# .-@ A hydrogen atom is in the ground stat

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```					                                                                                                                                     PROBLEMS        I   971

atom, (b) the volume- of the nucleus, and' (c) the percentage of the
volume of the atom that is occupied by the nucleus.                                     the hydrogen atom the total of orbit is electron in orbit A is
rilf.)In the radius of orbit A. The radius energy ofBthe sixteen times greater
~
2. In a Rutherford scattering experiment a target nucleus has a di-             - 3AO eV. What is the total energy of the electron in orbit B?
ameter of lA X 10-14 m. The incoming Cl particle has a mass of               * 15.     ssm A wavelength of 410.2 nm is emitted by the hydrogen
6.64 X 10-27 kg. What is the kinetic energy of an Cl particle that has            atoms in a high-voltage discharge tube. What are the initial and final
a de Broglie wavelength equal to the diameter of the target nucleus?              values of the quantum number n for the energy level transition that
Ignore relativistic effects.                                                      produces this wavelength?
3. The nucleus of a hydrogen atom is a single proton, which has a            *@        The energy of the n = 2 Bohr orbit is - 30.6 eV for an unidenti-
radius of about 1.0 X IO-IY m. The single electron in a hydrogen                  fied ionized atom in which only one electron moves about the nu-
atom normally orbits the nucleus at a distance of 5.3 X 10-11 m.                  cleus. What is the radius of the n = 5 orbit for this species?
What is the ratio of the density of the hydrogen nucleus to the den-         * 17.     ssm For atomic hydrogen, the Paschen series of lines occurs
sity of the complete hydrogen atom?                                               when nf = 3, whereas the Brackett series occurs when nf = 4 in
4. Review Conceptual Example I and use the information therein as                 Equation 30.14. Using this equation, show that the ranges of wave-
an aid in working this problem. Suppose you're building a scale                   lengths in these two series overlap.
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
*@         Interactive LearningWare 30.1 at www.wiley.comlcolIege/cutnell
reviews the concepts that play roles in this problem. A hydrogen
miles away (I mi = 1.61 X 105 cm) should the electron be placed?                  atom emits a photon that has momentum with a magnitude of
* 5. ssm The nucleus of a copper atom contains 29 protons and has a                5A52 X 10-27 kg' m/so This photon is emitted because the electron
radius of 4.8 X 10-15 m. How much work (in electron volts) is done                in the atom falls from a higher energy level into the n = I leveL
by the electric force as a proton is brought from infinity, where it is           What is the quantum number of the level from which the electron
at rest, to the "surface" of a copper nucleus?                                    falls? Use a value of 6.626 X 10-34 J. s for Planck's constant.
* 6. There are Z protons in the nucleus of an atom, where Z is the           **    19. A diffraction grating is used in the first order to separate the
atomic number of the element. An Cl particle carries a charge of                  wavelengths in the Balmer series of atomic hydrogen. (Section 27.7
+2e. In a scattering experiment, an Cl particle, heading directly to-             discusses diffraction gratings.) The grating and an observation
ward a nucleus in a metal foil, will come to a halt when all the parti-           screen (see Figure 27.32) are separated by a distance of 81.0 cm.
cle's kinetic energy is converted to electric potential energy. In such           You may assume that () is small, so sin () = () when radian measure
a situation, how close will an Cl particle with a kinetic energy of               is used for (). How many lines per centimeter should the grating
5.0 X 10-13 J come to a gold nucleus (Z = 79)?                                    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?
Section 30.2 Line Spectra                                                   ** 20.     A certain species of ionized atoms produces an emission line
Section 30.3 The Bohr Model of the Hydrogen         Atom                          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
7. ssm www Concept Simulation 30.1 at www.wiley.comlcollege/
series in which the shortest wavelength is 22.79 nm and the longest
cutnell reviews the concepts on which the solution to this problem
wavelength is 41.02 nm. Find the next-to-the-Iongest wavelength in
depends. The electron in a hydrogen atom is in the first excited state,           the series of lines.
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             Section 30.5 The Quantum      Mechanical   Picture
orbit of a triply ionized beryllium atom (BeH, Z = 4) to the energy               of the Hydrogen Atom
of the nth orbit of a hydrogen atom (H).
"@        The orbital quantum number for the electron in a hydrogen atom
9. A singly ionized helium atom (He+) has only one electron in or-                IS £  = 5. What is the smallest possible value (algebraically) for the
bit about the nucleus. What is the radius of the ion when it is in the            total energy of this electron? Give your answer in electron volts.
22. A hydrogen atom is in its second excited state.     Determine, ac-
~ond      excited state?                                                           cording to quantum mechanics, (a) the total energy       (in eV) of the
(!!J: (a) What is the minimum energy (in electron volts) that is re-
quired to remove the electron from the ground state of a singly ion-             atom, (b) the magnitude of the maximum angular          momentum the
ized helium atom (He+, Z = 2)? (b) What is the ionization energy                 electron can have in this state, and (c) the maximum    value that the z
for He+?
11. ssm Find the energy (in joules) of the photon that is emitted           , 23. ssm www The principal quantum number for an electron in an
when the electron in a hydrogen atom undergoes a transition from            ~mponent n = 6, and the magnetic quantum number is me = 2. What
a om is     Lz of the angular momentum can have.
the n = 7 energy level to produce a line in the Paschen series .              possible values for the orbital quantum number £ could this electron
have?
.-@ A
makes
hydrogen atom is in the ground state. It absorbs energy and
a transition to the n = 3 excited state. The atom returns to the         24. The maximum value for the magnetic quantum number in state
ground state by emitting two photons. What are their wavelengths?                 A is me = 2, while in state B it is me = I. What is the ratio LA/LB of
13. Consider the Bohr energy expression (Equation 30.13) as it ap-                the magnitudes of the orbital angular momenta of an electron in
these two states?
plies to singly ionized helium He+ (Z = 2) and doubly ionized
lithium Li2+ (Z = 3). This expression predicts equal electron ener-          * 25.     Interactive Solution 30.25 at www.wiley.comlcollege/cutnell
gies for these two species for certain values of the quantum number               offers one approach to problems of this type. For an electron in a hy-
n (the quantum number is different for each species). For quantum                 drogen atom, the z component of the angular momentum has a maxi-
numbers less than or equal to 9, what are the lowest three energies               mum value of L, = 4.22 X 10-34 J . s. Find the three smallest possi-
(in electron volts) for which the helium energy level is equal to the             ble values (algebraically) for the total energy (in electron volts) that
lithium energy level?                                                             this atom could have.
972   I    CHAPTER   30   I   THE   NATURE   OF THE ATOM

36. What is the minimum potential difference that must be applied
ff6J Review Conceptual Example 6 model and quantum mechanics
~    the hydrogen atom, the Bohr as backg;ound for this problem.              to an X-ray tube to knock a K-shell electron completely out of an
both give the same value for the energy of the nth state. However,          atom in a copper (2 = 29) target? Use the Bohr model as needed.
they do not give the same value for the orbital angular momentum L.        * 37. ssm An X-ray tube contains a silver (2 = 47) target. The high
(a) For n = I, determine the values of L [in units of h/(27T)] pre-         voltage in this tube is increased from zero. Using the Bohr model,
dicted by the Bohr model and quantum mechanics. (b) Repeat part             find the value of the voltage at which the Kcx X-ray just appears in
(a) for n = 3, noting that quantum mechanics permits more than one          the X-ray spectrum.
value of £ when the electron is in the n = 3 state.
* 38. Multiple-Concept   Example 9 reviews the concepts that are
** 27.   ssm www An electron is in the n    = 5 state.   What is the small-
important in this problem. An electron, traveling at a speed of
est possible value for the angle between the z component of the or-         6.00 X 107 m/s, strikes the target of an X-ray tube. Upon impact, the
bital angular momentum and the orbital angular momentum?                    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?

r.'
Section 30.6 The Pauli Exclusion Principle      and the Periodic
Table of the Elements                                                       Section 30.8 The Laser

28. Two of the three electrons in a lithium atom have quantum num-
bers of n = 1, £ = 0, me = 0, m, = +! and n = 1, £ = 0, me = 0,
R   tached retina back into place. The wavelength of the
m, =      -!.
What quantum numbers can the third electron have if the          39.         ssm beam is 514 nm, and the power is
laser wwwAlaserisusedineyesurgerytoweldade-1.5 W. During
atom is in (a) its ground state and (b) its first excited state?            surgery, the laser beam is turned on for 0.050 s. During this time,
29. In the style shown in Table 30.3, write down the ground-state           how many photons are emitted by the laser?
electronic configuration for arsenic As (2 = 33). Refer to Figure           40. ~       The dye laser used in the treatment of the port-wine stain
30.17 for the order in which the sub shells fill.
30. Figure 30.17 was constructed using the Pauli exclusion principle               ~8  585 nm. 30.30 (see Section 30.9) has a a wavelength
in Figure A carbon dioxide laser produces wavelength of
and indicates that the n = 1 shell holds 2 electrons, the n = 2 shell       of 1.06 X 10-5 m. What is the minimum number of photons that the
holds 8 electrons, and the n = 3 shell holds 18 electrons. These num-       carbon dioxide laser must produce to deliver at least as much or
bers can be obtained by adding the numbers given in'the figure for          more energy to a target as does a single photon from the dye laser?
the subshells contained within a given shell. How many electrons can        41. A pulsed laser emits light in a series of short pulses, each hav-
be put into the n = 5 shell, which is only partly shown in the figure?      ing a duration of 25.0 ms. The average power of each pulse is
31. Write down the fourteen sets of the four quantum numbers that           5.00 mW, and the wavelength of the light is 633 nm. Find (a) the
correspond to the electrons in a completely filled 4f subshell.             energy of each pulse and (b) the number of photons in each pulse.
* 32.

T'
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-              . an eye condition known as narrow-angle glaucoma, in
shell? Refer to Figure 30.17 for the order in which the subshells fill.     42.     •   A laser peripheral iridotomy eye can lead
which pressure buildup in the is a procedure tofor treating
loss of vi-
sion. A neodymium YAG laser (wavelength = 1064 nm) is used in
Section 30.7 X-Rays                                                          the procedure to punch a tiny hole in the peripheral iris, thereby
relieving the pressure buildup. In one application the laser delivers
33. ssm Molybdenum has an atomic number of 2 = 42. Using the                 4.1 X 10-3 J of energy to the iris in creating the hole. How many
Bohr model, estimate the wavelength of the Kcx X-ray.                        photons does the laser deliver?
34. Interactive LearningWare 30.2 at www.wiley.comlcollege/cutnell          * 43. Fusion is the process by which the sun produces energy. One
reviews the concepts that are pertinent to this problem. By using the        experimental    technique for creating controlled fusion utilizes
Bohr model, decide which element is likely to emit a Kcx X-ray with          a solid-state laser that emits a wavelength of 1060 nm and can pro-
a wavelength of 4.5 X 10-9 m.                                                duce a power of 1.0 X 1014 W for a pulse duration of 1.1 X 10-11 s.
35. Interactive Solution 30.35 at www.wiley.comlcollege/cutnell              In contrast, the helium/neon laser used at the checkout counter
provides one model for solving problems such as this one. An X-ray           in a bar-code scanner emits a wavelength of 633 nm and pro-
tube is being operated at a potential difference of 52.0 kV. What is         duces a power of about 1.0 X 10-3 W. How long (in days)
the Bremsstrahlung wavelength that corresponds to 35.0% of the ki-           would the helium/neon laser have to operate to produce the
netic energy with which an electron collides with the metal target in        same number of photons that the solid-state laser produces in
the tube?                                                                    1.1 X 1O-11 s?

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 ground-
state electronic configuration for cadmium Cd (2 = 48).
(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.

45. ssm www In the line spectrum of atomic hydrogen there is                 46. The atomic number of lead is 2 = 82. According to the Bohr
also a group of lines known as the Pfund series. These lines are pro-        model, what is the energy (in joules) of a Kcx X-ray photon?
duced when electrons, excited to high energy levels, make transi-            47. ssm When an electron makes a transition between energy levels
tions to the n = 5 level. Determine (a) the longest wavelength and           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:                              (30.12)
n

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 n, B e    3+   of the nth orbit of a beryllium atom

(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        -(2.18x 10-18 J) ZH =~=
"2   Z~            (4)2 = [lli
(1)2
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

(30.10)

10. REASONING
a. The total energy En for a single electron in the nth state is given by

Z2
En   = -(13.6 eV)-2                                   (30.13)
n

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) .
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 (2)' ] = 140.8 eVI
eV)

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)'
(00)                          [-(13.6 12
eV)(2)']_ -154.4 eVI

11,

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
A 2tr2mk2e4
1     h3e              ( nl
1   ni2 J
1         .

where 2tr2mk2e4 /(h3e) = 1.097 X 107 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

E=-=-----------=
he (6.63XlO-34 J.s)(3.00XI08                        rnls)   I1.98xl0-19      J
I
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                                                                                           (30.14)
1
,.1,=   ( 1.097x10 7 m _1)()2(1 "22-"32 =1.524xlO
1         1)                     6    m -1

,.1,=   !6.56XlO-7ml

n = 2 to n = 1                                                                                        (30.14)
~ =(1.097X107       m-1)(1)2(*_               )=8.228X106
212               m-I

A=I1.22X10-7         m!

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

Z2                   Z2
or
-(13.6 eV)    ~e
nHe
= -(13.6 eV)    ii
nLi

Simplifying, this becomes

4          9
or
2      2
nHe        nLi
Thus,

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
---    --~-   ---    -- --       ----

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.

36
-13.6 eV
-1.S1eV
-3.40eV
9 Energy
nLi                                          nHe
2
I.                                                             4

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
-I'   j.

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 n i = 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=
-13.6 eV = ~{(-30.6 eV)(2)2 = 3
Enn2          -13.6 eV

Therefore, the radius of the n = 5 orbit is

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

or    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

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

or    A = 4.051     X   10-6 m
v
~ =(l.097Xl07    m-I)(   41,- 5; J             Longest wavelength in
Brackett series

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:

h
j~.
p=-A
'~
',[i
oi
where his Planck's constant. Using Equation 30.14 to substitute for ~, we obtain
A
Chapter 30 Problems   1521

p=-=hR                  ---                                                       (1)
h
A            (1nf        nf
1   J

SOLUTION Rearranging Equation (1) gives

_1 __ 1 _ P             or          _1 __ 1         _L
nf   nf - hR                        nf - nf               hR
Thus, we find

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

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     ---                                     or      /4 = 656.3 nm
/4
1     ( nf
1    ni2 J
1      (       7 -1)( 22
1                     32
1   )

Next-to-Iongest:

or      ~ = 486.2 nm
_I =R( ~_~ nj J~(1.097XI07 m-I)(_l __1 )
~      nf                        22 42

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 n = -(13.6 eV)Z2/n2, to determine the smallest value for the total energy E n .

SOLUTION           The smallest value of E n (i.e., the most negative) occurs when n = 6. Thus,
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      can have is one less
.e

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

E3   =         eV)(1)2 = 1-1.51 eVI
(13.6 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)

Lz =m £ ~=(2)6.63XlO-34
2tr            2tr             ],s=12.11X10-34
.------         J.sl

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+         h
1) 2& ~   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:

1
E/1   = -(13.6 eV)2n                                   (30.13)

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
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

h
(30.16)
Lz    = mf   21C

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·

SOLUTION           Solving Equation 30.16 for mf gives

rn, ~ 21fL, ~ 21f(4.22X10-34 J.J,s)_4-
h        6.63xl0-34     S

As explained in the REASONING,                    this maximum       value for mf indicates that     £   = 4.
Therefore, a minimum value for n is

n mm =£+1=4+1=5
.                           or   n>5
-

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)~5 = 1-0.544 eVj

[n   = 6]                   E6    = -(13.6 eV)~6 = 1-0.378 eV[

[n=7]                        E7   = -(13.6 eV)~7 = 1-0.278 eV[

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                                            i
"1

,j
i
;
Equation 30.16 for Lz and Equation 30.15 for L, we have                                              1

!
"

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
~.e(.e+l)
re
~£+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

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