56157 28 ch28 p415-433 - City University of New Yorkuserhome.brooklyn.cuny.edu/kshum/documents/ISM_chapter28.pdf · 28 Atomic Physics Clicker Questions Question Q3.01 Description:

28Atomic Physics

Clicker Questions

Question Q3.01

Description: Distinguishing the Bohr theory’s postulates from related statements.

Question

Which of the following is not one of the Bohr postulates?

1. The laws of classical physics apply to orbital motion of the electrons but not during transitions from one orbit to another.

2. The possible energies that the electron can have are discrete and depend on the orbit radius.

3. Electrons emit electromagnetic radiation only during transitions between orbits and not while in an allowed orbit.

4. Permissible orbits have an angular momentum which is an integer multiple of h 2π .

5. The frequency of the radiation emitted during a transition is related to the energy difference between the orbits by f E h= Δ .

Commentary

Purpose: To check your understanding of the fundamental postulates of the Bohr theory (as distinct from their consequences), and stimulate discussion about what actually goes into the theory.

Discussion: Since students usually see the equations for the discrete energies and radii of the Bohr orbits more frequently than they encounter statements of the model’s postulates, they often believe these equa-tions themselves are the foundation of the model. An understanding of the postulates is essential to appreci-ating how Bohr’s theory relates to both classical and quantum physics.

Statements 1, 3, 4, and 5 are the basis of the theory, though you will often see them with different wording. Statement 2, on the other hand, is not fundamental; quantization of the orbit radii and electron energies is a consequence of the quantization of angular momentum (as described in Statement 4). Bohr’s assertion that angular momentum comes in discrete units of one “h-bar” (h 2π ) was suffi cient to predict the discrete energy levels observed in hydrogen.

The quantization condition is sometimes represented as requiring an integer number of electron wave-lengths to occur around the circumference of each orbit, where wavelength is related to energy via E hf= . This is a heuristic (“hand-waving”) argument designed to make the condition seem “reasonable” or “intui-tive,” however, and is not one of the formal postulates of the Bohr theory.

415

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416 Chapter 28

Key Points:

• Distinguishing a theory’s postulates from its derived statements and consequences helps you to understand the theory, know its limits, and appreciate its relationship to other physical theories.

• The Bohr model requires four fundamental postulates, “leaps” which cannot be proven or justifi ed. The value of the model comes entirely from its success in explaining physical phenomena.

• Bohr’s identifi cation of the quantization of angular momentum is a postulate; quantization of orbit radii and electron energies are consequences of that and of classical physics.

For Instructors Only

This is a simple question, but it gets at something students rarely pay attention to: the logical structure and foundation of a physical theory. They are usually far more concerned with the theory’s consequences, since these are used to solve the problems we give them.

If time permits, this question makes an excellent entrée into a general philosophical discussion about phys-ics, theories, model-building, and the role of observables.

QUICK QUIZZES

1. (b). The allowed energy levels in a one-electron atom may be expressed as E Z nn = − ( )2 213 6. eV , where Z is the atomic number. Thus, the ground state n =( )1 level in helium, with Z = 2, is lower than the ground state in hydrogen, with Z = 1.

2. (a) For n = 5, there are 5 allowed values of l, namely l = 0, 1, 2, 3, and 4.

(b) Since ml ranges from –l to +l in integer steps, the largest allowed value of l ( l = 4 in this case) permits the greatest range of values for ml. For n = 5, there are 9 possible values for ml: –4, –3, –2, –1, 0, +1, +2, +3, and +4.

(c) For each value of l, there are 2 1l + possible values of ml. Thus, there is 1 distinct pair with l = 0, 3 distinct pairs with l = 1, 5 distinct pairs with l = 2, 7 distinct pairs with l = 3, and 9 distinct pairs with l = 4. This yields a total of 25 distinct pairs of l and ml that are possible when n = 5.

3. (d). Krypton has a closed confi guration consisting of fi lled n n n= = =1 2 3, , and shells as well as fi lled 4s and 4p subshells. The fi lled n = 3 shell (the next to outer shell in krypton) has a total of 18 electrons, 2 in the 3s subshell, 6 in the 3p subshell and 10 in the 3d subshell.

ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. Wavelengths of the hydrogen spectrum are given by 1 1 12 2λ = −R n nf iH ( ), where the Rydberg constant is RH

1 m= × −1 097 373 2 107. . Thus, with nf = 3 and ni = 5,

1

1 097 373 2 101

3

1

57 80 107

2 25

λ= × −⎛

⎝⎜⎞⎠⎟ = ×−. . m 1 mm 1− ,

and λ = × = ×− −1 7 80 10 1 28 105 1 6( . ) . m m, so (a) is seen to be the correct choice.


Atomic Physics 417

2. The energy levels in a single electron atom having atomic number Z are

EZ

nn = −( )2

2

13 6. eV.

For beryllium, Z = 4, and for the ground state, n = 1. Thus,

E1

2

2

4 13 6

1218=

− ( ) = −. eV

eV

and the correct answer is choice (b).

3. With a principal quantum number of n = 3 , there are 3 possible values of the orbital quantum number, l = 0, 1, 2. There are a total of 2 2 1l +( ) possible quantum states for each value of l; 2 1l + possible values of the orbital magnetic quantum number ml, and 2 possible spin orien-tations ( )ms = ± 1

2 for each value of ml.Thus, there are 10 3d states ( , )having n = =3 2l , 6 3p states ( , )with n = =3 1l , and 2 3s states ( , )with n = =3 0l , giving a grand total of 10 6 2 18+ + = n = 3 states and the correct choice is (e).

4. There are 6 distinct possible downward transitions with 4 energy levels. These transitions are:4 1 2→ → → → → →, , 4 4 3, 3 1, 3 2, and 2 1. Thus, assuming that each transition has a unique photon energy, E E E Ei fphoton = = −Δ , associated with it, there are 6 different wavelengths λ = hc Ephoton the atom could emit and (e) is the correct choice.

5. The structure of the periodic table is the result of the Pauli exclusion principle, which states that no two electrons in an atom can ever have the same set of values for the set of quantum numbersn m ms, , , and l l . This principle is best summarized by choice (c).

6. All states associated with l = 2 are referred to as d states. Thus, all 10 possible quantum states hav-ing n = =3 2, l are called 3d states (see Question 3 above), and the correct answer is choice (c).

7. Of the electron confi gurations listed, (b) and (e) are not allowed. Choice (b) is not possible because the Pauli exclusion principle limits the number of electrons in any p subshell to a maxi-mum of 6. Choice (e) is impossible because the selection rules of quantum mechanics limit the maximum value of l to n − 1. Thus, a 2d state (n = =2, 2l ) cannot exist.

8. Since the electron is in some bound quantum state of the atom, the atom is not ionized and choice (a) is false. The fact that the electron is in a d state means that its orbital quantum number is l = 2, so choice (b) is false. Also, since the maximum value of l is n − 1, choice (e) is false. Finally, the ground state of hydrogen is a 1s state, so choice (d) is false, leaving (c) as the only true statement in the list of choices.

9. If it were possible for the spin quantum number to take on the four values ms = ± ±32 and 1

2 , the fi rst closed shell would occur for beryllium with 4 electrons in states of ( , , , )1 0 0 3

2 , ( , , , )1 0 0 12 ,

( , , , )1 0 0 12− , and ( , , , )1 0 0 3

2− . The correct answer is choice (c).

10. According to de Broglie’s interpretation of Bohr’s quantization postulate, the circumference of the n = 3 orbit would be exactly 3 electron wavelengths long. However, the de Broglie wave-length of the electron is given by

λπ

π πn

n e n e n n n

n nh

p

h

m

h

m r r

hr

n h

r

n n

n= = = = ( ) = =v v 2

2 2 22 2

2

2

2

2h h

m k en

m k ee e e e

⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

π

Thus, the wavelength of an electron in the n = 3 orbit is 3 times longer than the wavelength of the electron when in the n = 1 orbit, and the circumference of the n = 3 orbit must be 3 3 9( ) = times greater than that of the n = 1 orbit. Choice (c) is the correct answer for this question.


418 Chapter 28

ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS

2. Neon signs do not emit a continuous spectrum. They emit many discrete wavelengths, as could be determined by observing the light from the sign through a spectrometer. However, they do not emit all wavelengths. The specifi c wavelengths and intensities account for the color of the sign.

4. An atom does not have to be ionized to emit light. For example, hydrogen emits light when a transition carries an electron from a higher state to the n = 2 state.

6. Classically, the electron can occupy any energy state. That is, all energies would be allowed. Therefore, if the electron obeyed classical mechanics, its spectrum, which originates from transi-tions between states, would be continuous rather than discrete.

8. The de Broglie wavelength of macroscopic objects such as a baseball moving with a typical speed such as 30 m/s is very small and impossible to measure. That is, λ = h mv is a very small number for macroscopic objects. We are not able to observe diffraction effects because the wave-length is much smaller than any aperture through which the object could pass.

10. In both cases the answer is yes. Recall that the ionization energy of hydrogen is 13.6 eV. The elec-tron can absorb a photon of energy less than 13.6 eV by making a transition to some intermediate state such as one with n = 2. It can also absorb a photon of energy greater than 13.6 eV, but in doing so, the electron would be separated from the proton and have some residual kinetic energy.

12. It replaced the simple circular orbits in the Bohr theory with electron clouds. More important, quan-tum mechanics is consistent with Heisenberg’s uncertainty principle, which tells us about the limits of accuracy in making measurements. In quantum mechanics, we talk about the probabilistic nature of the outcome of a measurement of a system, a concept which is incompatible with the Bohr theory. Finally, the Bohr theory of the atom contains only one quantum number n, while quantum mechanics provides the basis for additional quantum numbers to explain the fi ner details of atomic structure.

14. Each of the given atoms has a single electron in an l = ( )0 or s state outside a fully closed-shell core, shielded from all but one unit of the nuclear charge. Since they reside in very similar environ-ments, one would expect these outer electrons to have nearly the same electrical potential energies and hence nearly the same ionization energies. This is in agreement with the given data values. Also, since the distance of the outer electron from the nuclear charge should tend to increase with Z (to allow for greater numbers of electrons in the core), one would expect the ionization energy to decrease somewhat as atomic number increases. This is also in agreement with the given data.

PROBLEM SOLUTIONS

28.1 (a) The wavelengths in the Lyman series of hydrogen are given by 1 1 1 2λ = −R nH ( ), where n = 2, 3, 4, . . . , and the Rydberg constant is RH

1 m= × −1 097 373 2 107. . This can also be written as λ = −( )( )1 12 2R n nH so the fi rst three wavelengths in this series are

λ1 7

2

2

1

1 097 373 2 10

2

2 11 215 10=

× −⎛⎝⎜

⎞⎠⎟

= ×−−

..

m 177 121 5 m nm= .

λ2 7

2

2

1

1 097 373 2 10

3

3 11 025 10=

× −⎛⎝⎜

⎞⎠⎟

= ×−−

..

m 177 102 5 m nm= .

λ3 7

2

2

1

1 097 373 2 10

4

4 19 720 10=

× −⎛⎝⎜

⎞⎠⎟

= ×−−

..

m 188 97 20 m nm= .

(b) These wavelengths are all in the far ultraviolet region of the spectrum.


Atomic Physics 419

28.2 (a) The wavelengths in the Paschen series of hydrogen are given by 1 1 3 12 2λ = −R nH ( ),

where n = 4, 5, 6, . . . , and the Rydberg constant is RH1 m= × −1 097 373 2 107. . This can

also be written as λ = −( )[ ( )]1 9 92 2R n nH so the fi rst three wavelengths in this series are

λ1 7

2

2

1

1 097 373 2 10

9 4

4 91 875=

× −⎡

⎣⎢

⎤

⎦⎥ = ×−.

( ).

m 1 110 1 8756− = m nm

λ2 7

2

2

1

1 097 373 2 10

9 5

5 91 281=

× −⎡

⎣⎢

⎤

⎦⎥ = ×−.

( ).

m 1 110 1 2816− = m nm

λ1 7

2

2

1

1 097 373 2 10

9 6

6 91 094=

× −⎡

⎣⎢

⎤

⎦⎥ = ×−.

( ).

m 1 110 1 0946− = m nm

(b) These wavelengths are all in the infrared region of the spectrum.

28.3 (a) From Coulomb’s law,

Fk q q

re=

N m C C2 21 22

9 19 28 99 10 1 60 10

=× ⋅( ) ×( )−. .

11 0 102 3 10

10 28

..

×( )= ×

−

−

m N

(b) The electrical potential energy is

PEk q q

re= =

× ⋅( ) − ×( )−1 2

9 198 99 10 1 60 10. . N m C C2 2 11 60 10

1 0 10

2 3 10

19

10

18

.

.

.

×( )×

= − ×

−

−

−

C

m

J1 eV

11.60 10 J eV19×

⎛⎝⎜

⎞⎠⎟ = −− 14

28.4 (a) From Coulomb’s law,

Fk q q

re= =

× ⋅( ) ×( )−1 22

9 19 28 99 10 1 60 10. . N m C C2 2

11 0 102 3 10

15 22

..

×( )= ×

− m N

(b) The electrical potential energy is

PEk q q

re= =

× ⋅( ) ×( )−1 2

9 198 99 10 1 60 10 1. .N m C C2 2 ..

.

.

60 10

1 0 10

2 3 10

19

15

13

×( )×

= + ×

−

−

−

C

m

J1 MeV

11.60 10 JMeV

13×⎛⎝⎜

⎞⎠⎟

= +− 1 4.

28.5 (a) The electrical force supplies the centripetal acceleration of the electron, so

mr

k e

rev2 2

2= or v = k e

mre

2

v =× ⋅( ) ×( )

×

−

−

8 99 10 1 60 10

9 11 10

9 19 2

3

. .

.

N m C C2 2

11 106

1 0 101 6 10

kg m m s( ) ×( ) = ×−.

.

continued on next page


420 Chapter 28

(b) No. vc

= ××

= × −1 6 10

3 00 105 3 10

6

83.

..

m s

m s << 1, so the electron is not relativistic.

(c) The de Broglie wavelength for the electron is λ = =h p h mv , or

λ = × ⋅×( ) ×( )

−

−

6 63 10

9 11 10 1 6 10

34

31 6

.

. .

J s

kg m s== × =−4 6 10 0 4610. . m nm

(d) Yes. The wavelength and the atom are roughly the same size.

28.6 Assuming a head-on collision, the a-particle comes to rest momentarily at the point of closest approach. From conservation of energy,

KE PE KE PEf f i i+ = +

or

02 79 2 79

+( )( ) = +

( )( )k e e

rKE

k e e

re

fi

e

i

With ri → ∞, this gives the distance of closest approach as

re

KEfe

i

= =× ⋅( ) × −

158 158 8 99 10 1 60 102 9 1k . .N m C2 2 99

2

14

5 0 10

4 5 10

C

MeV 1.60 J MeV

m

13

( )×( )

= ×

−

−

.

. == 45 fm

28.7 (a) r n an = 20 yields r2 4 0 052 9 0 212= ( ) =. . nm nm

(b) With the electrical force supplying the centripetal acceleration, m r k e re n n e nv2 2 2= ,

giving vn e e nk e m r= 2 and p m m k e rn e n e e n= =v 2 .

Thus,

pm k e

re e

2

2

2

31 99 11 10 8 99 10= =

×( ) × ⋅(−. . kg N m C2 2 )) ×( )×

= × ⋅

−

−

−

1 6 10

0 212 10

9 95 10

19 2

9

25

.

.

.

C

m

kg mm s

(c) L nh

Ln = ⎛⎝⎜

⎞⎠⎟ → = × ⋅⎛

⎝⎜⎞⎠⎟

=−

22

6 63 10

22

34

π π

J s.22 11 10 34. × ⋅− J s

(d) KE mp

me2 2

2 22 25 2

1

2 2

9 95 10

2 9 11= = =

× ⋅( )×

−

v.

.

kg m s

1105 44 10

103119

−−

−( ) = ××

⎛⎝⎜ kg

J1 eV

1.60 J19. ⎞⎞⎠⎟ = 3 40. eV

(e) PEk e e

re

22

9 198 99 10 1 60 10=

−( ) = −× ⋅( ) × −. . N m C 2 2 CC

m

J eV

( )×( )

= − × = −

−

−

2

9

18

0 212 10

1 09 10 6 80

.

. .

(f) E KE PE2 2 2 3 40 6 80 3 40= + = − −. . . eV eV= eV


Atomic Physics 421

28.8 (a) With the electrical force supplying the centripetal acceleration, m r k e re n n e nv2 2 2= ,

giving vn e e nk e m r= 2 , where r n a nn = =20

2 0 052 9( . ) nm .

Thus,

v1

2

1

9 198 99 10 1 60 10= =

× ⋅( ) ×( )−k e

m re

e

. . N m C C2 2 22

31 96

9 11 10 0 052 9 102 19 10

. ..

×( ) ×( ) = ×− − kg m m ss

(b) kgKE me1 1

2 311

2

1

29 11 10 2 19 10= = ×( ) ×−v . . 66

2

182 18 1010

m s

J1 eV

1.60 J19

( )= ×

×⎛⎝⎜

⎞⎠⎟

−−. == 13 6. eV

(c) PEk e e

re

11

9 198 99 10 1 60 10=

−( ) = −× ⋅( ) × −. . N m C 2 2 CC

m

J eV

( )×( )

= − × = −

−

−

2

9

18

0 052 9 10

4 35 10 27 2

.

. .

28.9 Since the electrical force supplies the centripetal acceleration,

m

r

k e

re n

n

e

n

v2 2

2= or vne n

k e

m r2 2

2

=

From L m r nn e n n= =v h , we have r n mn e n= h v , so

vv

ne

e nk e

m

m

n2 2

2

=⎛⎝⎜

⎞⎠⎟h

which reduces to vn e

k e n= 2 h .

28.10 (a) The Rydberg equation is1 1 12 2λ = −( )R n nf iH , or

λ =−

⎛

⎝⎜⎞

⎠⎟1

2 2

2 2R

n n

n ni f

i fH

With n ni f= =5 3 and ,

λ =×

( )( )−

⎡⎣⎢

⎤⎦⎥

= ×−

1

1 097 37 10

25 9

25 91 281 17..

m 1 00 1 2816− = m nm

(b) fc= = ×

×= ×−λ

3 00 10

1 281 102 34 10

8

914.

. m s

m Hz

(c) Ehc

photon

J s m s

1 28= =

× ⋅( ) ×( )−

λ6 63 10 3 00 1034 8. .

11 m

1 eV

1.60 J eV19× ×

⎛⎝⎜

⎞⎠⎟ =− −10 10

0 9709 .


422 Chapter 28

28.11 The energy of the emitted photon is

Ehc

photon

J s m s= =

× ⋅( ) ×( )−

λ6 626 10 2 998 1034 8. .

6656 m

ev

1.60 10 J eV19× ×

⎛⎝⎜

⎞⎠⎟ =− −10

11 899 .

This photon energy is also the difference in the electron’s energy in its initial and fi nal orbits. The energies of the electron in the various allowed orbits within the hydrogen atom are

Enn = − 13 6

2

. eV where n = 1 2 3, , , . . .

giving E E E E1 2 313 6 3 40 1 51= − = − = −. . . eV, eV, eV, 44 0 850= − . . . . eV,

Observe that E E Ephoton = −3 2 , so the transition was from the orbit to the orbitn n= =3 2 .

28.12 The change in the energy of the atom is

ΔE E En nf i

i f

= − = −⎛

⎝⎜⎞

⎠⎟13 6

1 12 2. eV

Transition I: ΔE = −⎛⎝⎜

⎞⎠⎟ =13 6

1

4

1

252 86. . eV eV (absorption)

Transition II: ΔE = −⎛⎝⎜

⎞⎠⎟ = −13 6

1

25

1

90 967. . eV eV (emission)

Transition III: ΔE = −⎛⎝⎜

⎞⎠⎟ = −13 6

1

49

1

160 572. . eV eV (emission)

Transition IV: ΔE = −⎛⎝⎜

⎞⎠⎟ =13 6

1

16

1

490 572. . eV eV (absorption)

(a) Since λ = =−

hc

E

hc

Ephoton Δ, transition II emits the shortest wavelength photon.

(b) The atom gains the most energy in transition I .

(c) The atom loses energy in transitions II and III .

28.13 The energy absorbed by the atom is

E E En nf i

i fphoton eV= − = −

⎛

⎝⎜⎞

⎠⎟13 6

1 12 2.

(a) Ephoton eV eV= −⎛⎝⎜

⎞⎠⎟ =13 6

1

2

1

52 862 2. .

(b) Ephoton eV eV= −⎛⎝⎜

⎞⎠⎟ =13 6

1

4

1

60 4722 2. .


Atomic Physics 423

28.14 (a) The energy absorbed is

ΔE E En nf i

i f

= − = −⎛

⎝⎜⎞

⎠⎟= −13 6

1 113 6

1

1

12 2. . eV eV

9912 1⎛

⎝⎜⎞⎠⎟ = . eV

(b) Three transitions are possible as the electron returns to the ground state. These transitions and the emitted photon energies are

n ni f= → =3 1 : ΔE = −⎛⎝⎜

⎞⎠⎟ =13 6

1

1

1

312 12 2. . eV eV

n ni f= → =3 2 : ΔE = −⎛⎝⎜

⎞⎠⎟ =13 6

1

2

1

31 892 2. . eV eV

n ni f= → =2 1 : ΔE = −⎛⎝⎜

⎞⎠⎟ =13 6

1

1

1

210 22 2. . eV eV

28.15 To ionize the atom, it is necessary that nf → ∞. The required energy is then

ΔE E En nf i

f i

= − = − −⎛

⎝⎜⎞

⎠⎟= − ∞13 6

1 113 6

12 2. . eV eV −−

⎛⎝⎜

⎞⎠⎟

=1 13 62 2n ni i

. eV

(a) If ni = 1, the required energy is ΔE = =13 6

113 62

..

eV eV

(b) If ni = 3, ΔE = =13 6

31 512

..

eV eV

28.16 The magnetic force supplies the centripetal acceleration, so

m

rq B

vv

2

= , or rm

qB= v

If angular momentum is quantized according to

L m r nn n n= =v 2 h , then mn

rnn

v = 2 h

and the allowed radii of the path are given by

rqB

n

rnn

=⎛⎝⎜

⎞⎠⎟

1 2 h or r

n

qBn = 2 h

28.17 (a) The energy emitted by the atom is

ΔE E E= − = − −⎛⎝⎜

⎞⎠⎟ =4 2 2 213 6

1

4

1

22 55. . eV eV

The wavelength of the photon produced is then

λγ

= = =× ⋅( ) ×( )−

hc

E

hc

EΔ6 63 10 3 00 10

2

34 8. .

.

J s m s

555 1 60 10

4 88 10 488

19

7

eV J eV

m nm

( ) ×( )

= × =

−

−

.

.



424 Chapter 28

(b) Since momentum must be conserved, the photon and the atom go in opposite directions with equal magnitude momenta. Thus, p m h= =atomv λ, or

v = = × ⋅×( )

−

−

h

matom

J s

kgλ6 63 10

1 67 10 4 88

34

27

.

. . ××( ) =−100 814

7 m m s.

28.18 (a) Starting from the n = 4 state, there are 6 possible transitions as the electron returns to the ground ( )n = 1 state. These transitions are: n n= → =4 1, n n= → =4 2, n n= → =4 3, n n= → =3 1, n n= → =3 2, and n n= → =2 1. Since there is a different change in energy associated with each of these transitions, there will be 6 different wavelengths observed in the emission spectrum of these atoms.

(b) The longest observed wavelength is produced by the transition involving the smallest change in energy. This is the n n= → =4 3 transition, and the wavelength is

λmax

. .=

−=

× ⋅( ) ×(−hc

E E4 3

34 86 626 10 2 998 10 J s m s))− −⎛

⎝⎜⎞⎠⎟

×⎛⎝⎜ −

13 61 1

3

1

2. eV4

eV

1.602 10 J2

19

⎞⎞⎠⎟

⎛⎝⎜

⎞⎠⎟−

1 nm

10 m9

or λmax .= ×1 88 103 nm .

Since this transition terminates on the n = 3 level, this is part of the Paschen series .

28.19 For minimum initial kinetic energy, KEtotal = 0 after collision. Hence, the two atoms must have equal and opposite momenta before impact. The atoms then have the same initial kinetic energy, and that energy is converted into excitation energy of the atom during the collision. Therefore,

KE m E Eatom atom eV= = − =1

210 22

2 1v .

or v =( ) =

( ) ×( −2 10 2 2 10 2 1 60 10 19

. . . eV eV J eV

atomm

))×

= ×−1.67 10 kg m s27 4 42 104.

28.20 (a) L m r mr

Tr= = ⎛

⎝⎜⎞⎠⎟

=×( ) ×

v2

2 7 36 10 3 84 1022 8

π

π . . kg m

s kg m s2( )

×= × ⋅

2

634

2 36 102 89 10

..

(b) nL L

h= = =

× ⋅( )× ⋅−h

2 2 2 89 10

6 63 10

34

34

π π .

.

kg m s

J

2

ss= ×2 74 1068.



Atomic Physics 425

(c) The gravitational force supplies the centripetal acceleration so

m

r

GM m

rEv2

2= , or r GM Ev2 =

Then, from L m r nn n n

= =v h or vn

n

n

mr= h

,

we have rn

mrGMn

nE

h⎛⎝⎜

⎞⎠⎟

=2

which gives r nGM m

n rnE

=⎛⎝⎜

⎞⎠⎟

=22

22

1

h

Therefore, when n increases by 1, the fractional change in the radius is

Δr

r

r r

r

n r n r

n r

n

n nn n

n

=−

=+( ) −

= + ≈+1

2

12

12

12

1 2 1 2

Δr

r≈

×= × −2

2 74 107 30 1068

69

..

28.21 (a) r n a n rn = = ⇒ = =20

23

20 052 9 3 0 052 9 0( . ) ( . ) . nm nm 4476 nm

(b) In the Bohr model, the circumference of an allowed orbits must be an integral multiple of the de Broglie wavelength for the electron in that orbit, or 2π λr nn = . Thus, the wavelength of the electron when in the n = 3 orbit in hydrogen is

λ π π= =

( ) =2

3

2 0 476

30 9973r .

. nm

nm

28.22 (a) The Coulomb force supplies the necessary centripetal force to hold the electron in orbit so

m r k e re n n e nv2 2 2= , or m k e r

e n e nv2 2= . But m KE

e n nv2 2= and k e r PEe n n

2 = − , where PEn is the electrical potential energy of the electron-proton system when the electron is in an orbit

of radius rn. We then have 2KE PEn n= − , or KE PEn n= − 12 .

(b) When the atom absorbs energy, E, and the electron moves to a higher level, both the kinetic and potential energy will change. Conservation of energy requires that E KE PE= +Δ Δ . But, from the result of part (a), Δ ΔKE PE= − 1

2 and we have

E PE PE PE= − + = +1

2

1

2Δ Δ Δ or ΔPE E= 2

(c) Δ ΔKE PE E= − = − ( )1

2

1

22 or ΔKE E= −

28.23 rn

Z m k e

n a

Zne e

=⎛⎝⎜

⎞⎠⎟

=2 2

2

20h

so ra

Z Z10 0 052 9= = . nm

(a) For He +, Z = 2 and r = =0 052 9 0 026 5. . . nm 2 nm

(b) For Li 2+, Z = 3 and r = =0 052 9 0 017 6. . . nm 3 nm

(c) For Be 3+, Z = 4 and r = =0 052 9 0 013 2. . . nm 4 nm


426 Chapter 28

28.24 (a) The energy levels in a single electron atom with nuclear charge +Ze are

E Z nn = − 2 213 6( . ) eV . For doubly-ionized lithium, Z = 3, giving E nn = −(122 2 eV) .

(b) E4

1227 63= − = − eV

4 eV2 .

(c) E2

12230 5= − = − eV

2 eV2 .

(d) E E Ei fphoton eV eV eV= − = − − −( ) =7 63 30 5 22 9. . .

Ephoton eV J

1 eV= ( ) ×⎛

⎝⎜⎞⎠⎟

=−

22 91 60 10

3 619

..

. 66 10 18× − J

(e) fE

h= = ×

× ⋅= ×

−

−photon

34

J

6.63 J s

3 66 10

105 52

18.. 11015 Hz

λ = = ××

= × =−c

f

3 00 10

105 43 10 54

88.

. m s

5.52 Hz m15 ..3 nm

(f) This wavelength is in the deep ultraviolet region of the spectrum.

28.25 From L m r ne n n= =v h and r n an = 20

we fi nd that p mn

r

n h

n a

h

a nn nn

= = = =vh ( )2

220 0

ππ

Thus, the de Broglie wavelength of the electron in the n th orbit is λ π= = ( )h p a nn 2 0 . For n = 4, this yields

λ π π= = ( ) =8 8 0 052 9 1 330a . . nm nm

28.26 (a) For standing waves in a string fi xed at both ends, Ln= λ2

or λ = 2 L

n. According to the de Broglie hypothesis, p

h=λ

Combining these expressions gives p mnh

L= =v

2

(b) Using E mp

m= =1

2 22

2

v , with p as found in (a) above:

En h

L mn E E

h

mLn = ( ) = =2 2

22

0 0

2

24 2 8 where


Atomic Physics 427

28.27 In the 3d subshell, n = 3 and l = 2. The 10 possible quantum states are

n = 3 l = 2 ml = + 2 ms = + 12

n = 3 l = 2 ml = + 2 ms = − 12

n = 3 l = 2 ml = +1 ms = + 12

n = 3 l = 2 ml = +1 ms = − 12

n = 3 l = 2 ml = + 0 ms = + 12

n = 3 l = 2 ml = + 0 ms = − 12

n = 3 l = 2 ml = −1 ms = + 12

n = 3 l = 2 ml = −1 ms = − 12

n = 3 l = 2 ml = − 2 ms = + 12

n = 3 l = 2 ml = − 2 ms = − 12

28.28 (a) For a given value of the principal quantum number n, the orbital quantum number l

varies from 0 to n − 1 in integer steps. Thus, if n = 4, there are 4 possible values of l : l = 0, 1, 2, and 3.

(b) For each possible value of the orbital quantum number l , the orbital magnetic quantum

number ml ranges from − l l to + in integer steps. When the principal quantum number

is n = 4 and the largest allowed value of the orbital quantum number is l = 3, there are 7

distinct possible values for ml. These values are: ml = − − − + + +3 1, ,2, 0, 1, 2, and 3.

28.29 The 3d subshell has n = 3 and l = 2. For l-mesons, we also have s = 1. Thus, there are 15 possible quantum states, as summarized in the table below.

n 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3l 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

ml +2 +2 +2 +1 +1 +1 0 0 0 −1 −1 −1 −2 −2 −2

ms +1 0 −1 +1 0 −1 +1 0 −1 +1 0 −1 +1 0 −1


428 Chapter 28

28.30 (a) The electronic confi guration for nitrogen ( )Z = 7 is 1 2 22 2 3s s p .

(b) The quantum numbers for the 7 electrons can be:

1s states n = 1 � = 0 m� = 0ms = + 1

2

ms = − 12

2s states n = 2 � = 0 m� = 0ms = + 1

2

ms = − 12

2 p states n = 2 � = 1

m� = −1ms = + 1

2

ms = − 12

m� = 0ms = + 1

2

ms = − 12

m� = 1ms = + 1

2

ms = − 12

28.31 (a) For Electron #1 and also for Electron #2, n = 3 and � = 1. The other quantum numbers for each of the 30 allowed states are listed in the tables below.

m� ms m� ms m� ms m� ms m� ms m� ms

Electron #1 +1 + 12 +1 + 1

2 +1 + 12 +1 − 1

2 +1 − 12 +1 − 1

2

Electron #2 +1 − 12 0 ± 1

2 −1 ± 12 +1 + 1

2 0 ± 12 −1 ± 1

2


Electron #1 0 + 12 0 + 1

2 0 + 12 0 − 1

2 0 − 12 0 − 1

2

Electron #2 +1 ± 12 0 − 1

2 −1 ± 12 +1 ± 1

2 0 + 12 −1 ± 1

2


Electron #1 −1 + 12 −1 + 1

2 −1 + 12 −1 − 1

2 −1 − 12 −1 − 1

2

Electron #2 +1 ± 12 0 ± 1

2 −1 − 12 +1 ± 1

2 0 ± 12 −1 + 1

2

There are 30 allowed states , since Electron #1 can have any of three possible values of m�

for both spin up and spin down, totaling six possible states. For each of these states, Elec-tron #2 can be in either of the remaining fi ve states.



Atomic Physics 429

(b) Were it not for the exclusion principal, there would be 36 possible states, six for each electron independently.

28.32 (a) For n = =1 0, l and there are 2 2 1( )l + states = =2 1( ) 2 sets of quantum numbers.

(b) For n = =2 0, l for 2 2 1( )l + states = + =2 0 1 2( ) sets and l = 1 for 2 2 1( )l + states = + =2 2 1 6( ) sets

total number of sets = 8

(c) For n = =3 0, l for 2 2 1( )l + states = + =2 0 1 2( ) sets and l = 1 for 2 2 1( )l + states = + =2 2 1 6( ) sets and l = 2for 2 2 1( )l + states = + =2 4 1 10( ) sets

total number of sets = 18

(d) For n = =4 0, l for 2 2 1( )l + states = + =2 0 1 2( ) sets and l = 1 for 2 2 1( )l + states = + =2 2 1 6( ) sets and l = 2 for 2 2 1( )l + states = + =2 4 1 10( ) sets and l = 3 for 2 2 1( )l + states = + =2 6 1 14( ) sets total number of sets = 32

(e) For n = =5 0, l for 2 2 1( )l + states = + =2 0 1 2( ) sets and l = 1 for 2 2 1( )l + states = + =2 2 1 6( ) sets and l = 2 for 2 2 1( )l + states = + =2 4 1 10( ) sets and l = 3 for 2 2 1( )l + states = + =2 6 1 14( ) sets and l = 4 for 2 2 1( )l + states = + =2 8 1 18( ) sets total number of sets = 50

For n n= =1 2 22: For n n= =2 2 82: For n n= =3 2 182: For n n= =4 2 322: For n n= =5 2 502:

Thus, the total number of sets of quantum states agrees with the 2 2n rule.

28.33 (a) Zirconium, with 40 electrons, has 4 electrons outside a closed krypton core. The krypton core, with 36 electrons, has all states up through the 4 p subshell fi lled. Normally, one would expect the next 4 electrons to go into the 4d subshell. However, an exception to the rule occurs at this point, and the 5s subshell fi lls (with 2 electrons) before the 4d subshell starts fi lling. The two remaining electrons in zirconium are in an incomplete 4d subshell. Thus, n = =4 2, and l for each of these electrons.

(b) For electrons in the 4d subshell, with l = 2, the possible values of ml are

ml = ± ±0 1 2, , and those for ms are ms = ± 1 2 .

(c) We have 40 electrons, so the electron confi guration is:

1 2 2 3 3 3 4 4 4 5 4 52 2 6 2 6 10 2 6 2 2 2s s p s p d s p d s d s = [Kr] 22


430 Chapter 28

28.34 The photon energy is E E EL Kphoton eV eV eV= − = − − −( ) =951 8 979 8 028 , and thewavelength is

λ = =

× ⋅( ) ×( )−hc

Ephoton

J s m s6 63 10 3 00 10

8 0

34 8. .

228 1 60 101 55 10 0 155

1910

eV J eV m( ) ×( ) = × =−

−

.. . nm

To produce the Kα line, an electron from the K-shell must be excited to the L-shell or higher. Thus, a minimum energy of 8 028 eV must be given to the atom. A minimum accelerating voltage of ∆V = =8 028 8 03 V kV. is required.

28.35 For nickel, Z = 28 and

E ZK ≈ − −( )( )

= − ( ) ( ) = −113 6

27 13 6 9 912

2

2.. .

eV

1 eV ×× 103 eV

E ZL ≈ − −( )( )

= − ( ) ( ) = −313 6

2513 6

42 1

2

2

2. ..

eV

2

eV33 103× eV

Thus, E E EL Kphoton keV keV ke= − = − − −( ) =2 13 9 91 7 78. . . VV

and

λ = =× ⋅( ) ×( )−

hc

Ephoton

J s m s6 63 10 3 00 10

7

34 8. .

.778 1 60 101 60 10 0 160

1610

keV J keV m

.. .

×( ) = × =−− nm

28.36 The energies in the K and M shells are

E ZK ≈ − −( )( )

113 62

2

. eV

1and E ZM ≈ − −( )

( )9

13 62

2

. eV

3

Thus, E E EZ

ZM Kphoton eV= − ≈ ( ) −−( ) + −( )⎡

⎣⎢⎢

⎤13 6

9

91

22

.⎦⎦⎥⎥

= ( ) −⎛⎝⎜

⎞⎠⎟13 6

8

982. eV Z

and Ehc

photon =λ

gives Zhc2 9

88

13 6= + ( )

⎡

⎣⎢

⎤

⎦⎥. eV λ

, or

Z ≈ +× ⋅( ) ×( )

(−

99 6 63 10 3 00 10

8 13 6

34 8. .

.

J s m s

eV)) ×( ) ×⎛⎝⎜

⎞⎠⎟ =− −0 101 10

1

1 60 109 19. . m

eV

J32 0.

The element is germanium .


Atomic Physics 431

28.37 The transitions that produce the three longest wavelengths in the K series are shown at the right. The energy of the K shell is EK = − 69 5. keV.

Thus, the energy of the L shell is

E Ehc

L K= +λ3

or EL = − +× ⋅( ) ×( )−

69 56 63 10 3 00 10

0

34 8

.. .

keV J s m s

..021 5 10 9× − m

= − + ××

⎛⎝⎜

−−69 5 9 25 10 15. . keV J

1 keV

1.60 10 J16

⎞⎞⎠⎟

= − + = −69 5 57 8 11 7. . . keV keV keV

Similarly, the energies of the M and N shells are

E Ehc

M K= +

= − +× ⋅( ) ×−

λ2

34

69 56 63 10 3 00 1

.. .

keV J s 00

0 020 9 10 1 60 101

8

9 16

m s

m J keV

( )×( ) ×( ) = −− −. .

00 0. keV

and

E Ehc

N K= +

= − +× ⋅( ) ×−

λ1

34

69 56 63 10 3 00 1

.. .

keV J s 00

0 018 5 10 1 60 102

8

9 16

m s

m J keV

( )×( ) ×( ) = −− −. .

..30 keV

The ionization energies of the L, M, and N shells are

1 keV, 10.0 keV, and 2.30 keV, respectiv1 7. eely

28.38 According to the Bohr model, the radii of the electron orbits in hydrogen are given by

r n an = 20 with a0

110 052 9 5 29 10= = × −. . nm m

Then, if rn ≈ = × −1 00 1 00 10 6. . m m µ , the quantum number is

n

r

an= = ×

×≈

−

−0

6

11

1 00 10

5 29 10137

.

.

m

m

28.39 (a) ∆E E E= − = − − −( ) =2 12 213 6 2 13 6 1 10 2. ) . ) . eV ( eV ( eVV

(b) The average kinetic energy of the atoms must equal or exceed the needed excitation energy, or 3

2 k T EB ≥ ∆ , which gives

TE

k≥

( ) =( ) ×( )

×

−2

3

2 10 2 1 60 10 19∆

B

eV J eV

3 1.38

. .

110 J K K

23−( ) = ×7 88 104.


432 Chapter 28

28.40 (a) L c t= ( ) = ×( ) ×( ) = ×−Δ 3 00 10 14 0 10 4 20 108 12. . . m s s −− =3 4 20 m mm.

(b) NE

E

E

hc= =pulse

photon

pulse

λ

=×( )( )

× ⋅( )−

−

694 3 10 3 00

6 63 10 3 00

9

34

. .

. .

m J

J s ××( ) = ×10

1 05 108

19

m s photons.

(c) nN

V

N

L d= = ( )π 2 4

=×( )

( ) ( )=

4 1 05 10

6 008

19

2

.

.

photons

4.20 mm mmπ..84 1016× photons mm3

28.41 (a) With one vacancy in the K shell, an electron in the L shell has one electron shielding it from the nuclear charge, so Z Zeff = − = − =1 24 1 23. The estimated energy the atom gives up during a transition from the L shell to the K shell is then

ΔE E EZ

n

Zi f

i

≈ − = −( ) − −

( )eff2

eff2 eV eV13 6 13 6

2

. .

nnZ

n nf f i2 2 213 6

1 1⎡

⎣⎢⎢

⎤

⎦⎥⎥

= ( ) −⎡

⎣⎢⎢

⎤

⎦⎥eff

2 eV.⎥⎥

or

ΔE ≈ ( ) ( ) −⎡

⎣⎢⎤⎦⎥

= × =23 13 61

1

1

25 40 10

2

2 23. . eV eV 55 40. keV

(b) With a vacancy in the K shell, we assume that Z − = − =2 24 2 22 electrons shield the outermost electron (in a 4s state) from the nuclear charge. Thus, for this outerelectron, Zeff = − =24 22 2 and the estimated energy required to remove this electronfrom the atom is

E E E E

Z

nf i ii

ionizationeff2 ev

= − = − ≈ − −( )⎡

013 6

2

.

⎣⎣⎢

⎤

⎦⎥ =

( ) =2 13 6

43 40

2

2

..

ev eV

(c) KE E E= − = − ≈Δ ionization keV eV 5.40 keV5 40 3 40. .

28.42 (a) The energy levels of a hydrogen-like ion whose charge number is Z are given by

E

Z

nn = −( )13 62

2. eV

For helium, Z = 2 and the energy levels are

En

nn = − =54 412

. eV , 2, 3, . . .

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

E E E= − = −−( )( )

( )=∞ 1

2

201 2

1

3.6 eV54 4. eV


Atomic Physics 433

28.43 (a) IA

E t

d= =

( )=

× ×( )− −P Δ Δπ π

2

3

4

4 3 00 10. J 1.00 10 s

3

9

00.0 10 m W m

6

2

×( )= ×

− 2154 24 10.

(b) E I A t= ( )Δ

= ×⎛

⎝⎜⎞⎠⎟ ×( )⎡

⎣⎢⎤⎦⎥

−4 24 104

0 600 1015 9 2. .

W

m m2

π11 00 10 1 20 109 12. .×( ) = ×− − s J

28.44 (a) Given that the de Broglie wavelength is λ = 2 0a , the momentum is p h h a= =λ 2 0. The kinetic energy of this non-relativistic electron is

KEp

m

h

m ae e

= =2 2

022 8

=× ⋅( ) ×( )

×

− −6 63 10 1

8 9 11 1

34 2.

.

J s eV 1.60 10 J19

00 0 052 9 10135

31 9 2− −( ) ×( )=

kg m eV

.

(b) The kinetic energy of this electron is ≈ 10 times the magnitude of the ground state energy of the hydrogen atom, which is –13.6 eV.

28.45 In the Bohr model,

fE

h

E E

h

h

m k e

n n

n n

e e

= = −

= − −−( )

⎛

⎝⎜⎞

−Δ 1

2 4

2 2 2

1

2

1 1

1h ⎠⎠⎟⎡

⎣⎢⎢

⎤

⎦⎥⎥

=−( )

−⎡

⎣⎢⎢

⎤

⎦⎥

4

2

1

1

12 2 4

3 2 2

π m k e

h n ne e

⎥⎥

which reduces to fm k e

h

n

n ne e= −

−( )⎛

⎝⎜⎞

⎠⎟2 2 1

1

2 2 4

3 2 2

π

28.46 Ehc

photon

J s m s= =

× ⋅( ) ×( )−

λ6 626 10 2 998 1034 8. .

λλ λ1 602 10 10

1 24019 9. ×( )( ) = ⋅ =− − J eV m nm

eV nm ΔEE

For: λ = 310 0. nm, ΔE = 4 000. eV

λ = 400 0. nm, ΔE = 3 100. eV

and λ = 1 378 nm, ΔE = 0 900 0. eV

The ionization energy is 4.100 eV. The energy level diagram having the smallest number of levels and consistent with these energy differences is shown below.


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