV 110 MeV
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545
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
37. We start with Eq. 38-13 for the energy level as a function of n. If we solve for n, we have the
number of levels with energies between 0 and E. Taking the differential of that expression will give
the number of levels with energies between E and dE. Finally, we multiply by 2 since there can be 2
electrons (with opposite spins) in each energy level.
E n2 h2 n 8ml2 E dn 8ml2 1 dE
8ml2 h2 h2 2 E
8ml2 1 dE 8ml2
dn h2 2 h2E
gl 2 dE 2 dE E
38. We use Eq. 40-14, with E EF.
f 1 1 11 1
eE EF / kT e0 1 11 2
The result is independent of the value of T.
39. (a) Eq. 38-13 gives the energy levels as a function of n, the number of levels. Since there are 2
electrons in every energy level, n = N / 2. The Fermi energy will be the highest energy level
occupied with N electrons.
n2 h2 2 h2 h2N 2 h2N 2
8ml2 8ml2 32ml2 32ml2
En 1 N EF
2
(b) The smallest amount of energy that this metal can absorb is the spacing between energy levels.
E En1 En h2 n 12 n2 h2 2n 1 h2 N 1
8ml2 8ml2 8ml2
(c) We calculate the limit requested.
E h2 N 1 4
EF 8ml2 N
h2N 2
32ml2
For large N, this is a very small change in energy. Thus a very small change in energy will
allow an electron to change energy levels, and so the metal conducts very easily.
40. (a) We use Eq. 40-14, with the data as given in the problem.
E EF
kT
0.12 eV 1.60 1019 J eV 4.74848
k 1.38 1023 J K 293K
f 1 1 11 8.590 103 8.6 103
eE EF / kT e4.74848 1 116.409
This is reasonable. Very few states this far above the Fermi energy are occupied at this
relatively low temperature.
(b) Use a similar calculation to part (a).
E EF
kT
0.12 eV 1.60 1019 J eV 4.74848
k 1.38 1023 J K 293K
f 1 1 1 1 0.991409 0.99
eEEF / kT 1 e4.74848 1.008665
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546
Chapter 40 Molecules and Solids
(c) Since the probability that the state is occupied is 0.991409, the probability that the state is
unoccupied is 1 0.991409 8.591103 8.6 103 . This is the same as part (a).
41. We consider the cube to be a three-dimensional infinite well, with a width of l in each dimension.
We apply the boundary conditions as in Section 38-8 separately to each dimension. Each dimension
gives a quantum number which we label as n1, n2, and n3. We then have a contribution to the
energy of the bound particle from each quantum number, as in Eq. 38-13.
E n12 h2 h2 n32 h2 h2
E1 E2 E3 8ml2 n22 8ml2 8ml2 8ml2 n12 n22 n32 , n1, n2,n3 1,2,3,
Specifying the three quantum numbers gives a state and the corresponding energy.
Choosing axes as specified in the problem, the equation of a sphere of radius R in that coordinate
system is R2 n12 n22 n32. Each state “contained” in that sphere could be indicated by a cube of
side length 1, and each state can have two electrons (two spin states). The “volume” of that sphere is
1 of a full sphere. From that we calculate the number of states in one octant, and then g(E).
8
3/2 8ml2 3 / 2
3 3
N 2 R3
1 4 n12 n22 n32 h2 E
8 3
1 dN 1 8ml2 3/ 2 3 E1/2 8m 3 / 2 8 2 m3/ 2
V dE l3 2 2 h2 h3
g E 3 h2 E1/ 2 E1/ 2
42. The photon with the minimum frequency for conduction must have an energy equal to the energy
gap.
Eg
hf hc 6.631034 Js 3.00 108 m/s 2.14 eV
1.60 1019 J/eV 580 109m
43. The photon with the longest wavelength or minimum frequency for conduction must have an energy
equal to the energy gap:
c hc hc 6.63 1034 Js 3.00 108 m/s 1.09 106 m 1.09 m.
f hf Eg
1.60 1019 J/eV 1.14eV
44. The energy of the photon must be greater than or equal to the energy gap. Thus the longest
wavelength that will excite an electron is
c hc hc 6.63 1034 Js 3.00 108 m/s 1.7 106 m 1.7 m
f hf Eg
1.60 1019 J/eV 0.72eV
Thus the wavelength range is 1.7m .
45. (a) In the 2s shell of an atom, l 0, so there are two states: ms 1 . When N atoms form bands,
2
each atom provides 2 states, so the total number of states in the band is 2N.
(b) In the 2p shell of an atom, l 1, so there are three states from the ml values: m 0, 1; each
of which has two states from the ms values: ms 1 , for a total of 6 states. When N atoms
2
form bands, each atom provides 6 states, so the total number of states in the band is 6N.
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547
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
(c) In the 3p shell of an atom, l 1, so there are three states from the ml values: ml 0, 1; each
of which has two states from the ms values: ms 1 , for a total of 6 states. When N atoms
2
form bands, each atom provides 6 states, so the total number of states in the band is 6N.
(d) In general, for a value of l, there are 2l 1 states from the ml values: ml 0, 1, ... , l. For
each of these there are two states from the ms values: ms 1 , for a total of 22l 1 states.
2
When N atoms form bands, each atom provides 22l 1 states, so the total number of states
in the band is 2N 2l 1.
46. The minimum energy provided to an electron must be equal to the energy gap. Divide the total
available energy by the energy gap to estimate the maximum number of electrons that can be made
to jump.
N
hf 760 103eV 1.1 106
Eg
0.72 eV
47. Calculate the number of conduction electrons in a mole of pure silicon. Also calculate the additional
conduction electrons provided by the doping, and then take the ratio of those two numbers of
conduction electrons.
NSi
28.09 103 kg mol 1.206 1011 electrons mole
2330 kg m3 1016 electrons m3
Ndoping
6.02 1023atoms 5.017 1017 added conduction electrons.
1.2 106
Ndoping
NSi
5.017 1017 4.16 106 4 106
1.206 1011
48. The wavelength is found from the energy gap.
c hc hc 6.63 1034 Js 3.00 108 m/s 7.8 107 m 0.78 m
f hf Eg
1.60 1019 J/eV 1.6eV
49. The photon will have an energy equal to the energy gap:
Eg
hf hc 6.63 1034 Js 3.00 108 m/s 1.8eV
1.60 1019 J/eV 680 109 m
50. From the current-voltage characteristic graph in Figure 40-38, we see that a current of 12 mA means
a voltage of about 0.68 V across the diode. The battery voltage is the sum of the voltages across the
diode and the resistor.
Vbattery Vdiode VR 0.68 V 0.012 A860 11V
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548
Chapter 40 Molecules and Solids
51. The battery voltage is the sum of the voltages across the diode
and the resistor.
Vbattery Vdiode VR ;
2.0 V Vdiode I 150 Ω I Vdiode 2.0 V
150 150
This is the equation for a straight line which passes through
the points (0 V, 13.3 mA) and (0.8 V, 8 mA). The line has a
y-intercept of 13.3 mA and a slope of 6.67 mA/V. If we
assume the operating voltage of the diode is about 0.7 V, then
the current is about 8.6 mA. There is some approximation
involved in the answer.
52. We have copied the graph for V > 0 and rotated it so that it shows V as a function of I. This is the
first diagram below. The resistance is the slope of that first graph. The slope, and thus the
resistance, is very high for low currents, and decreases for larger currents, approaching 0. As an
approximate value, we see that the voltage changes from about 0.55 V to 0.65 V as the current goes
from 0 to 10 mA. That makes the resistance about 10 ohms when the current is about 5 mA. The
second diagram is a sketch of the resistance.
0.8 V volts R
0.6 20
0.4 15
10
0.2 5
I mA
10 20 30 10 20 30
I (mA)
53. (a) For a half-wave rectifier without a capacitor, the current is zero for half the time. We
approximate the average current as half of the full rms current.
I av V1 rms 1 120 V 1.7 mA
2R 2 35 k
(b) For a full-wave rectifier without a capacitor, the current is positive all the time. We
approximate the average current as equal to the full rms current.
I av Vrms 120 V 3.4 mA
R 35 k
54. The band gap is the energy corresponding to the emitted wavelength.
E hc
6.631034 Js 3.00 108 m s 0.96eV
1.3106 m 1.60 1019 J eV
55. There will be a current in the resistor while the ac voltage varies from 0.6 V to 9.0 V rms. Because
the 0.6 V is small, the voltage across the resistor will be almost sinusoidal, so the rms voltage across
the resistor will be close to 9.0 V 0.6 V 8.4 V.
(a) For a half-wave rectifier without a capacitor, the current is zero for half the time. We ignore
the short time it takes for the voltage to increase from 0 to 0.6 V, and so current is flowing in
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549
Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual
the resistor for about half the time. We approximate the average current as half of the full rms
current.
I av V1 rms 1 8.4 V 35 mA
2R 2 0.120 kΩ
(b) For a full-wave rectifier without a capacitor, the current is positive all the time. We ignore the
short times it takes for the voltage to increase from 0 to 0.6 V, and so current is flowing in the
resistor all the time. We approximate the average current as the full rms current.
I av Vrms 8.4 V 70 mA
R 0.120 kΩ
56. (a) The time constant for the circuit is 1 RC1 28 103 35 106 F 0.98s. As seen in
Figure 40-40(c), there are two peaks per cycle. The period of the rectified voltage is
T s1 0.0083s. Because 1 T , the voltage across the capacitor will be essentially
120
constant during a cycle, so the average voltage is the same as the peak voltage. The average
current is basically constant.
Vavg Vpeak 2Vrms 2 120 V
R R R
28 103
Iavg 6.1mA
(b) With a different capacitor, the time constant for the circuit changes.
2 RC2 28 103 0.10 106 F 0.0028s
Now the period of the rectified voltage is about 3 time constants, and so the voltage will
decrease to about 5% e3 of the peak value during each half-cycle. We approximate the
voltage as dropping linearly from its peak value to 0 over each half-cycle, and so take the
average voltage as half the peak voltage.
Vavg 2Vrms 2 120 V
R R
28 103
Iavg 1 1 3.0 mA
2 2
57. By Ohm’s law, the output (collector) current times the output resistor will be the output voltage.
RC Vout Vout 0.35 V
iC iB 1.0 106 A
Vout iCRC 95 3684 3700
58. By Ohm’s law, the output (collector) current times the output resistor will be the output voltage.
Vout iCRC iBRC 85 2.0 106 A 4300 0.73V
59. By Ohm’s law, the output (collector) current times the output resistor will be the output voltage.
Vout iCR iC Vout VV input 650.080 V 2.08 104 A 0.21mA
R R
25, 000
60. (a) The voltage gain is the collector ac voltage divided by the base ac voltage.
V VC iC RC I RC 75 7.8 k 153.9 150
VB iB RB RB 3.8 k
(b) The power amplification is the output power divided by the input power.
P iCVC I V 75153.9 11,543 12, 000
iBVB
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