UNIT 2 PROBLEM SET ANSWERS
1.
a) Refer to the notes for the figure of the electromagnetic spectrum. The trend in increasing wavelength is:
x-ray < ultraviolet < visible < infrared < microwave < radio waves.
b) Frequency is inversely proportional to wavelength, so frequency has the opposite trend:
radio < microwave < infrared < visible < ultraviolet < x-ray.
c) Energy is directly proportional to frequency. Therefore, the trend in increasing energy matches the trend in
increasing frequency: radio < microwave < infrared < visible < ultraviolet < x-ray.
2.
Wavelength is related to frequency through the equation c = . Recall that a Hz is a reciprocal second, or 1/s = s–1.
3.00 108 m/s
c
(m) =
=
= 312.5 = 313 m

 1  103 Hz   1 s 1 
 960.kHz  
 

 1 kHz   1 Hz 
 1 nm 
3.00 108 m/s 

9
c
 1  10 m 
 (nm) =
=
= 3.125×1011 = 3.13×1011 nm

 1  103 Hz  1 s 1 
 960. kHz  



 1 kHz  1 Hz 


1Å
3.00  108 m/s 

10
c
 1  10 m  = 3.125×1012 = 3.13×1012 Å
 (Å) =
=

 1  103 Hz   1 s 1 
 960. kHz  
 

 1 kHz   1 Hz 
3.
(m) =
c
=

3.00  108 m/s
= 3208556 = 3.21 m
 1  106 Hz   1 s 1 
 93.5 MHz  
 

 1 MHz   1 Hz 
 1 nm 
3.00  108 m/s 

1  109 m 
c

 (nm) =
=
= 3.208556×109 = 3.21×109 nm

 1  106 Hz   1 s 1 
 93.5 MHz  
 

 1 MHz   1 Hz 


1Å
3.00  108 m / s 

10
1

10
m

 = 3.208556×1010 = 3.21×1010 Å
 (Å) =
=

 1  106 Hz  1 s 1 
 93.5 MHz  



 1 MHz  1 Hz 
c
4.
Frequency is related to energy through the equation Ephoton = h. Note that 1 Hz = 1 s–1.
Ephoton = (6.626×10–34 J•s)(3.6×1010 s–1) = 2.385×10–23 = 2.4×10–23 J
5.
= 1.529×10–15 = 1.5×10–15 J
 1  10 m 
1.3 Å 

1Å


Since energy is directly proportional to frequency (Ephoton = h) and frequency and wavelength are inversely related
( = c/), it follows that energy is inversely related to wavelength. As wavelength decreases, energy increases. In
terms of increasing energy the order is red < yellow < blue.
6.
7.
Ephoton =
hc

=
 6.626  10
34

J • s 3.00 108 m/ s

10
Since energy is directly proportional to frequency (Ephoton = h), the smaller the frequency, the smaller the energy.
The order is UV > IR > microwave.
1
8.
a) The least energetic photon has the longest wavelength (242 nm).
3.00  108 m/s
c
= =
= 1.239669×1015 = 1.24×1015 s–1

 1  10 9 m 
242 nm 
 1 nm 


Ephoton =
hc

 6.626  10
=
34

J • s 3.00  108 m/s
 = 8.2140×10
–19
 1  10 m 
242 nm 

 1 nm 
b) The most energetic photon has the shortest wavelength (2200 Å).


1Å
3.00  108 m/s 

10
c
 1  10 m  = 1.3636×1015 = 1.4×1015 s–1
= =
2200 Å

Ephoton =
hc
hc

 6.626  10
=
9
34

J • s 3.00  108 m/s
 1  10 m 
2200 Å 

1Å


 6.626  10
=
34
10
J • s)(3.00  108 m/s

= 8.21×10–19 J
= 9.03545×10–19 = 9.0×10–19 J

9.
= 2.8397×10–19 J (unrounded)
 1  10 m 
700. nm 

 1nm 
This is the value for each photon, that is, J/photon.
Number of photons = (2.0×10–17 J)/(2.8397×10–19 J/photon) = 70.4296 = 70. photons
10.
Ephoton =
Ephoton =
a)  =
b)  =
11.

hc

thus  =
hc
Ephoton
hc
Ephoton
a) Ek =
=
=
=
9
hc
Ephoton
 6.626  10
34
 6.626  10
34




 1 nm 
J • s 3.00  108 m/s 

 1  109 m  = 432.130 = 432 nm
4.60  1019 J
 1 nm 
J • s 3.00  108 m/s 

 1  109 m  = 286.4265 = 286 nm
6.94  1019 J
1
mu 2
2


1
9.11  1031 kg 6.40  105 m/s
2

2

1J

2 2
 1 kg • m /s



= 1.865728×10–19 = 1.87×10–19 J
b) In this example, the incident light is higher than the threshold frequency, so the kinetic energy of the electron,
Ek, must be subtracted from the total energy of the incident light to yield the work function, .
hc
 6.626  10
34

J • s 3.00  108 m/s

= 5.55096×10–19 J (unrounded)
 1  10 m 
 358.1 nm  

 1 nm 
 = Ephoton − Ek = (5.55096×10–19 J) − (1.86552×10–19 J) = 3.68544×10–19 = 3.69×10–19 J
Ephoton =

=
9
2
12.
(a) Use Bohr’s equation for the difference in energy levels, then find the wavelength.
 1
1 
E   A Z 2  2  2 
 nf ni 
 1
1 
 (2.179  1018 J)(1) 2  2  2 
(2)
(5)


19
 4.576  10 J
 Ephoton = |E| = 4.576  10−19 J


hc
E
(6.626  10 34 J s)(3.00  108 m/s)
4.576  10 19 J
 1 nm 
 4.34  10 7 m 

9
 1 10 m 
 434 nm
(b) Multiply E by Avogadro’s number.
E = −4.576  10−19 J×6.022×1023 mol−1 = −2.75567×105 = −2.756×105 J/mol
The value is negative because light is emitted.
13.
(a) Use Bohr’s equation for the difference in energy levels, then find the wavelength.
 1
1 
E   A Z 2  2  2 
 nf ni 
 1
1 
 (2.179  10 18 J)(1) 2  2  2 
 (3) (1) 
18
 1.93689  10 J
 Ephoton = |E| = 1.93689  10−18 J


hc
E
(6.626  10 34 J s)(3.00  108 m/s)
1.93689  1018 J
 1Å

 1.02628  10 7 m 


 1  10 m 
 1026.28 Å
 1030 Å
(b) Multiply E by Avogadro’s number.
E = 1.93689  10−18 J×6.022×1023 mol−1 = 1.166395×106 = 1.166×106 J/mol
The value is positive because light is absorbed.
3
14.
For the infrared series of the H atom, the final state is the second excited state, which means that nf = 3. The least
energetic spectral line in this series would represent an electron moving from the next highest energy level, ni = 4.
 1
1 
E   A Z 2  2  2 
 nf ni 
 1
1 
 (2.179  10 18 J)(1) 2  2  2 
(3)
(4)


 1.05924  10 19 J
 Ephoton = |E| = 1.05924  10−19 J


hc
E
(6.626  10 34 J s)(3.00  108 m/s)
1.05924  10 19 J
 1 nm 
 1.8766  106 m 


 1  10 m 
 1876.6 nm
 1880 nm
15.
For the visible series of the H atom, the final state is the second excited state, which means that nf = 2. The least
energetic spectral line in this series would represent an electron moving from the next highest energy level, ni = 3.
 1
1 
E   A Z 2  2  2 
 nf ni 
 1
1 
 (2.179  10 18 J)(1) 2  2  2 
 (2) (3) 
19
 3.02639  10 J
 Ephoton = |E| = 3.02639  10−19 J


hc
E
(6.626  10 34 J s)(3.00  108 m/s)
3.02639  10 19 J
 1 nm 
 6.56822 10 7 m 


 1  10 m 
 656.822 nm
 657 nm
16.
E 

hc

(6.626  10 34 J s)(3.00  108 m/s)
 1  10 m 
97.20 nm 

 1 nm 
 2.04506  1018 J
E = 2.04506×10−18 J (positive since absorption occurs)
4
 1
1 
E   A Z 2  2  2 
n
n
 f
i 
 1
1 
2.04506  10 18 J  (2.179  10 18 J)(1) 2  2  2 
n
(1)
 f

1
0.0614686  2
nf
nf2  16.2685
nf  4.03  4 (round to the nearest whole number)
17.
a)
E 

hc

(6.626  10 34 J s)(3.00  108 m/s)
 1 10 m 
94.91 nm 

 1 nm 
 2.09441 10 18 J
E = 2.09441×10−18 J (positive since absorption occurs)
 1
1 
E   A Z 2  2  2 
n
n
 f
i 
 1
1 
2.09441  1018 J  (2.179  1018 J)(1) 2  2  2 
n
(1)
 f

1
0.038821  2
nf
nf2  25.760
nf  5.08  5 (round to the nearest whole number)
b)
E 
hc

(6.626  10 34 J s)(3.00  108 m/s)

 1  10  m 
1281 nm 

 1 nm 
 1.55176  10 19 J
E = −1.55176×10−19 J (negative since emission occurs)
 1
1 
E   A Z 2  2  2 
 nf ni 
 1
1 
1.55176  10 19 J  (2.179  10 18 J)(1) 2  2  2 
 nf (5) 
1
0.1112143  2
nf
nf2  8.9916
nf  3.00  3 (round to the nearest whole number)
5
c)
 1
1 
E   A Z 2  2  2 
 nf ni 
 1
1 
 (2.179  10 18 J)(1) 2  2  2 
(1)
(3)


18
 1.93689  10 J
Ephoton = |E| = 1.93689  10−18 J
hc

E

(6.626  10 34 J s)(3.00  108 m/s)
1.93689 10 18 J
 1 nm 
 1.02628  10 7 m 


 1  10 m 
 102.628 nm
 103 nm
18.
In the ground state, ni = 1.
 1
1 
E   A Z 2  2  2 
 nf ni 
 1
1 
 (2.179  1018 J) (5)2 
 2
2
(

)
(1)


 (2.179  1018 J) (5) 2  0  1
 5.4475  1017 J
This is energy to ionize one of the ions. For one mole and converting to kJ:
E  5.4475  10 17 J  6.022  10 23 mol 1
 3.28048  107 J mol 1
19.
 1 kJ 
 3.280  107 J mol 1 

 1000 J 
 3.280  104 kJ mol 1
 1 1 
E   A Z 2  2  2 
 nf ni 
 1
1 
 (2.179  1018 J) (2) 2 
 2
2
 ( ) (3) 
1

 (2.179  10 18 J) (2) 2  0  
9

 9.6844  1019 J
hc

E

(6.626  10 34 J s)(3.00  108 m/s)
9.6844  10 19 J
 1 nm 
 2.05258  10 7 m 


 1  10 m 
 205.258 nm
 205 nm
6
20.
The highest-energy line corresponds to the shortest wavelength photon. It also corresponds to ionization of the
electron with nf = .
hc
E 

(6.626  10 34 J s)(3.00  108 m/s)

 1 10  m 
3282 nm 

 1 nm 
 6.05667  1020 J
E = 6.05667×10−20 J (positive since absorption occurs)
 1
1 
E   A Z 2  2  2 
 nf ni 
 1
1 
6.05667  10 20 J  (2.179  10 18 J)(1) 2  2  2 
ni 


1 
6.05667  10 20 J  (2.179  10 18 J)(1) 2  0  2 
ni 

1
0.0277957   2
ni
ni2  35.9768
ni  6.00  6 (round to the nearest whole number)
21.
The highest-energy line corresponds to ionizing the electron from the ground state. Therefore, the transition is from
n = 1 to n = . The energy of the photon is:
E = h = (6.626×10–34 J∙s)(2.961×1016 Hz) (1 s–1/1 Hz) = 1.9619586×10–17 J (unrounded)
E = 1.9619586×10–17 J (positive since absorption occurs)
 1 1 
E   A Z 2  2  2 
 nf ni 
 1
1 
1.9619586  10 17 J  (2.179  10 18 J) Z 2 
 2
2
 () (1) 
17
18
2
1.9619586  10 J  (2.179  10 J) Z  0  1
Z 2  9.0039
Z  3.00  3 (round to nearest whole number)
Find the Z = 3 on the period table (Li). Note, however, that it is a one-electron ion of Li in this question, not the Li
atom itself (the equations we use above hold only for one-electron species). Therefore, the ion is Li2+.
22.
 1 1 
E   A Z 2  2  2 
 nf ni 
Ionization involves complete removal of the electron, so nf =  (assuming that the electron has no kinetic energy
after the ionization). The above equation applies to one ion, but we are given the ionization energy for one mole of
the ions (they are often tabulated per mole). Therefore, we must divide this by Avogadro’s number to get the value
for one ion.
 1
1310 103 J mol1
1 
 (2.179  1018 J) Z 2 
 2
23
1
2
6.022 10 mol
 () (5) 
1 

2.175 10 18 J  (2.179  1018 J) Z 2  0  
25 

2
Z  24.95
Z  5.00  5 (round to nearest whole number)
The element with atomic number 5 is B. Neutral B atoms have 5 electrons, but for boron to have a hydrogen-like
ion, it must have only one electron. Thus, the ion must be B4+.
7
23.
Z can be calculated from the photon energy and the energy change in a one-electron species:
hc
Ephoton 

(6.626  10 34 J s)(3.00  108 m/s)

(30.4  109 m)
 6.539  1018 J
E = −6.53910−18 J (emission)
 1
1 
E   A Z 2  2  2 
 nf ni 
 1 1 
6.539  1018 J  (2.179  10 18 J) Z 2  2  2 
2 4 
 1 1 
3.00092  Z 2  2  2 
2 4 
Z 2  16.005
Z  4.00  4 (round to the nearest whole number)
Species: Be3+
24.
Use the de Broglie equation.
 1 kg m 2 /s 2 
J • s 

1J


 = h/mu =

 1 kg  
mi   1.609 km  1  103 m   1 h 
 220 lb  

  19.6  


h   1 mi   1 km   3600 s 
 2.205 lb  
 6.626 × 10
34

= 7.58102×10–37 = 7.6×10–37 m
 1 kg m 2 /s 2 
J • s 

1J


 = h/mu =


 1  103 m   1 h


mi
1
kg
1.609
km


6.6 × 1024 g  3.4 × 107






h   1  103 g   1 mi   1 km   3600

 6.626 × 10
25.

34




s
= 6.60656×10–15 = 6.6×10–15 m
26.
 = h/mu
 1 kg m 2 /s 2 
J • s 

1J


u = h/m =
= 2.1717×10–26 = 2.2×10–26 m/s

10
 1 kg 
 1  10 m 
 56.5 g  

3 
  5400 Å  
1Å
 1  10 g 


 6.626 × 10
27.
34

 = h/mu
 1 kg m 2 /s 2 
J • s 

1J


u = h/m =
= 4.666197×10–23 = 4.67×10–23 m/s
 1 kg   1  1012 m 
142 g 100. pm  
 

3 
 1  10 g   1 pm 
 6.626 × 10
34

8