42

CHAPTER OUTLINE42.1 Atomic Spectra of Gases42.2 Early Models of the Atom42.3 Bohr’s Model of the Hydrogen Atom42.4 The Quantum Model of the Hydrogen Atom42.5 The Wave Functions of Hydrogen42.6 Physical Interpretation of the Quantum Numbers42.7 The Exclusion Principle and the Periodic Table

Visible and X-ray

42.8 More on Atomic Spectra:

Spontaneous and42.9 Stimulated Transitions42.10 Lasers

Atomic Physics

ANSWERS TO QUESTIONS

Q42.1 Neon signs emit light in a bright-line spectrum, rather than in acontinuous spectrum. There are many discrete wavelengthswhich correspond to transitions among the various energylevels of the neon atom. This also accounts for the particularcolor of the light emitted from a neon sign. You can see theseparate colors if you look at a section of the sign through adiffraction grating, or at its reflection in a compact disk. Aspectroscope lets you read their wavelengths.

Q42.2 One assumption is natural from the standpoint of classicalphysics: The electron feels an electric force of attraction to thenucleus, causing the centripetal acceleration to hold it in orbit.The other assumptions are in sharp contrast to the behavior ofordinary-size objects: The electron’s angular momentum mustbe one of a set of certain special allowed values. During thetime when it is in one of these quantized orbits, the electronemits no electromagnetic radiation. The atom radiates a photonwhen the electron makes a quantum jump from one orbit to alower one.

Q42.3 If an electron moved like a hockey puck, it could have any arbitrary frequency of revolution aroundan atomic nucleus. If it behaved like a charge in a radio antenna, it would radiate light withfrequency equal to its own frequency of oscillation. Thus, the electron in hydrogen atoms wouldemit a continuous spectrum, electromagnetic waves of all frequencies smeared together.

Q42.4 (a) Yes—provided that the energy of the photon is precisely enough to put the electron into oneof the allowed energy states. Strangely—more precisely non-classically—enough, if theenergy of the photon is not sufficient to put the electron into a particular excited energylevel, the photon will not interact with the atom at all!

(b) Yes—a photon of any energy greater than 13.6 eV will ionize the atom. Any “extra” energywill go into kinetic energy of the newly liberated electron.

Q42.5 An atomic electron does not possess enough kinetic energy to escape from its electrical attraction tothe nucleus. Positive ionization energy must be injected to pull the electron out to a very largeseparation from the nucleus, a condition for which we define the energy of the atom to be zero. Theatom is a bound system. All this is summarized by saying that the total energy of an atom isnegative.

515

516 Atomic Physics

Q42.6 From Equations 42.7, 42.8 and 42.9, we have − = − = + − = +Ek e

rk e

rk e

rK Ue e e

e

2 2 2

2 2. Then K E= and

U Ee = −2 .

Q42.7 Bohr modeled the electron as moving in a perfect circle, with zero uncertainity in its radialcoordinate. Then its radial velocity is always zero with zero uncertainty. Bohr’s theory violates theuncertainty principle by making the uncertainty product ∆ ∆r pr be zero, less than the minimum

allowable 2

.

Q42.8 Fundamentally, three quantum numbers describe an orbital wave function because we live in three-dimensional space. They arise mathematically from boundary conditions on the wave function,expressed as a product of a function of r, a function of θ, and a function of φ.

Q42.9 Bohr’s theory pictures the electron as moving in a flat circle like a classical particle described byF ma∑ = . Schrödinger’s theory pictures the electron as a cloud of probability amplitude in the

three-dimensional space around the hydrogen nucleus, with its motion described by a waveequation. In the Bohr model, the ground-state angular momentum is 1 ; in the Schrödinger modelthe ground-state angular momentum is zero. Both models predict that the electron’s energy is

limited to discrete energy levels, given by −13 606

2. eVn

with n = 1 2 3, , .

Q42.10 The term electron cloud refers to the unpredictable location of an electron around an atomic nucleus.It is a cloud of probability amplitude. An electron in an s subshell has a spherically symmetricprobability distribution. Electrons in p, d, and f subshells have directionality to their distribution. Theshape of these electron clouds influences how atoms form molecules and chemical compounds.

Q42.11 The direction of the magnetic moment due to an orbiting charge is given by the right hand rule, butassumes a positive charge. Since the electron is negatively charged, its magnetic moment is in theopposite direction to its angular momentum.

Q42.12 Practically speaking, no. Ions have a net charge and the magnetic force q v B×a f would deflect thebeam, making it difficult to separate the atoms with different orientations of magnetic moments.

Q42.13 The deflecting force on an atom with a magnetic moment is proportional to the gradient of themagnetic field. Thus, atoms with oppositely directed magnetic moments would be deflected inopposite directions in an inhomogeneous magnetic field.

Q42.14 If the exclusion principle were not valid, the elements and their chemical behavior would be grosslydifferent because every electron would end up in the lowest energy level of the atom. All matterwould be nearly alike in its chemistry and composition, since the shell structures of all elementswould be identical. Most materials would have a much higher density. The spectra of atoms andmolecules would be very simple, and there would be very little color in the world.

Q42.15 The Stern-Gerlach experiment with hydrogen atoms shows that the component of an electron’s spinangular momentum along an applied magnetic field can have only one of two allowed values. Sodoes electron spin resonance on atoms with one unpaired electron.

Chapter 42 517

Q42.16 The three elements have similar electronic configurations. Each has filled inner shells plus oneelectron in an s orbital. Their single outer electrons largely determine their chemical interactionswith other atoms.

Q42.17 When a photon interacts with an atom, the atom’s orbital angular momentum changes, thus thephoton must carry orbital angular momentum. Since the allowed transitions of an atom arerestricted to a change in angular momentum of ∆ = ±1 , the photon must have spin 1.

Q42.18 In a neutral helium atom, one electron can be modeled as moving in an electric field created by thenucleus and the other electron. According to Gauss’s law, if the electron is above the ground state itmoves in the electric field of a net charge of + − = +2 1 1e e e We say the nuclear charge is screened bythe inner electron. The electron in a He+ ion moves in the field of the unscreened nuclear charge of2 protons. Then the potential energy function for the electron is about double that of one electron inthe neutral atom.

Q42.19 At low density, the gas consists of essentially separate atoms. As the density increases, the atomsinteract with each other. This has the effect of giving different atoms levels at slightly differentenergies, at any one instant. The collection of atoms can then emit photons in lines or bands,narrower or wider, depending on the density.

Q42.20 An atom is a quantum system described by a wave function. The electric force of attraction to thenucleus imposes a constraint on the electrons. The physical constraint implies mathematicalboundary conditions on the wave functions, with consequent quantization so that only certain wavefunctions are allowed to exist. The Schrödinger equation assigns a definite energy to each allowedwave function. Each wave function is spread out in space, describing an electron with no definiteposition. If you like analogies, think of a classical standing wave on a string fixed at both ends. Itsposition is spread out to fill the whole string, but its frequency is one of a certain set of quantizedvalues.

Q42.21 Each of the electrons must have at least one quantum number differentfrom the quantum numbers of each of the other electrons. They can differ(in ms ) by being spin-up or spin-down. They can also differ (in ) inangular momentum and in the general shape of the wave function. Thoseelectrons with = 1 can differ (in m ) in orientation of angularmomentum—look at Figure Q42.21.

FIG. Q42.21

Q42.22 The Mosely graph shows that the reciprocal square root of the wavelength of Kα characteristic x-rays is a linear function of atomic number. Then measuring this wavelength for a new chemicalelement reveals its location on the graph, including its atomic number.

Q42.23 No. Laser light is collimated. The energy generally travels in the same direction. The intensity of alaser beam stays remarkably constant, independent of the distance it has traveled.

Q42.24 Stimulated emission coerces atoms to emit photons along a specific axis, rather than in the randomdirections of spontaneously emitted photons. The photons that are emitted through stimulation canbe made to accumulate over time. The fraction allowed to escape constitutes the intense, collimated,and coherent laser beam. If this process relied solely on spontaneous emission, the emitted photonswould not exit the laser tube or crystal in the same direction. Neither would they be coherent withone another.

518 Atomic Physics

Q42.25 (a) The terms “I define” and “this part of the universe” seem vague, in contrast to the precisionof the rest of the statement. But the statement is true in the sense of being experimentallyverifiable. The way to test the orientation of the magnetic moment of an electron is to applya magnetic field to it. When that is done for any electron, it has precisely a 50% chance ofbeing either spin-up or spin-down. Its spin magnetic moment vector must make one of two

allowed angles with the applied magnetic field. They are given by cosθ = =SS

z 1 2

3 2 and

cosθ =−1 2

3 2. You can calculate as many digits of the two angles allowed by “space

quantization” as you wish.

(b) This statement may be true. There is no reason to suppose that an ant can comprehend thecosmos, and no reason to suppose that a human can comprehend all of it. Our experiencewith macroscopic objects does not prepare us to understand quantum particles. On theother hand, what seems strange to us now may be the common knowledge of tomorrow.Looking back at the past 150 years of physics, great strides in understanding the Universe—from the quantum to the galactic scale—have been made. Think of trying to explain thephotoelectric effect using Newtonian mechanics. What seems strange sometimes just has anunderlying structure that has not yet been described fully. On the other hand still, it hasbeen demonstrated that a “hidden-variable” theory, that would model quantum uncertaintyas caused by some determinate but fluctuating quantity, cannot agree with experiment.

SOLUTIONS TO PROBLEMS

Section 42.1 Atomic Spectra of Gases

P42.1 (a) Lyman series1

112λ

= −FHGIKJR

ni

ni = 2 3 4, , , …

1 194 96 10

1 097 10 11

97

2λ=

×= × −

FHGIKJ−.

.e jni

ni = 5

(b) Paschen series:1 1

31

2 2λ= −FHG

IKJR

ni

ni = 4 5 6, , , …

The shortest wavelength for this series corresponds to ni = ∞ for ionization

11 097 10

19

172λ

= × −FHG

IKJ.

ni

For ni = ∞ , this gives λ = 820 nm

This is larger than 94.96 nm, so this wave lengthcannot be associated with the Paschen series .

Balmer series:1 1

21

2 2λ= −FHG

IKJR

ni

ni = 3 4 5, , , …

11 097 10

14

172λ

= × −FHG

IKJ.

ni

with ni = ∞ for ionization, λmin = 365 nm

Once again the shorter given wavelength cannot be associated with the Balmer series .

Chapter 42 519

P42.2 (a) λminmax

=hc

E

Lyman (n f = 1): λmin .= =⋅

=hcE1

1 24091 2

eV nm13.6 eV

nm (Ultraviolet)

Balmer (n f = 2 ): λmin .= =

⋅=

hcE2

1 24013 6

365 eV nm

1 4 eV nmb g (UV)

Paschen (n f = 3 ): λmin .= = =… 3 91 2 8212 nm nma f (Infrared)

Bracket (n f = 4 ): λmin .= = =… 4 91 2 1 4602 nm nma f (IR)

(b) Ehc

maxmin

=λ

Lyman: E Emax .= =13 6 1 eV c hBalmer: E Emax .= =3 40 2 eV c hPaschen: E Emax .= =1 51 3 eV c hBrackett: E Emax .= =0 850 4 eV c h

Section 42.2 Early Models of the Atom

P42.3 (a) For a classical atom, the centripetal acceleration is

avr

er m

Ee

rm v e

r

e

e

= =∈

= −∈

+ = −∈

2

0

2

2

2

0

2 2

0

14

4 2 8

π

π π

sodEdt

er

drdt

e ac

ec

er me

=∈

=−∈

=−∈ ∈

FHG

IKJ

2

02

0

2 2

3

2

03

2

02

2

81

6 6 4π π π π.

Therefore,drdt

er m ce

= −∈

4

202 2 2 312π

.FIG. P42.3

(b) − ∈ =× −z z12 2

02 2 2 3

2 00 10

4

010

π r m c dr e dte

T

. m

0 123

8 46 102

02 2 3

4

3

0

2 00 1010

10

π ∈= = ×

×−

−

m ce

rTe

.

. s

Since atoms last a lot longer than 0.8 ns, the classical laws (fortunately!) do not hold forsystems of atomic size.

520 Atomic Physics

P42.4 (a) The point of closest approach is found when

E K Uk q q

re= + = +0 α Au

or rk e e

Ee

min =2 79a fa f

rmin

. .

. ..=

× ⋅ ×

×= ×

−

−−

8 99 10 158 1 60 10

4 00 1 60 105 68 10

9 19 2

1314

N m C C

MeV J MeV m

2 2e ja fe ja fe j

.

(b) The maximum force exerted on the alpha particle is

Fk q q

re

maxmin

. .

..= =

× ⋅ ×

×=

−

−

α Au2 2 N m C C

m N2

9 19 2

14 2

8 99 10 158 1 60 10

5 68 1011 3

e ja fe je j

away from the

nucleus.

Section 42.3 Bohr’s Model of the Hydrogen Atom

P42.5 (a) vk em r

e

e1

2

1=

where r a12

0111 0 005 29 5 29 10= = = × −a f . . nm m

v1

9 19 2

31 116

8 99 10 1 60 10

9 11 10 5 29 102 19 10=

× ⋅ ×

× ×= ×

−

− −

. .

. ..

N m C C

kg m m s

2 2e je je je j

(b) K m ve1 12 31 6 2 181

212

9 11 10 2 19 10 2 18 10 13 6= = × × = × =− −. . . . kg m s J eVe je j

(c) Uk ere

1

2

1

9 19 2

1118

8 99 10 1 60 10

5 29 104 35 10 27 2= − = −

× ⋅ ×

×= − × = −

−

−−

. .

.. .

N m C C

m J eV

2 2e je j

P42.6 ∆En ni f

= −FHG

IKJ

13 61 12 2. eVa f

Where for ∆E > 0 we have absorption and for ∆E < 0 we have emission.

(i) for ni = 2 and n f = 5 , ∆E = 2 86. eV (absorption)

(ii) for ni = 5 and n f = 3 , ∆E = −0 967. eV (emission)

(iii) for ni = 7 and n f = 4 , ∆E = −0 572. eV (emission)

(iv) for ni = 4 and n f = 7 , ∆E = 0 572. eV (absorption)

(a) Ehc

=λ

so the shortest wavelength is emitted in transition ii .

(b) The atom gains most energy in transition i .

(c) The atom loses energy in transitions ii and iii .

Chapter 42 521

P42.7 (a) r22 20 052 9 2 0 212= =. . nm nmb ga f

(b) m vm k e

ree e

2

2

2

31 9 19 2

9

9 11 10 8 99 10 1 60 10

0 212 10= =

× × ⋅ ×

×

− −

−

. . .

.

kg N m C C

m

2 2e je je j

m ve 2259 95 10= × ⋅−. kg m s

(c) L m v re2 2 225 9 349 95 10 0 212 10 2 11 10= = × ⋅ × = × ⋅− − −. . . kg m s m kg m s2e je j

(d) K m vm v

mee

e2 2

2 22 25 2

31191

2 2

9 95 10

2 9 11 105 43 10 3 40= = =

× ⋅

×= × =

−

−−b g e j

e j.

.. .

kg m s

kg J eV

(e) Uk ere

2

2

2

9 19 2

918

8 99 10 1 60 10

0 212 101 09 10 6 80= − = −

× ⋅ ×

×= − × = −

−

−−

. .

.. .

N m C C

m J eV

2 2e je j

(f) E K U2 2 2 3 40 6 80 3 40= + = − = −. . . eV eV eV

P42.8 We use Enn =

−13 62

. eV.

To ionize the atom when the electron is in the n th level,

it is necessary to add an amount of energy given by E Enn= − =

13 62

. eV.

(a) Thus, in the ground state where n = 1, we have E = 13 6. eV .

(b) In the n = 3 level, E = =13 6

1 51.

. eV

9 eV .

P42.9 (b)1 1 1

1 097 101

21

62 27 1

2 2λ= −FHG

IKJ= × −FHG

IKJ

−Rn nf i

. me j so λ = 410 nm

(a) Ehc

= =× ⋅ ×

×= × =

−

−−

λ

6 626 10 3 00 10

410 104 85 10 3 03

34 8

919

. .. .

J s m s

m J eV

e je j

(c) fc

= =××

= ×−λ3 00 10410 10

7 32 108

914.

. Hz

522 Atomic Physics

P42.10 Starting with12 2

22

m vk e

ree=

we have vk em r

e

e

22

=

and using rn

m k ene e

=2 2

2

gives vk e

m n m k en

e

e e e

22

2 2 2=e j

or vk enne=

2

.

P42.11 Each atom gives up its kinetic energy in emitting a photon,

so12

6 626 10 3 00 10

1 216 101 63 102

34 8

718mv

hc= =

× ⋅ ×

×= ×

−

−−

λ

. .

..

J s m s

m J

e je je j

v = ×4 42 104. m s .

P42.12 The batch of excited atoms must make these six transitions to get back to state one: 2 1→ , and also3 2→ and 3 1→ , and also 4 3→ and 4 2→ and 4 1→ . Thus, the incoming light must have justenough energy to produce the 1 4→ transition. It must be the third line of the Lyman series in theabsorption spectrum of hydrogen. The absorbing atom changes from energy

Ei = − = −13 6

13 6.

. eV

1 eV2 to E f = − = −

13 60 850

..

eV4

eV2 ,

so the incoming photons have wavelength

λ =−

=× ⋅ ×

− − − ×

FHG

IKJ = × =

−

−−hc

E Ef i

6 626 10 3 00 10

0 850 13 61 00

9 75 10 97 534 8

198

. .

. ..

. . J s m s

eV eV eV

1.60 10 J m nm

e je ja f .

P42.13 (a) The energy levels of a hydrogen-like ion whose charge numberis Z are given by

EZnn = −13 6

2

2. eVa f .

Thus for Helium Z = 2a f, the energy levels are

En

nn = − =54 4

1 2 32.

, , , eV … .

(b) For He+ , Z = 2 , so we see that the ionization energy (theenergy required to take the electron from the n = 1 to the n = ∞state) is

E E E= − = −−

=∞ 1

2

2013 6 2

154 4

..

eV eV

a fa fa f .

FIG. P42.13

Chapter 42 523

*P42.14 (a)1 1 12

2 2λ= −

FHG

IKJ

Z Rn nH

f i

. The shortest wavelength, λ s , corresponds to ni = ∞ , and the longest

wavelength, λ , to n ni f= + 1 .

1 2

2λ s

H

f

Z Rn

= (1)

1 1 1

11

12

2 2

2

2

2

λ= −

+

L

NMMM

O

QPPP= −

+

FHGIKJ

L

NMM

O

QPPZ R

n n

Z Rn

n

nHf f

H

f

f

fd i(2)

Divide (1) and (2): λλ

s f

f

n

n= −

+

FHGIKJ1

1

2

∴+

= − = − =n

nf

f

s

11 1

22 80 800

λλ

..

nm63.3 nm

∴ =n f 4

From (1): Zn

Rf

s H= =

× ×=

− −

2 2

9 7 1

4

22 8 10 1 097 108 00

λ . ..

m me je j.

Hence the ion is O7+ .

(b) λ = × −+

LNMM

OQPP

RS|T|

UV|W|

−

−

7 020 8 1014

1

48 1

2 2

1

. me j a fk , k = 1 2 3, , , …

Setting k = 2 3 4, , gives λ = 41 0. nm, 33.8 nm, 30.4 nm .

P42.15 (a) The speed of the moon in its orbit is vr

T= =

×

×= ×

2 2 3 84 10

2 36 101 02 10

8

63π π .

..

m

s m s

e j.

So, L mvr= = × × × = × ⋅7 36 10 1 02 10 3 84 10 2 89 1022 3 8 34. . . . kg m s m kg m s2e je je j .

(b) We have L n=

or nL

= =× ⋅

× ⋅= ×−

2 89 101 055 10

2 74 1034

3468.

..

kg m s J s

2

.

(c) We have n L mvr mGM

rre= = = FHG

IKJ

1 2

,

so rm GM

n Rne

= =2

22 2 and

∆rr

n R n R

n Rnn

=+ −

=+1 2 12 2

2 2

a f

which is approximately equal to 2

7 30 10 69

n= × −. .

524 Atomic Physics

Section 42.4 The Quantum Model of the Hydrogen Atom

P42.16 The reduced mass of positronium is less than hydrogen, so the photon energy will be less forpositronium than for hydrogen. This means that the wavelength of the emitted photon will belonger than 656.3 nm. On the other hand, helium has about the same reduced mass but more chargethan hydrogen, so its transition energy will be larger, corresponding to a wavelength shorter than656.3 nm.

All the factors in the given equation are constant for this problem except for the reduced massand the nuclear charge. Therefore, the wavelength corresponding to the energy difference for thetransition can be found simply from the ratio of mass and charge variables.

For hydrogen, µ =+

≈m m

m mmp e

p ee . The photon energy is ∆E E E= −3 2 .

Its wavelength is λ = 656 3. nm, where λ = =cf

hcE∆

.

(a) For positronium, µ =+

=m m

m mme e

e e

e

2

so the energy of each level is one half as large as in hydrogen, which we could call“protonium”. The photon energy is inversely proportional to its wavelength , so forpositronium,

λ µ32 2 656 3 1 31= =. . nm ma f (in the infrared region).

(b) For He+ , µ ≈ me , q e1 = , and q e2 2= ,

so the transition energy is 2 42 = times larger than hydrogen.

Then, λ 32656

4164= FHG

IKJ = nm nm (in the ultraviolet region).

P42.17 (a) ∆ ∆x p ≥2

so if ∆x r= , ∆pr

≥2

.

(b) Choosing ∆pr

≈ , Kpm

p

m m re e e

= ≈ =2 2 2

22 2 2

∆b g

Uk ere=

− 2

, so E K Um r

k ere

e= + ≈ −2

2

2

2.

(c) To minimize E,

dEdr m r

k er

rm k e

ae

e

e e

= − + = → = =2

3

2

2

2

2 00 (the Bohr radius).

Then, Em

m k ek e

m k e m k e

e

e ee

e e e e=FHG

IKJ −

FHG

IKJ = − = −

2 2

2

22

2

2

2 4

22 213 6. eV .

Chapter 42 525

Section 42.5 The Wave Functions of Hydrogen

P42.18 ψπ

1

03

10

sr ar

aea f = − (Eq. 42.22)

P rra

esr a

1

2

03

240a f = − (Eq. 42.25)

FIG. P42.18

P42.19 (a) ψ π ψ ππ

2 2 2

0 03

2 2

0

4 41

0dV r dra

r e drr az z z= =FHGIKJ

∞−

∞

Using integral tables, ψ 2

02

2 20

02

0 02

022

22

210dV

ae r a r

aa

ar az = − + +FHG

IKJ

LNMM

OQPP = −FHGIKJ −FHGIKJ =

−

∞

so the wave function as given is normalized.

(b) P r dra

r e dra aa

ar a

a

a

0 0

0

0

0

0

0

2 32 2

2

3 2

03

2 2

2

3 2

4 41

2→−= =

FHGIKJz zπ ψ π

π

Again, using integral tables,

Pa

e r a ra

ae

ae

aa a

r a

a

a

0 00

0

02

22 17

45

40 497

02

2 20

02

2

3 2

02

3 02

1 02

2 3 2→− − −= − + +FHG

IKJ

LNMM

OQPP = −

FHGIKJ −FHGIKJ

LNMM

OQPP = . .

P42.20 ψ = −13

1

2 03 2

0

2 0

a

ra

e r a

b gso P r r

ra

err a= = −4 4

242 2 2

2

05

0π ψ π .

SetdPdr a

r e ra

er a r a= + −FHGIKJ

LNMM

OQPP=− −4

244

10

05

3 4

0

0 0π

.

Solving for r, this is a maximum at r a= 4 0 .

P42.21 ψπ

= −1

03

0

ae r a 2 2 2

05 0

0

rddr r a

era

r aψ

πψ=

−= −−

ddr a

ea

r a2

207

02

1 10

ψ

πψ= =− − −

FHG

IKJ −

∈=

2

02

0

2

021 2

4m a rae

rE

eψ

πψ ψ

But am ee

0

20

2

4=

∈πb g

so −∈

=e

aE

2

0 08π

or Ek e

ae= −

2

02.

This is true, so the Schrödinger equation is satisfied.

526 Atomic Physics

P42.22 The hydrogen ground-state radial probability density is

P r rra

rasa f = = −FHGIKJ4

4 221

22

03

0π ψ exp .

The number of observations at 2 0a is, by proportion

NP a

P a

a

a

ee

ea a

a a= = = =−

−−1 000

2

21 000

2

21 000 16 7970

0

02

02

43

0 0

0 0

b gb g

b gb g

a f times .

Section 42.6 Physical Interpretation of the Quantum Numbers

Note: Problems 17 and 25 in Chapter 29 and Problem 68 in Chapter 30 can be assigned with this section.

P42.23 (a) In the 3d subshell, n = 3 and = 2 ,

we have n 3 3 3 3 3 3 3 3 3 32 2 2 2 2 2 2 2 2 2

m +2 +2 +1 +1 0 0 –1 –1 –2 –2ms +1/2 –1/2 +1/2 –1/2 +1/2 –1/2 +1/2 –1/2 +1/2 –1/2

(A total of 10 states)

(b) In the 3p subshell, n = 3 and = 1 ,

we have n 3 3 3 3 3 31 1 1 1 1 1

m +1 +1 +0 +0 –1 –1ms +1/2 –1/2 +1/2 –1/2 +1/2 –1/2

(A total of 6 states)

P42.24 (a) For the d state, = 2 , L = = × ⋅−6 2 58 10 34. J s .

(b) For the f state, = 3 , L = + = = × ⋅−1 12 3 65 10 34a f . J s .

P42.25 L = + 1a f : 4 714 10 16 626 10

234

34

..

× = +×F

HGIKJ

−−

a fπ

+ =×

×= × ≈ = +

−

−1

4 714 10 2

6 626 101 998 10 20 4 4 1

4 2 2

34 21a f e j a f

e ja f.

..

π

so = 4 .

Chapter 42 527

P42.26 The 5th excited state has n = 6 , energy−

= −13 6

0 378.

. eV

36 eV .

The atom loses this much energy:hcλ=

× ⋅ ×

× ×=

−

− −

6 626 10 3 00 10

1 090 10 1 60 101 14

34 8

9 19

. .

..

J s m s

m J eV eV

e je je je j

to end up with energy − − = −0 378 1 14 1 52. . . eV eV eV

which is the energy in state 3: − = −13 6

1 51.

. eV

3 eV3 .

While n = 3 , can be as large as 2, giving angular momentum + =1 6a f .

P42.27 (a) n = 1 : For n = 1 , = 0 , m = 0 , ms = ±12

n m ms

1 0 0 –1/21 0 0 +1/2

Yields 2 sets; 2 2 1 22 2n = =a f(b) n = 2 : For n = 2 ,

we have

n m ms

2 0 0 ±1/22 1 –1 ±1/22 1 0 ±1/22 1 1 ±1/2

yields 8 sets; 2 2 2 82 2n = =a fNote that the number is twice the number of m values. Also, for each there are 2 1+a fdifferent m values. Finally, can take on values ranging from 0 to n −1 .

So the general expression is number = +−

∑ 2 2 10

1a f

n.

The series is an arithmetic progression: 2 6 10 14+ + + …

the sum of which is number = + −n

a n d2

2 1a f

where a = 2 , d = 4 : number = + − =n

n n2

4 1 4 2 2a f .

(c) n = 3 : 2 1 2 3 2 5 2 6 10 18a f a f a f+ + = + + = 2 2 3 182 2n = =a f

(d) n = 4 : 2 1 2 3 2 5 2 7 32a f a f a f a f+ + + = 2 2 4 322 2n = =a f

(e) n = 5 : 32 2 9 32 18 50+ = + =a f 2 2 5 502 2n = =a f

528 Atomic Physics

P42.28 For a 3d state, n = 3 and = 2 .

Therefore, L = + = = × ⋅−1 6 2 58 10 34a f . J s

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

so Lz can have the values and − −2 0 2, , , .

Using the relation cosθ =LL

z

we find the possible values of θ 145 114 90 0 65 9 35 3° ° ° ° °, , . , . , . and .

P42.29 (a) Density of a proton: ρπ

= =×

×= ×

−

−

mV

1 67 10

1 00 103 99 10

27

15 317.

..

kg

4 3 m kg m3

b g e j.

(b) Size of model electron: rm

=FHGIKJ =

×

×

FHGG

IKJJ = ×

−−3

4

3 9 11 10

4 3 99 108 17 10

1 3 31

17

1 3

17

π ρ π

.

..

kg

kg m m

3

e je j

.

(c) Moment of inertia: I mr= = × × = × ⋅− − −25

25

9 11 10 8 17 10 2 43 102 31 17 2 63. . . kg m kg m2e je j

L IIvrz = = =ω

2.

Therefore, vrI

= =× ⋅ ×

× × ⋅= ×

− −

−2

6 626 10 8 17 10

2 2 2 43 101 77 10

34 17

6312

. .

..

J s m

kg m m s

2

e je je jπ

.

(d) This is 5 91 103. × times larger than the speed of light.

P42.30 In the N shell, n = 4 . For n = 4 , can take on values of 0, 1, 2, and 3. For each value of , m can be− to in integral steps. Thus, the maximum value for m is 3. Since L mz = , the maximum

value for Lz is Lz = 3 .

P42.31 The 3d subshell has = 2 , and n = 3 . Also, we have s = 1 .

Therefore, we can have n m s ms= = = − − = = −3 2 2 1 0 1 2 1 1 0 1, ; , , , , ; ; , ,and

leading to the following table:

n 3 3 3 3 3 3 3 3 3 3 3 3 3 3 32 2 2 2 2 2 2 2 2 2 2 2 2 2 2

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

Chapter 42 529

Section 42.7 The Exclusion Principle and the Periodic Table

P42.32 (a) 1 2 22 2 4s s p

(b) For the 1s electrons, n = 1 , = 0 , m = 0 , ms = +12

and −12

.

For the two 2s electrons, n = 2 , = 0 , m = 0 , ms = +12

and −12

.

For the four 2p electrons, n = 2 ; = 1 ; m = −1 , 0, or 1; and ms = +12

or −12

.

P42.33 The 4s subshell fills first , for potassium and calcium, before the 3d subshell starts to fill for

scandium through zinc. Thus, we would first suppose that Ar 3 44 2d s would have lower energy

than Ar 3 45 1d s . But the latter has more unpaired spins, six instead of four, and Hund’s rule

suggests that this could give the latter configuration lower energy. In fact it must, for Ar 3 45 1d s isthe ground state for chromium.

P42.34 Electronic configuration: Sodium to Argon

1 2 22 2 6s s p +3 1s → Na11

+3 2s → Mg12

+3 32 1s p → Al13

+3 32 2s p → Si14

+3 32 3s p → P15

+3 32 4s p → S16

+3 32 5s p → Cl17

+3 32 6s p → Ar18

1 2 2 3 3 42 2 6 2 6 1s s p s p s → K19

*P42.35 In the table of electronic configurations in the text, or on a periodic table, we look for the elementwhose last electron is in a 3p state and which has three electrons outside a closed shell. Its electronconfiguration then ends in 3 32 1s p . The element is aluminum .

530 Atomic Physics

P42.36 (a) For electron one and also for electron two, n = 3 and = 1 . The possible states are listedhere in columns giving the other quantum numbers:

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

one ms

12

12

12

12

12

−12

−12

−12

−12

−12

12

12

12

12

12

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

two ms−

12

12

−12

12

−12

12

12

−12

12

−12

12

−12

−12

12

−12

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

one ms−

12

−12

−12

−12

−12

12

12

12

12

12

−12

−12

−12

−12

−12

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

two ms

12

−12

12

12

−12

12

−12

12

−12

−12

12

−12

12

−12

12

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

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

electron independently.

P42.37 (a) n + 1 2 3 4 5 6 7subshell 1s 2s 2p , 3s 3p , 4s 3d , 4p , 5s 4d , 5p , 6s 4f , 5d , 6p , 7s

(b) Z = 15 : Filled subshells: 1 2 2 3s s p s, , ,(12 electrons)

Valence subshell: 3 electrons in 3p subshellPrediction: Valence = +3 or –5Element is phosphorus, Valence = +3 or –5 (Prediction correct)

Z = 47 : Filled subshells: 1 2 2 3 3 4 3 4 5s s p s p s d p s, , , , , , , ,

(38 electrons)Outer subshell: 9 electrons in 4d subshellPrediction: Valence = −1Element is silver, (Prediction fails) Valence is +1

Z = 86 : Filled subshells: 1 2 2 3 3 4 3 4 5 4 5 6s s p s p s d p s d p s, , , , , , , , , , , ,4 5 6f d p, ,(86 electrons)

Prediction Outer subshell is full: inert gasElement is radon, inert (Prediction correct)

P42.38 Listing subshells in the order of filling, we have for element 110,

1 2 2 3 3 4 3 4 5 4 5 6 4 5 6 7 5 62 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 8s s p s p s d p s d p s f d p s f d .

In order of increasing principal quantum number, this is

1 2 2 3 3 3 4 4 4 4 5 5 5 5 6 6 6 72 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 8 2s s p s p d s p d f s p d f s p d s .

Chapter 42 531

*P42.39 In the ground state of sodium, the outermost electron is in an s state. This state is sphericallysymmetric, so it generates no magnetic field by orbital motion, and has the same energy no matterwhether the electron is spin-up or spin-down. The energies of the states 3pA and 3pB above 3s are

hfhc

1 = λ and hf

hc2

2=λ

.

The energy difference is

21 1

1 2µ

λ λBB hc= −FHG

IKJ

so Bhc

B= −FHG

IKJ =

× ⋅ ×

× ×−

×FHG

IKJ

−

− − −21 1 6 63 10 3 10

2 9 27 10

1588 995 10

1589 592 101 2

34 8

24 9 9µ λ λ

.

. . .

J s m s

J T m m

e je je j

B = 18 4. T .

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

P42.40 (a) n m= = =3 0 0, ,

n m= = = −3 1 1 0 1, , , ,

For n m= = = − −3 2 2 1 0 1 2, , , , , ,

(b) ψ 300 corresponds to EZ E

n300

20

2

2

2

2 13 6

36 05= − = − = −

..

a f eV .

ψ ψ ψ31 1 310 311− , , have the same energy since n is the same.

ψ ψ ψ ψ ψ32 2 32 1 320 321 322− −, , , , have the same energy since n is the same.

All states are degenerate.

P42.41 Ehc

e V= =λ

∆ :6 626 10 3 00 10

10 0 101 60 10

34 8

919

. .

..

× ⋅ ×

×= ×

−

−−

J s m s

m

e je je j e j∆V

∆V = 124 V

P42.42 Some electrons can give all their kinetic energy K e Ve = ∆ to the creation of a single photon of x-radiation, with

hfhc

e V

hce V V V

= =

= =× ⋅ ×

×=

⋅−

−

λ

λ

∆

∆ ∆ ∆

6 626 1 10 2 997 9 10

1 602 2 10

1 24034 8

19

. .

.

J s m s

C

nm Ve je je j

532 Atomic Physics

P42.43 Following Example 42.9 Eγ = − = × = × −34

42 1 13 6 1 71 10 2 74 102 4 15a f a f. . . eV eV J

f = ×4 14 1018. Hz

and λ = 0 072 5. nm .

P42.44 Ehc

= =⋅

=⋅

λ λ λ1 240 1 240 eV nm keV nm.

For λ1 0 018 5= . nm , E = 67 11. keV

λ 2 0 020 9= . nm , E = 59 4. keV

λ 3 0 021 5= . nm , E = 57 7. keV

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

FIG. P42.44

L shell keV= 11 8. M shell keV= 10 1. N shell keV= 2 39. .

P42.45 The Kβ x-rays are emitted when there is a vacancy in the ( n = 1 ) K shell and an electron from the

( n = 3 ) M shell falls down to fill it. Then this electron is shielded by nine electrons originally and byone in its final state.

hc Z Z

Z ZZ Z

Z

λ= −

−+

−

× ⋅ ×

× ×= − + − + − +

FHG

IKJ

× = −FHG

IKJ

−

− −

13 6 9

3

13 6 1

1

6 626 10 3 00 10

0 152 10 1 60 1013 6

918

9819

2 1

8 17 10 13 68

98

2

2

2

2

34 8

9 19

22

32

. .

. .

. ..

. .

a f a f

e je je je j

a f

a f

eV eV

J s m s

m J eV eV

eV eV

so 6018

98

2

= −Z

and Z = 26 Iron .

Section 42.9 Spontaneous and Stimulated Transitions

Section 42.10 Lasers

P42.46 The photon energy is E Ehc

4 3 20 66 18 70 1 96− = − = =. . .a f eV eVλ

λ =× ⋅ ×

×=

−

−

6 626 10 3 00 10

1 96 1 60 10633

34 8

19

. .

. .

J s m s

J nm

e je je j

.

P42.47 fEh e

= =× ⋅

×FHG

IKJ ⋅FHG

IKJ = ×−

−−0 117

6 630 101 60 10

2 82 1034

1913 1.

..

. eV

J s C 1 J

1 V C s

λ µ= =×

×=−

cf

3 00 102 82 10

10 68

13 1

..

. m s s

m , infrared

Chapter 42 533

P42.48 (a) I =×

× ×LNM

OQP= ×

−

− −

3 00 10

1 00 10 15 0 104 24 10

3

9 6 215

.

. ..

J

s m W m2e j

e j e jπ

(b) 3 00 100 600 10

30 0 101 20 10 7 503

9 2

6 212.

.

.. .×

×

×= × =−

−

−

− J m

m J MeVe j e j

e j

P42.49 E t= = × × =−P ∆ 1 00 10 1 00 10 0 010 06 8. . . W s Je je j

E hfhc

γ λ= = =

× ×

×= ×

−

−−

6 626 10 3 00 10

694 3 102 86 10

34 8

919

. .

..

e je j J J

NE

E= =

×= ×−

γ

0 010 02 86 10

3 49 101916.

.. photons

*P42.50 (a)NN

N e

N ee eg

E k

gE k

E E k hc kB

B

B B3

2

300

300300 300

3

2

3 2= = =− ⋅

− ⋅− − ⋅ − ⋅

K

K K K

b g

b gb g b g b gλ

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

NN

e

NN

e

3

2

6 63 10 3 10 632 8 10 1.38 10 300

3

2

75 9 33

34 8 9 23

1 07 10

=

= = ×

− × ⋅ × × ×

− −

− − −. .

. .

J s m s m J K Ke je j e je ja f

(b)NN

eu E E k Tu B= − −b g

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

Since E E Ehc

u − = =photon λNN

eu hc k TB= − λ .

Thus, we require 1 02. = −e hc k TBλ

or ln ..

. .1 02

6 63 10 3 10

632 8 10 1 38 10

34 8

9 23a f e je j

e je j= −

× ⋅ ×

× ×

−

−

J s m s

m J K T

T = −×

= − ×2 28 10

1 021 15 10

46.

ln ..a f K .

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

(c) Because E Eu − > 0 , and in any real equilibrium state T > 0 ,

e E E k Tu B− − <b g 1 and N Nu < .

Thus, a population inversion cannot happen in thermal equilibrium.

534 Atomic Physics

*P42.51 (a) The light in the cavity is incident perpendicularly on themirrors, although the diagram shows a large angle ofincidence for clarity. We ignore the variation of the indexof refraction with wavelength. To minimize reflection ata vacuum wavelength of 632.8 nm, the net phasedifference between rays (1) and (2) should be 180°. Thereis automatically a 180° shift in one of the two rays uponreflection, so the extra distance traveled by ray (2) shouldbe one whole wavelength:

1 2

SiO2

FIG. P42.51

2

2632 8

217

tn

tn

=

= = =

λ

λ . nm2 1.458

nma f

(b) The total phase difference should be 360°, including contributions of 180° by reflection and180° by extra distance traveled

22

4543

93 1

tn

tn

=

= = =

λ

λ nm4 1.458

nma f .

Additional Problems

*P42.52 (a) Using the same procedure that was used in the Bohr model of the hydrogen atom, we applyNewton’s second law to the Earth. We simply replace the Coulomb force by thegravitational force exerted by the Sun on the Earth and find

GM M

rM

vr

S EE2

2

= (1)

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

M vr n nE = = …1 2 3, , ,b g .Solving for v gives

vn

M rE= . (2)

Substituting (2) into (1), we find

rn

GM MS E

=2 2

2 . (3)

continued on next page

Chapter 42 535

(b) Solving (3) for n gives

n GM rM

SE= . (4)

Taking MS = ×1 99 1030. kg , and ME = ×5 98 1024. kg , r = ×1 496 1011. m ,

G = × −6 67 10 11. Nm kg2 2 , and = × −1 055 10 34. Js , we find

n = ×2 53 1074. .

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

rn

GM MnS E

=2 2

2 and rn

GM MnS E

+ =+

1

2 2

2

1a f.

Hence, the separation between these two orbits is

∆rGM M

n nGM M

nS E S E

= + − = +2

22 2

2

21 2 1a f a f .Since n is very large, we can neglect the number 1 in the parentheses and express theseparation as

∆rGM M

nS E

≈ = × −2

2632 1 18 10a f . m .

This number is much smaller than the radius of an atomic nucleus ~10 15− me j , so the

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

*P42.53 (a) ∆Ee Bme

= =× × ⋅

×

⋅⋅ ⋅FHG

IKJ

⋅

⋅FHGIKJ = ×

=

− −

−−

1 60 10 5 26

2 9 11 109 75 10

609

19 34

3123

. .

..

C 6.63 10 J s T

kg

N sT C m

kg mN s

J

eV

2

e ja fe jπ

µ

(b) k TB = × × = × =− − −1 38 10 80 10 1 10 10 6 9023 3 24. . . J K K J eVe je j µ

(c) fEh

= =×× ⋅

= ×−∆ 9 75 10

1 47 1023

11..

J6.63 10 J s

Hz-34

λ = =×

×= × −c

f3 10

1 47 102 04 10

8

113 m s

Hz m

..

*P42.54 (a) Probability = ′ ′ = ′ ′∞

− ′∞z zP r dr

ar e drs

r

r a

r1

03

2 240a f = −

′+

′+

FHG

IKJ

LNMM

OQPP

− ′

∞2 2

12

02

0

2 0ra

ra

e r a

r

,

using integration by parts, or Example 42.5

= + +FHG

IKJ

−2 21

2

02

0

2 0r

ar

ae r a

continued on next page

536 Atomic Physics

(b)1.2

1

0.8

0.6

0.4

0.2

00 1 2 3 4 5

Probability Curve for Hydrogen

r /a0

FIG. P42.66

(c) The probability of finding the electron inside or outside the sphere of radius r is 12

.

∴ + +FHG

IKJ =−2 2

112

2

02

0

2 0r

ar

ae r a or z z ez2 2 2+ + = where z

ra

=2

0

One can home in on a solution to this transcendental equation for r on a calculator, theresult being r a= 1 34 0. to three digits.

P42.55 Let r represent the distance between the electron and the positron. The two move in a circle of

radius r2

around their center of mass with opposite velocities. The total angular momentum of the

electron-positron system is quantized to according to

Lmvr mvr

nn = + =2 2

where n = 1 2 3, , , … .

For each particle, F ma∑ = expands tok er

mvr

e2

2

2

2= .

We can eliminate vnmr

= to findk e

rmnm r

e2 2 2

2 22

= .

So the separation distances are rn

mk ea n n

e

= = = × −22 1 06 10

2 2

2 02 10 2. me j .

The orbital radii are r

a n2 0

2= , the same as for the electron in hydrogen.

The energy can be calculated from E K U mv mvk e

re= + = + −

12

12

2 22

.

Since mvk e

re2

2

2= , E

k er

k er

k er

k ea n n

e e e e= − = − =−

= −2 2 2 2

02 22 2 4

6 80. eV.

P42.56 (a) The energy difference between these two states is equal to the energy that is absorbed.

Thus, E E E= − =−

−−

= = × −2 1

1813 64

13 61

10 2 1 63 10. .

. . eV eV

eV Ja f a f

.

(b) E k TB=32

or TEkB

= =×

×= ×

−

−

23

2 1 63 10

3 1 38 107 88 10

18

234

.

..

J

J K K

e je j

.

Chapter 42 537

P42.57 hf Emk eh n n

e= =−

−FHG

IKJ

∆4

21

1

12 2 4

2 2 2π

a f

fmk eh

n

n ne=

−

−

FHG

IKJ

2 2 1

1

2 2 4

3 2 2

π

a f

As n approaches infinity, we have f approaching2 22 2 4

3 3π mk e

h ne

The classical frequency is fv

rk em re= =

21

212

3 2π π

where rn hmk ee

=2 2

24π

Using this equation to eliminate r from the expression for f, fmk eh n

e=2 22 2 4

3 3π

P42.58 (a) The energy of the ground state is: Ehc

11 240

8 16= − = −⋅

= −λ series limit

eV nm152.0 nm

eV. .

From the wavelength of the Lyman α line: E Ehc

2 11 240

6 12− = =⋅

=λ

nm eV202.6 nm

eV.

E E2 1 6 12 2 04= + = −. . eV eV .

The wavelength of the Lyman β line gives: E E3 11 240

7 26− =⋅

= nm eV

170.9 nm eV.

so E3 0 902= − . eV .

Next, using the Lyman γ line gives: E E4 11 240

7 65− =⋅

= nm eV

162.1 nm eV.

and E4 0 508= − . eV .

From the Lyman δ line, E E5 11 240

7 83− =⋅

= nm eV

158.3 nm eV.

so E5 0 325= − . eV .

(b) For the Balmer series,hc

E Eiλ= − 2 , or λ =

⋅−

1 240

2

nm eVE Ei

.

For the α line, E Ei = 3 and so λ a =⋅

− − −=

1 2400 902 2 04

1 090 nm eV

eV eV nm

. .a f a f .

Similarly, the wavelengths of the β line, γ line, and the short wavelength limit are found tobe: 811 nm , 724 nm , and 609 nm .

continued

538 Atomic Physics

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

0 600 202 6 122. . nm nma f = , 0 600 170 9 103. . nm nma f = , 0 600 162 1 97 3. . . nm nma f = ,

0 600 158 3 95 0. . . nm nma f = , and 0 600 152 0 91 2. . . nm nma f =

These are seen to be the wavelengths of the α, β, γ, and δ lines as well as the shortwavelength limit for the Lyman series in Hydrogen.

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

′=

′=

−

+=

ff

v c

v cλλ

1

10 600

b gb g . yielding v c= 0 471. .

P42.59 The wave function for the 2s state is given by Eq. 42.26: ψπ2

0

3 2

0

214 2

12 0

sr ar

ara

ea f = FHGIKJ −LNMOQP

− .

(a) Taking r a= = × −0

100 529 10. m

we find ψπ2 0 10

3 21 2 14 3 21

4 21

0 529 102 1 1 57 10s a eb g =

×FHG

IKJ − = ×−

− −

..

m m .

(b) ψ 2 02 14 3 2 2 28 31 57 10 2 47 10s ab g e j= × = ×− −. . m m

(c) Using Equation 42.24 and the results to (b) gives P a a as s2 0 02

2 02 8 14 8 69 10b g b g= = × −π ψ . m .

*P42.60 From Figure 42.20, a typical ionization energy is 8 eV. For internal energy to ionize most of theatoms we require

32

8k TB = eV : T =× ×

×

−

−

2 8 1 60 10

3 1 38 10

19

23

.

.~

J

J K between 10 K and 10 K4 5e j

e j.

P42.61 (a) 3 00 10 14 0 10 4 208 12. . .× × =− m s s mme je j

(b) Ehc

= = × −

λ2 86 10 19. J

N =×

= ×−3 00

1 05 101919.

. J

2.86 10 J photons

(c) V = =4 20 3 00 1192. . mm mm mm3a f a fπ

n =×

= × −1 05 10119

8 82 1019

16 3.. mm

Chapter 42 539

P42.62 (a) The length of the pulse is ∆ ∆L c t= .

(b) The energy of each photon is Ehc

γ λ= so N

EE

Ehc

= =γ

λ.

(c) V Ld

= ∆ π2

4n

NV c t d

Ehc

= =FHG

IKJFHGIKJ

42∆ π

λ

*P42.63 The fermions are described by the exclusion principle. Two of them, one spin-up and one spin-down, will be in the ground energy level, with

d LNN = =12λ , λ = =2L

hp

, and phL

=2

K mvpm

hmL

= = =12 2 8

22 2

2 .

The third must be in the next higher level, with

dL

NN = =2 2

λ, λ = L , and p

hL

= Kpm

hmL

= =2 2

22 2.

The total energy is thenhmL

hmL

hmL

hmL

2

2

2

2

2

2

2

28 8 23

4+ + = .

P42.64 ∆∆

zat F

mt

dB dz

mx

vz z z= =FHGIKJ = F

HGIKJ

2

0

2

0

2

212 2

µ b gand µ z

e

em

=2

dBdz

m z v m

x e

dBdz

z e

z

= =× × ×

× × ⋅

=

− − −

− −

2 2 2 108 1 66 10 10 10 2 9 11 10

1 00 1 60 10 1 05 10

0 389

02

2

27 3 4 31

19 34

∆

∆

a f b g a fe je je je je je je j

. .

. . .

.

kg m m s kg

m C J s

T m

2 2

2

P42.65 We use ψ π2 03 1 2

0

214

2 2 0s

r ar ara

ea f e j= −FHGIKJ

− − .

By Equation 42.24, P r rra

ra

e r aa f = =FHGIKJ −FHGIKJ

−418

22 22

03

0

2

0π ψ .

(a)dP r

drr

ara

ra a

ra

ra

ra a

e r aa f= −

FHGIKJ −

FHGIKJ −FHGIKJ − −FHGIKJFHGIKJ

LNMM

OQPP =−1

82

22 1

2 21

003

0

2 2

03

0 0

2

03

0

2

0

0

or 18

2 2 22

2 003

0 0 0 0 0

0r

ara

ra

ra

ra

ra

e r aFHGIKJ −FHGIKJ −FHGIKJ − − −

FHGIKJ

LNMM

OQPP

=− .

The roots of dPdr

= 0 at r = 0 , r a= 2 0 and r = ∞ are minima with P ra f = 0 .

continued

540 Atomic Physics

Therefore we require ..... = −FHGIKJ +FHGIKJ =4

60

0 0

2r

ara

with solutions r a= ±3 5 0e j .

We substitute the last two roots into P ra f to determine the most probable value:

When r a a= − =3 5 0 763 90 0e j . , P ra

a f = 0 051 9

0

..

When r a a= + =3 5 5 2360 0e j . , P ra

a f = 0 191

0

..

Therefore, the most probable value of r is 3 5 5 2360 0+ =e ja a. .

(b) P r drra

ra

e drr aa f0

2

03

0

2

0

18

2 0

∞−

∞z z= FHGIKJ −FHGIKJ

Let ura

=0

, dr a du= 0 ,

P r dr u u u e dr u u u e du u u u eu u ua f e j e j e j0

2 2

0

4 3 2

0

4 2

0

18

4 418

4 418

4 8 8 1∞

−∞

−∞

−∞

z z z= − + = − + = − + + + =

This is as required for normalization.

P42.66 Ehc

E= =⋅

=λ λ

1 240 eV nm∆

λ1 310= nm, so ∆E1 4 00= . eV

λ 2 400= nm, ∆E2 3 10= . eV

λ 3 1 378= nm, ∆E3 0 900= . eV

and the ionization energy = 4 10. eV .FIG. P42.66

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

P42.67 With one vacancy in the K shell, excess energy

∆E Z≈ − − −FHGIKJ =1 13 6

12

11

5 4022 2a f a f. . eV keV .

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

Eionization eV

eV≈ =2 13 6

43 40

2

2

..

a f.

Note the experimental ionization energy is 6.76 eV.

K E E= − ≈∆ ionization keV5 39. .

Chapter 42 541

P42.68 (a) The configuration we may model as SN NS has higher energy than SN SN . The

higher energy state has antiparallel magnetic moments, so it has parallel spins of the

oppositely charged particles.

(b) Ehc

= = × =−

λµ9 42 10 5 8925. . J eV

(c) ∆ ∆E t ≈2

so ∆E ≈× ⋅

× ×

FHG

IKJ = ×

−

−−1 055 10

3 16 10

1 001 04 10

34

7 1930.

.

..

J s

2 10 yr s yr

eV1.60 10 J

eV7e je j

P42.69 Pra

e dr z e dzr a

a

z= =−∞

−∞

z z4 12

2

03

2

2 50

2

5 00

0

0. .

where zr

a≡

2

0

P z z e ez= − + + = − + + + = FHGIKJ =−

∞−1

22 2

12

012

25 0 10 0 2 00372

0 006 74 0 1252

5 00

5e j a f b g.

. . . . .

P42.70 (a) One molecule’s share of volume

Al: V = =×

FHG

IKJ

×FHG

IKJ = ×

−−mass per molecule

density g mol

molecules mol m

g m

3327 0

6 02 101 00 10

2 701 66 1023

629.

..

..

V3 10 12 55 10= × − −. m~10 nm .

U: V =×

FHG

IKJ

×FHG

IKJ = ×

−−238 1 00 10

18 92 09 10

629 g

6.02 10 molecules m

g m23

33.

..

V3 10 12 76 10 10= × − −. ~ m nm .

(b) The outermost electron in any atom sees the nuclear charge screened by all the electronsbelow it. If we can visualize a single outermost electron, it moves in the electric field of netcharge, + − − = +Ze Z e e1a f , the charge of a single proton, as felt by the electron in hydrogen.So the Bohr radius sets the scale for the outside diameter of every atom. An innermostelectron, on the other hand, sees the nuclear charge unscreened, and the scale size of its (K-

shell) orbit is aZ

0 .

P42.71 ∆E B hf= =2µB

so 2 9 27 10 0 350 6 626 1024 34. . .× = × ⋅− − J T T J se ja f e j f

and f = ×9 79 109. Hz .

542 Atomic Physics

P42.72 ψ π251 2

0

3 2

0

2

0

214

21

2 20 0=FHGIKJ −FHGIKJ = −

FHGIKJ

− − −a fa

ra

e Ara

er a r a ddr

Aea

ra

r aψ= − +

FHG

IKJ

− 2

0 02

02

2

ddr

Aea

ra

r a2

2

2

02

0

0 32 4

ψ=FHG

IKJ −FHG

IKJ

−

Substituting into Schrödinger’s equation and dividing by Ae r a− 2 0 , we will have a solution if

− + + + − = −54 8

2 22

2

02

2

0

2

03

2

0

2

0m ak ea

rm a m a r

k er

EErae

e

e e

e .

Now with am e ke e

0

2

2= , this reduces to

− −FHGIKJ = −FHGIKJ

m e k ra

Era

e e4 2

20 08

2 2 .

This is true, so ψ 25 is a solution to the Schrödinger equation, provided E E= = −14

3 401 . eV.

P42.73 (a) Suppose the atoms move in the +x direction. The absorption of a photon by an atom is acompletely inelastic collision, described by

mvh

mvi fi i i+ − =λ e j so v v

hmf i− = −λ

.

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

av v

th

m tf i=−

= − −× ⋅

×−

−

− − −∆ ∆λ~

.~

6 626 10

500 10 1010

34

25 9 86 J s

10 kg m s m s2

e je je j.

(b) With constant average acceleration,

v v a xf i2 2 2= + ∆ 0 10 2 103 2 6~ m s m s2e j e j+ − ∆x

so ∆x ~ ~10

101

3 2

6

m s

m s m2

e j.

P42.741 4 1 4 4 1

2

12

03

2

0 03

2

0 03

02

0

0 0

rra

er

dra

re dra a a

r a a r= = = =−∞

−∞

z z b g b g

We compare this to 1 1

3 22

30 0r a a= = , and find that the average reciprocal value is NOT the

reciprocal of the average value.

Chapter 42 543

ANSWERS TO EVEN PROBLEMS

P42.2 (a) 91.2 nm, 365 nm, 821 nm, 1 46. mµ ; P42.38 1 2 2 3 3 3 4 4 42 2 6 2 6 10 2 6 10s s p s p d s p d4 5 5 5 5 6 6 6 714 2 6 10 14 2 6 8 2f s p d f s p d s(b) 13.6 eV, 3.40 eV, 1.51 eV, 0.850 eV

P42.4 (a) 56.8 fm; (b) 11.3 N away from thenucleus

P42.40 (a) = 0 with m = 0 ; = 1 with m = 1, 0,or 1; and = 2 with m = −2 , –1, 0, 1, 2;(b) –6.05 eV

P42.6 (a) ii; (b) i; (c) ii and iii

P42.42 see the solutionP42.8 (a) 13.6 eV; (b)1.51 eV

P42.44 L shell 11.8 keV, M shell 10.1 keV, N shell2.39 keV, see the solutionP42.10 see the solution

P42.12 97.5 nm P42.46 see the solution

P42.14 (a) O7+ ; (b) 41.0 nm, 33.8 nm, 30.4 nm P42.48 (a) 4 24. PW m2 ; (b) 1 20 7 50. . pJ MeV=

P42.16 (a) 1 31. mµ ; (b) 164 nm P42.50 (a) 1 07 10 33. × − ; (b) − ×1 15 106. K ;(c) negative temperatures do not describesystems in thermal equilibriumP42.18 see the solution

P42.52 (a) see the solution; (b) 2 53 1074. × ;P42.20 4 0a(c) 1 18 10 63. × − m , unobservably small

P42.22 797 times

P42.54 (a) Probability = + +FHG

IKJ

−2 21

2

02

0

2 0r

ar

ae r a ;P42.24 (a) 6 2 58 10 34= × ⋅−. J s;

(b) 12 3 65 10 34= × ⋅−. J s (b) see the solution; (c) 1 34 0. a

P42.26 6 P42.56 (a) 10 2 1 63. . eV aJ= ; (b) 7 88 104. × K

P42.28 6 ; −2 , − , 0, , 2 ; 145°, 114°, 90.0°,65.9°, 35.3°

P42.58 (a) –8.16 eV, –2.04 eV, –0.902 eV, –0.508 eV,–0.325 eV;(b) 1 090 nm, 811 nm, 724 nm, 609 nm;

P42.30 3 (c) see the solution; (d) The spectrumcould be that of hydrogen, Doppler shiftedby motion away from us at speed 0 471. c .P42.32 (a) 1 2 22 2 4s s p ;

(b) n

m

ms

1 1 2 2 2 2 2 20 0 0 0 1 1 1 10 0 0 0 1 1 0 112

12

12

12

12

12

12

12

−

− − −

P42.60 between 104 K and 105 K

P42.62 (a) c t∆ ; (b) Ehcλ

; (c) 4

2 2E

t d hcλ

π∆

P42.64 0.389 T/mP42.34 see the solution

P42.66 Energy levels at 0, –0.100 eV, –1.00 eV, and–4.10 eVP42.36 (a) see the solution;

(b) 36 states instead of 30

544 Atomic Physics

P42.68 (a) parallel spins; (b) 5 89. eVµ ; P42.72 see the solution(c) 1 04 10 30. × − eV

P42.741

0a, no

P42.70 (a) diameter ~10 1− nm for both;(b) A K-shell electron moves in an orbit

with size on the order of aZ

0