CHAPTER 7 | Electrons in Atoms and Periodic Properties
7.37. Collect and Organize
The kinetic energy of an ejected photon is given by the equation
KEelectron  h – 
where  is the work function or the energy threshold required to eject the electron from the metal.
Analyze
In this problem, we solve for :
  h – KEelectron
where h = 6.626  10–34 J  s,  = frequency of the light used to eject the electron, and KEelectron = 5.34 
10–19 J. We can find  of the irradiating light from  = c/ where c = 3.00  108 m/s and we are given
 = 162 nm (1.62  10–7 m).
Solve

3.00  108 m/s 
   6.626  10–34 J  s 
– 5.34  10–19 J  6.93  10–19 J
–7

1.62  10 m 

Think about It
This does not seem to be a large amount of energy to emit one electron, but remember if we were to
consider a mole of electrons to be ejected we would need (6.93  10–19 J)  (6.022  1023/mol) =
4.17  105 J/mol or 417 kilojoules per mol. This is the ionization energy in kilojoules per mole for
this metal.
7.38. Collect and Organize
The work function equation to answer this problem is
KEelectron  h – 
We are asked to determine the wavelength corresponding to the work function to eject one electron.
Analyze
The kinetic energy for the electron ejected by the maximum wavelength must be zero because no
additional energy is supplied to the electron by the incident radiation other than ejecting it. Therefore,
the frequency of the light to eject the electron would be
0  h –  or   /h
The work function is the energy required to eject an electron. This is equivalent to the first ionization
energy (given as 6.24  10–19 J). The wavelength can then be found from  using  = c/.
Solve
6.24  10 –19 J
 9.417  1014 s –1
6.626  10 –34 J  s
3.00  108 m/s

 3.19  10 –7 m or 319 nm
14 –1
9.417  10 s

Think about It
This wavelength (319 nm) is in the UV region of the electromagnetic spectrum.
7.42. Collect and Organize
We can calculate the energy (and from that the wavelength) of electrons ejected from silicon and
titanium when a 250 nm wavelength light source shines on these surfaces to determine which
electrons have the longest wavelength.
344
Electrons in Atoms and Periodic Properties | 345
Analyze
We are given the work functions for each of the elements (Ti = 6.94  10–19 J, Si = 7.24  10–19 J).
Knowing the wavelength of the incident radiation (2.50  10–7 m), we can calculate the frequency of
the incident radiation using  = c/. Knowing , we can then calculate the kinetic energy of the
electrons ejected from the metals using
KE  h – 
We can use the calculated kinetic energies in E = hc/ to determine the wavelengths.
Solve
Frequency of the incident radiation:

For titanium:
3.00  108 m/s
 1.20  1015 s –1
2.50  10 –7 m


KE  6.626  10–34 J  s  1.20  1015 s –1 – 6.94  10–19 J  1.01  10–19 J

For silicon:
hc 6.626  10 J  s  3.00  10 m/s

 1.97  10–6 m or 1970 nm
E
1.01  10–19 J
–34
8


KE  6.626  10 –34 J  s  1.20  1015 s –1 – 7.24  10 –19 J  7.11  10 –20 J
hc 6.626  10 J  s  3.00  10 m/s

 2.80  10–6 m or 2800 nm
E
7.11  10–20 J
Silicon will emit electrons with a longer wavelength than titanium.
–34

8
Think about It
This answer makes sense because silicon’s work function is greater; electrons ejected using the same
wavelength of light will have lower (kinetic) energies and thus longer wavelengths for silicon
compared to titanium.
7.55. Collect and Organize
To calculate the wavelength emitted when an electron in hydrogen undergoes a transition from n = 4
to n = 3, we can use the Rydberg equation. The region of the electromagnetic spectrum corresponding
to that wavelength can be found from Figure 7.1.
Analyze
In the Rydberg equation, n1 = 3 and n2 = 4.
 1
1
 [1.097  10–2 (nm) –1 ]  2  2 

 n1 n2 
1
Solve
1

1 1
 [1.097  10 –2 (nm) –1 ]  2  2  = 5.333  10 –4 (nm) –1
3 4 
  1875 nm
This wavelength occurs in the infrared region of the electromagnetic spectrum.
Think about It
Notice that in the Rydberg equation, n2 is a higher orbit number (n) than n1. In this way we don’t get
the nonsensical result of a negative wavelength.
7.56. Collect and Organize
To calculate the frequency of light emitted from an electron that undergoes a transition from n = 5 to
n = 3 we can use the Rydberg equation to calculate the wavelength of the emitted light, then convert
346 | Chapter 7
the wavelength into frequency. We can find the corresponding region of the electromagnetic spectrum
from Figure 7.1.
Analyze
In the Rydberg equation we need to use n1 = 3 and n2 = 5:
 1
1
 [1.097  10–2 (nm) –1 ]  2  2 

 n1 n2 
1
The conversion from  to  is given by
 = c/
Solve
 1 1
 [1.097  10 –2 (nm) –1 ]  2  2   7.801  10 –4 (nm) –1

3 5 
1
 = 1282 nm
3.00  108 m/s

 2.34  1014 s –1
1.282  10 –6 m
This transition occurs in the infrared region of the electromagnetic spectrum.
Think about It
When calculating frequency, don’t forget to convert nanometers to meters.
7.60. Collect and Organize
Given the wavelength of a transition in hydrogen as 92.3 nm (ultraviolet region), we are to determine
the nature of this transition.
Analyze
Figure 7.14 shows the transitions observed in a hydrogen atom. Since 92.3 nm is in the ultraviolet
region, this transition is part of the Lyman series. To determine n2 we can use the following equation:
 1
1
1
 [1.097  10–2 (nm) –1 ]  2  2 

n 
n
1
2
where  = 92.3 nm and n1 = 1. The longest wavelength that a ground-state hydrogen atom can absorb
is associated with the n = 1 to n = 2 transition. All n = 1 transitions to higher n levels (3, 4, 5, etc.)
require higher energy or shorter wavelengths.
Solve
(a) Because this line is probably in the Lyman series, the transition is associated with the emission of
light from an electron in an excited state relaxing to the ground state (n = 1).
(b) Here, ni = n2.
1 1
1
 [1.097  10 –2 (nm) –1 ]  2  2 
92.3 nm
1 n 
2
1 1

 0.988
12 n22
1
 0.012
n22
n2  9
(c) The longest wavelength (lowest energy) transition from the ground state (n = 1) is n = 1 to n = 2.
 1 1
E  2.18  10–18 J  2 – 2   1.64  10–18 J
2 1 
Electrons in Atoms and Periodic Properties | 347
Think about It
Remember, long wavelengths are associated with low energy.
7.75. Collect and Organize
As the principal quantum number, n, increases so too does the number of orbitals available at the n
level.
Analyze
The number of orbitals at each level n is n2.
Solve
(a) For n = 1, there is only 1 orbital (an s orbital).
(b) For n = 2, there are 4 orbitals (one s and three p orbitals).
(c) For n = 3, there are 9 orbitals (one s, three p, and five d orbitals).
(d) For n = 4, there are 16 orbitals (one s, three p, five d, and seven f orbitals).
(e) For n = 5, there are 25 orbitals (one s, three p, five d, seven f, and nine g orbitals). This totals 55
orbitals in the atom.
Think about It
Notice that in each subshell, there is an odd number of orbitals and that the number of orbitals in a
particular subshell is 2 + 1 where = 0 for s orbitals, 1 for p orbitals, 2 for d orbitals, and 3 for f
orbitals.
7.76. Collect and Organize
For different combinations of n,
that particular orbital set.
, and m we are to determine the number of orbitals in an atom for
Analyze
The principal quantum number, n, defines the distance from the nucleus; as n increases, the number of
orbitals increases in the shell. The angular momentum quantum number, , defines the type of orbital
(s, p, d, f ); the number of orbitals in each subshell is given by (2 + 1). The magnetic quantum
number, m , describes the orientation of each of the orbitals in a subshell.
Solve
(a) When = 2, there are (2  2 + 1) = 5 orbitals
(b) When = 1, there are (2  1 + 1) = 3 orbitals
(c) Because only one m value is listed, this set of quantum numbers describes 1 orbital.
Think about It
Notice that in a there are five d orbitals and in c there are three p orbitals. This is true for every value
of n.
7.80. Collect and Organize
For each set of quantum numbers of n and
, we are to write the orbital designation.
Analyze
The principal quantum number gives the shell number that is just expressed as the number. The
angular momentum quantum number, however, is given a letter designation ( = 0 is an s orbital, =
1 is a p orbital, = 2 is a d orbital, and = 3 is an f orbital).
Solve
(a) 2p is the orbital for n = 2,
(b) 5f is the orbital for n = 5,
= 1.
= 3.
348 | Chapter 7
(c) 3d is the orbital for n = 3,
(d) 4f is the orbital for n = 4,
= 2.
= 3.
Think about It
The letter designation for the shape of the orbital provides a shorthand designation of the orbital as a
number plus a letter. This is easier than describing the orbital with two numbers (n and ).
7.81. Collect and Organize
Given values for the quantum numbers n, , and m , we are to determine the number of electrons
that could occupy the orbitals described by these quantum numbers.
Analyze
The principal quantum number gives us the shell of the orbitals. This then gives the allowed values of
(n – 1) which in turn describe the type of orbital (s, p, d, or f ). The m quantum number gives us
the orientation of the orbital and its allowed values (– , – + 1, . . . , – 1, ), which gives us the
number of orbitals available for that subshell. Each orbital can accommodate two electrons.
Solve
(a) The set of quantum numbers n = 2,
= 0 describes a 2s orbital that can be occupied by two
electrons.
(b) The set of quantum numbers n = 3, = 1, m = 0 describes one of the 3p orbitals that can be
occupied by two electrons.
(c) The set of quantum numbers n = 4, = 2 describes the set of 4d orbitals. There are five d orbitals
in the subshell so 10 electrons can occupy this orbital set.
(d) The set of quantum numbers n = 1, = 0, m = 0 describes the 1s orbital that can be occupied by
two electrons.
Think about It
Remember that there are one s, three p, five d, seven f, and nine g orbitals in shells for which these are
allowed.
7.82. Collect and Organize
Given values for the quantum numbers n, , and m , we are to determine the number of electrons
that could occupy the orbitals described by these quantum numbers.
Analyze
The principal quantum number gives us the shell of the orbitals. This then gives the allowed values of
(n – 1) which in turn describe the type of orbital (s, p, d, or f ). The m quantum number gives us
the orientation of the orbital and its allowed values (– , – , + 1, . . . , – 1, ), which gives us the
number of orbitals available for that subshell. Each orbital can accommodate two electrons.
Solve
(a) The set of quantum numbers n = 3, = 2 describes the 3d orbitals. There are five d orbitals in the
subshell, so 10 electrons can occupy this orbital set.
(b) The set of quantum numbers n = 5, = 4 describes the 5g orbitals. There are nine g orbitals ( m
= –4, –3, –2, –1, 0, 1, 2, 3, 4) in the subshell, so 18 electrons can occupy this orbital set.
(c) The set of quantum numbers n = 3,
= 0 describes the 3s orbital that can be occupied by 2
electrons.
(d) The set of quantum numbers n = 4, = 1, m = 1 describes one of the 4p orbitals that can be
occupied by 2 electrons.
Think about It
Remember that there is one s, three p, five d, seven f, and nine g orbitals in shells for which these are
allowed. And in part b we see that there are nine g orbitals.
7.83. Collect and Organize
Electrons in Atoms and Periodic Properties | 349
Given values for the quantum numbers n,
allowed.
, m , and ms we are to determine which combinations are
Analyze
The principal quantum number (n) can take on whole numbers starting with 1 (n = 1, 2, 3, 4, . . . ).
The angular momentum quantum numbers ( ) possible for a given n value are n – 1, n – 2, . . . , 0.
The magnetic quantum numbers ( m ) allowed for a given are – , – + 1, . . . , – 1, . Allowed
values for ms are + 12 or – 12 .
Solve
(a) For n = 1, the only allowed value of and m is 0; the combination n = 1, = 1, m = 0, ms =
+ 12 is not allowed because  1 for n = 1.
(b) For n = 3, the allowed values of are 0, 1, 2 and when = 0 the allowed value of m is 0; this
combination of n = 3, = 0, m = 0, ms = – 12 is allowed.
(c) For n = 1, the only allowed value for and m is 0; the combination n = 1, = 0, m = 1,
ms = – 12 is not allowed because m  1 when = 0.
(d) For n = 2, the allowed values of are 0 and 1 and when = 1 the allowed value of m = –1, 0, 1;
this combination of n = 2, = 1, m = 2, ms = + 12 is not allowed because m  2 when = 1.
Think about It
For the allowed combination of quantum numbers, part b describes a 3s orbital.
7.84. Collect and Organize
Given values for the quantum numbers n,
allowed.
, m , and ms we are to determine which combinations are
Analyze
The principal quantum number (n) can take on whole numbers starting with 1 (n = 1, 2, 3, 4, . . . ).
The angular momentum quantum numbers ( ) possible for a given n value are n – 1, n – 2, . . . , 0.
The magnetic quantum numbers ( m ) allowed for a given are – , – + 1, . . . , – 1, . Allowed
values for ms are + 12 or – 12 .
Solve
(a) For n = 3, the allowed values of are 2, 1, and 0 and for = 2 the allowed values of m are –2,
–1, 0, 1, and 2; this combination of n = 3, = 2, m = 0, ms = – 12 is allowed.
(b) For n = 5, the allowed values of are 4, 3, 2, 1, and 0 and for = 4 the allowed values of m are
–4, –3, –2, –1, 0, 1, 2, 3, and 4; this combination of n = 5, = 4, m = 4, ms = + 12 is allowed.
(c) For n = 3, the allowed values of
are 2, 1, and 0 and for
= 0 the allowed value of m is 0;
because m cannot equal 1 when = 0, this combination of n = 3, = 0, m = 1, ms = + 12 is not
allowed.
(d) For n = 4, the allowed values of
are 3, 2, 1, and 0; because
cannot equal 4 for n = 4, this
combination of n = 4, = 4, m = 1, ms = – 12 is not allowed.
Think about It
For the allowed combinations of quantum numbers, part a describes a 3d orbital and part b describes a
5g orbital.
on energy of Mg would be higher still.