Download Ch 28 Solutions

QUANTUM MECHANICS OF ATOMS
28
Responses to Questions
1.
(a)
(b)
A matter wave Ψ does not need a medium, but a wave on a string does. The square of the wave
function Ψ for a matter wave describes the probability of finding a particle within a certain
spatial range. The function for a wave on a string describes the displacement of a piece of the
string from its equilibrium position as a function of location and time.
Neither an EM nor a matter wave needs a medium in which to exist. The EM wave describes the
amplitudes and directions of the electric and magnetic fields as a function of location and time.
An EM wave represents a vector field and can be polarized. A matter wave is a scalar and cannot
be polarized.
2.
According to Bohr’s theory, each electron in an atom travels in a circular orbit and has a precise
position and momentum at any point in time. This view is inconsistent with the postulates of quantum
mechanics and the uncertainty principle, which does not allow both the position and momentum to be
known precisely. According to quantum mechanics, the “orbitals” of electrons do not have precise
radii, but describe the probability of finding an electron in a given spatial range.
3.
As a particle becomes more massive, the uncertainty of its momentum (Δp = mΔυ ) becomes larger,
which reduces the uncertainty of its position (due to Δx ≥ /Δp ). Thus, with a small uncertainty of its
position, we can do a better job of predicting its future position.
4.
Because of the very small value of , the uncertainties in both a baseball’s momentum and position can
be very small compared to typical macroscopic positions and momenta without being close to the limit
imposed by the uncertainty principle. For instance, the uncertainty in the baseball’s position could be
10−10 m, and the uncertainty in the baseball’s momentum 10−10 kg ⋅ m/s, and still easily satisfy the
uncertainty principle. Yet we never try to measure the position or momentum of a baseball to that
precision. For an electron, however, typical values of uncertainty in position and momentum can be close
to the uncertainty principle limit, so the relative uncertainty can be much higher for that small object.
5.
No. According to the uncertainty principle, if the needle were balanced, then the position of the center of
mass would be known exactly, and there would have to be some uncertainty in its momentum. The center
of mass of the needle could not have a zero momentum and therefore would fall over. If the initial
momentum of the center of mass of the needle were exactly zero, then there would be uncertainty in its
position, and the needle could not be perfectly balanced (with the center of mass over the tip).
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28-1
28-2
Chapter 28
6.
No. The hot soup will warm the thermometer when they come in contact. Thus the final temperature of
the thermometer/soup system will be less than the initial temperature of the soup. Accordingly, the
final thermometer reading will be lower than the original temperature of the soup. This is a situation
where making a measurement affects the physical situation and causes uncertainty in the measurement.
7.
No. However, the greater the precision of the measurement of position, the greater the uncertainty in
the measurement of the momentum of the object will be.
8.
If you knew the position precisely, then you would know nothing about the momentum.
9.
Yes, some of the air escapes the tire (and goes into the tire gauge) in the act of measuring the pressure.
It is impossible to avoid this escape. The act of measuring the air pressure in a tire therefore actually
changes the pressure, although not by much since very little air escapes compared to the total amount
of air in the tire. This is similar to the uncertainty principle, in which one of the two factors limiting the
precision of measurement is the interaction between the object being observed, or measured, and the
observing instrument. By making the measurement, you have affected the state of the system.
10.
Yes, this is consistent with the uncertainty principle for energy (ΔE Δt ≥ ). The ground-state energy
can be very precisely known because the electrons remain in that state for a very long time. Thus, as
Δt → ∞, ΔE → 0. However, the electrons do not remain in the excited states for very long, thus the
Δt in this case is relatively small, making the ΔE , the energy width, relatively large.
11.
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 most
probable location for the electron is the Bohr radius, but it can also be found closer to the nucleus (and
farther away). See Fig. 28–6 for a visual representation of this question.
12.
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.
13.
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.
14.
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:
H 1
1s1
He 2
Li 3
Be 4
B 5
2s1
2p1
2p 2
2p3
C 6
N 7
O 8
F 9
3p1
3p 2
3p3
S 16
Cl 17
Ar 18
1
3s
Ne 10
Na 11
Mg 12
Al 13
1
1
2
3
4s
3d
K 19
Ca 20
1
5s
3d
3d
Si 14
3d
4
P 15
3d
5
1
4p
4p
2
4p3
4d 1
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Quantum Mechanics of Atoms
28-3
15.
(a) and (c) are allowed for atoms in an excited state. (b) is not allowed. Only six electrons are allowed
in the 2p state.
16.
(a)
Group VII. Outer electrons are 2s 2 2p5 : fluorine.
(b)
Group II. Outer electrons are 3s 2 : magnesium.
(c)
Group VIII. Outer electrons are 3s 2 3p 6 : argon.
(d)
Group I. Outer electron is 4s1: potassium.
17.
Both chlorine and iodine are in the same column of the Periodic Table (column VII). Thus, both of
their outer electron configurations are p5 (5 electrons in their outer p shell), which makes them both
halogens. Halogens lack one electron from a filled shell, so their outer electron orbit shapes are similar
and can readily accept an additional electron from another atom. This makes their chemical reaction
properties extremely similar.
18.
Both sodium and potassium are in the same column of the Periodic Table (column I). Thus, both of
their outer electron configurations are s1 (1 electron in their outer s shell), which makes them both
alkali metals. Alkali metals have one lone electron in their outermost shell, so their outer electron orbit
shapes are similar and they can readily share this electron with another atom (especially since this
outermost electron spends a considerable amount of time very far away from the nucleus). This makes
their chemical reaction properties extremely similar.
19.
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 little effect on the chemical properties of these elements.
20.
Neon is a “closed shell” atom—a noble gas. All of its shells are completely full and spherically
symmetric, so its electrons are not readily ionized. Sodium is a Group I atom (an alkali metal) that has
one lone electron in the outermost shell, and this electron spends a considerable amount of time far
away from the nucleus and also shielded from the nucleus. The result is that sodium can be ionized
much easier than neon, which the two ionization energies demonstrate very well.
21.
The difference in energy between n = 1 and n = 2 levels is much bigger than between any other
combination of energy levels. Thus, electron transitions between the n = 1 and n = 2 levels produce
photons of higher energy and frequency, which means that these photons have shorter wavelengths
(since frequency and wavelength are inversely proportional to each other).
22.
The Bohr model does violate the uncertainty principle, specifically by saying that an electron in a
given orbit has a precise radius and a precise momentum. The uncertainty principle says that knowing
both the location and the momentum precisely is not possible.
23.
Noble gases are nonreactive because they have completely filled shells or subshells. In that
configuration, other electrons are not attracted, nor are electrons readily lost. Thus they do not react
(share or exchange electrons) with other elements. Alkali metals all have a single outer s electron. This
outer electron is easily removed and can spend much of its time around another atom, forming a
molecule. Thus alkali metals are highly reactive.
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28-4
Chapter 28
24.
Spontaneous emission occurs when an electron is in an excited state of an atom and it spontaneously
(with no external stimulus) drops back down to a lower energy level. To do this, it emits a photon to
carry away the excess energy. Stimulated emission occurs when an electron is in an excited state of an
atom but a photon strikes the atom and causes or stimulates the electron to make its transition to a
lower energy level sooner than it would have done so spontaneously. The stimulating photon has to
have the same energy as the difference in energy levels of the transition. The result is two photons of
the same frequency—the original one and the one due to the emission.
25.
Laser light is coherent (all of the photons are in the same phase) and ordinary light is not coherent (the
photons have random phases relative to each other). Laser light is nearly a perfect plane wave, while
ordinary light is spherically symmetric (which means that the intensity of laser light remains nearly
constant as it moves away from the source, while the intensity of ordinary light drops off as 1/r 2 ).
Laser light is always monochromatic, while ordinary light can be monochromatic, but it usually is not.
However, both travel at c and both display the wave–particle duality of photons. Both are created when
electrons fall to lower energy levels and emit photons. Both are electromagnetic waves.
26.
The 0.0005-W laser beam’s intensity is nearly constant as it travels away from its source since it is
approximately a plane wave. The street lamp’s intensity drops off as 1/r 2 as it travels away from its
source since it is a spherically symmetric wave. So, at a distance, the power from the lamp light
actually reaching the camera is much less than 1000 W. Also, the street lamp’s intensity is spread out
over many wavelengths (many of which are probably not in the visible range), whereas the laser’s
intensity is all at one (assumed visible) wavelength.
27.
No, the intensity of light from a laser beam does not drop off as the inverse square of the distance.
Laser light is much closer to being a plane wave than a spherically symmetric wave, so its intensity is
nearly constant along the entire beam. It does spread slightly over long distances, so it is not a perfect
plane wave.
28.
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 a “cutoff ” wavelength in the X-ray spectrum. An increase in the number of
electrons will not change the cutoff 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.
29.
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.
30.
To figure out which lines in an X-ray spectrum correspond to which transitions, you would use the
Bohr model to estimate the differences in the sizes of the energy level jumps that the falling electrons
will make. Then, you just need to match the n = 2 → n = 1 energy to the K α line, the n = 3 → n = 1
energy to the K β line, the n = 3 → n = 2 jump energy to the Lα line, and so on.
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Quantum Mechanics of Atoms
28-5
Responses to MisConceptual Questions
1.
(b)
A common error is to add up the coefficients (principal quantum numbers), which would give an
answer of 15. However, the exponents represent the number of electrons in each subshell. In this
case there are 19 electrons.
2.
(b)
The orbital quantum numbers are represented by the subshell letters: s = 0 and p = 1. Since
only s and p subshells are present, only the orbital quantum numbers 0 and 1 are in the electron
configuration.
3.
(a)
In considering electrons, if only the particle property of electrons is considered, then the electron
is viewed as a point particle passing through the slit. The electron also has wave properties and
therefore undergoes diffraction as it passes through the slit.
4.
(c)
A common misconception is that all of the electrons have their lowest quantum numbers when in
the ground state. However, due to the Pauli exclusion principle, no two electrons can have the
same quantum numbers. The ground state of an atom is when each electron occupies the lowest
energy state available to it without violating the exclusion principle.
5.
(e)
The Pauli exclusion principle applies to all electrons within the same atom. No two electrons in
an atom can have the same set of four quantum numbers. Electrons from different atoms can
have the same set of quantum numbers (for example, two separate atoms in the same container or
two atoms from different parts of the Periodic Table).
6.
(c)
The Heisenberg uncertainty principle deals with our inability to know the position and
momentum of a particle precisely. The more precisely you can determine the position, the less
precisely you can determine the momentum, and vice versa.
7.
(a)
In an atom, multiple electrons will be in the same shell and can therefore have the same value
of n. Electrons in different shells can have the same value of m . Electrons in different atoms
can also have the same set of four quantum numbers. The Pauli exclusion principle does not
allow electrons within the same atom to have the same set of four quantum numbers.
8.
(e)
The Heisenberg uncertainty principle prohibits the exact measurement of position and velocity at
the same time; the product of their uncertainties will always be greater than a constant value.
9.
(c)
The uncertainty principle allows for a precise measurement of the position, provided that there is
a large uncertainty in the momentum. It also allows for a precise measurement of the momentum,
provided that the simultaneous measurement of the position has a large uncertainty. It does not
allow for the precise measurement of position and momentum at the same time.
10.
(d)
Laser light is monochromatic (all photons have nearly identical frequencies), is coherent (all
photons have the same phase), moves as a beam, and is created by a population inversion. A
laser beam usually is intense, but it does not have to be brighter than other sources of light.
Solutions to Problems
1.
We find the wavelength of the neutron from Eq. 27–8, using the classical relationship between
momentum and kinetic energy for the low energy neutrons. The peaks of the interference pattern are
given by Eq. 24–2a and Fig. 24–10. For small angles, we have sin θ = tan θ.
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28-6
Chapter 28
h
=
p
λ=
h
2mn KE
sin θ = tan θ
→
λ
h
Δx =
=
d
d sin θ = mλ ,
;
mλ x
=
d
d 2mn KE
=
m = 1, 2, … ;
→ x=
mλ
,
d
x = tan θ
m = 1, 2, … →
(6.63 × 10−34 J ⋅ s)(1.0 m)
(4.0 × 10−4 m) 2(1.67 × 10−27 kg)(0.025 eV)(1.60 × 10−19 J/eV)
= 4.5 × 10−7 m
2.
We find the wavelength of a pellet from Eq. 27–8. The half-angle for the central circle of the
1.22λ
diffraction pattern is given in Section 25–7 as sin θ =
, where D is the diameter of the opening.
D
Assuming the angle is small, the diameter of the spread of the bullet beam is d = 2 tan θ = 2 sin θ .
h
h
=
;
p mυ
λ=
=
d = 2 tan θ = 2 sin θ = 2
1.22λ
1.22h
=2
D
Dmυ
→
Dmυ d (3.0 × 10−3 m)(2.0 × 10−3 kg)(120 m/s)(0.010 m)
=
= 4.5 × 1027 m
2.44h
2.44(6.63 × 10−34 J ⋅ s)
This is more than 1011 light-years.
3.
The uncertainty in the velocity is given, so the uncertainty in the momentum can be calculated. Use
Eq. 28–1 to find the minimum uncertainty in the position.
Δx ≥
4.
Δp
=
m Δυ
kg)(1200 m/s)
= 5.3 × 10−11 m
Δx
→ Δυ ≥
m Δx
=
(1.055 × 10−34 J ⋅ s)
(9.11× 10
−31
kg)(2.4 × 10
−8
m)
= 4825 m/s ≈ 4.8 × 103 m/s
Δt
=
(1.055 × 10−34 J ⋅ s)
(1× 10
−8
s)
⎛
1 eV
= 1.055 × 10−26 J ⎜
⎜ 1.60 × 10−19
⎝
⎞
−8
−8
⎟ = 6.59 × 10 eV ≈ 7 × 10 eV
J ⎟⎠
We find the lifetime of the particle from Eq. 28–2.
Δt ≥
7.
(1.67 × 10
−27
The minimum uncertainty in the energy is found from Eq. 28–2.
ΔE ≥
6.
(1.055 × 10−34 J ⋅ s)
The uncertainty in position is given. Use Eq. 28–1 to find the uncertainty in the momentum.
Δp = mΔυ ≥
5.
=
ΔE
=
(1.055 × 10−34 J ⋅ s)
(2.5 GeV)(1.60 × 10−10 J/GeV)
= 2.6 × 10−25 s
We find the uncertainty in the energy of the muon from Eq. 28–2, and then find the uncertainty in the
mass.
ΔE ≥
Δm ≥
Δt
2
; ΔE = ( Δm ) c 2
c Δt
=
→
(1.055 × 10−34 J ⋅ s)
2
c (2.20 × 10
−6
J ⎞⎛
1 eV
⎛
= ⎜ 4.7955 × 10−29 2 ⎟ ⎜
⎜
c ⎠ ⎝ 1.60 × 10−19
s) ⎝
⎞
−10
eV/c 2
⎟ = 3.00 × 10
J ⎟⎠
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Quantum Mechanics of Atoms
8.
We find the uncertainty in the energy of the free neutron from Eq. 28–2, and then the mass uncertainty
from Eq. 26–7. We assume that the lifetime of the neutron is good to 2 significant figures. The current
experimental lifetime of the neutron is between 881 and 882 seconds.
ΔE ≥
9.
28-7
Δt
ΔE = (Δm)c 2
;
→ Δm ≥
2
c Δt
=
(1.055 × 10−34 J ⋅ s)
8
2
(3.00 × 10 m/s) (880 s)
= 1.3 × 10−54 kg
The uncertainty in the position is found from the uncertainty in the velocity and Eq. 28–1. The
uncertainty in velocity is found by multiplying the velocity by its precision.
Δxelectron ≥
Δxbaseball ≥
Δp
Δp
=
=
m Δυ
m Δυ
=
=
(1.055 × 10−34 J ⋅ s)
(9.11× 10
−31
−4
kg)(120 m/s)(6.5 × 10 )
(1.055 × 10−34 J ⋅ s)
(0.14 kg)(120 m/s)(6.5 × 10−4 )
= 1.485 ×10−3 m ≈ 1.5 × 10−3 m
= 9.661× 10−33 m ≈ 9.7 × 10−33 m
Δxelectron melectron Δυ mbaseball
(0.14 kg)
=
=
=
= 1.5 × 1029
Δxbaseball
melectron (9.11× 10−31 kg)
mbaseball Δυ
The uncertainty for the electron is greater by a factor of 1.5 × 1029.
10.
The uncertainty in the energy is found from the lifetime and the uncertainty principle.
ΔE =
Δt
=
(1.055 × 10−34 J ⋅ s) ⎛
1 eV
⎜⎜
−6
(12 × 10 s) ⎝ 1.60 × 10−19
⎞
−11
⎟ = 5.49 × 10 eV
J ⎟⎠
ΔE 5.49 ×10−11 eV
=
= 1.0 × 10−17
6
E
5.5 × 10 eV
11.
The uncertainty in position is given. Use Eq. 28–1 to find the uncertainty in the momentum.
Δp = mΔυ ≥
Δx
→ Δυ ≥
mΔx
=
(1.055 × 10−34 J ⋅ s)
(9.11× 10
−31
kg)(15 × 10
−9
m)
= 7720 m/s ≈ 7.7 × 103 m/s
The speed is low enough that the electron may be treated with classical mechanics instead of
relativistic mechanics. We calculate the minimum kinetic energy from the speed calculated above.
KE
⎛
1 eV
= 12 mυ 2 = 12 (9.11× 10−31 kg)(7720 m/s) 2 = 2.715 × 10−23 J ⎜
⎜ 1.602 × 10−19
⎝
⎞
−4
⎟ = 1.694 × 10 eV
J ⎟⎠
= 2.7 × 10−23 J, or 1.7 × 10−4 eV
12.
We use the radius as the uncertainty in position for the neutron. We find the uncertainty in the
momentum from Eq. 28–1. If we assume that the lowest value for the momentum is the least
uncertainty, then we can estimate the lowest possible kinetic energy (nonrelativistic) as the following:
2
2
⎛ 1.055 × 10−34 kg ⋅ m/s ⎞
⎛ ⎞
⎜⎜
⎟⎟
1.2 × 10−15 m
(Δp) 2 ⎜⎝ Δx ⎟⎠
⎝
⎠ = 2.314 × 10−12
E=
=
=
−
27
2m
2m
2(1.67 × 10
kg)
⎛ 1 MeV
J⎜
⎜ 1.60 × 10−13
⎝
⎞
⎟
J ⎟⎠
= 14.46 MeV ≈ 14 MeV
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28-8
13.
Chapter 28
The momentum of the electron is found from the energy by the classical relationship.
p = [ 2m( KE ) ] 2 = [2(9.11× 10−31 kg)(5.00 keV)(1.60 × 10−16 J/keV)] 2 = 3.817 × 10−23 kg ⋅ m/s
1
1
If the uncertainty in the energy is 1.00%, then the uncertainty in momentum can be found using the
binomial expansion. Then use Eq. 28–1 to find the uncertainty in position.
1
1
1
p′ = [2m(1.0100 KE )] 2 = [2m(KE )] 2 (1 + 0.0100) 2 ≈ p[1 + 12 (0.0100)]
Δp = p ′ − p = p
Δx ≥
14.
Δp
=
( 12 ) (0.0100) = (3.8127 ×10−23 kg ⋅ m/s) ( 12 ) (0.0100) = 1.909 ×10−25 kg ⋅ m/s
(1.055 × 10−34 J ⋅ s)
(1.909 × 10−25 kg ⋅ m/s)
= 5.53 × 10−10 m
Use the radius as the uncertainty in position for the electron. We find the uncertainty in the momentum
from Eq. 28–1, and then find the energy associated with that momentum from Eq. 26–9.
Δp ≥
Δx
=
(1.055 × 10−34 J ⋅ s)
(1.0 × 10−15 m)
= 1.055 × 10−19 kg ⋅ m/s
If we assume that the lowest value for the momentum is the least uncertainty, then we can estimate the
lowest possible energy.
E = ( p 2 c 2 + m 2 c 4 )1/2 = [(Δp) 2 c 2 + m 2 c 4 ]1/2
= [(1.055 × 10−19 kg ⋅ m/s) 2 (3.00 × 108 m/s) 2 + (9.11× 10−31 kg) 2 (3.00 × 108 m/s) 4 ]1/2
⎛ 1 MeV
= 3.175 × 10−11 J ⎜
⎜ 1.60 × 10−13
⎝
⎞
⎟ ≈ 200 MeV
J ⎟⎠
ranges from 0 to n − 1. Thus, for n = 6,
15.
The value of
16.
The value of m can range from − to + . Thus, for
= 0, 1, 2, 3, 4, 5 .
= 3, m = −3, − 2, − 1, 0, 1, 2, 3 .
The possible values of ms are − 12 , + 12 .
17.
The number of electrons in the subshell is determined by the value of . For each , the value of m
can range from − to + , which is 2 + 1 values. For each m value, there are two values of ms .
Thus the total number of states for a given is N = 2(2 + 1).
N = 2(2 + 1) = 2[2(3) + 1] = 14 electrons
18.
The value of ranges from 0 to n − 1. Thus for n = 4, = 0, 1, 2, 3. For each , the value of m can
range from − to + , or 2 + 1 values. For each m , there are two values of ms . Thus the number of
states for each is 2(2 + 1). The number of states is therefore given by the following:
N = 2(0 + 1) + 2(2 + 1) + 2(4 + 1) + 2(6 + 1) = 32 states
We start with
= 0, and list the quantum numbers in the order (n, , m , ms ).
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Quantum Mechanics of Atoms
28-9
(4, 0, 0, − 12 ), (4, 0, 0, + 12 ), (4, 1, –1, − 12 ), (4, 1, –1, + 12 ), (4, 1, 0, − 12 ), (4, 1, 0, + 12 ),
(4, 1, 1, − 12 ), (4, 1, 1, + 12 ), (4, 2, – 2, − 12 ), (4, 2, – 2, + 12 ), (4, 2, –1, − 12 ), (4, 2, –1, + 12 ),
(4, 2, 0, − 12 ), (4, 2, 0, + 12 ), (4, 2, 1, − 12 ), (4, 2, 1, + 12 ), (4, 2, 2, − 12 ), (4, 2, 2, + 12 ),
(4, 3, – 3, − 12 ), (4, 3, – 3, + 12 ), (4, 3, – 2, − 12 ), (4, 3, – 2, + 12 ), (4, 3, −1, − 12 ), (4, 3, – 1, + 12 ),
(4, 3, 0, − 12 ), (4, 3, 0, + 12 ), (4, 3, 1, − 12 ), (4, 3, 1, + 12 ), (4, 3, 2, − 12 ), (4, 3, 2, + 12 ),
(4, 3, 3, − 12 ), (4, 3, 3, + 12 )
19.
(a)
For carbon, Z = 6. We start with the n = 1 shell, and list the quantum numbers in the order
(n, , m , ms ).
(1, 0, 0, − 12 ), (1, 0, 0, + 12 ), (2, 0, 0, − 12 ), (2, 0, 0, + 12 ), (2, 1, − 1, − 12 ) , (2, 1, − 1,1 12 )
Note that, without additional information, there are other possibilities for the last two electrons.
The third quantum number for the last two electrons could be –1, 0, or 1.
(b)
For aluminum, Z = 13. We start with the n = 1 shell, and list the quantum numbers in the order
(n, , m , ms ).
(1, 0, 0, − 12 ), (1, 0, 0, + 12 ), (2, 0, 0, − 12 ), (2, 0, 0, + 12 ), (2, 1, −1, − 12 ), (2, 1, − 1, + 12 ),
(2, 1, 0, − 12 ), (2, 1, 0, + 12 ), (2, 1, 1, − 12 ), (2, 1, 1, + 12 ), (3, 0, 0, − 12 ), (3, 0, 0, + 12 ), (3, 1, − 1, − 12 )
Note that, without additional information, there are other possibilities for the last electron. The
third quantum number could be –2, –1, 0, 1, or 2, and the fourth quantum number could be either
+ 12 or − 12 .
20.
For oxygen, Z = 8. We start with the n = 1 shell, and list the quantum numbers in the order
(n, , m , ms ).
(1, 0, 0, − 12 ), (1, 0, 0, + 12 ), (2, 0, 0, − 12 ), (2, 0, 0, + 12 ),
(2, 1, − 1, − 12 ), (2, 1, −1, + 12 ), (2, 1, 0, − 12 ), (2, 1, 0, + 12 )
Note that, without additional information, there are two other possibilities that could substitute for any
of the last four electrons. The third quantum number could be changed to 1.
21.
The magnitude of the angular momentum depends only on .
L=
22.
( + 1)
=
12
= 12(1.055 × 10−34 J ⋅ s) = 3.65 × 10−34 J ⋅ s
The value of m can range from − to + , so we have m = −4 , − 3, − 2, −1, 0, 1, 2, 3, 4 .
The value of
can range from 0 to n − 1. Thus we have n ≥ 5 .
There are two values of ms : ms = − 12 , + 12 .
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28-10
23.
Chapter 28
The value of
can range from 0 to n − 1. Thus we have n ≥ 4 .
The value of m can range from − to + , so
≥ 3 , and
≤ n − 1. So 3 ≤ ≤ n − 1.
There are two values of ms : ms = − 12 , + 12 .
24.
The “g” subshell has = 4. The value of m can range from − to + , so there are 9 values of m :
m = − 4, − 3, − 2, −1, 0, 1, 2, 3, 4 . Each of the 9 m values represents two states, so the “g” subshell
can hold 18 electrons .
25.
The configurations can be written with the subshells in order by energy (4s before 3d, for example) or
with the subshells in principal number order (3d before 4s, for example, as in Table 28–4). Both are
shown here.
(a) Z = 26 is iron.
1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 6 4 s 2 or 1s 2 2 s 2 2 p 6 3s 2 3 p 6 4s 2 3d 6
(b)
Z = 34 is selenium.
1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 4 or 1s 2 2 s 2 2 p 6 3s 2 3 p 6 4s 2 3d 10 4 p 4
(c)
Z = 38 is strontium.
1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 6 5s 2 or 1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p 6 5s 2
26.
The configurations can be written with the subshells in order by energy (4s before 3d, for example) or
with the subshells in principal number order (3d before 4s, for example, as in Table 28–4). Both are
shown here.
(a) Silver has Z = 47.
1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 6 4d 10 5s1 or 1s 2 2s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p 6 5s1 4d 10
(b)
Gold has Z = 79.
1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 6 4d 10 4 f 14 5s 2 5 p 6 5d 10 6 s1 or
1s 2 2s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p 6 5s 2 4d 10 5 p 6 4 f 14 6s1 5d 10
(c)
Uranium has Z = 92.
1s 2 2s 2 2 p 6 3s 2 3 p 6 3d 10 4 s 2 4 p 6 4d 10 4 f 14 5s 2 5 p 6 5d 10 6s 2 6 p 6 5 f 3 6d 1 7 s 2 or
1s 2 2s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p 6 5s 2 4d 10 5 p 6 4 f 14 6s 2 5d 10 6 p 6 7 s 2 5 f 3 6d 1
27.
(a)
The principal quantum number is n = 5 .
(b)
The energy of the state is found from Eq. 27–15b.
E7 = −
(13.6 eV)
n
2
=−
(13.6 eV)
52
= −0.544 eV
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Quantum Mechanics of Atoms
(c)
The d subshell has
L=
(d)
= 2 . The magnitude of the angular momentum depends on
( + 1) =
only.
6
the value of m can range from −
For each
28-11
to + .
m = −2, − 1, 0, 1, 2
28.
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. Use the energy of the
hydrogen atom, Eq. 27–15b.
E2 = −
(13.6 eV)
n
2
=−
(13.6 eV)
22
= −3.4 eV
We predict the binding energy to be 3.4 eV. Our assumption of complete shielding of the nucleus by
the 2s electrons is evidently not correct. The partial shielding means the net charge of the “nucleus” is
higher than +1e, so it holds the outer electron more tightly, requiring more energy to remove it.
29.
In a filled subshell, there are an even number of electrons. Those electrons have m values of
− , − + ,… , − 1, 0,1,… , −1, . For each of those electrons, Lz = m . When all of those Lz values
are calculated and summed, the total is − + (− + 1) + + ( − 1) + = 0. Half of the electrons
have spin angular momentum of “up,” and the other half have spin angular momentum of “down.” The
sum of those spin angular momenta is also 0. Thus the total angular momentum is 0.
30.
(a)
The 4 p → 3 p transition is forbidden, because Δ = 0 ≠ ±1.
(b)
The 3 p → 1s transition is allowed, because Δ = −1.
(c)
The 4d
→ 2d transition is forbidden, because Δ = 0 ≠ ±1.
(d)
The 5d
→ 3s transition is forbidden, because Δ = −2 ≠ ±1.
(e)
The 4s → 2 p transition is allowed, because Δ = +1.
31.
Since the electron is in its lowest energy state, we must have the lowest possible value of n. Since
m = 2, the smallest possible value of is = 2 , and the smallest possible value of n is n = 3 .
32.
Photon emission means a jump to a lower state, so n = 1, 2, 3, 4, or 5. For a d subshell,
because Δ = ±1, the new value of must be 1 or 3.
(a)
= 2, and
= 1 corresponds to a p subshell, and = 3 corresponds to an f subshell. Keeping in mind that
0 ≤ ≤ n − 1, we find the following possible destination states: 2 p, 3 p, 4 p, 5 p, 4 f , 5 f . In the
format requested, these states are: (2, 1), (3, 1), (4, 1), (5, 1), (4, 3), (5, 3) .
(b)
In a hydrogen atom, has no appreciable effect on energy, so for energy purposes there are four
possible destination states, corresponding to n = 2, 3, 4, and 5. Thus, there are four different
photon wavelengths corresponding to four possible changes in energy.
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28-12
33.
Chapter 28
The smallest wavelength X-ray has the most energy, which is the maximum KE of the electron in the
tube. A 28.5-kV potential difference means that the electrons have 28.5 keV of KE.
λ=
hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
= 4.362 × 10−11 m = 0.0436 nm
E (1.60 × 10−19 J/eV)(28.5 × 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.
34.
The shortest wavelength X-ray has the most energy, which is the maximum kinetic energy of the
electron in the tube:
E=
hc
λ
=
(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
(1.60 × 10−19 J/eV)(0.035 × 10−9 m)
= 3.552 ×104 eV ≈ 36 keV
Thus the operating voltage of the tube is 36 kV .
35.
The energy of the photon with the shortest wavelength must equal the maximum kinetic energy of an
electron. We assume that V is in volts.
E = hf 0 =
λ0 =
36.
hc
λ0
= eV
→
hc (6.63 ×10−34 J ⋅ s)(3.00 ×108 m/s)(109 nm/m) 1243 nm 1240 nm
=
=
≈
eV
V
V
(1.60 × 10−19 C)(V V)
The maximum energy is emitted when the shortest wavelength is emitted. That appears to be about
0.025 nm. The minimum energy is emitted when the longest wavelength is emitted. That appears to be
about 0.2 nm. Use Eq. 27–4.
Emax = hf max =
hc
λmin
=
(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
(0.025 × 10
−9
(1 eV)
(1.60 × 10−19 J)
m)
= 49, 725 eV
≈ 5.0 × 104 eV
Emin = hf min =
37.
hc
λmax
=
(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
(0.2 × 10
m)
(1 eV)
(1.60 × 10−19 J)
= 6215 eV ≈ 6 × 103 eV
We follow Example 28–7, using the Bohr formula, Eq. 27–16, with Z replaced by Z – 1.
⎛ 2π 2 e 4 mk 2 ⎞
1
2⎛ 1
=⎜
⎟⎟ ( Z − 1) ⎜ 2 − 2
3
⎜
λ ⎝
h c
n
⎝ n′
⎠
1
= 5.562 × 109 m −1
38.
−9
→ λ=
1 ⎞
⎞
−1
7
2⎛ 1
⎟ = (1.097 × 10 m )(27 − 1) ⎜ 2 − 2 ⎟
⎝1 2 ⎠
⎠
1
5.562 × 109 m −1
= 1.798 × 10−10 m
We follow Example 28–7, using the Bohr formula, Eq. 27–16, with Z replaced by Z – 1.
⎛ 2π 2 e 4 mk 2 ⎞
1
2⎛ 1
=⎜
⎟⎟ ( Z − 1) ⎜ 2 − 2
3
⎜
λ ⎝
h c
n
⎝ n′
⎠
1
= 9.523 × 109 m −1 → λ =
1 ⎞
⎞
−1
7
2⎛ 1
⎟ = (1.097 × 10 m )(26 − 1) ⎜ 2 − 2 ⎟
3 ⎠
⎠
⎝2
1
9.523 × 109 m −1
= 1.050 × 10−9 m
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Quantum Mechanics of Atoms
39.
28-13
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. 27–10 and 27–15b.
1 ⎞
⎛ 1
hf = ΔE = (13.6 eV)( Z − 1) 2 ⎜ 2 − 2 ⎟
n ⎠
⎝ n′
hc
=
ΔE
λ=
hc
1 ⎞
⎛ 1
(13.6 eV)( Z − 1) 2 ⎜ 2 − 2 ⎟
′
n ⎠
⎝n
=
(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
⎛1 1 ⎞
(13.6 eV)(412 ) ⎜ 2 − 2 ⎟ (1.60 ×10−19 J/ev)
⎝1 3 ⎠
= 6.12 × 10−11 m = 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 with 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.
40.
The wavelength of the Kα line is calculated for molybdenum in Example 28–7. We use that same
procedure. Note that the wavelength is inversely proportional to ( Z − 1) 2 .
λunknown
( Z Fe − 1) 2
=
λFe
( Z unknown − 1) 2
⎡
194 pm ⎤
→ Z unknown = ⎢(26 − 1)
⎥ + 1 = 24
229 pm ⎦
⎣
The unknown material has Z = 24, so it is chromium.
41.
The energy of a pulse is the power of the pulse times the duration in time.
E = P Δt = (0.68 W)(25 ×10−3 s) = 0.017 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. 27–4.
E Eλ
(0.017 J)(640 × 10−9 m)
=
=
= 5.5 × 1016 photons
hf
hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
N=
42.
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 lightbulb is the surface area of a sphere.
(a)
I=
(b)
I=
P
πr
2
=
P
4π r
2
0.50 × 10−3 W
π (1.5 × 10
=
−3
m)
100 W
4π (2.0 m) 2
2
= 70.74 W/m 2 ≈ 71 W/m 2
= 1.989 W/m 2 ≈ 2.0 W/m 2
The laser beam is more intense by a factor of
43.
70.74 W/m 2
1.989 W/m 2
= 35.57 ≈ 36 .
Transition from the E3 ′ state to the E2 ′ state releases photons with energy 1.96 eV, as shown in
Fig. 28–20. The wavelength is determined from the energy.
λ=
hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
= 6.34 × 10−7 m = 634 nm
−
19
ΔE
(1.60 × 10
J/eV)(1.96 eV)
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28-14
44.
Chapter 28
The angular half-width of the beam is given in
r
1.22λ
, where d is the diameter
Eq. 25–7, θ1/2 =
D
θ
d
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 = rθ , for small angles. See the diagram.
θ1/2 =
(a)
1.22 λ 1.22(694 × 10−9 m)
=
= 2.822 × 10−4 rad → θ = 5.644 × 10−4 rad ≈ 5.6 × 10−4 rad
d
(3.0 × 10−3 m)
The diameter of the beam when it reaches the satellite is as follows:
D = rθ = (340 × 103 m)(5.644 × 10−4 rad) = 190 m
(b)
The diameter of the beam when it reaches the Moon is as follows:
D = rθ = (384 × 106 m)(5.644 × 10−4 rad) = 2.2 × 105 m
45.
We use Eq. 28–3 to find l and the unnumbered equation following it to find m .
L=
2
( + 1)
46.
(a)
→ m =
=
Lz
L2
2
=
(6.84 × 10−34 J ⋅ s) 2
(1.055 × 10−34 J ⋅ s) 2
= 42
−1 ± 1 − 4(1)(−42) −1 ± 13
=
= 6, −7 →
2
2
=
2.11× 10−34 J ⋅ s
1.055 × 10−34 J ⋅ s
=6
= 2
= 6, we must have n ≥ 7 .
The minimum uncertainty in the energy is found from Eq. 28–2.
ΔE ≥
(b)
( + 1) =
+ − 42 = 0 →
Lz = m
Since
→
Δt
=
(1.055 × 10−34 J ⋅ s)
(1× 10
−8
s)
⎛
1 eV
= 1.055 ×10 −26 J ⎜
⎜ 1.60 × 10−19
⎝
⎞
−8
−7
⎟⎟ = 6.59 ×10 eV ≈ 10 eV
J⎠
The transition energy can be found from the Bohr hydrogen atom energies, Eq. 27–15b, with
Z = 1 for hydrogen.
En = −(13.6 eV)
Z2
n2
⎡
12 ⎤ ⎡
12 ⎤
→ E2 − E1 = ⎢ −(13.6 eV) 2 ⎥ − ⎢ −(13.6 eV) 2 ⎥ = 10.2 eV
2 ⎦⎥ ⎣⎢
1 ⎦⎥
⎣⎢
6.59 ×10−8 eV
ΔE
=
= 6.46 × 10−9 ≈ 10−8
E2 − E1
10.2 eV
The uncertainty in the energy is a very small fraction of the actual transition energy.
(c)
The wavelength is given by Eq. 27–4.
E = hf =
hc
λ
→ λ=
hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
= 1.22 × 10−7 m = 122 nm
E
⎛ 1.60 × 10−19 J ⎞
(10.2 eV) ⎜
⎟⎟
⎜
eV
⎝
⎠
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Quantum Mechanics of Atoms
28-15
Take the width in the line to be the uncertainty in the line’s value, based on the uncertainty in the
energy. We will use Eq. 27–4 with the binomial expansion.
47.
hc
hc ⎛ ΔE ⎞
= ⎜1 +
⎟
Δ
E
E ⎠
⎛
⎞ E⎝
E ⎜1 +
⎟
E ⎠
⎝
−1
hc ⎛ ΔE ⎞
⎛ ΔE ⎞
⎜1 −
⎟ = λ ⎜1 −
⎟ →
E⎝
E ⎠
E ⎠
⎝
λ + Δλ =
hc
=
( E + ΔE )
| Δλ | = λ
ΔE
6.59 × 10−8 eV
= (122 nm)
= 8 × 10−7 nm
10.2 eV
E
≈
can range from 0 to n − 1. Thus for n = 6, we have 0 ≤ ≤ 5. The magnitude of the
The value of
angular momentum is given by Eq. 28–3, L =
( + 1) .
Lmin = 0; Lmax = 30
48.
(a)
We find the wavelength from Eq. 27–8.
λ=
(b)
h
h
(6.63 × 10−34 J ⋅ s)
=
=
= 3.683 × 10−34 m ≈ 3.7 × 10−34 m
p mυ (0.012 kg)(150 m/s)
Use Eq. 28–1 to find the uncertainty in momentum.
Δpx ≥
49.
(1.055 × 10−34 J ⋅ s)
= 1.758 × 10−32 kg ⋅ m/s ≈ 1.8 × 10−32 kg ⋅ m/s
(0.0060 m)
Use Eq. 28–1 (the uncertainty principle) and Eq. 27–8 (the de Broglie wavelength).
Δp ≥
λ=
50.
Δx
=
(a)
Δx
=
(1.055 × 10−34 J ⋅ s)
(2.0 × 10−8 m)
= 5.275 × 10−27 kg ⋅ m/s ≈ 5.3 × 10−27 kg ⋅ m/s
h
(6.63 × 10−34 J ⋅ s)
=
= 1.257 ×10−7 m ≈ 1.3 × 10−7 m
p 5.275 × 10−27 kg ⋅ m/s
Boron has Z = 4, with an electron configuration of 1s 2 2 s 2 , so the outermost electron has
n = 2. We use the Bohr result with an effective Z. We might naively expect to get Z eff = 1,
indicating that the other three electrons shield the outer electron from the nucleus, or Z eff = 2,
indicating that only the inner two electrons accomplish the shielding.
E2 = −
(13.6 eV)( Z eff ) 2
n2
→ − 8.26 eV = −
(13.6 eV)( Z eff ) 2
22
→ Z eff = 1.56
This indicates that the other electron in the n = 2 shell does partially shield the electron that is to
be removed.
(b)
We find the average radius from Eq. 27–14.
r=
n 2 r1 22 (0.529 × 10−10 m)
=
= 1.36 × 10−10 m
Z eff
(1.56)
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28-16
51.
Chapter 28
Use Eq. 27–14, which says that the radius of a Bohr orbit is inversely proportional to the atomic
number. Also use Eq. 27–15b, which says that the energy of a Bohr orbit is proportional to the square
of the atomic number. The energy to remove the electron is the opposite of the total energy. A simple
approximation is used in which the electron “feels” the full effect of all 92 protons in the nucleus.
rn =
n2
1
(0.529 × 10−10 m) = (0.529 × 10−10 m) = 5.75 × 10−13 m
Z
92
| En | = (13.6 eV)
52.
Z2
n
= (13.6 eV)
2
922
2
1
= 1.15 ×105 eV = 115 keV
We find the wavelength of the protons from their kinetic energy, using nonrelativistic mechanics and
Eq. 27–8, and then use Eq. 24–21 for two-slit interference, with a small-angle approximation. If the
protons were accelerated by a 480-V potential difference, then they will have 480 eV of kinetic energy.
λ=
h
=
p
sin θ = tan θ
Δx =
λ
d
=
h
2mp KE
→
; d sin θ = mλ ,
mλ x
=
d
h
d 2mp KE
=
→ x=
m = 1, 2, … ;
mλ
,
d
x = tan θ
m = 1, 2, … →
(6.63 × 10−34 J ⋅ s)(28 m)
(7.0 × 10−4 m) 2(1.67 × 10−27 kg)(480 eV)(1.60 × 10−19 J/ev)
= 5.2 × 10−8 m
53.
An h subshell has
= 5. For a given
value, m ranges from − to + , taking on 2 + 1 different
values. For each m , there are two values of ms . Thus the number of states for a given
value is
2(2 + 1). Thus there are 2(2 + 1) = 2(11) = 22 possible electron states.
54.
(a)
For each , the value of m can range from − to + , or 2 + 1 values. For each of these, there
are two values of ms . Thus the total number of states in a subshell is N = 2(2 + 1) .
(b)
55.
= 0, 1, 2, 3, 4, 5, and 6, N = 2, 6, 10, 14, 18, 22, and 26 , respectively.
For
For a given n, 0 ≤ ≤ n − 1. Since for each , the number of possible states is 2(2 + 1), as shown in
Problem 54, the number of possible states for a given n is as follows:
n −1
Ä 2(2
=0
n −1
n −1
⎛ n(n − 1) ⎞
2
+ 1) = 4 Ä + Ä 2 = 4 ⎜
⎟ + 2n = 2n
2
⎝
⎠
=0
=0
The summation formulas can be found (and proven) in a pre-calculus mathematics textbook.
56.
We find the wavelength of the electrons from their kinetic energy, using nonrelativistic mechanics.
Classical formulas are justified because the kinetic energy of 45 keV = 0.045 MeV is much less than
the rest energy of 0.511 MeV. Eq. 27–8 is used to find the wavelength, and Eq. 24–21 is used for the
two-slit interference, with a small-angle approximation.
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Quantum Mechanics of Atoms
λ=
h
=
p
h
2me KE
sin θ = tan θ
→
λ
h
Δx =
=
d
; d sin θ = mλ ,
mλ x
=
d
d 2me KE
=
28-17
m = 1, 2, … ; x = tan θ
→ x=
mλ
,
d
m = 1, 2, … →
(6.63 × 10−34 J ⋅ s)(0.450 m)
(2.0 × 10−6 m) 2(9.11× 10−31 kg)(45 × 103 eV)(1.60 × 10−19 J/ev)
= 1.30 × 10−6 m ≈ 1.3 μ m
57.
In the Bohr model, LBohr = n
or
58.
h
= 2 . In quantum mechanics, Lqm =
2π
= 1, so that Lqm = 0(0 + 1) = 0 or Lqm = 1(1 + 1) =
2
( + 1) . For n = 2,
=0
.
The fractional uncertainty in the energy is found from the lifetime and the uncertainty principle,
Eq. 28–2. Also use Eq. 27–4.
ΔE =
Δt
=
h
hc
; E = hf =
λ
2πΔt
h
λ
ΔE 2πΔt
500 × 10−9 m
=
=
=
= 2.65 × 10−8 ≈ 3 × 10−8
hc
E
2π cΔt 2π (3.00 × 108 m/s)(10 × 10−9 s)
λ
Use the binomial expansion to find the wavelength’s fractional uncertainty.
59.
hc
=
( E + ΔE )
| Δλ | = λ
ΔE
E
−1
≈
hc ⎛ ΔE ⎞
⎛ ΔE ⎞
⎜1 −
⎟ = λ ⎜1 −
⎟ →
E⎝
E ⎠
E ⎠
⎝
Find the uncertainty in the position from Eq. 28–1.
Δx =
60.
hc
hc ⎛ ΔE ⎞
= ⎜1 +
⎟
E ⎠
⎛ ΔE ⎞ E ⎝
E ⎜1 +
⎟
E ⎠
⎝
Δλ
ΔE
→ |
= 3 × 10−8
|=
λ
E
λ + Δλ =
mΔυ
=
1.055 × 10−34 J ⋅ s
= 3.7 × 10−37 m
(1300 kg)(0.22 m/s)
The difference in measured energies is found from Eq. 27–4.
ΔE = E2 − E1 =
hc
λ2
−
hc
λ1
⎛
⎞
1
1
= (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s) ⎜
−
= 1.670 × 10−21 J
⎜ 487 × 10−9 m 489 × 10−9 m ⎟⎟
⎝
⎠
The lifetime of the excited state is determined from Eq. 28–2.
ΔE Δt ≥
→ Δt ≥
ΔE
=
1.055 × 10−34 J ⋅ s
1.670 × 10−21 J
= 6.317 × 10−14 s ≈ 6.32 × 10−14 s
This solution assumes that the entire spread in energy is the uncertainty.
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28-18
61.
Chapter 28
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.
→ eV = 12 mυ 2
PE initial = KE final
ΔxΔp =
h
2π
Δpproton
Δpelectron
→ Δp =
→ υ=
h
h
2eV
; λ= =
=
m
p mυ
h
2eV
m
m
=
h
2meV
= Δx
2π h
Δx
2π h
Δxproton
Δx
=
= electron =
2π h
Δxproton
Δxelectron
h
2melectron eV
=
h
2mproton eV
mproton
melectron
=
1.67 × 10−27 kg
9.11× 10−31 kg
= 42.8
62.
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, there would be 86 elements.
63.
The de Broglie wavelength is determined by Eq. 27–8.
λ=
h
h
=
p mυ
→ υ=
h
6.63 × 10−34 J ⋅ s
=
= 1.95 × 10−35 m/s ≈ 2.0 × 10−35 m/s
mλ (68.0 kg)(0.50 m)
Since λ is comparable to the width of a typical doorway, yes, you would notice diffraction effects.
However, assuming that walking through the doorway requires travel through a distance of 0.2 m, the
time Δt required is huge.
υΔt = d → Δt =
64.
d
υ
=
0.20 m
1.95 × 10
−35
m/s
= 1.0 × 1034 s ≈ 3 × 1026 yr
1 ⎞
⎛ 1
∝ Z 2 ⎜ 2 − 2 ⎟ , we seek Z and n′X so that for every fourth value of nX
n ⎠
⎝ n′
there is an nH such that the Lyman formula equals the formula for the unknown element.
Since from Eq. 27–16,
1
λ
⎛ 1
1 1
1 ⎞ Z2 Z2
− 2 = Z2 ⎜ 2 − 2 ⎟ = 2 − 2
⎜ n′
⎟
1 nH
⎝ X nX ⎠ n′X nX
Consider atoms (with Z > 1) for which Z 2 /n′2X = 1 → n′X = Z , and for which
nX2 = Z 2 nH2 → nX = ZnH for every fourth line. This last condition is obviously met when Z = 4
and n′X = 4. The resulting spectral lines for nX = 4, 8, 12, … match the hydrogen Lyman series lines
for n = 1, 2, 3, … . The element is beryllium.
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Quantum Mechanics of Atoms
65.
28-19
The K α line is from the n = 2 to n = 1 transition. We use the energies of the hydrogen atom with Z
replaced by Z − 1, as in Example 28–7, using Eq. 27–16.
⎛ 2π 2 e 4 mk 2 ⎞
1 ⎞
2⎛ 1
=⎜
⎟⎟ ( Z − 1) ⎜ 2 − 2 ⎟ →
3
⎜
λ ⎝
h c
n ⎠
⎝ n′
⎠
1
−1
1 ⎛ 2π 2 e 4 mk 2 ⎞ ⎛ 1
1 ⎞
Z = 1+
⎜⎜
⎟⎟ ⎜ 2 − 2 ⎟
3
λ⎝
h c
n ⎠
⎠ ⎝ n′
= 1 + 28.09 ≈ 29
−1
= 1+
1
(0.154 × 10−9
⎛1 1 ⎞
(1.097 × 107 m −1 ) −1 ⎜ 2 − 2 ⎟
⎝1 2 ⎠
m)
−1
The element is copper.
Solutions to Search and Learn Problems
1.
We use Eq. 27–11, Bohr’s quantum condition, with n = 1 for the ground state.
mυ rn = n
ΔxΔp ≈
h
2π
→ mυ r1 =
→ Δx ≈
Δp
=
→ mυ = p =
/r1
r1
= Δp
= r1
The uncertainty in position is comparable to the Bohr radius.
2.
The periodicity of the Periodic Table depends on the number and arrangement of the electrons in the
atom. The Pauli exclusion principle requires that each electron have distinct values for the principal
quantum number, angular orbital quantum number, magnetic quantum number, and spin quantum
number. As electrons are added to the atom, the electrons fill the lowest available energy states. Atoms
with the same number of electrons in the outer shells have similar properties, which accounts for the
periodicity of the Periodic Table. The quantization of the angular momentum determines the number of
angular momentum states allowed in each shell. These are represented by the letters s, p, d, f, etc. The
magnitude of the angular momentum also limits the number of magnetic quantum states available for
each orbital angular state. Finally, the electron spin allows for two electrons (one spin up and one spin
down) to exist in each magnetic quantum state. So the Pauli exclusion principle, quantization of
angular momentum, direction of angular momentum, and spin all play a role in determining the
periodicity of the Periodic Table.
3.
(a)
(b)
4.
(a)
The quantum-mechanical theory retained the aspect that electrons only exist in discrete states of
definite energy levels. It also retained the feature that an electron absorbs or emits a photon when
it changes from one energy state to another.
In the Bohr model the electrons moved in circular orbits around the nucleus. In the quantummechanical model the electrons occupy orbitals about the nucleus and do not have exact, welldefined orbits.
We treat the Earth as a particle in rotation about the Sun. The angular momentum of a particle is
given in Eq. 27–11 as L = mυ rn , where r is the orbit radius. We equate this to the quantummechanical expression in Eq. 28–3. We anticipate that the quantum number will be very large, so
( + 1) as .
we approximate
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28-20
Chapter 28
L = M Earthυ rSun- = M
Earth
=
(b)
1
2π r
r = [ ( + 1)] 2 =
T
→
M Earth 2π r 2 (5.98 × 1024 kg)2π (1.496 × 1011 m) 2
=
= 2.5255 × 1074 ≈ 2.53 × 1074
T
(1.055 × 10−34 J ⋅ s)(3.156 × 107 s)
There are 2 + 1 values of m for a value of , so the number of orientations is as follows:
N = 2 + 1 = 2(2.5255 × 1074 ) + 1 = 5.051× 1074 ≈ 5.05 × 1074
5.
The “location” of the beam is uncertain in the transverse (y) 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, so we obtain a value for the transverse
momentum.
Δp y Δy ≥
→ Δp y ≥
Δy
=
D
→
py ≈
D
The momentum in the forward direction is related to the
h
wavelength of the light by px = . See the diagram to
λ
D
θ
px
py
relate the momentum to the angle.
θ≈
6.
py
px
≈
/D
λ
λ
λ
; “spread” = 2θ =
=
≈
h /λ 2π D
πD D
For noble gases, the electrons fill all available states in the outermost shell. Helium has two electrons
which fill the 1s shell. For the other noble gases, there are six electrons in the p subshells, filling each of
those shells. For each of these gases, the next electron would have to be added to an s subshell at much
higher energy. Since the noble gases have a filled shell, they do not react chemically with other elements.
Lithium, which has three electrons and is the first alkali metal, has its third electron in the 2s shell.
This shell has significantly higher energy than the 1s shell. All of the alkali metals have one electron in
their outer s subshell and therefore react well with halogens.
Halogens all have five electrons in their outer p subshell. They react well with atoms that have a donor
electron in their outer shell, like the alkali metals.
7.
The uncertainty in the electron position is Δx = r1. The minimum uncertainty in the velocity, Δυ , can
be found by solving Eq. 28–1, where the uncertainty in the momentum is the product of the electron
mass and the uncertainty in the velocity.
ΔxΔpmin = r1 (mΔυ ) ≥
Δυ ≥
8.
mr1
=
1.055 × 10−34 J ⋅ s
(9.11× 10−31 kg)(0.529 × 10−10 m)
= 2.19 × 106 m/s
According to the uncertainty principle, it is impossible to measure a system without disturbing, or
affecting, the system. The more precisely you measure the position, the greater the disturbance in the
momentum, and hence the greater the uncertainty in the momentum. The more precisely you measure
the momentum, the greater the disturbance in the position, and hence the greater the uncertainty in the
position. When measuring position and momentum, there is a minimum value for the product of the
uncertainties in momentum and position.
© Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.