Chapter 29
Easy
P29.5
(a)
Longest wavelength implies lowest frequency and smallest energy:
n3
the atom falls from
n2
to
losing energy 
13.6 eV
3
2

13.6 eV
22
 1.89 eV
E
h
f 
The photon frequency is
Balmer Series
FIG. P29.5
and its wavelength is



6.626  1034 J  s 3.00  108 m s 

eV
c hc
 


19
f E
 1.60  10 J 
1.89 eV 
  656 nm
(b)
The biggest energy loss is for an atom to fall from an ionized configuration,
n
to the
n  2 state .
It loses energy

13.6 eV 13.6 eV

 3.40 eV

22
to emit light of wavelength

P29.15
(a)



6.626  1034 J  s 3.00  108 m s
hc

 365 nm .
E
3.40 eV  1.60  1019 J eV


In the 3d subshell, n  3 and   2 ,
we have n

m 
ms
3
3
3
3
3
3
3
3
3
3
2
2
2
2
2
2
2
2
2
2
+2
+2
+1
+1
0
0
–1
–1
–2
–2
+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

m 
3
1
+1
3
1
+1
3
1
+0
3
1
+0
3
1
–1
3
1
–1
ms
+1/2 –1/2 +1/2 –1/2 +1/2 –1/2
(A total of 6 states)
P29.16
6  2.58  1034 J  s .
(a)
For the d state,   2 ,
L
(b)
For the f state,   3 ,
L     1 
12  3.65  1034 J  s .
For a 3d state, n  3 and   2 .
P29.20
Therefore,
L     1 
6  2.58  1034 J  s
m  can have the values
–2, –1, 0, 1,
and 2
so
Lz can have the values  2 ,   , 0,  and 2
Using the relation
cos  
.
Lz
L
we find the possible values of 
145, 114, 90.0, 65.9, and 35.3
P29.24
(a)
(b)
.
1s 2 2s 2 2 p 4
For the 1s electrons,
For the two 2s electrons,
For the four 2p electrons,
n  1 ,   0 , m  0 ,
ms  
and

n  2 ,   0 , m  0 ,
ms  
and

n  2 ;   1 ; m   1 , 0, or 1; and
ms  
or

1
2
1
.
2
1
2
1
.
2
1
2
1
.
2
P29.27 In the table of electronic configurations in the text, or on a periodic table, we look
for the element whose last electron is in a 3p state and which has three electrons outside a
closed shell. Its electron configuration then ends in 3s 2 3 p1 . The element is aluminum .
Medium
*P29.3
(a)
The speed of the moon in its orbit is
v


8
2 r 2 3.84  10 m

 1.02  103 m s .
T
2.36  106 s
So,




L  mvr  7.36  10 22 kg 1.02  103 m s 3.84  108 m  2.89  1034 kg  m 2 s
.
(b)
We have L  n 
or
(c)
L 2.89  1034 kg  m 2 s
 2.74  1068 .
n 
34

1.055

10
J

s

 GM e 
We have n   L  mvr  m 
 r 

12
r,
2
r n  1 R  n 2 R 2n  1


r 2
n 2  Rn 2 and
r
n 2R
n2
m GM e

2
so
which is approximately equal to
P29.6
(a)
2
 7.30  1069 .
n
The photon has energy 2.28 eV.
13.6 eV
 3.40 eV is required to ionize a hydrogen atom from state n  2 .
22
So while the photon cannot ionize a hydrogen atom pre-excited to n  2 , it can
13.6 eV
ionize a hydrogen atom in the n  3 state, with energy 
 1.51 eV .
32
And
(b)
The electron thus freed can have kinetic energy
1
K e  2.28 eV  1.51 eV  0.769 eV  m e v 2 . Therefore,
2
v


2 0.769  1.60  1019 J
9.11  10
31
kg
 520 km s .
The 5th excited state has n  6 , energy
P29.18
13.6 eV
 0.378 eV .
36
The atom loses this much energy:
hc

6.626  10

1 090  10

m 1.60  10
34
9
  1.14 eV
J eV 
J  s 3.00  108 m s
19
to end up with energy
0.378 eV  1.14 eV  1.52 eV
which is the energy in state 3:

13.6 eV
33
 1.51 eV .
While n  3 ,  can be as large as 2, giving angular momentum    1 

P29.19
(a)
n 1:
For n  1 ,   0 , m   0 , m s  
n
1
1

m
ms
0
0
0
0
–1/2
+1/2
6 .
1
2
Yields 2 sets; 2n 2  2 1  2
2
(b)
n  2:
For n  2 ,
we have
n
2
2
2
2

m
0
1
1
1
0
–1
0
1
ms
1/2
1/2
1/2
1/2
yields 8 sets;
2n 2  2 2   8
2
Note that the number is twice the number of m  values. Also, for each  there
are 2  1 different m  values. Finally,  can take on values ranging from 0
to n  1 .
So the general expression is
number 
n 1
 2 2  1 .
0
The series is an arithmetic progression:
2  6  10  14
P29.28
the sum of which is
number 
n
2 a  n  1d 
2
where a  2 , d  4 :
number 
n
4  n  1 4   2n 2 .
2
(c)
n  3:
2 1  2 3   2 5   2  6  10  18
2n 2  2 3   18
(d)
n  4:
2 1  2 3   2 5   2 7   32
2n 2  2 4   32
(e)
n  5:
32  2 9   32  18  50
2n 2  2 5   50
(a)
For electron one and also for electron two, n  3 and   1 . The possible states
are listed here in columns giving the other quantum numbers:
1
1
1
1
1
one
1
2
1
2
1
2
1
2
1
2
1
0
0
electro m 
n
two
ms
electro m 
n
one
ms

1
2
0

0

1
2
0
1
two
1
2
1

1
2
0
1
2
1

1
1
0
1
2
1
2
1
2
1
2
0

0
1
1
1
0
0
0
0
0
1
2
1
2
1
2
1
2
1
2
–1 –1
1
1
0
1
2
1
2
1
1
1
1
1




2
2
2
2
2
–1 –1
1
1
1
1
1




2
2
2
2
2
electro m 
n
ms
1
2
2
2
electro m 
n
ms
2
0

1
2
1
2

1
2
1
2
1
2
–1 –1
1
1
0
0
–1
1
1
2
1
2
1
1

2
2
1
2

1
2
1
2


1

1
2
0
1
2
There are thirty allowed states, since electron one can have any of three
possible values for m  for both spin up and spin down, amounting to six
states, and the second electron can have 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.
1
2

1
2
1
1
1
1
1




2
2
2
2
2
1
2

1
1

2
2
–1 –1 –1 –1 –1 –1 –1 –1 –1 –1
1
2
1
2

–1 –1
0

1
2
–1
1
2
P29.30 Listing subshells in the order of filling, we have for element 110,
1s 2 2s 2 2 p6 3s 2 3 p6 4 s 2 3d10 4 p6 5s 2 4 d10 5 p6 6s 2 4 f 14 5d10 6 p6 7 s 2 5 f 14 6d 8 .
In order of increasing principal quantum number, this is
1s 2 2s 2 2p6 3s 2 3p6 3d10 4s 2 4p6 4d10 4 f 14 5s 2 5p6 5d10 5 f 14 6s 2 6p6 6d 8 7s 2 .
*P29.42
(a)
1 
 1
The energy emitted by the atom is E  E4  E2  13.6 eV  2  2   2.55 eV .
4
2 
The wavelength of the photon produced is then

(b)



6.626  1034 J  s 3.00  108 m s
hc hc


 4.87  107 m  487 nm
19
E
E
2.55
eV
1.60

10
J
eV




Since momentum must be conserved, the photon and the atom go in opposite
h
directions with equal magnitude momenta. Thus, p  matom v 
or

v
h
matom 

6.626  1034 J  s
1.67  10
27

kg 4.87  107 m

 0.814 m s .
Hard
*P29.40
(a)
The energy of the ground state is:
E1  
hc
series limit

1 240 eV  nm
 8.16 eV .
152.0 nm
From the wavelength of the Lyman  line:
E2  E1 
hc


1 240 nm  eV
 6.12 eV
202.6 nm
E2  E1  6.12 eV  2.04 eV .
The wavelength of the Lyman  line gives: E3  E1 
1 240 nm  eV
 7.26 eV
170.9 nm
so
E3  0.902 eV .
Next, using the Lyman  line gives:
E4  E1 
and
E4  0.508 eV .
From the Lyman  line,
E5  E1 
1 240 nm  eV
 7.65 eV
162.1 nm
1 240 nm  eV
 7.83 eV
158.3 nm
(b)
so
E5  0.325 eV .
For the Balmer series,
hc

 Ei  E2 , or  
1 240 nm  eV
.
Ei  E2
For the  lin e, Ei  E3 and so
a 
1 240 nm  eV
 1 090 nm .

0.902
eV   2.04 eV 

Similarly, the wavelengths of the  line,  line, and the short wavelength
limit are found to be: 811 nm , 724 nm , and 609 nm .
(c)
Computing 60.0% of the wavelengths of the spectral lines shown on the
energy-level diagram gives:
0.600 202.6 nm   122 nm , 0.600 170.9 nm   103 nm ,
0.600 162.1 nm   97.3 nm , 0.600 158.3 nm   95.0 nm , and
0.600 152.0 nm   91.2 nm
These are seen to be the wavelengths of the  ,  ,  , and  lines as well as the
short wavelength limit for the Lyman series in Hydrogen.
(d)
The observed wavelengths could be the result of Doppler shift when the source
moves away from the Earth. The required speed of the source is found from
1  v c 
f 


 0.600

1  v c 
f
yielding
v  0.471c .