atom, (b) the volume- of the nucleus, and' (c) the percentage of the ...

atom, (b) the volume- of the nucleus, and' (c) the percentage of the
volume of the atom that is occupied by the nucleus.
2. In a Rutherford scattering experiment a target nucleus has a diameter
of lA X 10-14 m. The incoming Cl particle has a mass of
6.64 X 10-27 kg. What is the kinetic energy of an Cl particle that has
a de Broglie wavelength equal to the diameter of the target nucleus?
Ignore relativistic effects.
3. The nucleus of a hydrogen atom is a single proton, which has a
radius of about 1.0 X IO-IY m. The single electron in a hydrogen
atom normally orbits the nucleus at a distance of 5.3 X 10-11 m.
What is the ratio of the density of the hydrogen nucleus to the density
of the complete hydrogen atom?
4. Review Conceptual Example I and use the information therein as
an aid in working this problem. Suppose you're building a scale
model of the hydrogen atom, and the nucleus is represented by a ball
of radius 3.2 cm (somewhat smaller than a baseball). How many
miles away (I mi = 1.61 X 105 cm) should the electron be placed?
* 5. ssm The nucleus of a copper atom contains 29 protons and has a
radius of 4.8 X 10-15 m. How much work (in electron volts) is done
by the electric force as a proton is brought from infinity, where it is
at rest, to the "surface" of a copper nucleus?
* 6. There are Z protons in the nucleus of an atom, where Z is the
atomic number of the element. An Cl particle carries a charge of
+2e. In a scattering experiment, an Cl particle, heading directly toward
a nucleus in a metal foil, will come to a halt when all the particle's
kinetic energy is converted to electric potential energy. In such
a situation, how close will an Cl particle with a kinetic energy of
5.0 X 10-13 J come to a gold nucleus (Z = 79)?
Section 30.2 Line Spectra
Section 30.3 The Bohr Model of the Hydrogen Atom
7. ssm www Concept Simulation 30.1 at www.wiley.comlcollege/
cutnell reviews the concepts on which the solution to this problem
depends. The electron in a hydrogen atom is in the first excited state,
when the electron acquires an additional 2.86 eV of energy. What is
the quantum number n of the state'into which the electron moves?
8. Using the Bohr model, determine the ratio of the energy of the nth
orbit of a triply ionized beryllium atom (BeH, Z = 4) to the energy
of the nth orbit of a hydrogen atom (H).
9. A singly ionized helium atom (He+) has only one electron in orbit
about the nucleus. What is the radius of the ion when it is in the
~ond excited state?
(!!J: (a) What is the minimum energy (in electron volts) that is required
to remove the electron from the ground state of a singly ionized
helium atom (He+, Z = 2)? (b) What is the ionization energy
for He+?
11. ssm Find the energy (in joules) of the photon that is emitted
when the electron in a hydrogen atom undergoes a transition from
the n = 7 energy level to produce a line in the Paschen series .
.-@ A hydrogen atom is in the ground state. It absorbs energy and
makes a transition to the n = 3 excited state. The atom returns to the
ground state by emitting two photons. What are their wavelengths?
13. Consider the Bohr energy expression (Equation 30.13) as it applies
to singly ionized helium He+ (Z = 2) and doubly ionized
lithium Li2+ (Z = 3). This expression predicts equal electron ener­
gies for these two species for certain values of the quantum number
n (the quantum number is different for each species). For quantum
numbers less than or equal to 9, what are the lowest three energies
(in electron volts) for which the helium energy level is equal to the
lithium energy level?
PROBLEMS I 971
rilf.)In the hydrogen atom the radius of orbit B is sixteen times greater
~ the radius of orbit A. The total energy of the electron in orbit A is
- 3AO eV. What is the total energy of the electron in orbit B?
* 15. ssm A wavelength of 410.2 nm is emitted by the hydrogen
atoms in a high-voltage discharge tube. What are the initial and final
values of the quantum number n for the energy level transition that
produces this wavelength?
*@ The energy of the n = 2 Bohr orbit is - 30.6 eV for an unidentified
ionized atom in which only one electron moves about the nucleus.
What is the radius of the n = 5 orbit for this species?
* 17. ssm For atomic hydrogen, the Paschen series of lines occurs
when nf = 3, whereas the Brackett series occurs when nf = 4 in
Equation 30.14. Using this equation, show that the ranges of wavelengths
in these two series overlap.
*@ Interactive LearningWare 30.1 at www.wiley.comlcolIege/cutnell
reviews the concepts that play roles in this problem. A hydrogen
atom emits a photon that has momentum with a magnitude of
5A52 X 10-27 kg' m/so This photon is emitted because the electron
in the atom falls from a higher energy level into the n = I leveL
What is the quantum number of the level from which the electron
falls? Use a value of 6.626 X 10-34 J. s for Planck's constant.
** 19. A diffraction grating is used in the first order to separate the
wavelengths in the Balmer series of atomic hydrogen. (Section 27.7
discusses diffraction gratings.) The grating and an observation
screen (see Figure 27.32) are separated by a distance of 81.0 cm.
You may assume that ()is small, so sin ()= ()when radian measure
is used for (). How many lines per centimeter should the grating
have so that the longest and the next-to-the-Iongest wavelengths in
the series will be separated by 3.00 cm on the screen?
** 20. A certain species of ionized atoms produces an emission line
spectrum according to the Bohr model, but the number of protons Z
in the nucleus is unknown. A group of lines in the spectrum forms a
series in which the shortest wavelength is 22.79 nm and the longest
wavelength is 41.02 nm. Find the next-to-the-Iongest wavelength in
the series of lines.
"@ The
Section 30.5 The Quantum Mechanical Picture
of the Hydrogen Atom
orbital quantum number for the electron in a hydrogen atom
IS £ = 5. What is the smallest possible value (algebraically) for the
total energy of this electron? Give your answer in electron volts.
22. A hydrogen atom is in its second excited state. Determine, according
to quantum mechanics, (a) the total energy (in eV) of the
atom, (b) the magnitude of the maximum angular momentum the
electron can have in this state, and (c) the maximum value that the z
, 23. ssm www The principal quantum number for an electron in an
~mponent a om is n = Lz6, ofand the the angular magnetic momentum quantumcannumber have. is me = 2. What
possible values for the orbital quantum
have?
number £ could this electron
24. The maximum value for the magnetic quantum number in state
A is me = 2, while in state B it is me = I. What is the ratio LA/LB of
the magnitudes
these two states?
of the orbital angular momenta of an electron in
* 25. Interactive Solution 30.25 at www.wiley.comlcollege/cutnell
offers one approach to problems of this type. For an electron in a hydrogen
atom, the z component of the angular momentum has a maximum
value of L, = 4.22 X 10-34 J . s. Find the three smallest possible
values (algebraically)
this atom could have.
for the total energy (in electron volts) that

972 I CHAPTER 30 I THE NATURE OF THE ATOM
ff6J Review Conceptual Example 6 as backg;ound for this problem.
~ the hydrogen atom, the Bohr model and quantum mechanics
both give the same value for the energy of the nth state. However,
they do not give the same value for the orbital angular momentum L.
(a) For n = I, determine the values of L [in units of h/(27T)] predicted
by the Bohr model and quantum mechanics. (b) Repeat part
(a) for n = 3, noting that quantum mechanics permits more than one
value of £ when the electron is in the n = 3 state.
** 27. ssm www An electron is in the n = 5 state. What is the small­
est possible value for the angle between the z component of the orbital
angular momentum and the orbital angular momentum?
Section 30.6 The Pauli Exclusion Principle and the Periodic
Table of the Elements
28. Two of the three electrons in a lithium atom have quantum numbers
of n = 1, £ = 0, me = 0, m, = +! and n = 1, £ = 0, me = 0,
m, = -!. What quantum numbers can the third electron have if the
atom is in (a) its ground state and (b) its first excited state?
29. In the style shown in Table 30.3, write down the ground-state
electronic configuration for arsenic As (2 = 33). Refer to Figure
30.17 for the order in which the sub shells fill.
30. Figure 30.17 was constructed using the Pauli exclusion principle
and indicates that the n = 1 shell holds 2 electrons, the n = 2 shell
holds 8 electrons, and the n = 3 shell holds 18 electrons. These num­
bers can be obtained by adding the numbers given in'the figure for
the subshells contained within a given shell. How many electrons can
be put into the n = 5 shell, which is only partly shown in the figure?
31. Write down the fourteen sets of the four quantum numbers that
correspond to the electrons in a completely filled 4f subshell.
* 32. What is the atom with the smallest atomic number that contains
the same number of electrons in its s subshells as it does in its d sub­
shell? Refer to Figure 30.17 for the order in which the subshells fill.
Section 30.7 X-Rays
33. ssm Molybdenum has an atomic number of 2 = 42. Using the
Bohr model, estimate the wavelength of the Kcx X-ray.
34. Interactive LearningWare 30.2 at www.wiley.comlcollege/cutnell
reviews the concepts that are pertinent to this problem. By using the
Bohr model, decide which element is likely to emit a Kcx X-ray with
a wavelength of 4.5 X 10-9 m.
35. Interactive Solution 30.35 at www.wiley.comlcollege/cutnell
provides one model for solving problems such as this one. An X-ray
tube is being operated at a potential difference of 52.0 kV. What is
the Bremsstrahlung wavelength that corresponds to 35.0% of the kinetic
energy with which an electron collides with the metal target in
the tube?
IADDITIONAL PROBLEMS
44. Referring to Figure 30.17 for the order in which the sub shells
fill and following the style used in Table 30.3, determine the groundstate
electronic configuration for cadmium Cd (2 = 48).
45. ssm www In the line spectrum of atomic hydrogen there is
also a group of lines known as the Pfund series. These lines are produced
when electrons, excited to high energy levels, make transitions
to the n = 5 level. Determine (a) the longest wavelength and
36. What is the minimum potential difference that must be applied
to an X-ray tube to knock a K-shell electron completely out of an
atom in a copper (2 = 29) target? Use the Bohr model as needed.
* 37. ssm An X-ray tube contains a silver (2 = 47) target. The high
voltage in this tube is increased from zero. Using the Bohr model,
find the value of the voltage at which the Kcx X-ray just appears in
the X-ray spectrum.
* 38. Multiple-Concept Example 9 reviews the concepts that are
important in this problem. An electron, traveling at a speed of
6.00 X 107 m/s, strikes the target of an X-ray tube. Upon impact, the
electron decelerates to one-quarter of its original speed, emitting an
X-ray in the process. What is the wavelength of the X-ray photon?
39.r.'
Section 30.8 The Laser
R tached retina back into place. The wavelength of the
laser ssm wwwAlaserisusedineyesurgerytoweldade­
beam is 514 nm, and the power is 1.5 W. During
surgery, the laser beam is turned on for 0.050 s. During this time,
how many photons are emitted by the laser?
40. ~ The dye laser used in the treatment of the port-wine stain
8~ in 585Figure nm. A30.30 carbon (seedioxide Sectionlaser 30.9) produces has a wavelength a wavelengthof
of 1.06 X 10-5 m. What is the minimum number of photons that the
carbon dioxide laser must produce to deliver at least as much or
more energy to a target as does a single photon from the dye laser?
41. A pulsed laser emits light in a series of short pulses, each having
a duration of 25.0 ms. The average power of each pulse is
5.00 mW, and the wavelength of the light is 633 nm. Find (a) the
energy of each pulse and (b) the number of photons in each pulse.
. an eye condition known as narrow-angle glaucoma, in
42. T'A • which laserpressure peripheral buildup iridotomy in the is eye a procedure can lead tofor loss treating of vision.
A neodymium YAG laser (wavelength = 1064 nm) is used in
the procedure to punch a tiny hole in the peripheral iris, thereby
relieving the pressure buildup. In one application the laser delivers
4.1 X 10-3 J of energy to the iris in creating the hole. How many
photons does the laser deliver?
* 43. Fusion is the process by which the sun produces energy. One
experimental technique for creating controlled fusion utilizes
a solid-state laser that emits a wavelength of 1060 nm and can produce
a power of 1.0 X 1014 W for a pulse duration of 1.1 X 10-11 s.
In contrast, the helium/neon laser used at the checkout counter
in a bar-code scanner emits a wavelength of 633 nm and produces
a power of about 1.0 X 10-3 W. How long (in days)
would the helium/neon laser have to operate to produce the
same number of photons that the solid-state laser produces in
1.1 X 1O-11 s?
(b) the shortest wavelength in this series. (c) Refer to Figure 24.10
and identify the region of the electromagnetic spectrum in which
these lines are found.
46. The atomic number of lead is 2 = 82. According to the Bohr
model, what is the energy (in joules) of a Kcx X-ray photon?
47. ssm When an electron makes a transition between energy levels
of an atom, there are no restrictions on the initial and final values of

Chapter 30 Problems 1515
8. REASONING According to the Bohr model, the energy (in joules) of the nth orbit of an
atom containing a single electron is
En = -(2.18x10-18 J) Z:
n
(30.12)
where Z is the atomic number of the atom. The ratio of the energies of the two atoms can be
obtained directly by using this relation.
SOLUTION Taking the ratio of the energy E B 3+ of the nth orbit of a beryllium atom
n, e
(ZBe3+ = 4) to the energy En, H of the nth orbit of a hydrogen (ZH = I) atom gives
En, B,;' -(2.18xlO-18 J) Z~,;.
= 2 Z2
En.H "2 =~= (4)2
-(2.18x 10-18 J) ZH Z~ (1)2 =[lli
n2
9. REASONING The atomic number for helium is Z = 2. The ground state is the n = 1 state,
the first excited state is the n = 2 state, and the second excited state is the n = 3 state. With
Z= 2 and n = 3, we can use Equation 30.10 to find the radius of the ion.
SOLUTION The radius ofthe second excited state is
10. REASONING
a. The total energy En for a single electron in the nth state is given by
Z2
En = -(13.6 eV)-2 n
(30.10)
(30.13)
where Z = 2 for helium. The minimum amount of energy required to remove the electron
from the ground state (n = 1) is that needed to move the electron into the state for which
n = 2. This amount equals the difference between the two energy levels.
b. The ionization energy defined as the minimum amount of energy required to remove the
electron from the n = 1 orbit to the highest possible excited state (n = 00) .

11,
1516 THE NATURE OF THE ATOM
SOLUTION
a. The minimum amount of energy required to remove the electron from the ground state
(n = 1) and move it into the state for which n = 2 is
Minimum energy = E, _ El = -(13.6 22 eV)(2)' - [-(13.6 12 eV) (2)' ] = 140.8 eVI
b. The ionization energy is the difference between the ground-state energy (n = 1) and the
energy in the highest possible excited state (n = 00) . Thus,
Ionization energy ~ E__ El = -(13.6 e~) (2)' [-(13.6 eV)(2)']_
(00) 12 -154.4 eVI
11. I SSMI REASONING According to Equation 30.14, the wavelength A emitted by the
hydrogen atom when it makes a transition from the level with nj to the level with nf is given
by
-=---- (Z2) --- wIth nj,nf =1,2,3, ... and nj >nf
A1 2tr2mk2e4 h3e ( nl 1 ni2 1 J .
where 2tr2mk2e4 /(h3e) = 1.097 X107 m-I and Z= 1 for hydrogen. Once the wavelength
for the particular transition in question is determined, Equation 29.2 (E = hf = he / A) can
be used to find the energy of the emitted photon.
SOLUTION In the Paschen series, nf= 3. Using the above expression with Z= 1, nj = 7
and nf= 3, we find that
~ = (1.097 X 107 m-I)(I') (3~- 71,) or A=1.005xlO-6 m
The photon energy is
he (6.63XlO-34 J.s)(3.00XI08 rnls) I I
E=-=-----------= 1.98xl0-19 J
A 1.005xl0-6 m '------

Chapter 30 Problems 1517
12. REASONING Since the atom emits two photons as it returns to the ground state, one is
emitted when the electron falls from n = 3 to n = 2, and the other is emitted when it
subsequently drops from n = 2 to n = 1. The wavelengths of the photons emitted during
these transitions are given by Equation 30.14 with the appropriate values for the initial and
final numbers, ni and ne
SOLUTION The wavelengths of the photons are
n =3 ton =2 ,.1,= 1 ( 1.097x10 7 m_1)()2(1 1 "22-"32 1) =1.524xlO 6 m-1
n = 2 to n = 1
,.1,= !6.56XlO-7ml
~ =(1.097X107 m-1)(1)2(*_ 212)=8.228X106 m-I
A=I1.22X10-7 m!
(30.14)
(30.14)
13. REASONING The Bohr expression as it applies to anyone-electron species of atomic
number Z, is given by Equation 30.13: En = -(13.6 eV)(Z2 / n2). For certain values of the
quantum number n, this expression predicts equal electron energies for singly ionized
helium He + (Z = 2) and doubly ionized lithium Li + (Z = 3). As stated in the problem, the
quantum number n is different for the equal energy states for each species.
SOLUTION For, equal energies, we can write
Simplifying, this becomes
Thus,
or
Z2 Z2
-(13.6 eV) ~e = -(13.6 eV) ii
nHe nLi
or
4 9
2 2
nHe nLi
Therefore, the value of the helium energy level is equal to the lithium energy level for any
value of nHe that is two-thirds of nLi· For quantum numbers less than or equal to 9, an

I.
-I' j.
-3.40eV 9-1.S1eV
3-13.6 6nLiEnergy
eV
--- --~- --- -- -- ----
1518 THE NATURE OF THE ATOM
equality in energy levels will occur for nHe = 2,4, 6 corresponding to nLi = 3, 6, 9. The
results are summarized in the following table.
24
nHe
14. REASONING In the Bohr model of the hydrogen atom the total energy En of the electron is
given in electron volts by Equation 30.13 and the orbital radius rn is given in meters by
Equation 30.10:
Solving the radius equation for n2 and substituting the result into the energy equation gives
En = -13.6 = (-13.6)(S.29X10-11)
rnl(S.29X10-11) rn
Thus, the energy is inversely proportional to the radius, and it is on this fact that we base our
solution.
SOLUTION We know that the radius of orbit B is sixteen times greater than the radius of
orbit A. Since the total energy is inversely proportional to the radius, it follows that the total
energy of the electron in orbit B is one-sixteenth of the total energy in orbit A:
EA _ -3.40 eV = 1-0.213 eVI
EB=16.0- 16.0
IS. I SSM I REASONING A wavelength of 410.2 nm is emitted by the hydrogen atoms in a
high-voltage discharge tube. This transition lies in the visible region (380-7S0 nm) of the
hydrogen spectrum. Thus, we can conclude that the transition is in the Balmer series and,
therefore, that nf = 2. The value of ni can be found using Equation 30.14, according to which
the Ba1mer series transitions are given by
11 = 3, 4, S, ...

Chapter 30 Problems 1519
This expression may be solved for ni for the energy transition that produces the given
wavelength.
SOLUTION Solving for ni' we find that
Therefore, the initial and final states are identified by I ni = 6 and nf = 2 I·
16. REASONING The energy levels and radii of a hydrogenic species of atomic number Z are
given by Equations30.13 and 30.10, respectively: En=-(13.6eV)(Z2/n2) and
rn = (5.29 x 10-11 m)( n2 /Z) . We can use Equation 30.13 to find the value of Z for the
unidentified ionized atom and then calculate the radius of the n = 5 orbit using
Equation 30.10.
SOLUTION Solving Equation 30.13 for atomic number Z of the unknown species, we have
Z=
Therefore, the radius of the n = 5 orbit is
-13.6 Enn2 eV = ~ {(-30.6 -13.6eV)(2)2 eV = 3
17. I SSMI REASONING AND SOLUTION For the Paschen series, nf = 3. The range of
wavelengths occurs for values of nj = 4 to ni= 00. Using Equation 30.14, we find that the
shortest wavelength occurs for nj = 00 and is given by
/L=8.204x10-7 m
~---~v~---~
Shortest wavelength in
Paschen series

1520 THE NATURE OF THE ATOM
The longest wavelength in the Paschen series occurs for ni= 4 and is given by
A = 1.875 X 10-6 m
~---~v·~---~
Longest wavelength in
Paschen series
For the Brackett series, nf= 4. The range of wavelengths occurs for values of ni= 5 to
ni= 00, Using Equation 30.14, we find that the shortest wavelength occurs for ni= 00 and is
given by
~ =(l.097Xl07 m-I)( 41,- 5; J
or
or
A = 4.051 X 10-6 m
v
Longest wavelength in
Brackett series
A=1.459x10-6 m
v
Shortest wavelength in
Brackett series
The longest wavelength in the Brackett series occurs for ni= 5 and is given by
Since the longest wavelength in the Paschen series falls within the Brackett series, the
wavelengths of the two series overlap.
18. REASONING· To obtain the quantum number of the higher level from which the electron
falls, we will use Equation 30.14 for the reciprocal of the wavelength A of the photon:
where R is the Rydberg constant and nf and ni' respectively, are the quantum numbers of the
final and initial levels. Although we are not directly given the wavelength, we do have a
value for the magnitude p of the photon's momentum, and the momentum and the
wavelength are related according to Equation 29.6:
j~.
'~
',[i
o i where his Planck's constant.
h
p=- A
Using Equation 30.14 to substitute for ~, we obtain
A

SOLUTION Rearranging Equation (1) gives
Thus, we find
Chapter 30 Problems 1521
p=-=hR --- (1)
Ah (1nf nf 1 J
_1 __ 1 _ P
nf nf - hR
_1 __ 1 _L
nf - nf hR
1 1 P _ 1 5.452x10-27 kg·m/s -0.2499 or n =!lI
nj2 = ni - hR -12- (6.626X10-34 J.s)(1.097X107 m-l) 1
or
19. REASONING AND SOLUTION We need to use Equation 30.14 to find the spacing
between the longest and next-to-the longest wavelengths in the Ba1mer series. In order to do
this, we need to first find these two wavelengths.
Longest:
-=R --- = 1.097x10 m ---
/41 ( nf 1 ni2 1 J ( 7 -1)( 22 1 32 1 )
Next-to-Iongest:
_I ~ =R( ~_~ nf nj J~(1.097XI07 m-I)(_l 22 __1 42 )
or /4 = 656.3 nm
or ~ = 486.2 nm
Equation 27.7 states that sin B= mAid. Using the small angle approximation, we have
sin B"'" tan B"'" B = Y
L
so that y/L = mAid. The position of the fringe due to the longest wavelength is Yl = mAlL/d.
For the next-to-Iongest, Y2 = m~L/d. The difference in the positions on the screen is,
therefore, Yl - Y2 = (mL/d)(Al -~) which gives
d= mL(/4 -~)
Yj- Y2

Chapter 30 Problems 1523
21. REASONING The orbital quantum number .e has values of 0, 1, 2, ....., (n -1), according
to the discussion in Section 30.5. Since.e = 5, we can conclude, therefore, that n ;? 6. This
knowledge about the principal quantum number n can be used with Equation 30.13,
E = -(13.6 eV)Z2/n2, to determine the smallest value for the total energy E .
n n
SOLUTION The smallest value of E (i.e., the most negative) occurs when n = 6. Thus,
n
using Z = 1 for hydrogen, we find
Z2 12
En = -(13.6 eV)~ = -(13.6 eV)"62 = 1-0.378 eV I
22. REASONING
a. The ground state is the n = 1 state, the first excited state is the n = 2 state, and the second
excited state is the n = 3 state. The total energy (in eV) of a hydrogen atom in the n = 3
state is given by Equation 30.13.
b. According to quantum mechanics, the magnitude L of the angular momentum is given by
Equation 30.15 as L=~.e(.e+l)(h/21r), where .e is the orbital quantum number. The
discussion in Section 30.5 indicates that the maximum value that .e can have is one less
than the principal quantum number, so that .e max = n - 1.
c. Equation 30.16 gives the z-component Lz of the angular momentum as Lz = m£ (h / 21r) ,
where me is the magnetic quantum number. According to the discussion in Section 30.5,
the maximum value that m£ can attain is when it is equal to the orbital quantum number,
which is .e max'
SOLUTION
a. The total energy of the hydrogen atom is given by Equation 30.13. Using n = 3, we have
(13.6 eV)(1)2 = 1-1.51 eVI
E3 = 32
b. The maximum orbital quantum number is .e max = n - 1 = 3 - 1 = 2. The maximum
angular momentum Lmax has a magnitude given by Equation 30.15:

1524 THE NATURE OF THE ATOM
c. The maximum value for the z-component Lz of the angular momentum is
(withmf =J!max=2)
L =m ~=(2)6.63XlO-34 ],s=12.11X10-34 J.sl
z £ 2tr 2tr .------
23. ISSMII wwwl REASONING The values that I can have depend on the value of n, and
only the following integers are allowed: I = 0, 1, 2, ... (n -1). The values that ml can
have depend on the value of I , with only the following positive and negative integers being
permitted: mf = -J!, ... -2, -1,0, +1, +2, .. .+e.
SOLUTION Thus, when n = 6, the possible values of I are 0, 1,2,3, 4,5. Now when
mf = 2 , the possible values of I are 2, 3, 4, 5, ... These two series of integers overlap for
the integers 2, 3, 4, and 5. Therefore, the possible values for the orbital quantum number I
that this electron could have are 11 =
2, 3, 4, 51.
24. REASONING The maximum value for the magnetic quantum number is ml = I ; thus, in
state A, I = 2, while in state B, I = 1. According to the quantum mechanical theory of
angular momentum, the magnitude of the orbital angular momentum for a state of given I is
L = -JJ!(J! + 1) (h / 2tr) (Equation 30.15). This expression can be used to form the ratio
LA / ~ of the magnitudes of the orbital angular momenta for the two states.
SOLUTION Using Equation 30.15, we find that
h
LA = .j2(2+ 1) 2& h ~ V2 @" ~v'3~11.7321
LB -J1(1 + 1) 2tr
25. REASONING The total energy En for a hydrogen atom in the quantum mechanical picture
is the same as in the Bohr model and is given by Equation 30.13:
E/1 = -(13.6 eV)2 n
Thus, we need to determine values for the principal quantum number n if we are to calculate
the three smallest possible values for E. Since the maximum value of the orbital quantum
number J! is n - 1, we can obtain a minimum value for n as nmin = J! + 1. But how to obtain
1
(30.13)

Chapter 30 Problems 1525
£? It can be obtained, because the problem statement gives the maximum value of Lz, the z
component of the angular momentum. According to Equation 30.16, Lz is
where mf is the magnetic quantum number and his Planck's constant. For a given value of £
the allowed values for mf are as follows: -£, .. " -2, -1, 0, +1, +2, ... , +£. Thus, the
maximum value ofmf is £, and we can use Equation 30.16 to calculate the maximum value of
mf from the maximum value given for Lz·
h
Lz = mf 21C
SOLUTION Solving Equation 30.16 for mf gives
rn, ~ 21fL, ~ 21f(4.22X10-34 J,s)_4
h 6.63xl0-34 J. S -
(30.16)
As explained in the REASONING, this maximum value for mf indicates that £ = 4.
Therefore, a minimum value for n is
n . =£+1=4+1=5 or n>5
mm -
This means that the three energies we seek correspond to n = 5, n = 6, and n = 7. Using
Equation 30.13, we find them to be
[n = 5] E5 = -(13.6 eV)~ = 1-0.544 eVj
5
[n = 6] E6 = -(13.6 eV)~ = 1-0.378 eV[
6
[n=7] E7 = -(13.6 eV)~ = 1-0.278 eV[
7
26. REASONING AND SOLUTION
a. For the angular momentum, Bohr's value is given by Equation 30.8, with n = 1,
According to quantum theory, the angular momentum is given by Equation 30.15. For
n=I,£=O

1526 THE NATURE OF THE ATOM
b. For n = 3; Bohr theory gives
while quantum mechanics gives
[n = 3, .e = 0]
[n=3,.e=1]
[n = 3, .e = 2]
27. I SSMII wwwl REASONING Let e denote the angle between
the angular momentum L and its z-component Lz· We can see
from the figure at the right that Lz = L cos e . Using
Equation 30.16 for Lz and Equation 30.15 for L, we have
The smallest value for e corresponds to the largest value of cos e. For a given value of I ,
the largest value for cos e corresponds to the largest value for rn,. But the largest possible
value for rnl is rnl = I . Therefore, we find that
cos e =.e re
~.e(.e+l) ~£+1
SOLUTION The smallest value for e corresponds to the largest value for I. When the
electron is in the n = 5 state, the largest allowed value of I is I = 4 ; therefore, we see that
or
e = cos-1 (J4i5) = I 26.6° I
i
"1
,j
i;
1
!
"