Chapter 28
Atomic Physics
Problem Solutions
28.1
(a) The w avelengths in the Lym an series of hyd rogen are given by 1   RH (1 1 n2 ) ,
w here n  2, 3, 4, . . . , and the Ryd berg constant is RH  1.097 373 2 107 m1 . This can


also be w ritten as   (1 RH ) n2 n2  1 so the first three w avelengths in this series are
1 
 22 
1
7
  1.215  10 m  121.5 nm
7
1  2
1.097 373 2  10 m  2  1 
2 
 32 
1
7

  1.025  10 m  102.5 nm
1.097 373 2  107 m 1  32  1 
3 
 42 
1
8

  9.720  10 m  97.20 nm
1.097 373 2  107 m 1  42  1 
(b) These w avelengths are all in the far ultraviolet region of the spectrum .
28.2
(a) The w avelengths in the Paschen series of hyd rogen are given by 1   RH (1 32  1 n2 ) ,
w here n  4, 5, 6, . . . , and the Ryd berg constant is RH  1.097 373 2 107 m1 . This can
also be w ritten as   (1 RH )[9n2 (n2  9)] so the first three w avelengths in this series
are
1 
 9(4) 2 
1
6
  1.875  10 m  1875 nm
7
1  2
1.097 373 2  10 m  4  9 
2 
 9(5) 2 
1
6
  1.281  10 m  1 281 nm
7
1  2
1.097 373 2  10 m  5  9 
1 
 9(6) 2 
1
6
  1.094  10 m  1 094 nm
7
1  2
1.097 373 2  10 m  6  9 
544
A tomic Physics
(b) These w avelengths are all in the infrared region of the spectrum .
28.3
(a) From Coulom b’ s law ,
F=
ke q1q2
r
2
8.99 10


N  m 2 C2 1.60  1019 C
9
1.0 10
10
m


2
 2.3  108 N
2
(b) The electrical potential energy is
PE 



8.99  109 N  m 2 C 2 1.60  1019 C 1.60  1019 C
ke q1q2

r
1.0  1010 m

1 eV


  2.3  1018 J 
  14 eV
-19
 1.60  10 J 
28.4
(a) From Coulom b’ s law ,


8.99  109 N  m2 C2 1.60 1019 C
k qq
F  e 12 2 
2
r
1.0  1015 m



2
 2.3  102 N
(b) The electrical potential energy is



8.99  109 N  m 2 C 2 1.60  1019 C 1.60  10 19 C
ke q1q2
PE 

r
1.0  1015 m
 1 MeV
 2.3  1013 J 
-13
 1.60  10
28.5


  1.4 MeV
J
(a) The electrical force supplies the centripetal acceleration of the electron, so
m
v 2 ke e 2
 2
r
r
(b)
No .
ke e 2
mr
8.99 10 N  m C 1.60 10 C 
 9.1110 kg 1.0 10 m 
9
v
v
or
2
31
19
2
10
2
 1.6  106 m s
v 1.6  106 m s

 5.3  103 << 1 , so the electron is not relativistic.
8
c 3.00  10 m s
545
546
CH APTER 28
(c) The d e Broglie w avelength for the electron is   h p  h mv , or

(d )
28.6
6.63 1034 J  s
9.1110
31

kg 1.6 10 m s
6

 4.6 1010 m  0.46 nm
Yes . The w avelength and the atom are roughly the sam e size.
Assum ing a head -on collision, the -particle com es to rest m om entarily at the point of
closest approach. From conservation of energy,
KE f  PE f  KEi  PEi
or
0
ke  2e  79e 
rf
 KEi 
ke  2e  79e 
ri
With ri   , this gives the d istance of closest approach as


9
2
2
19
158ke e2 158 8.99 10 N  m C 1.60 10 C
rf 

KEi
5.0 MeV 1.60 10-13 J MeV
 4.5  10
28.7
(a)
14



2
m  45 fm
rn  n2 a0 yield s r2  4  0.052 9 nm  0.212 nm
(b) With the electrical force supplying the centripetal acceleration, mevn2 rn  kee2 rn2 ,
giving vn  ke e2 me rn and pn  me vn  me ke e2 rn .
Thus,
me ke e 2
p2 

r2
 9.1110
31


kg 8.99  109 N  m 2 C 2 1.6  10 19 C

2
0.212  109 m
 9.95  1025 kg  m s
(c)
 6.63  1034 J  s 
 h 
34
Ln  n 
  2.11  10 J  s
  L2  2 
2
 2 



9.95  1025 kg  m s
p22
1 2
(d ) KE2  mv2 

2
2me
2 9.111031 kg



2
1 eV


 5.44  1019 J 
  3.40 eV
-19
 1.60  10 J 
A tomic Physics
(e)
PE2 
ke  e  e
r2
8.99 10

9

N  m 2 C2 1.60  1019 C
 0.212 10
9
m


2
 1.09  1018 J   6.80 eV
(f)
28.8
E2  KE2  PE2  3.40 eV  6.80 eV=  3.40 eV
(a) With the electrical force supplying the centripetal acceleration, mevn2 rn  kee2 rn2 ,
giving vn  ke e2 me rn , w here rn  n2 a0  n2 (0.052 9 nm) .
Thus,
k e2
v1  e 
me r1
8.99 10 N  m C 1.60 10 C 
 9.1110 kg  0.052 9 10 m 
9
2
19
2
31


(c)
PE1 
r1
8.99 10

9
 2.19  106 m s
9
1
1
(b) KE1  me v12  9.11 1031 kg 2.19 106 m s
2
2
1 eV


 2.18  1018 J 
  13.6 eV
-19
 1.60  10 J 
ke  e  e
2


2
N  m 2 C2 1.60  1019 C
 0.052 9 10
9
m


2
  4.35  1018 J   27.2 eV
28.9
Since the electrical force supplies the centripetal acceleration,
me vn2 ke e 2
 2
rn
rn
or
vn2 
k2 e 2
me rn
From Ln  me rnvn  n , w e have rn  n mevn , so
vn2 
k2 e 2  me vn 


me  n 
w hich red uces to vn  ke e2 n .
547
548
28.10
CH APTER 28


(a) The Ryd berg equation is 1   RH 1 n2f  1 ni2 , or
2 2
1  ni n f


RH  ni2  n2f



With ni  5 and n f  3 ,

(b)
(c)
28.11
f 
c

  25 9 
1
6
  1.28110 m  1 281 nm
7
1 
1.097 37 10 m  25  9 

Ephoton 
3.00  108 m s
 2.34  1014 Hz
1 281  10 9 m
hc

 6.63 10


34

J  s 3.00 108 m s 
1 eV

9
-19
1 28110 m
 1.60 10

  0.970 eV
J
The energy of the em itted photon is
Ephoton 
hc

 6.626 10
34



J  s 2.998 108 m s 
1 ev

9
19
656 10 m
 1.60 10

  1.89 eV
J
This photon energy is also the d ifference in the electron’ s energy in its initial and final
orbits. The energies of the electron in the various allow ed orbits w ithin the hyd rogen atom
are
En  
13.6 eV
n2
w here
n  1, 2, ,3,
giving E1  13.6 eV, E2  3.40 eV, E3  1.51 eV, E4  0.850 eV,
Observe that Ephoton  E3  E2 , so the transition w as from the n  3 orbit to the n  2 orbit .
A tomic Physics
28.12
The change in the energy of the atom is
 1
1 
E  E f  Ei  13.6 eV  2  2 
 ni n f 


Transition I:
1 1 
E  13.6 eV     2.86 eV (absorption)
 4 25 
Transition II:
 1 1
E  13.6 eV     0.967 eV (em ission)
 25 9 
Transition III:
1
 1
E  13.6 eV      0.572 eV (em ission)
 49 16 
Transition IV:
1 
 1
E  13.6 eV     0.572 eV (absorption)
16
49


(a) Since  
hc
Ephoton

hc
, transition II em its the shortest w avelength photon.
E
(b) The atom gains the m ost energy in transition I
(c) The atom loses energy in transitions II and III
28.13
The energy absorbed by the atom is
 1
1 
Ephoton  E f  Ei  13.6 eV  2  2 
 ni n f 


(a)
1
 1
Ephoton  13.6 eV  2  2   2.86 eV
2 5 
1 
 1
(b) Ephoton  13.6 eV  2  2   0.472 eV
6 
4
28.14
(a) The energy absorbed is
 1
1 
1 1 
E  E f  Ei  13.6 eV  2  2   13.6 eV     12.1 eV
 ni n f 
1 9 


549
550
CH APTER 28
(b) Three transitions are possible as the electron returns to the ground state. These
transitions and the em itted photon energies are
1 1
ni  3  n f  1 : E  13.6 eV  2  2   12.1 eV
1 3 
ni  3  n f  2 :
1
 1
E  13.6 eV  2  2
3
2

  1.89 eV

1 1 
ni  2  n f  1: E  13.6 eV  2  2   10.2 eV
1 2 
28.15
To ionize the atom , it is necessary that n f   . The required energy is then
 1
 1 1  13.6 eV
1
E  E f  Ei  13.6 eV  2  2   13.6 eV   2  
 n f ni 
ni2
  ni 


(a) If ni  1 , the required energy is E 
(b) If ni  3 , E 
28.16
13.6 eV
 13.6 eV
12
13.6 eV
 1.51 eV
32
The m agnetic force supplies the centripetal acceleration, so
mv 2
mv
 qvB , or r 
qB
r
If angular m om entum is quantized accord ing to
Ln  mvn rn  2n , then mvn 
2n
rn
and the allow ed rad ii of the path are given by
rn 
1  2n 

 or
qB  rn 
rn 
2n
qB
A tomic Physics
28.17
551
(a) The energy em itted by the atom is
1 
 1
E  E4  E2  13.6 eV  2  2   2.55 eV
2 
4
The w avelength of the photon prod uced is then



6.63  1034 J  s 3.00  108 m s
hc hc


E E
 2.55 eV  1.60  1019 J eV



 4.88  107 m  488 nm
(b) Since m om entum m ust be conserved , the photon and the atom go in opposite
d irections w ith equal m agnitud e m om enta. Thus, p  matomv  h  or
v
28.18
h
matom

6.63 1034 J  s
1.67 10
27

kg 4.88 107 m

 0.814 m s
(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  4  n  1, n  4  n  2,
n  4  n  3, n  3  n  1, n  3  n  2, and n  2  n  1. Since there is a d ifferent
change in energy associated w ith each of these transitions there w ill be
6 different wavelengths observed in the em ission spectrum of these atom s.
(b) The longest observed w avelength is prod uced by the transition involving the sm allest
change in energy. This is the n  4  n  3 transition, and the w avelength is
max 



6.626  1034 J  s 2.998  108 m s 
hc
1 eV


-19
E4  E3
 1 1
 1.602  10
13.6 eV  2  2 
4 3 
 1 nm 


J  10-9 m 
or max  1.88 103 nm .
Since this transition term inates on the n  3 level, this is part of the Paschen series .
552
28.19
CH APTER 28
For m inim um initial kinetic energy, KEtotal  0 after collision. H ence, the tw o atom s m ust
have equal and opposite m om enta before im pact. The atom s then have the sam e initial
kinetic energy, and that energy is converted into excitation energy of the atom d uring the
collision. Therefore,
1
KEatom  matom v 2  E2  E1  10.2 eV
2
or
28.20
(a)
2 10.2 eV 
v
matom

2 10.2 eV  1.60 1019 J eV

1.67 10
27
kg

4.42  104 m s
 2 r 
L  mvr  m 
r
 T 

(b) n 


2  7.36  1022 kg 3.84  108 m
2.36  106 s
L



2
 2.89  1034 kg  m 2 s

34
2
2 L 2 2.89  10 kg  m s

 2.74  1068
h
6.63  1034 J  s
(c) The gravitational force supplies the centripetal accelerat ion so
mv 2 GM E m

, or rv2  GM E
2
r
r
Then, from Ln  mvn rn  n or vn 
n
,
mrn
2
2

 n 
2
r

n
 n 2 r1
w e have rn 
w
hich
gives

GM


n
E
2 
GM
m
mr
E


 n
Therefore, w hen n increases by 1, the fractional change in the rad ius is
2
r rn1  rn  n  1 r1  n r1 2n  1 2


 2 
r
rn
n
n2 r1
n
2
r
2

 7.30  1069
r
2.74  1068
28.21
(a)
rn  n2 a0  n2 (0.052 9 nm)  r3  32 (0.052 9 nm)  0.476 nm
A tomic Physics
553
(b) In the Bohr m od el, the circum ference of an allow ed orbits m ust be an integral
m ultiple of the d e Broglie w avelength for the electron in that orbit, or 2 rn  n .
Thus, the w avelength of the electron w hen in the n  3 orbit in hyd rogen is

28.22
2 r3 2  0.476 nm 

 0.997 nm
3
3
(a) The Coulom b force supplies the necessary centripetal force to hold the electron in
orbit so mevn2 rn  kee2 rn2 , or mevn2  kee2 rn . But mevn2  2KEn and kee2 rn  PEn ,
w here PEn is the electrical potential energy of the electron -proton system w hen the
electron is in an orbit of rad ius rn . We then have 2KEn  PEn , or KEn   12 PEn .
(b) When the atom absorbs energy, E, and the electron m oves to a higher level, both the
kinetic and p otential energy w ill change. Conservation of energy requires that
E  KE  PE . But, from the result of part (a), KE   12 PE and w e have
(c)
28.23
28.24
rn 
1
1
E   PE  PE   PE
2
2
or
PE  2E
1
1
KE   PE    2 E 
2
2
or
KE  E
2
 n2 a0
n2 


Z  me ke e2 
Z
so
r1 
a0 0.052 9 nm

Z
Z
(a) For He+ ,
Z  2 and r  0.052 9 nm 2  0.026 5 nm .
(b) For Li2+ ,
Z  3 and r  0.052 9 nm 3  0.017 6 nm .
(c) For Be3+ ,
Z  4 and r  0.052 9 nm 4  0.013 2 nm .
(a) The energy levels in a single electron atom w ith nuclear charge Ze are
En   Z 2 (13.6 eV) n2 . For d oubly-ionized Lithium , Z  3 , giving En  (122 eV) n2 .
(b) E4 
122 eV
 7.63 eV
42
E2 
122 eV
 30.5 eV
22
(c)
554
CH APTER 28
(d ) Ephoton  Ei  E f  7.63 eV   30.5 eV   22.9 eV
 1.60  1019 J 
18
Ephoton   22.9 eV  
  3.66  10 J
1
eV


(e)
f 

(f)
28.25
Ephoton
h

3.66  1018 J
 5.52  1015 Hz
-34
6.63  10 J  s
c 3.00  108 m s

 5.43  108 m  54.3 nm
f 5.52  1015 Hz
This w avelength is in the deep ultraviolet region of the spectrum .
From L  mevn rn  n and rn  n2 a0
w e find that pn  mvn 
n
n(h 2 )
h


2
rn
2 a0 n
n a0
Thus, the d e Broglie w avelength of the electron in the n th orbit is   h pn   2 a0  n .
For n  4 , this yield s
  8 a0  8  0.052 9 nm  1.33 nm
28.26
(a) For stand ing w aves in a string fixed at both end s, L 
or  
n
2
2L
h
. Accord ing to the d e Broglie hypothesis, p 
n

Com bining these expressions gives p  mv 
nh
2L
1
p2
(b) Using E  mv 2 
, w ith p as found in (a) above:
2
2m
En 
28.27
n2 h2
h2
2

n
E
where
E

0
0
4 L2  2m 
8mL2
In the 3d subshell, n  3 and
2.
The 10 possible quantum states are
A tomic Physics
28.28
n3
2
m  2
ms   12
n3
2
m  2
ms   12
n3
2
m  1
ms   12
n3
2
m  1
ms   12
n3
2
m  0
ms   12
n3
2
m  0
ms   12
n3
2
m  1
ms   12
n3
2
m  1
ms   12
n3
2
m  2
ms   12
n3
2
m  2
ms   12
555
(a) For a given value of the principle quantum num ber n, the orbital quantu m num ber
varies from 0 to n 1 in integer steps. Thus, if n  4 , there are 4 possible values of
:
 0, 1, 2, and 3
(b) For each possible value of the orbital quantum num ber , the orbital m agnetic
quantum num ber m ranges from  to + in integer steps. When the principle
quantum num ber is n  4 and the largest allow ed value of the orbital quantum
num ber is  3 , there are 7 d istinct possible values for m . These values are:
m  3,  2, 1, 0, +1, +2, and +3
28.29
The 3d subshell has n  3 and  2 . For -m esons, w e also have s  1 . Thus, there are 15
possible quantum states as sum m arized in the table below .
 2(6  1)  14
m
ms
3
3
2
2
+2 +2
+1 0
3
2
+2
1
3
3
2
2
+1 +1
+1 0
3
2
+1
1
3
2
0
+1
3
2
0
0
3
2
0
1
3
2
1
+1
3
2
1
0
3
2
1
1
3
2
2
+1
3
2
2
0
3
2
2
1
556
28.30
CH APTER 28
(a) The electronic configuration for nitrogen (Z  7) is 1s 2 2s 2 2 p3 .
(b) The quantum num bers for the 7 electrons can be:
1s states
n 1
2(2  1)
2s states
n2
2(2  1)
ms   12
 2(4  1)  10
ms   12
ms   12
m 0
ms   12
ms   12
m  1
ms   12
ms   12
n  3: 2n2 n184: 2n2  32
m 0
2 p states
ms   12
ms   12
m 1
28.31
ms   12
(a) For Electron #1 and also for Electron #2, n  3 and  1 . The other quantum num bers
for each of the 30 allow ed states are listed in the tables below .
2
m
m
2 13.6 eV
2
2 1 2

E6K3
8528
979
 Z8d
028

1s2 2s2E2 p6 3EsL2 3 p
d 10951
4s2 4eV
p64Zd
s2E=KeV
[Kr]4
5s eV 2
ms
m
ms
1
+1

1
2
+1

m
ms
EM    Z  9  1
+1  2
2
Electron #1
+1

Electron #2
+1
 12
0
 12
1
m
ms
 12
m
ms
 12
m
Electron #1
Electron #2
0
+1
1
2

0
0
1
2
32.0

ms
 12
+1

+1
Germanium 1
2
0
1
 12
ms
 12
m
ms
 12
m
0
1
2
+1
m
0

ms
 12
0

13.6 eV
 3
2


4 1.05 1019 photons
6.63 1034 J  s 3.00 108 m s
hc
16
3

E E mm

0

8.82 10 photons
1 10
 1210-19 J eV
+1 limit 12
0152010
1 2 1
m
1 1.60
2
 4.200mm   6.00mm

557
A tomic Physics

m
ms
m
ms
m
ms
Electron #1
1
Electron #2
+1
m

10 J  s  3.00 10 hcm s   6.63 10 J  s  3.00  10 hc
m s  6.626 10
 6.631.09


 609 nm
E 
 
10 nm
 E 0J eV
 1 eV    2.04
 1 eV  1.60
E1 10
1 2.04 eV  1.60
1  10 J eV1.602 10

0.904





34

34
8
8
3
short

  310.0 nmE  4.000 eV
  400.0 nm 1
0
1
+1
 -19
-19
2
2
0
 12
1
 12
There are 30 allowed states , since Electron #1 can have any of three possible values of
E  I A t  for both spin up and spin d ow n, totaling six possible states. For each of
these states, Electron #2 can be in either of the rem aining five states.
(b) Were it not for the exclusion principle, there w ould be 36 possible states, six for each
electron ind epend ently.
28.32
(a) For n  1,  0 and there are 2(2  1) states  2(1)  2 sets of quantum
num bers
(b) For n  2,  0 for 2(2  1) states  2(0  1)  2 sets
 1 for 2(2  1) states  2(2  1)  6 sets
and
total num ber of sets = 8
(c) For n  3,
and
and

34
6.63
s 3.00 108 m
hc10 hcJ  hc
hc



E5  E1 8.18eV

0.325
 eV long10
E 10Ei E
E3 f1.60
4
E2 10-19 J
1583
m
 0 for 2(2  1) states  2(0  1)  2 sets
 1 for 2(2  1) states  2(2  1)  6 sets
 2 for 2(2  1) states  2(4  1)  10 sets
total num ber of sets = 18
(d ) For n  4,  0 for 2(2
 1 for 2(2
and
 2 for 2(2
and
and
 3 for 2(2
 1) states  2(0  1)  2 sets
 1) states  2(2  1)  6 sets
 1) states  2(4  1)  10 sets
 1) states  2(6  1)  14 sets
total num ber of sets = 32
34
558
CH APTER 28
(e) For n  5,
and
and
and
and
 0 for 2(2
 1 for 2(2
 2 for 2(2
 3 for 2(2
 4 for 2(2
 1) states  2(0  1)  2 sets
 1) states  2(2  1)  6 sets
 1) states  2(4  1)  10 sets
 1) states  2(6  1)  14 sets
 1) states  2(8  1)  18 sets
total num ber of sets = 50
For n  1: 2n2  2
For n  2: 2n2  8
For n  3: 2n2  18
For n  4: 2n2  32
For n  5: 2n2  50
Thus, the total num ber of sets of quantum states agrees w ith the 2n2 rule.
28.33
(a) Zirconium , w ith 40 electrons, has 4 electrons outsid e a closed Krypton core. The
Krypton core, w ith 36 electrons, has all states up through the 4 p subshell filled .
N orm ally, one w ould expect the next 4 electrons to go into the 4d subshell. H ow ever,
an exception to the rule occurs at this point, and the 5s subshell fills (w ith 2
electrons) before the 4d subshell starts filling. The tw o rem aining electrons in
Zirconium are in an incom plete 4d subshell. Thus, n  4, and  2 for each of these
electrons.
(b) For electrons in the 4d subshell, w ith
 2 , the possible values of m are
m  0,  1,  2 and those for ms are ms   1 2
(c) We have 40 electrons, so the electron configuration is:
1s 2 2s 2 2 p6 3s 2 3 p6 3d 10 4s 2 4 p6 4d 2 5s 2 = [Kr]4d 2 5s 2
28.34
The photon energy is Ephoton  EL  EK  951 eV   8 979 eV   8 028 eV , and the
w avelength is

hc
Ephoton
 6.63 10

34

J  s 3.00 108 m s
8 028 eV  1.60 10
19
J eV

  1.55 10
10
m  0.155 nm
To prod uce the K line, an electron from the K-shell m ust be excited to the L-shell or
higher. Thus, a m inim um energy of 8 028 eV m ust be given to the atom . A m inim um
accelerating voltage of V  8 028 V  8.03 kV is required .
A tomic Physics
28.35
For nickel, Z  28 and
EK    Z  1
2
EL    Z  3
13.6 eV
1
2
2
13.6 eV
 2
2
   27  13.6 eV    9.91 103 eV
2
   25
2
13.6 eV   2.13 103 eV
4
Thus, Ephoton  EL  EK  2.13 keV   9.91 keV   7.78 keV
and

28.36
hc
Ephoton
 6.63 10

34

J  s 3.00 108 m s

7.78 keV 1.60 10
16
J keV

  1.60 10
10
m  0.160 nm
The energies in the K and M shells are
EK    Z  1
Thus,
Ephoton
and Ephoton 
2
13.6 eV
1
2
and EM    Z  9 
2
13.6 eV
 3
2
  Z  9 2

2
8

 EM  EK  13.6 eV  
  Z  1   13.6 eV   Z 2  8 
9
9





9
hc
gives Z 2  8 
 , or

8  13.6 eV   
hc



9 6.63 1034 J  s 3.00 108 m s 
1 eV

Z  9

  32.0
19
9
8 13.6 eV  0.101 10 m
 1.60 10 J 
The elem ent is Germanium .


559
560
28.37
CH APTER 28
The transitions that prod uce the three
longest w avelengths in the K series are
show n at the right. The energy of the K
shell is EK   69.5 keV .
Thus, the energy of the L shell is
EL  EK 
or
hc
3
EL   69.5 keV 
 6.63 10

34
J  s 3.00 108 m s
0.0215 10
9

m
 1 keV

  69.5 keV  9.25  1015 J 

-16
 1.60  10 J 
  69.5 keV  57.8 keV  11.7 keV
Sim ilarly, the energies of the M and N shells are
EM  E K 
hc
2
  69.5 keV 
 6.63 10
34
 0.020 9 10
9

m 1.60  10
J  s 3.00  108 m s
16


 10.0 keV

  2.30 keV
J keV
and
E N  EK 
hc
1
  69.5 keV 
 6.63 10
34
 0.018 5 10
9

m 1.60  10
J  s 3.00  108 m s
16
The ionization energies of the L, M, and N shells are
11.7 keV, 10.0 keV, and 2.30 keV respectively

J keV
A tomic Physics
28.38
Accord ing to the Bohr m od el, the rad ii of the electron orbits in hyd rogen are given
by
rn  n2 a0 w ith a0  0.052 9 nm  5.29 1011 m
Then, if rn  1.00 m  1.00 106 m , the quantum num ber is
n
28.39
(a)
rn
1.00 106 m

 137
a0
5.29 1011 m


E  E2  E1 13.6 eV (2)2  13.6 eV (1)2  10.2 eV
(b) The average kinetic energy of the atom s m ust equal or exceed the need ed
excitation energy, or 32 kBT  E w hich gives
T
28.40
2  E 
3kB


2 10.2 eV  1.60 1019 J eV

3 1.38 10
-23

JK


  7.88 10
4
K

(a)
L  c  t   3.00 108 m s 14.0 1012 s  4.20 103 m  4.20 mm
(b)
N

(c)
n
Epulse

Ephoton
Epulse
hc 
 694.3 10
 6.63 10
34



m  3.00 J 

J  s 3.00 10 m s
N
N

V L  d2 4

9
8

 1.05 1019 photons

4 1.05 1019 photons

 4.20 mm   6.00 mm 
2
 8.82  1016 photons mm3
561
562
28.41
CH APTER 28
(a) With one vacancy in the K shell, an electron in the L shell has one electron shield ing it
from the nuclear charge, so Zeff  Z  1  24  1 23. The estim ated energy the atom
gives up d uring a transition from the L shell to the K shell is then
E  Ei  E f  
2
Zeff
13.6 eV    Zeff2 13.6 eV    Z 2 13.6 eV  1  1 
 2 2 


eff 
ni2
n2f


 n f ni 
or
2
1 1 
E   23 13.6 eV   2  2   5.40  103 eV  5.40 keV
1 2 
(b) With a vacancy in the K shell, w e assum e that Z  2  24  2  22 electrons shield the
outerm ost electron (in a 4s state) from the nuclear charge. Thus, for this outer
electron, Zeff  24  22  2 and the estim ated energy required to rem ove this electron
from the atom is
 Z 2 13.6 ev   22 13.6 ev 
Eionization  E f  Ei  0  Ei    eff 2
 3.40 eV

ni
42


(c)
28.42
KE  E  Eionization  5.40 keV  3.40 eV  5.40 keV
(a) The energy levels of a hyd rogen -like ion w hose
charge num ber is Z are given by
En   13.6 eV 
Z2
n2
For H elium , Z  2 and the energy levels are
En  
54.4 eV
n2
n  1, 2, 3, . . .
(b) For He , Z  2 , so w e see that the ionization energy (the energy required to take the
electron from the n  1 to the n   state) is
 13.6 eV  2
E  E  E1  0 
2
1
28.43
(a)

2
 54.4 eV

3
-9
P  E t  4 3.00  10 J 1.00 10 s
I 

 4.24  1015 W m2 
2
2
-6
A d 4
 30.0 10 m


A tomic Physics
563
(b) E  I A t 
2
W  

   4.24  1015 2    0.600  109 m   1.00  109 s   1.20  1012 J 
m  4


28.44
(a) Given that the d e Broglie w avelength is   2a0 , the m om entum is p  h   h 2 a 0 .
The kinetic energy of this non -relativistic electron is
KE 
p2
h2

2 me 8 me a 20
 6.63 10

8  9.11 10
34
 1 eV 1.60 10 J   135 eV 
kg  0.052 9 10 m 
J s
31
2
-19
9
2
(b) The kinetic energy of this electron is  10 times the m agnitud e of the ground state
energy of the hyd rogen atom w hich is –13.6 eV.
28.45
In the Bohr m od el,
f 

E En  En 1

h
h
1  me ke2 e 4  1
1   4 2 me ke2 e 4


 

h  2 2  n 2  n  12  
2 h3


w hich red uces to
f 
2 2 me ke2 e4  2n  1 


  n  12 n 2 
h3


 1
1
 2

2
n 
  n  1
564
28.46
CH APTER 28
Ephoton 
hc

 6.626 10
 1.602 10
34


  1 240 eV  nm  E

m nm 
J  s 2.998 108 m s
19

J eV 10
9
For:
  310.0 nm , E  4.000 eV
  400.0 nm , E  3.100 eV
and
  1378 nm , E  0.900 0 eV
The ionization energy is 4.100 eV. The energy level d iagram having the few est num ber of
levels and consistent w ith these energy d ifferences is show n below .