4
The Particle Nature of Matter
4-14
(a)
n 2 2
rn 
2 ; where n
  1, 2 , 3, 
 m e ke
rn  n
2
1.055  10 34 Js
2
2  0.052 9 nm n
31
9
2
2
19
9.11  10 kg 9.0  19 Nm C 1.6  10 C 
2
For n  1: rn  0.052 9 nm
For n  2: rn  0.212 1 nm
For n  3: rn  0.477 2 nm
12
(b)
 ke2 
From Equation 4.26, v  

me r 


2 
 
 8.99  109 Nm 2 C 2 1.60  10 19 C
v 1  
 9.11  10 31 kg 0.052 9  10 9 m




12
 2.19  10 6 m s
v 2  1.09  106 m s
v 3  7.28  10
4-15
5
m s
(c)
As c  3.0  108 m s , v  c and no relativistic correction is necessary.
(a)
The energy levels of a hydrogen-like ion whose charge number is 2 is given by
2
54.4 eV
Z
E n  13.6 eV 2 
for Z  2 . He 
2
n
n
 
0
E3  6.04 eV
E2  13.6 eV
So E1  54.4 eV
E2  13.6 eV
E3  6.04 eV , etc.
E1  54.4 eV
(b)
For He  , Z  2 so we see that the ionization energy (the energy required to take the
2
54.4 eV
2
electron from the state n  1 to the state n   is E  13.6 eV  2 
for
2
1
n
Z  2 . He 
 
4-16
For Li 2 , Z  3 from Equation 4.36
0
E3  13.6 eV
En  
13.6Z 2
2
n eV

122.4
2
n eV
E2  30.6 eV
So E1  122.4 eV
E2  30.6 eV
E3  13.6 eV , etc.
E1  122.4 eV
4-17
 n 2   2 
n 2 2

r
 Z 
2
2  ; n  1
Zm
ke
m
ke




e
e




2

34
1.055  10
Js
1 
r  
Z  9.11  10 31 kg 9  10 9 Nm 2 C2 1.6  10 19 C


4-18


2

11
m
 5.30  10


Z


(a)
For He  , Z  2 , r 
5.30  10 11 m
 2.65 10 11 m  0.026 5 nm
2
(b)
For Li 2 , Z  3 , r 
5.30  10 11 m
 1.77  10 11 m  0.017 7 nm
3
(c)
For Be3 , Z  4 , r 
5.30  10 11 m
 1.32  10 11 m  0.013 2 nm
4
(a)
1 
 1
E  13.6 eV 2  2   12.1 eV
1
3 
(b)
1 1 
Either E  12.1 eV or E  13.6 eV  2   10.2 eV and
1 2 
 1
1 
E  13.6 eV 2  2   1.89 eV .
2
3 
4-21
(a)
For the Paschen series;
 1
1 
 R 2  2 ;  max  1 874.606 nm . For minimum wavelength, n i   ,
3
4 
9
 1
1 
 R 2  ;  min   820.140 nm .
3
 
R
1
ni  4 ,
1
 min
(b)
hc
 min
hc
 min
4-22
E  K U 
 1
1 
 R 2  2  ; the maximum wavelength corresponds to

3
n i 
1
 max


1.6  10
J eV
hc
1 874.606 nm
19

hc
820.140
nm 

1.6  10
19
J eV
 0.662 7 nm
 1.515 nm
2
mv 2 1  ke2
mv 2 ke2
1  ke  U

. But
  
. Thus E   
 , so
2  r 
2  r
2
2
r
 2
U  2E  2 13.6 eV   27.2 eV and K  E U  13.6 eV  27.2 eV  13.6 eV .
4-23
(a)
r1  0.052 9 nm n  0.052 9 nm (when n  1 )
(b)
 ke2 
m e v  m e 

me r 
2
12


 9.1  10 31 kg 9  109 Nm 2 C 2
m e  

5.29  10 11 m

M e v  1.99  10 24 kg m s
4-25



12




 1.6  10 19 C



(c)
L  m e vr  1.99  10 24 kg m s 5.29  10 11 m , L  1.05  10 34 kg m 2 s  

(d)
K  E  13.6 eV
(e)
U  2K  27.2 eV
(f)
E  K U  13.6 eV
(a)
1
1


 16
 1
1 
9
14
E  hf  13.6 eV  2  2  or f  13.6 eV
  1.60  10 Hz
15
4.14

10
eV
s
n f n i 


(b)
T 
2 rn
v
12
so frev
r3  3 a0 and
2
12
 ke2 
 ke2 
1
v
 
. Using v  
, frev  

 . For n  3 ,
T 2  rn
me rn 
mrn 
frev
8.99  10 9

Nm
9.11 10
2

C 2 1.60  10 19 C
 

2 3.149 5.29 10 11 m 
31
kg 9  5.29 10 11 m

2
12
frev  2.44  1014 Hz n  3
frev  1.03  1014 Hz n  4 
Thus the photon frequency is about halfway between the two frequencies of the
revolution.
4-32
(a)
1H



1


 


9.109 390  10 31 kg 1.672 63  10 27 kg
meM


me  M
9.109 390  10 31 kg  1.672 63  10 27 kg

9.109 390  10 31 kg 1.672 63  10 27 kg   9.104 431  10 31 kg
0.000 910 939 0  10 27 kg  1.672 63  10 27 kg 

me
 1
k 
n f
2

31
kg 
1  9.104 431 6  10
1 
7
1  1

 2 

 1.097 315 3  10 m 
31
2
2 
2
n i   9.109 390  10
3 
kg 


1H  656.469 1 nm
4-33
(b)
Similarly, we find  2H  656.292 5 nm .
(c)
 3H  656.232 5 nm


2
1.05 10 34 Js
n 2 a0
n 2 2
2
rn 



1

 3.1  10 15 m . Note
2
Z
9
2
19
mke2 Z
207 m e  8.99  10 Nm C 1.60  10 C 82

that this means the muon grazes the nuclear surface, and so experiments with muonic atoms
give information about the nuclear charge distribution.

ke Z  2a n  ke Z

2
En


2
2
2
0
2
2
2 n mke
2

2
2 4

mk e Z
2 2
2 n
2
Using m  207me , n  1 , and Z  82 yields E1  18.9 MeV . (Using the reduced mass makes no
difference in the answers to three significant figures.)
4-35
2 2
meM
me
me
n 

since m e  M . In general, rn 
2 , so for positronium Z  1 ,  
me  M
2
2
Z  ke

2 n 2 m e  2
and rpositronium  
ke  2a0 n 2  2rhydrogen . Similarly,




1
2



E hydrogen
6.80 eV

.
Epositronium 
2
2
n

4-37
hf   E 
4 2 me k 2 e 4
2 h2
 1
2  2 m e k 2 e 4  2n 1
1 
f


,




2 2
n  12 n 2 


h3


n 1 n

 as n   ,


 
1 2
2  2 m e k 2 e 4  2 
 1 ke2   1 
v
.
The
revolution
frequency
is

f

f  
 


 3 
  3 2  where
3
2 r 2 
h
 m e  r 

 n
12
 1 ke2 
r 2
substituting
for
r,
f



2  

 me 
4 m e ke2
n 2 h2
8 3 m ke3 m k 1 2  4 2 m k 2 e 4
e
e
e


.
 

n 3 h3
h3 n 3

