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CHAPTER 1
THE QUANTUM THEORY OF THE
SUBMICROSCOPIC WORLD
1.1
1.2
(a)
 
(b)
 
c

c



3.00  108 m s1
8.6  1013 s1
3.00  108 m s1
566  10
9
 3.5  106 m  3.5  103 nm
 5.30  1014 s1  5.30  1014 Hz
m
(a)
Strategy: We are given the wavelength of an electromagnetic wave and asked to calculate the frequency.
Rearranging Equation (1.2) of the text and replacing u with c (the speed of light) gives:
 
c

Solution: Because the speed of light is given in meters per second, it is convenient to first convert
wavelength to units of meters. Recall that 1 nm  1  109 m. We write:
456 nm 
1  109 m
 456  109 m  4.56  107 m
1 nm
Substituting in the wavelength and the speed of light ( 3.00  108 mŹs1 ), the frequency is:
 
c


3.00  108 m s 1
4.56  107 m
 6.58  1014 s1 or 6.58  1014 Hz
Check: The answer shows that 6.58  1014 waves pass a fixed point every second. This very high frequency
is in accordance with the very high speed of light.
(b)
Strategy: We are given the frequency of an electromagnetic wave and asked to calculate the wavelength.
Rearranging Equation (1.2) of the text and replacing u with c (the speed of light) gives:
 
c

Solution: Substituting in the frequency and the speed of light ( 3.00  108 mŹs1 ) into the above equation,
the wavelength is:
 
c


3.00  108 m s 1
2.45  109 s 1
 0.122 m
The problem asks for the wavelength in units of nanometers. Recall that 1 nm  1  109 m.
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
  0.122 m 
1.3
 
c


1 nm
1  10
9
195
 1.22  108 nm
m
3.00  108 m s 1
9192631770 s 1
 3.26  102 m  3.26  107 nm
This radiation falls in the microwave region of the spectrum. (See Figure 1.6 of the text.)
1.4
1.5
1.6
The wavelength is:
 
1m
 6.05780211  107 m
1,650,763.73 wavelengths
 
c

=
E  h 
3.00  108 m s 1
6.05780211  10
hc


7
 4.95  1014 s 1 or 4.95  1014 Hz
m
(6.63  1034 J s)(3.00  108 m s1 )
624  10
9
 3.19  1019 J
m
(a)
Strategy: We are given the frequency of an electromagnetic wave and asked to calculate the wavelength.
Rearranging Equation (1.2) of the text and replacing u with c (the speed of light) gives:
 
c

Solution: Substituting in the frequency and the speed of light (3.00  108 m s-1) into the above equation, the
wavelength is:
 
3.00  108 m s 1
14 1
7.5  10 s
 4.0  107 m  4.0  102 nm
Check: The wavelength of 400 nm calculated is in the blue region of the visible spectrum as expected.
(b)
Strategy: We are given the frequency of an electromagnetic wave and asked to calculate its energy.
Equation (1.3) of the text relates the energy and frequency of an electromagnetic wave.
E  h
Solution: Substituting in the frequency and Planck's constant (6.63  1034 J s) into the above equation, the
energy of a single photon associated with this frequency is:


E  h  (6.63  1034 J s) 7.5  1014 s1  5.0  1019 J
Check: We expect the energy of a single photon to be a very small energy as calculated above, 5.0  1019 J.
196
1.7
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
(a)
 
c


3.00  108 m s1
4 1
6.0  10 s
 5.0  103 m  5.0  1012 nm
The radiation does not fall in the visible region; it is radio radiation. (See Figure 1.6 of the text.)
1.8
(b)
E  h  (6.63  1034 J s)(6.0  104 s-1)  4.0  1029 J
(c)
Converting to J mol-1: E 
4.0  1029 J 6.022  1023 photons

 2.4  105 J mol 1
1 photon
1 mol
The energy given in this problem is for 1 mole of photons. To apply E  h, we must divide the energy by
Avogadro’s number. The energy of one photon is:
E 
1.0  103 kJ
1 mol
1000 J


 1.7  1018 J
23
1 mol
1 kJ
6.022  10 photons
The wavelength of this photon can be found using the relationship, E 


(6.63  1034 J s) 3.00  108 m s1
hc
 

E
1.7  1018 J

 1.2  107 m 
hc

.
1 nm
1  10
9
 1.2  102 nm
m
The radiation is in the ultraviolet region (see Figure 1.6 of the text).
1.9
1.10
hc
E  h 
 


( 6.63  1034 J s )( 3.00  108 m s 1 )
( 0.154  10
9
 1.29  1015 J
m)
c
(a)

3.00  108 m s1
 
 3.70  107 m  3.70  102 nm
14 1
8.11  10 s
(b)
Checking Figure 1.6 of the text, you should find that the visible region of the spectrum runs from 400 to
700 nm. 370 nm is in the ultraviolet region of the spectrum.
(c)
E  h. Substitute the frequency () into this equation to solve for the energy of one quantum
associated with this frequency.


E  h  ( 6.63  1034 J s ) 8.11  1014 s1  5.38  1019 J
1.11
First, we need to calculate the energy of one 600 nm photon. Then, we can determine how many photons are
needed to provide 4.0  1017 J of energy.
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
The energy of one 600 nm photon is:
E 
hc


(6.63  1034 J s)(3.00  108 m s1 )
600  10
9
 3.32  1019 J
m
The number of photons needed to produce 4.0  1017 J of energy is:
(4.0  1017 J) 
1.12
1 photon
3.32  10
19
 1.2  102 photons
J
The heat needed to raise the temperature of 150 mL of water from 20C to 100C is:
heat  (150 ml)(4.184 J ml-1 C-1)(100  20)C  5.0  104 J
The microwave will need to supply more energy than this because only 92.0% of microwave energy is
converted to thermal energy of water. The energy that needs to be supplied by the microwave is:
5.0  104 J
 5.4  104 J
0.920
The energy supplied by one photon with a wavelength of 1.22  108 nm (0.122 m) is:
E 
hc


(6.63  1034 J s)(3.00  108 m s1 )
 1.63  1024 J
(0.122 m)
The number of photons needed to supply 5.4  104 J of energy is:
(5.4  104 J) 
1.13
1 photon
1.63  1024 J
 3.3  1028 photons
This problem must be worked to four significant figure accuracy. We use 6.6256  1034 Js for Planck’s
constant and 2.998  108 mŹs1 for the speed of light. First calculate the energy of each of the photons.
E 
E 
hc

hc



(6.6256  1034 J s)(2.998  108 m s1 )
589.0  10
9
m
(6.6256  1034 J s)(2.998  108 m s1 )
589.6  10
9
 3.372  1019 J
 3.369  1019 J
m
For one photon the energy difference is:
E  (3.372  1019 J)  (3.369  1019 J)  3  1022 J
For one mole of photons the energy difference is:
3  1022 J 6.022  1023 photons

 2  102 J mol 1
1 photon
1 mol
197
198
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
1.14
The radiation emitted by the star is measured as a function of wavelength (). The wavelength corresponding
to the maximum intensity is then plugged in to Wien’s law (Tmax = 1.44  10-2 K m) to determine the surface
temperature.
1.15
Wien’s law: Tmax = 1.44  10-2 Km.
Solving for temperature (T) yields:
T
1.44  10 2 Km

100nm: ultraviolet
T
1.44  10 2 K m
 1.44  10 5 K
100  10 9 m
300nm: ultraviolet
T
1.44  10 2 K m
 4.80  10 4 K
300  10 9 m
500nm: visible (green)
T
1.44  10 2 K m
 2.88  10 4 K
500  10 9 m
800nm: infrared
T
1.44  10 2 K m
 1.80  10 4 K
800  10 9 m
Most night vision goggles work by converting low energy infrared radiation (which our eyes can not detect) to
higher energy visible radiation.
1.16
Because the same number of photons are being delivered by both lasers and because both lasers produce
photons with enough energy to eject electrons from the metal, each laser will eject the same number of
electrons. The electrons ejected by the blue laser will have higher kinetic energy because the photons from
the blue laser have higher energy.
1.17
In the photoelectric effect, light of sufficient energy shining on a metal surface causes electrons to be ejected
(photoelectrons). Since the electrons are charged particles, the metal surface becomes positively charged as
more electrons are lost. After a long enough period of time, the positive surface charge becomes large enough
to increase the amount of energy needed to eject electrons from the metal.
1.18
(a) The maximum wavelength is obtained by solving Equation 1.4 when the kinetic energy of the ejected
electrons is equal to zero. In this case, Equation 1.4, together with the relation   c /  , gives
E
(b)
hc


 6.626  10
hc

J s  3.00  108 m s 1 
520  109 m
  h  (kinetic Energy)
(kinetic Energy) 
34
 
 3.82  1019 J
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
 6.626  10
(kinetic Energy) 
34
J s  3.00  108 m s 1 
9
450  10 m
199
 3.8227  10 19 J  5.9  1020 J
Please remember to carry extra significant figures during the calculation and then round at the
very end to the appropriate number of significant figures.
1.19 The minimum (threshold) frequency is obtained by solving Equation 1.4 when the kinetic energy of the ejected
electrons is equal to zero. In this case, Equation 1.4 gives
E  h   6.626  1034 J s  8.54  1014 s 1   5.66  1019 J
1.20
The emitted light could be analyzed by passing it through a prism.
1.21
Light emitted by fluorescent materials always has lower energy than the light striking the fluorescent
substance. Absorption of visible light could not give rise to emitted ultraviolet light because the latter has
higher energy.
The reverse process, ultraviolet light producing visible light by fluorescence, is very common. Certain brands
of laundry detergents contain materials called “optical brighteners” which, for example, can make a white
shirt look much whiter and brighter than a similar shirt washed in ordinary detergent.
1.22
Excited atoms of the chemical elements emit the same characteristic frequencies or lines in a terrestrial
laboratory, in the sun, or in a star many light-years distant from earth.
1.23
(a)
The energy difference between states E1 and E4 is:
E4  E1  (1.0  1019 J)  (15  1019 J)  14  1019 J

(b)
hc
(6.63  1034 J s)(3.00  108 m s1 )

 1.4  107 m  140 nm
19
E
14  10 J
The energy difference between the states E2 and E3 is:
E3  E2  (5.0  1019 J)  (10.0  1019 J)  5  1019 J
(c)
The energy difference between the states E1 and E3 is:
E1  E3  (15  1019 J)  (5.0  1019 J)  10  1019 J
Ignoring the negative sign of E, the wavelength is found as in part (a).
 
1.24
hc
(6.63  1034 J s)(3.00  108 m s1 )

 2.0  107 m  200 nm
19
E
10  10 J
We use more accurate values of h and c for this problem.
200
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
E 
1.25
hc

=
( 6.6256  1034 J s )( 2.998  108 m s1 )
656.3  109 m
 3.027  1019 J
In this problem ni  3 and nf  5.
 1
 1
1 
1
E  RH 


 1.55  1019 J
  (2.18  1018 J) 
2
2
2 
 n 2

nf 
3
5
i
The sign of E means that this is energy associated with an absorption process.
 
hc
(6.626  1034 J s)(2.998  108 m s1 )

 1.28  106 m  1.28  103 nm
19
E
1.55  10 J
Is the sign of the energy change consistent with the sign conventions for exo- and endothermic processes?
1.26
Strategy: We are given the initial and final states in the emission process. We can calculate the energy of
the emitted photon using Equation (1.5) of the text. Then, from this energy, we can solve for the frequency of
the photon, and from the frequency we can solve for the wavelength. The value of Rydberg's constant is
2.18  1018 J.
Solution: From Equation (1.5) we write:
 1
1
E  RH 


 n2
nf2 
i
 1
1
E  (2.18  1018 J) 


2
4
22 
E  4.09  1019 J
The negative sign for E indicates that this is energy associated with an emission process. To calculate the
frequency, we will omit the minus sign for E because the frequency of the photon must be positive. We
know that
E  h
Rearranging this equation and substituting in the known values,
 
E
4.09  1019 J
=
 6.17  1014 s1 or 6.17  1014 Hz
h
6.63  1034 J s
We also know that  
 
3.00  108 m s1
14 1
6.17  10 s
c

. Substituting the frequency calculated above into this equation gives:
 4.86  107 m  486 nm
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
201
Check: This wavelength is in the visible region of the electromagnetic region (see Figure 1.6 of the text).
This is consistent with the transition from ni  4 to nf  2 gives rise to a spectral line in the Balmer series (see
Figure 1.15 of the text).
1.27
The Balmer series corresponds to transitions to the n  2 level.
For He:
E  R
He
 

 1
1



2
 n2
n
i
f 
hc
(6.63  1034 J s)(3.00  108 m s1 )

E
E
For the transition, n  6  2
 1
1
E  (8.72  1018 J) 

  1.94  1018 J
2 
 62
2
 
1.99  1025 J m
1.94  1018 J
 1.03  107 m  103 nm
For the transition, n  6  2, E  1.94  1018 J
  103 nm
For the transition, n  5  2, E  1.83  1018 J
  109 nm
For the transition, n  4  2, E  1.64  1018 J
  121 nm
For the transition, n  3  2, E  1.21  1018 J
  164 nm
For H, the calculations are identical to those above, except the Rydberg constant for H is 2.18  1018 J.
For the transition, n  6  2, E  4.84  1019 J
  411 nm
For the transition, n  5  2, E  4.58  1019 J
  434 nm
For the transition, n  4  2, E  4.09  1019 J
  487 nm
For the transition, n  3  2, E  3.03  1019 J
  657 nm
All the Balmer transitions for He are in the ultraviolet region; whereas, the transitions for H are all in the
visible region. Note the negative sign for energy indicating that a photon has been emitted.
1.28
Equation 1.14:
rn 
0 h2 n2
Z  me e 2
for n = 2: r2 
(8.8542  1012 C2 J –1 m –1 )(6.626  1034 J s)2 2 
 2.117x1010 m = 211.7 pm
1 (9.109  1031 kg)(1.602  1019 C)2
for n = 3: r3 
(8.8542  1012 C2 J –1 m –1 )(6.626  1034 J s)2 3
 4.764x1010 m = 476.4 pm
1 (9.109  1031 kg)(1.602  1019 C)2
2
2
202
1.29
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
ni  236, nf  235
 1
1 
E  (2.18  1018 J) 

  3.34  1025 J
2 
 2362
235
 
hc
(6.63  1034 J s)(3.00  108 m s1 )

 0.596 m
E
3.34  1025 J
This wavelength is in the microwave region. (See Figure 1.6 of the text.)
1.30
  Z 2 RH
1
1

n12 n22
  1  3.289832496 1015 s –1 
2
1.31
1 1
  3.289832496x1015 s –1  3.289832496x1015 Hz
 1
First convert the parameters to the appropriate units and then use the following equation:  
(a)
 1.50  108 cm   m   1.50  106 m 
6
1

  100 cm   
  1.50  10 m s
s
s

(b)
h
mu
6.626  10 –34 J s
 4.85  10 –10 m
(9.1094  10 –31 kg)(1.50  10 6 m s–1 )
 60.0 g   kg 
 0.0600 kg  6.00  10 2 kg


1   1000 g 
 1500 cm   m   15.00 m 
1
1
1

 
 
  15.00 m s  1.500  10 m s
s
100 cm  
s

6.626  10 –34 J s
 7.36  10 -34 m
(6.00  10 –2 kg)(1.500  101 m s –1 )
h
6.63  1034 J s

 5.65  1010 m  0.565 nm
mu
(1.675  1027 kg)(7.00  102 m s1 )
1.32
 
1.33
Strategy: We are given the mass and the speed of the proton and asked to calculate the wavelength. We
need the de Broglie equation, which is Equation 1.20 of the text. Note that because the units of Planck's
constant are J s, m must be in kg and u must be in m s1 (1 J  1 kg m2 s2).
Solution: Using Equation 1.20 we write:
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
 
203
h
mu


6.63  1034 kg m2 s1
h
 

 1.37  1015 m
mu
(1.673  1027 kg)(2.90  108 ms1)
The problem asks to express the wavelength in nanometers.
  (1.37  1015 m) 
1.34
1 nm
1  10
9
 1.37  106 nm
m
Converting the velocity to units of ms1:
120 mi 1.61 km 1000 m
1 hr



 53.7 m s1
1 hr
1 mi
1 km
3600 s
 
1.35
h
6.63  1034 J s

 9.96  1034 m  9.96  1032 cm
1
mu
(0.0124 kg)(53.7 m s )
First, we convert mph to m s1.
35 mi 1.61 km 1000 m
1h



 16 m s1
1h
1 mi
1 km
3600 s
 
1.36


6.63  1034 kg m2 s1
h
=
 1.7  1032 m  1.7  1023 nm
3
1
mu
(2.5  10 kg)(16 ms )
From the Heisenberg uncertainty principle (Equation 1.22) we have
x  p 
h
2
For the minimum uncertainty we can change the greater than sign to an equal sign.
p 
h
1.055 1034 J s

 1 10 –24 kg m s –1
2 x
2 (4  10 –11m)
Because p = m u, the uncertainty in the velocity is given by
u 
p 1  10 –24 kg m s –1

 1  10 6 m s –1
m 9.1095  10 –31 kg
The minimum uncertainty in the velocity is 1  106 m s–1.
1.37
Below are the probability distributions for the first three energy levels for a one-dimensional particle in a box,
as shown in Figure 1.24. The average value for x for all of the states for a one-dimensional particle in a box is
L/2 (the middle of the box). This is true even for those states (like the second state) that have no probability
of being exactly in the middle of the box.
204
1.38
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
The difference between the energy levels n and (n + 1) for a one-dimensional particle in a box is given by
Equation 1.31:
Enn1 
h2
(2n  1) .
8mL2
All we have to do is to plug the numbers into the equation remembering that 1 Å=10-10 m.
Enn1 
(6.626  10 34 J s)2 [2(2)  1]
 1.20  10 18 J
8(9.109  10 31 kg)[5.00  10 10 m]2
Energy is equal to Planck’s constant times frequency. We can rearrange this equation to solve for frequency.

1.39
E
1.20  10 18 J

 1.81  1015 s1  1.81  1015 Hz
h 6.626  10 34 J s
From Equation 1.15 we can derive an equation for the change in energy going from the n=1 to n=2 states for a
hydrogen atom:
E 
3Z 2 e4 me
.
32h 2  02
From Equation 1.31 we can derive an equation for the change in energy going from the n=1 to n=2 states for a
particle in a box:
E 
3h 2
.
8mL2
We can combine the two equations and solve for L.
3Z 2 e4 me
3h 2

32h 2  02
8mL2
L
4h 4  02
2h 2 
2(6.626  10 34 J s)2 (8.8542  10 12 C2 J –1 m –1 )
 2 0 
 3.326  10 10 m
2 4 2
Z e me Ze me
1(1.602  1019 C)2 (9.109  1031 kg)
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
205
The radius of the ground state of the hydrogen atom (the Bohr radius) is 5.29  10-11 m (see Example 1.4).
This leads to a diameter of 1.06  10-10 m, which is only a third the size of L that we calculated. So, it seems
like a crude estimate.
1.40
The average value for x will be greater than L/2. The larger the electric field, the larger the average value of
x.
1.41
(a)
The number given in front of the designation of the subshell is the principal quantum number, so in this
case n  3. For s orbitals, l  0. ml can have integer values from l to l, therefore, ml  0. The
electron spin quantum number, ms, can be either 1/2 or −1/2.
Following the same reasoning as part (a)
(b)
4p: n  4; l  1; ml  −1, 0, or 1; ms  1/2, −1/2
(c)
3d: n  3; l  2; ml  −2, −1, 0, 1, or 2; ms  1/2, −1/2
An orbital in a subshell can have any of the allowed values of the magnetic quantum number for that subshell.
All the orbitals in a subshell have exactly the same energy.
1.42
(a)
(b)
(c)
2p: n  2, l  1, ml  1, 0, or 1
6s: n  6, l  0, ml  0 (only allowed value)
5d: n  5, l  2, ml  2, 1, 0, 1, or 2
An orbital in a subshell can have any of the allowed values of the magnetic quantum number for that subshell.
All the orbitals in a subshell have exactly the same energy.
1.43
The allowed values of l are 0, 1, 2, 3, and 4. These correspond to the 5s, 5p, 5d, 5f, and 5g subshells. These
subshells each have one, three, five, seven, and nine orbitals, respectively.
206
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
1.44
For n  6, the allowed values of l are 0, 1, 2, 3, 4, and 5 [l  0 to (n  1), integer values]. These l values
correspond to the 6s, 6p, 6d, 6f, 6g, and 6h subshells. These subshells each have 1, 3, 5, 7, 9, and 11 orbitals,
respectively (number of orbitals  2l  1).
1.45
There can be a maximum of two electrons occupying one orbital.
(a)
two
(b)
six
(c)
ten
(d)
fourteen
What rule of nature demands a maximum of two electrons per orbital? Do they have the same energy? How
are they different? Would five 4d orbitals hold as many electrons as five 3d orbitals? In other words, does
the principal quantum number n affect the number of electrons in a given subshell?
1.46
A 2s orbital is larger than a 1s orbital. Both have the same spherical shape. The 1s orbital is lower in energy
than the 2s. The 1s orbital does not have a node while the 2s orbital does.
1.47
The two orbitals are very similar in shape, though the 3py is larger and has a higher energy. The 3py orbital
will have 1 radial node, in contrast to the 2px orbital, which has no radial nodes. The two orbitals also differ in
their orientation: 3py is oriented parallel to the y-axis and 2px is oriented parallel to the x-axis. Can you assign
a specific value of the magnetic quantum number to these orbitals? What are the allowed values of the
magnetic quantum number for the 2p subshell?
1.48
We can modify Equation 1.15 to determine the ionization energies of single electron atoms. Ionization is the
process of removing an electron (exciting it to a infinitely high energy state).
Eionization 
Z 2 e4 me  1 1  Z 2 e4 me  1 


8h2 20  n12   8h2 20  n12 
He+ has two protons in the nucleus and hence a +2 nuclear charge, making Z=2.
Eionization 
Z 2 e4 me
8h2  02
 1 
1
2 2 (1.602  1019 C)4 (9.109  1031 kg)
18

 n2  8(6.626  10 34 J s)2 (8.8542  10 12 C2 J –1 m –1 )2  2   8.715  10 J
1




 1
We need to convert our answer to kJ mol–1.
 8.715  1018 J   1 kJ   6.022  1023 
3
-1
Eionization  
  1000  
  5.248  10 kJ mol
1
mol

Li2+ has three protons in the nucleus and hence a +3 nuclear charge, making Z = 3.
Eionization 
Z 2 e4 me
8h2  02
 1 
1
32 (1.602  1019 C)4 (9.109  1031 kg)
17

 n2  8(6.626  1034 J s)2 (8.8542  10 12 C2 J –1 m –1 )2  2   1.961  10 J
 1 
 1
We need to convert our answer to kJ mol–1.
 1.961  1017
Eionization  
1

J   1 kJ   6.022  1023 
4
-1
  1000  
  1.181  10 kJ mol
mol
207
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
1.49
Equation 1.15 yields the energy of the photon absorbed for different transitions. For an emission, Equation
1.15 can be written:
Ephoton 
Z 2 e4 me
8h2  02
 1
1 
 n2  n2 
 final
initial 
Energy equals Planck’s constant times the speed of light divided by the wavelength: E 
hc

.
We can use this relationship to modify the above equation.
hc


Z 2 e4 me
8h 2  02
 1
1 
 n2  n2 
 final
initial 
We can now solve for one over ninitial.
1
ninitial

1
2
final
n

8h 3 02 c

Z 2 e4 me 


34
3
12 2
–1
–1 2
8
-1
1 8(6.626  10 J s) (8.8542  10 C J m ) 3.0  10 m s
1

 0.20 
2
2
19
4
31
9
2
5
1 (1.602  10 C) (9.109  10 kg) 434  10 m


Hence, ninitial is 5.
1.50
1.51
(a)
False. n  2 is the first excited state.
(b)
False. In the n  4 state, the electron is (on average) further from the nucleus and hence easier to
remove.
(c)
True.
(d)
False. The n  4 to n  1 transition is a higher energy transition, which corresponds to a shorter
wavelength.
(e)
True.
The energy given in the problem is the energy of 1 mole of gamma rays. We need to convert this to the
energy of one gamma ray, and then we can calculate the wavelength and frequency of this gamma ray.
1.29  1011 J
1 mol

 2.14  1013 J
23
1 mol
6.022  10 gamma rays
Now, we can calculate the wavelength and frequency from this energy.
E 
hc

 
hc
(6.63  1034 J s)(3.00  108 m s1 )

 9.29  1013 m  0.929 pm
13
E
2.14  10 J
and
E  h
208
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
 
1.52
We first calculate the wavelength, then we find the color using Figure 1.6 of the text.
 
1.53
E
2.14  1013 J

 3.23  1020 s1  3.23  1020 Hz
h
6.63  1034 J s
(a)
hc
(6.63  1034 J s)(3.00  108 m s1 )

 4.63  107 m  463 nm, which is blue.
19
E
4.30  10 J
First, we can calculate the energy of a single photon with a wavelength of 633 nm.
E 
hc


(6.63  1034 J s)(3.00  108 m s1 )
633  109 m
 3.14  1019 J
The number of photons produced in a 0.376 J pulse is:
0.376 J 
(b)
1 photon
3.14  10
19
 1.20  1018 photons
J
Since a 1 W  1 J s-1, the power delivered per a 1.00  109 s pulse is:
0.376 J
1.00  109 s
 3.76  108 J s1  3.76  108 W
Compare this with the power delivered by a 100-W light bulb!
1.54
The energy required to heat the water is:
mCsT  (368 g)(4.184 J g-1 C-1)(5.00C)  7.70  103 J,
where m is the mass, Cs the specific heat of water, and T is the change in temperature.
The energy of a photon with a wavelength of 1.06  104 nm can be calculated as follows:
E  h 
hc
(6.63  1034 J s)(3.00  108 m s1 )

 1.88  1020 J
5

1.06  10 m
The number of photons required is:
(7.70  103 J) 
1.55
1 photon
1.88  10
20
 4.10  1023 photons
J
First, let’s find the energy needed to photodissociate one water molecule.
285.8 kJ
1 mol

 4.746  1022 kJ  4.746  1019 J
23
1 mol
6.022  10 molecules
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
209
The maximum wavelength of a photon that would provide the above energy is:
 
hc
(6.63  1034 J s)(3.00  108 m s1 )

 4.19  107 m  419 nm
19
E
4.746  10 J
This wavelength is in the visible region of the electromagnetic spectrum. Since water is continuously being
struck by visible radiation without decomposition, it seems unlikely that photodissociation of water by this
method is feasible.
1.56
For the Lyman series nf = 1. The longest wavelength emission corresponds to the smallest change in energy
(ni  2). By multiplying both sides of Equation 1.5 by Planck’s constant the following equation is obtained:
 1
 1
1 
1
E  h  hRH 

    1.64  1018 J
  (2.18  1018 J) 
2
2
2
 n
2
nf 
12 
i
 
hc
(6.63  1034 J s)(3.00  108 m s1 )

 1.21  107 m  121 nm
18
E
1.64  10 J
For the Balmer series nf = 2, the shortest wavelength emission corresponds to the largest change in energy (ni
 ).
 1
 1
1 
1
E  h  hRH 


  5.45  1019 J
  (2.18  1018 J) 
2
2
2
2 
 n


nf 

2
i
 
hc
(6.63  1034 J s)(3.00  108 m s1 )

 3.65  107 m  365 nm
19
E
5.45  10 J
The longest wavelength emission for the Lyman series (121 nm) is shorter than the shortest wavelength
emission for the Balmer series (365 nm). Hence, the two series do not overlap.
1.57
Since 1 W  1 J s-1, the energy output of the 75-W light bulb in 1 second is 75 J. The actual energy converted
to visible light is 15 percent of this value (0.15  75 = 11 J).
First, we need to calculate the energy of one 550 nm photon. Then, we can determine how many photons are
needed to provide 11 J of energy.
The energy of one 550 nm photon is:
E 
hc
(6.63  1034 J s)(3.00  108 m s1 )

 3.62  1019 J
9

550  10 m
The number of photons needed to produce 11 J of energy is:
11 J 
1 photon
3.62  1019 J
 3.0  1019 photons
210
1.58
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
The energy needed per photon for the process is:
248  103 J
1 mol

 4.12  1019 J
1 mol
6.022  1023 photons
 
hc
(6.63  1034 J s)(3.00  108 m s1 )

 4.83  107 m  483 nm
19
E
(4.12  10 J)
Any wavelength shorter than 483 nm will also promote this reaction. Once a person goes indoors, the reverse
reaction Ag  Cl  AgCl takes place.
1.59
Starting with Equation 1.4,
h   
1
m u2 ,
2 e
we can set the kinetic energy equal to eV,
h    eV ,
divide both sides by h to yield:

 e
 V.
h h
To use the equation above to determine h and  we plot  as a function of V. The slope will be equal to e/h
and the y-intercept will be equal to  /h. Before we can plot, we have to calculate  from  using v=c, where
c is the speed of light.

nm)
405.0
435.5
480.0
520.0
577.7
650.0
V (volt)
1.475
1.268
1.027
0.886
0.667
0.381
 (x1014 sec-1)
7.402
6.884
6.245
5.765
5.189
4.612
h
e
1.602  1019 J V 1

 6.125x10-34 Js
slope 2.6153  1014 s 1 V 1
7.5 10 14
Y = M0 + M1*X
M0 3.5314e+14
M1 2.6153e+14
7 1014
R
   y  intercept   6.125  10 34 Js 
 2.163  10-19 J
1.60
From Equation 1.17:
6.5 10 14
-1
Frequency (sec )
  3.5314  1014 s 1  6.125  10 34 Js 
0.99794
6 1014
5.5 10 14
5 1014
4.5 10 14
0.2
0.4
0.6
0.8
1
Voltage (V)
1.2
1.4
1.6
211
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
RH 
e 4 me
(1.60218  1019 C) 4 (9.10939  1031 kg)

3.28986  1015 s –1 .
8h3 02 8(6.62608  1034 J s)3 (8.85419  1012 C2 J –1 m –1 ) 2
For the hydrogen atom the mass of the nucleus is just equal to the mass of a proton.

 9.10939  1031 kg 1.67262  1027 kg   9.10443  1031 kg
me mN

me  mN  9.10939  1031 kg   1.67262  1027 kg 
RH' 
e4 
(1.60218  1019 C) 4 (9.10443  1031 kg)

3.28806  1015 s –1
3 2
8h  0 8(6.62608  1034 J s)3 (8.85419  1012 C2 J –1 m –1 ) 2
Percentage change 
RH  RH'
3.28986  1015 s –1  3.28806  1015 s –1
 100 
 100  0.0547%
RH
3.28986  1015 s –1
For the helium-4 isotope the mass of the nucleus is equal to 6.64610-27 kg.
9.10939  1031 kg  6.646  10-27 kg 

me mN


 9.10814  1031 kg
me  mN  9.10939  1031 kg    6.646  10-27 kg 
RH' 
e4 
(1.60218  1019 C) 4 (9.10814  1031 kg)

3.28940  1015 s –1
8h3 02 8(6.62608  1034 J s)3 (8.85419  1012 C2 J –1 m –1 ) 2
Percentage change 
RH  RH'
3.28986  1015 s –1  3.28940  1015 s –1
 100 
 100  0.014%
RH
3.28986  1015 s –1
That the percent change for a He+ is smaller illustrates that the bigger the difference in the two masses (the
mass of the electron and of the nucleus), the closer the reduced mass is to the lighter mass (the mass of the
electron).
1.61
(a)
For an s orbital,
zero.
is zero. Hence, the orbital angular momentum for an electron in an s orbital is
Bohr’s model only allowed for orbits with angular momentum equal to an integer multiple of

h
where is defined as Plank’s constant divided by two times pi  h 

2 

(b)
1.62
,
The Fe14+ in the corona is presumed to be formed by the ionization of Fe 13+, which takes 3.5104 kJ mol-1 of
energy. The energy for the ionization comes from collisions. Hence, for there to be Fe 14+ in the corona, we
can assume that the average kinetic energy of the corona gases has to be atleast equal to the ionization energy
of Fe13+.
3RT
 3.5  104 kJ mol-1
2
212
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
T
2  3.5  104 kJ mol-1 
3R

2  3.5  104 kJ mol-1 
3  8.31447  103 kJ K -1mol-1 
 2.8  106 K
The actual temperature can be, and most probably is, higher than this.
1.63
Since the energy corresponding to a photon of wavelength 1 equals the energy of photon of wavelength 2
plus the energy of photon of wavelength 3, then the equation must relate the wavelength to energy.
energy of photon 1  (energy of photon 2  energy of photon 3)
hc
Since E 
hc
1

hc
2


, then:
hc
3
Dividing by hc:








1.64
From the Heisenberg uncertainty principle (Equation 1.22) we have
x  p 
h
2
For the minimum uncertainty we can change the greater than sign to an equal sign. We can solve for the
uncertainty in momentum (p) by dividing both sides by the uncertainty in the position (x).
p 
h
2 x
The uncertainty of the oxygen molecules position in any one direction is equal to the diameter of the alveoli
(5.0 x 10-5 m).
p 
2 x

1.055  1034 J s
 1.1  1030 kg m s 1
2 (5.0  10 –5 m)
Momentum is equal to mass times velocity. If we assume that the there is no uncertainty in the mass, then the
uncertainty in the momentum is equal to mass times the uncertainty in the velocity (p = m u). By dividing
both sides by the mass, we see that the uncertainty in the velocity is equal to the uncertainty in momentum
divided by the mass.
u 
p 1.1  10 –30 kg m s –1

 2.0  10 –5 m s –1
m
5.3  10 –26 kg
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
Because we had started by changing the greater than sign in Heisenberg’s equation to an equal sign, the
minimum uncertainty in the velocity is 2.0  10-5 m s-1.
1.65
It takes:
(5.0  102 g ice) 
334 J
 1.67  105 J to melt 5.0  102 g of ice.
1 g ice
Energy of a photon with a wavelength of 660 nm:
E 
hc


(6.63  1034 J s)(3.00  108 m s1 )
 3.01  1019 J
660  109 m
Number of photons needed to melt 5.0  102 g of ice:
(1.67  105 J ) 
1 photon
 5.5  1023 photons
19
3.01  10 J
The number of water molecules is:
1 mol H 2O
6.022  1023 H2O molecules
(5.0  102 g H 2O) 

 1.7  1025 H 2O molecules
18.02 g H 2O
1 mol H 2O
The number of water molecules converted from ice to water by one photon is:
1.7  1025 H 2O molecules
5.5  1023 photons
1.66
 31 H 2O molecules per photon
Energy of a photon at 360 nm:
E  h 
hc


(6.63  1034 J s)(3.00  108 m s1 )
 5.53  1019 J
360  109 m
Area of exposed body in cm2:
2
 1 cm 
0.45 m 2  
 4.5  103 cm 2
 1  102 m 
The number of photons absorbed by the body in 2 hours is:
0.5 
2.0  1016 photons
7200 s
 (4.5  103 cm2 ) 
 3.2  1023 photons
2
2 hr
cm s
The factor of 0.5 is used above because only 50% of the radiation is absorbed.
3.2  1023 photons with a wavelength of 360 nm correspond to an energy of:
(3.2  1023 photons) 
5.53  1019 J
 1.8  105 J
1 photon
213
214
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
1.67
The anti-atom of hydrogen should show the same characteristics as a hydrogen atom. Should an anti-atom of
hydrogen collide with a hydrogen atom, they would be annihilated and energy would be given off.
1.68

h 
Looking at the de Broglie equation   
, the mass of an N2 molecule (in kg) and the velocity of an N2
mu 

molecule (in m/s) is needed to calculate the de Broglie wavelength of N2.
First, calculate the root-mean-square velocity of N2.
M(N2)  28.02 g mol1  0.02802 kg mol1


(3) 8.314 J mol1 K 1 300 K
urms =

1
0.02802 kg mol
1


516.8 m s1
Second, calculate the mass of one N2 molecule in kilograms.
28.02 g N 2
1 mol N 2
1 kg


 4.653  1026 kg
1 mol N 2
6.022  1023 N 2 molecules 1000 g
Now, substitute the mass of an N2 molecule and the root-mean-square velocity into the de Broglie equation to
solve for the de Broglie wavelength of an N2 molecule.
 
h
(6.63  1034 J s-1 )

 2.76  1011 m
26
1
mu
(4.653  10 kg)(516.8 m s )
1.69
Based on the selection rule, which states that l  1, only (b) and (d) are allowed transitions.
1.70
The kinetic energy acquired by the electrons is equal to the voltage times the charge on the electron. After
calculating the kinetic energy, we can calculate the velocity of the electrons (KE  1/2mu2). Finally, we can
calculate the wavelength associated with the electrons using the de Broglie equation.
KE  (5.00  103 V) 
1.602  1019 J
 8.01  1016 J
1V
We can now calculate the velocity of the electrons.
KE 
1 2
mu
2
8.01  1016 J 
1
(9.1094  1031 kg)u 2
2
u  4.19  107 m s-1
Finally, we can calculate the wavelength associated with the electrons using the de Broglie equation.
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
 

1.71
(a)
215
h
mu
(6.63  1034 J s)
 1.74  1011 m  17.4 pm
(9.1094  1031 kg)(4.19  107 m s1 )
We use the Heisenberg Uncertainty Principle to determine the uncertainty in knowing the position of the
electron.
xp 
h
4
We use the equal sign in the uncertainty equation to calculate the minimum uncertainty values.
p  mu, which gives:
x(mu) 
x 
h
4
h
6.63  1034 J s

 1  109 m
4 mu
4 (9.1094  1031 kg)[(0.01)(5  106 m s1 )]
Note that because of the unit Joule in Planck’s constant, mass must be in kilograms and velocity must be
in m s-1.
The uncertainly in the position of the electron is much larger than that radius of the atom. Thus, we have
no idea where the electron is in the atom.
(b)
We again start with the Heisenberg Uncertainly Principle to calculate the uncertainty in the baseball’s
position.
x 
h
4p
x 
6.63  1034 J s
 7.9  109 m
4 (1.0  107 )(6.7 kg m s1 )
This uncertainty in position of the baseball is such a small number as to be of no consequence.
1.72
(a)
The average kinetic energy of 1 mole of an ideal gas is 3/2RT. Converting to the average kinetic energy
per atom, we have:
3
1 mol
(8.314 J mol1 K 1 )(298 K) 
 6.171  1021 J atom 1
2
6.022  1023 atoms
(b)
To calculate the energy difference between the n  1 and n  2 levels in the hydrogen atom, we modify
Equation 1.5 of the text.
 1
1
1
1
E  RH  2  2   (2.180  1018 J)  2  2 
1
nf 
 
 ni
216
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
E  1.635  1018 J
(c)
For a collision to excite an electron in a hydrogen atom from the n  1 to n  2 level,
KE  E
3
RT
2
 1.635  1018 J
NA
T 
2 (1.635  1018 J)(6.022  1023 mol1 )
 7.90  104 K
3
(8.314 J mol1 K 1 )
This is an extremely high temperature. Other means of exciting H atoms must be used to be practical.
1.73
To calculate the energy to remove at electron from the n  1 state and the n  5 state in the Li2 ion, we use the
following equation:
 1
1
E  RH Z 2  2  2  .
nf 
 ni
For ni  1, nf  , and Z  3, we have:
1
1
E  (2.18  1018 J)(3)2  2  2   1.96  1017 J
1
 
For ni  5, nf  , and Z  3, we have:
1
1 
E  (2.18  1018 J)(3)2  2  2   7.85  1019 J
5
 
To calculate the wavelength of the emitted photon in the electronic transition from n  5 to n  1, we first
calculate E and then calculate the wavelength.
 1
1
1
1
E  RH Z 2  2  2   (2.18  1018 J)(3)2  2  2    1.88  1017 J
5
nf 
1 
 ni
We ignore the minus sign for E in calculating .
 
hc
(6.63  1034 J s1 )(3.00  108 m s1 )

E
1.88  1017 J
  1.06  108 m  10.6 nm
1.74
The minimum value of the uncertainty is found using xp 
x(mu) 
h
4
h
and p  mu. Solving for u:
4
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
217
h
4 mx
u 
For the electron with a mass of 9.109 × 10 31 kg and taking x as the radius of the nucleus, we find:
6.63  1034 J s
 5.8  1010 m s1
31
15
4 (9.109  10 kg)(1.0  10 m)
u 
This value for the uncertainty is impossible, as it far exceeds the speed of light. Consequently, it is impossible
to confine an electron within a nucleus.
Repeating the calculation for the proton with a mass of 1.673 × 10 27 kg gives:
6.63  1034 J s
 3.2  107 m s1
27
15
4 (1.673  10 kg)(1.0  10 m)
u 
While still a large value, the uncertainty is less than the speed of light, and the confinement of a proton to the
nucleus does not represent a physical impossibility. The large value does indicate the necessity of using
quantum mechanics to describe nucleons in the nucleus, just as quantum mechanics must be used for electrons
in atoms and molecules.
1.75
Below are the solutions for the particle in a one dimensional box (Equation 1.35).
 2
 n (x)   
 L
1/2
 n x 
sin 
 L 
n  1, 2,3,....
The equation below yields the average value of x, for a particle in a one dimensional box.
L
L
2
 2
 n x 
 2 L
 n x 
f   x  (x) dx   x   sin2 
dx

xsin2 
dx




 L
 L 
 L 0
 L 
0
0
We can use the following relation obtained from a table of integrals:
 xsin ax dx 
2
 
x 2 x sin(2ax) cos 2ax


4
4a
8a 2

 2n x 
 2n x  
 2 x sin 

cos 

2
 2
 2 x
 L 
 L  
2  n x 
f   x  (x) dx     xsin 
dx     


2
 n 
 L 0
 L 
 L 4
 n  
0

4
8

 L  

 L 


L
L
We now evaluate this for x equal to L and 0.
218
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD




 2



 2 L
1
1
  2 
 L
f   
   

2 
2 
 L 4
 n    L    n   2

8
8

  L  
 L  



The result above shows us that for all states for a particle in a box, the average value of x is L/2. This makes
sense, because the squared wavefunctions are all symmetric about x = L/2.
The equation below yields the average value of x2, for a particle in a one dimensional box.
L
L
2
 2
 n x 
 2  L 2 2  n x 
f   x 2  (x) dx   x 2   sin2 
dx

 L   x sin  L  dx
 L
 L 
0
0
0
We can use the following relation obtained from a table of integrals:
 
2
2
 x sin ax dx 

  
2 2
x 3 x cos(2ax) 2a x  1 sin 2ax


6
4a2
8a3
2




 2n x   2  n  x 2  1 sin  2n x  

x
cos




L
 L    L 

 L 
2
 2 L
 n x 
 2   x3


f   x 2  (x) dx     x 2 sin 2 
dx



 L  6
2
3
 L 0
 L 


 n 
 n 
0
4
8




 L 
 L 


We now evaluate this for x equal to L and 0.


 3


 2 L
L
1

2 1
 
f   

L
2 
 L 6
 3 2 n
 n  


4



L






2


 
for n = 1 the average value of x2 is 0.283 L2
for n = 2 the average value of x2 is 0.308 L2
for n = 3 the average value of x2 is 0.328 L2

1.76


The average value of r using Equation 1.41: r   rP r dr   r 3 R(r) dr
0
2
0
3
2Zr
  Z  3/2  Zr 
 Z   3  ao
ao
3
3

r   r R10 (r) dr   r 2   e
dr  4    r e dr
  a0 

 a0  0
0
0



2
From an integral table:

2
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD

x e
n  ax
dx 
0
n!
a n1
To use the above integral to solve our problem, n = 3, x = r, and a 
Z
r  4 
 a0 
219
3 
r e
3

2Zr
ao
0
Z
dr  4  
 a0 
3
2Z
,
ao
 6ao4  3ao 3

 ao .

4 
 16Z  2Z 2
Hence, the average value for r for a 1s orbital is
3
a . Remember that Z is the nuclear charge and is equal to 1
2 o
for a hydrogen atom.
To calculate the most probable value of r, we set the derivative of the radial probability function to zero and
solve for r.
2
2
2Zr
  Z  3/2  Zr 
 Z  2  ao
ao




P r  r R r   r 2  e
 4  r e
  a0 

 a0 




2
2
2
2

Z
Z
d
a
P r  4   r 2e o  4  
dr
a
 0
 a0 

2Zr
2
2
2Zr
2Zr
  2Zr
Z 
2Z 2  ao 
Z 2   ao
 2re ao 

r e
 8   r  r  e
ao
ao 
 a0  


ao
 ao the derivative is zero, which corresponds to the most likely value of r. This is consistent
Z
with ao being the Bohr radius.
When r 
For the 2s orbital



The average value of r: r   rP r dr   r 3 R(r) dr
0
2
0
2
Zr
 2 Z 
Zr  Š 2ao 
r   r 3 R10 (r) dr   r 3 
2

e
dr
ao 
 4  a0  

0
0


3
Zr
Zr
Zr
Š
Š
1 Z  
4Z
Z 2 5 Š ao 
ao
ao
3
4

r    4  r e dr 
 r e dr  a2  r e dr 
8  a0  
a
o
o




2
3/2
Using the same integral solution as above, we can solve for the three integrals in the square brackets

4 r e
3
Š
Zr
ao
dr  4
0
4Z
ao
4
r e
Š
Zr
ao
6ao4
Z4
dr 

24ao4
Z4
5
4
4Z  24ao  96ao
 5  4
ao  Z 
Z
220
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
6
4
Z 2 5 Š ao
Z 2  120ao  120ao
r
e
dr




ao2 
ao2  Z 6 
Z4
Zr
When we plug in the results for the integrals into the above equation we get that the average distance from the
nucleus for a 2s orbital is six times the Bohr radius. Notice that this is four times the average distance from
the nucleus for a 1s orbital.
1 Z 
r   
8  a0 
3
 24ao4 96ao4 120ao4  1  Z 
 4  4 
  
Z
Z 4  8  a0 
 Z
3
 48ao4 
ao
 4   6  6ao
Z
 Z 
To calculate the most probable value of r, we can set the derivative of the radial probability function to zero
and solve for r.
2
3
Zr
Zr
 2  Z  3/2 
Zr  Š 2ao 
1 Z   2 
4Zr Z 2 r 2  Š ao 




P r  r R r   r
2  e
   r 4
 2 e 
ao 
8  a0   
ao
 4  a0  

ao 




3
Zr
Zr
Zr
Š
1 Z  
4Zr 3 Š ao Z 2 r 4 Š ao 
a
P r     4r 2 e o 
e  2 e 
8  a0  
ao
ao



2

2
2

2
Zr
Zr
Zr
d
1  Z  d  2 Š ao 4Zr 3 Š ao Z 2 r 4 Š ao 
 4r e 
P r   
e  2 e 
dr
8  a0  dr 
ao
ao



Zr
d
1 Z   
Zr 2  4Z  2 Zr 3  Z 2  3 Zr 4   Š ao
P r     4  2r 

3r


4r

e

dr
8  a0   
ao  ao 
ao  ao2 
ao  

2
2
d
1 Z  
16Zr 2 8Z 2 r 3 Z 3r 4  Š ao
P r    8r 

 3 e
dr
8  a0  
ao
ao2
ao 

Zr
Zr
3
2
2
 Ša
d
1  Z   rZ 3   8a 16ao r ao r
3
P r     3   3o 


r
e
 o
dr
8  a0   ao   Z
Z
Z2


2
The above polynomial can be solved computationally to determine three roots. The largest root, which
corresponds to the most probable value is r 
3  5a  3  5a .
Z
o
o
1.77
1/2
1/2
L

 n x  
 m x  
 n x 
 m x 
2
 2 

 2 

sin
sin
dx

sin 
sin 
dx
0  L 

 L    L 
 L  

L0
 L 
 L 







L
Using a table of integrals, it can be shown that the last integral listed above is zero if n does not equal m, and
is L/2 if n equals m. Hence, the wavefunctions for a one-dimensional particle in a box are orthogonal.
1.78
The five lowest energy levels for a two-dimensional particle in a box correspond to the following sets of
quantum numbers: (nx=1, ny=1), (nx=1, ny=2), (nx=2, ny=1), (nx=2, ny=2), (nx=3, ny=1) , (nx=1, ny=3) , (nx=3,
ny=2) , (nx=2, ny=3). These wavefunctions are plotted below with the lowest energy wavefunction at the
bottom. Notice that the states (nx=1, ny=2) and (nx=2, ny=1) have the same energy as do the states (nx=1,
ny=3) and (nx=3, ny=1) and the states (nx=3, ny=2) and (nx=2, ny=3).
CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD
221