CHAPTER 7 QUANTUM THEORY AND ATOMIC STRUCTURE
The value for the speed of light will be 3.00 x 108 m/s except when more significant
figures are necessary, in which cases, 2.9979 x 108 m/s will be used.
7.1 All types of electromagnetic radiation travel as waves at the same speed. They differ
in both their frequency and
wavelength.
7.2 a) Figure 7.3 describes the electromagnetic spectrum by wavelength and
frequency. Wavelength increases from
left (10–2 nm) to right (1012 nm). The trend
in increasing wavelength is: x-ray < ultraviolet < visible < infrared <
microwave < radio waves.
b) Frequency is inversely proportional to wavelength according to equation 7.1, so
frequency has the opposite
trend: radio < microwave < infrared < visible <
ultraviolet < x-ray.
c) Energy is directly proportional to frequency according to equation 7.2.
Therefore, the trend in increasing energy
matches the trend in increasing frequency:
radio < microwave < infrared < visible < ultraviolet < x-ray. High- energy
electromagnetic radiation disrupts cell function. It makes sense that you want to limit
exposure to ultraviolet and x-ray radiation.
7.3 a) Refraction is the bending of light waves at the boundary of two media, as when
light travels from air into water.
b) Diffraction is the bending of light waves around an object, as when a wave
passes through a slit about as wide as its wavelength.
c) Dispersion is the separation of light into its component colors (wavelengths), as
when light passes through a prism.
d) Interference is the bending of light through a series of parallel slits to produce a
diffraction pattern of
brighter and darker spots.
Note: Refraction leads to a dispersion effect and diffraction leads to an
interference effect.
7.4 Evidence for the wave model is seen in the phenomena of diffraction and refraction.
Evidence for the particle model includes the photoelectric effect and blackbody
radiation.
7.5 In order to explain the formula he developed for the energy vs. wavelength data of
blackbody radiation, Max
Planck assumed that only certain quantities of energy,
7-1
called quanta, could be emitted or absorbed. The magnitude
were whole number multiples of the frequency: E = nh.
of these gains and losses
7.6 Radiation (light energy) occurs as quanta of electromagnetic radiation, where each
packet of energy is called a
photon. The energy associated with this photon is fixed
by its frequency, E = h. Since energy depends on frequency,
a threshold (minimum)
frequency is to be expected. A current will flow as soon as a photon of sufficient energy
reaches the metal plate, so there is no time lag.
7.7 Plan: Wavelength is related to frequency through the equation c = . Recall that
a Hz is a reciprocal second, or 1/s = s–1. Assume that the number “960” has three
significant figures. Solution:
c = 
(m) = c/ =
3.00 x 108 m/s
 103 Hz  s 1 
 960.kHz  



 1 kHz  Hz 
 (nm) = c/ =
 (Å) = c/ =
= 312.5 = 312 m
 1 nm 
3.00 x 108 m/s


3

1
 10 Hz  s   109 m 
 960. kHz  



 1k Hz  Hz 
 0.01 Å 
3.00 x 108 m/s


3

1
 10 Hz  s   1012 m 
 960. kHz  



 1 kHz  Hz 
= 3.125 x 1011 = 3.12 x 1011 nm
= 3.125 x 1012 = 3.12 x 1012 Å
7.8 Wavelength and frequency relate through the equation c = . Recall that a Hz is
a reciprocal second, or 1/s = s–1.
(m) = c/ =
3.00 x 108 m/s
 106 Hz  s 1 
 93.5 MHz  



 1 MHz  Hz 
 (nm) = c/ =
 (Å) = c/ =
= 3208556 = 3.21 m
3.00 x 108 m/s
 1 nm 


6

1
 10 Hz  s   109 m 
93.5
MHz

 



 1 MHz  Hz 
 0.01 Å 
3.00 x108 m / s


 106 Hz  s 1   1012 m 
 93.5 MHz  



 1 MHz  Hz 
= 3.208556 x 109 = 3.21 x 109 nm
= 3.208556 x 1010 = 3.21 x 1010 Å
7.9 Plan: Frequency is related to energy through the equation E = h. Note that 1 Hz
= 1 s–1.
Solution:
7-2
E = (6.626 x 10–34 J•s) (3.6 x 1010 s–1) = 2.385 x 10–23 = 2.4 x 10–23 J
7.10 E = hc/ =
 6.626 x 10
34


J • s 3.00 x 108 m/ s  0.01 Å 
 12 
1.3 Å
 10 m 
= 1.529 x 10–15 = 1.5 x 10–15 J
7.11 Since energy is directly proportional to frequency (E = 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.
7.12 Since energy is directly proportional to frequency (E = h): UV ( = 8.0 x 1015 s–
1
) > IR ( = 6.5 x 1013 s–1) >
microwave ( = 9.8 x 1011 s–1) or UV > IR >
microwave.
7.13 Plan: Frequency and wavelength can be calculated using the speed of light: c =
.
Solution:
 (nm) = c/ =
 1 nm 
2.99792 x 108 m/s


9

1
 10 Hz  s   109 m 
 22.235 GHz  



 1 GHz  Hz 
= 1.3482887 x 107 = 1.3483 x 107
nm
 (Å) = c/ =
 0.01 Å 
2.99792 x 108 m/s


 109 Hz  s 1   1012 m 
22.235
GHz

 



 1 GHz  Hz 
= 1.3482887 x 108 = 1.3483 x 108 Å
7.14 Frequency and wavelength can be calculated using the speed of light: c = .
a) = c/=
3.00 x 108 m / s  1 m 
 6 
9.6 m
 10 m 
b) (m) = c/ =

= 3.125 x 1013 = 3.1 x 1013 s–1
2.99792 x 108 m/s  1 m 


 s 1   106 m 
13
8.652 x 10 Hz 

 Hz 

= 3.465002 = 3.465 m
7.15 Frequency and energy are related by E = h, and wavelength and energy are
related by E = hc/.
(Hz) = E/h =
 106 eV  1.602 x 1019 J 



1 eV
 1 MeV 
  Hz 
 1 
34
6.626 x 10 J • s
s 
1.33 MeV  
7-3
= 3.2156 x 1020 = 3.22 x 1020
Hz
 (m) = hc/E =
 6.626 x 10
34

J • s 3.00 x 10 8 m/s

 106 eV  1.602 x 10 19 J 
1.33 MeV  



1 eV
 1 MeV 

= 9.32950 x 10–13 = 9.33 x 10–13 m
7.16 Plan: a) The least energetic photon has the longest wavelength (242 nm). b) The
most energetic photon has the shortest wavelength (2200 Å).
Solution:
a) = c/=
3.00 x 108 m/s  1 nm 
 9 
242 nm
 10 m 
E = hc/ =
 6.626 x 10
34
= 1.239669 x 1015 = 1.24 x 1015 s–1


J • s 3.00 x 10 8 m/s  1 nm 
 9 
242 nm
 10 m 
= 8.2140 x 10–19 = 8.21 x
10–19 J
b) = c/=
3.00 x 108 m/s  0.01 Å 
 12 
2200 Å
 10 m 
E = hc/ =
 6.626 x 10
34
= 1.3636 x 1015 = 1.4 x 1015 s–1


J • s 3.00 x 108 m/s  0.01 Å 
 12 
2200 Å
 10 m 
= 9.03545 x 10–19 = 9.0 x
10–19 J
7.17 “n” in the Rydberg equation is equal to a Bohr orbit of quantum number “n”
where n = 1, 2, 3, ...
7.18 Bohr’s key assumption was that the electron in an atom does not radiate energy
while in a stationary state, and the
electron can move to a different orbit by
absorbing or emitting a photon whose energy is equal to the difference in
energy
between two states. These differences in energy correspond to the wavelengths in the
known spectra for the hydrogen atoms. A solar system model does not allow for the
movement of electrons between levels.
7.19 An absorption spectrum is produced when atoms absorb certain wavelengths of
incoming light as electrons move
from lower to higher energy levels and results in
dark lines against a bright background. An emission spectrum is produced when atoms
that have been excited to higher energy emit photons as their electrons return to lower
energy levels and results in colored lines against a dark background. Bohr worked
with emission spectra.
7-4
7.20 The quantum number n is related to the energy level of the electron. An electron
absorbs energy to change from lower energy (low n) to higher energy (high n) giving
an absorption spectrum. An electron emits energy as it
drops from a higher energy
level to a lower one giving an emission spectrum.
a) absorption b) emission c) emission d) absorption
7.21 The Bohr model works for only a one-electron system. The additional attractions
and repulsions in many-electron
systems make it impossible to predict accurately
the spectral lines.
7.22 The Bohr model has successfully predicted the line spectra for the H atom and
Be3+ ion since both are one electron species. The energies could be predicted from En
=
  2.18 x 10
 Z2
n
18
J

where Z is the atomic number for the atom or ion. The line spectra
2
for H would not match the line spectra for Be3+ since the H nucleus contains one
proton while the Be3+ nucleus contains 4 protons (the Z values in the equation do
not match), thus the force of
attraction of the nucleus for the electron would be
greater in the beryllium ion than in the hydrogen atom. This
means that the pattern of
lines would be similar, but at different wavelengths.
7.23 Plan: Calculate wavelength by substituting the given values into equation 7.3,
where n1 = 2 and n2 = 5 because
n2 > n1. Although more significant figures could
be used, five significant figures are adequate for this calculation.
Solution:
 1
1
1 
 R 2  2 



 n1 n 2 
R = 1.096776 x 107 m–1
n1 = 2 n2 = 5
 1
1
1 
 R 2  2 



 n1 n 2 
 (nm) =
=
1.096776 x 10
7

1 
 1
m 1  2  2 
2 5 

 1 nm 
1

1 
 9 
 2303229.6 m  10 m 
= 2303229.6 m–1 (unrounded)
= 434.1729544 = 434.17 nm
7.24 Calculate wavelength by substituting the given values into equation 7.3, where n1
= 1 and n2 = 3 because n2 > n1. Although more significant figures could be used, five
significant figures are adequate for this calculation.
 1
1
1
 R 2  2


 n1 n 2



=
1.096776 x 10 m   11  31 
7
1
2
2
7-5
= 9749120 m–1 (unrounded)
 (Å) =

 0.01Å 
1

1 
 12 
 9749120 m  10 m 
= 1025.7336 = 1025.7 Å
7.25 Plan: The Rydberg equation is needed. For the infrared series of the H atom, n1
equals 3. The least energetic
spectral line in this series would represent an electron
moving from the next highest energy level, n2 = 4. Although
more significant figures
could be used, five significant figures are adequate for this calculation.
Solution:
 1
1
1 
 R 2  2 



 n1 n 2 
 (nm) =
=
1.096776 x 10
7

 1 1 
m 1  2  2 
3 4 

  1 nm 
1

1 
  9 
 533155 m   10 m 
= 533155 m–1 (unrounded)
= 1875.627 = 1875.6 nm
Check: Checking this wavelength with Figure 7.9, we find the line in the infrared
series with the greatest wavelength occurs at approximately 1850 nm.
7.26 The Rydberg equation is needed. For the visible series of the H atom, n1 equals 2.
The least energetic spectral line in this series would represent an electron moving from
the next highest energy level, n = 3. Although more
significant figures could be
used, five significant figures are adequate for this calculation.
Solution:
 1
1
1 
 R 2  2 



 n1 n 2 
 (nm) =
=
1.096776 x 10 m   21  31 
7
1
2

 1 nm 
1

1 
 9 
 1523300 m  10 m 
2
= 1523300 m–1 (unrounded)
= 656.4695 = 656.47 nm
7.27 Plan: To find the transition energy, apply equation 7.4 and multiply by
Avogadro’s number.
Solution:


 (Avogadro’s number)


23
 1 1   6.022 x 10 
5
2.18 x 1018 J  2  2  
 = -2.75687 x 10

mol
 2 5 

E =
 2.18 x 10 J   n 1
E =

18
2
final

1
2
n initial

= -2.76 x 105 J/mol
Check: The value is negative and so light is emitted.
7.28 To find the transition energy, apply equation 7.4 and multiply by Avogadro’s
number.
E =

 2.18 x 10 J   n 1
18

2
final

1
2
n initial

 (Avogadro’s

7-6
number)
E =

 2.18 x 10 J   31  11   6.022molx 10
18
2
2

23



= 1.1669 x 106 = 1.17 J/mol
7.29 Looking at an energy chart will help answer this question.
n=5 n=4
n=3
(d)
(a)
(c)
n=2
(b)
n=1
Frequency is proportional to energy so the smallest frequency will be d) n = 4 to n
= 3 and the largest frequency
b) n = 2 to n = 1. Transition a) n = 2 to n = 4 will be smaller than transition c) n =
2 to n = 5 since level 5 is a
higher energy than level 4. In order of increasing
frequency the transitions are d < a < c < b.
7.30 b > c > a > d
7.31 Plan: Use the Rydberg equation. A combination of E = hc/, and equation 7.4,
would also work.
Solution:
 = (97.20 nm) x (10–9 m/1 nm) = 9.720 x 10–8 m
1/ =
1.096776 x 10
7
 1
1
m1  2  2
n
 1 n2
1/(9.720 x 10–8 m) =
0.9380 =
1
n 22




1.096776 x 10
7
1 1 
m1  2  2 
1 n 
2 


1 1 
 2  2 
 1 n2 
= 1 - 0.9380 = 0.0620
= 16.1
n2 = 4
n 22
7.32  = (1281 nm) x (10–9 m/1 nm) = 1.281 x 10–6 m
7-7
1/ =


1.096776 x 10 m   n1  n1 
7
1

1/(1.281 x 10–6 m) =
0.07118 =
2
1
2
2

1.096776 x 10
7
 1 1
m1  2  2
n
 1 5




 1 1 
 2  2 
 n1 5 
1
= 0.07118 + 0.04000 = 0.11118
n12
n12 = 8.9944
n1 = 3
7.33 E = hc/ =
 6.626 x 10
34


J • s 3.00 x 10 8 m/s  1 nm 
 9 
 436 nm 
 10 m 
= 4.55917 x 10–19 = 4.56 x 10–19 J
7.34 Plan: The energy can be calculated from E = hc/.
Solution:
E = hc/ =
 6.626 x 10
34


J • s 3.00 x 10 8 m/s  1 nm 
 9 
 589 nm 
 10 m 
= 3.37487 x 10–19 = 3.37 x 10–19
J/photon
E=
 3.37487 x 10 19 J  6.022 x 10 23 photon



photon
1 einstein


  1 kJ 

  103 J 


= 203.23 = 203 kJ/einstein
7.35 If an electron occupies a circular orbit, only integral numbers of wavelengths (=
2nr) are allowed for acceptable
standing waves. A wave with a fractional number
of wavelengths is forbidden due to destructive interference with itself. In a musical
analogy to electron waves, the only acceptable guitar string wavelengths are those that are
an
integral multiple of twice the guitar string length (2 L).
7.36 de Broglie’s concept is supported by the diffraction properties of electrons
demonstrated in an electron
microscope.
7.37 Macroscopic objects have significant mass. A large m in the denominator of  =
h/mu will result in a very small wavelength. Macroscopic objects do exhibit a wavelike
motion, but the wavelength is too small for humans to see
it.
7-8
7.38 The Heisenberg uncertainty principle states that there is fundamental limit to the
accuracy of measurements. This
limit is not dependent on the precision of the
measuring instruments, but is inherent in nature.
7.39 Plan: Part a) uses the de Broglie equation, and part b) uses Heisenberg’s
relationship.
Solution:
a)  = h/mv =
 6.626 x 10

34
J • s  kg • m 2 /s 2

mi 
J
 220 lb  19.6  
h 

  2.205 lb  0.62 mi   1 km   3600 s 
 

  3  

  1 kg  1 km   10 m   1 h 
= 7.562675 x 10–37 = 7.6 x 10–37 m
b) x • mv  h
4
x  h/4 mv 
 6.626 x 10

34
J • s  kg • m 2 /s 2

J
 0.1 mi  
4  220 lb  

 h 
  2.205 lb  0.62 mi   1 km   3600 s 
 

 3 

  1 kg  1 km   10 m   1 h 
 1.17956 x 10–35  1 x 10–35 m
7.40 a)  = h/mv =
 6.626 x 10 J • s   kg • m /s

 6.6 x 10 g   3.4 x 10 mih   J
34
2
24
7
2
 103 g   0.62 mi   1 km   3600 s 

 1 kg   1 km   103 m   1 h 





= 6.59057 x 10–15 = 6.6 x 10–15 m
b)
x
mv 
h
4
x  h/4 mv 
 6.626 x 10 J • s 
 0.1 x 10
4  6.6 x 10 g  
h
34
24

 1.783166 x 10
–14
7.41  = h/mv
v = h/m =
 6.626 x 10
34
 kg • m 2 /s 2

7
J
mi  


 103 g   0.62 mi   1 km   3600 s 

 1 kg   1 km   103 m   1 h 





 2 x 10–14 m

J • s  kg • m2 /s 2

J
 56.5 g   5400 Å  
 103 g  0.01 Å 

 1 kg 
  1012 m 



= 2.1717 x 10–26 = 2.2 x 10–26
m/s
7.42  = h/mv
v = h/m =
 6.626 x 10
34

J • s  kg • m2 /s 2

J
142 g 100. pm  
 103 g   1 pm 

 1 kg   1012 m 


10–23 m/s
7-9
= 4.666197 x 10–23 = 4.67 x
7.43 Plan: The de Broglie wavelength equation will give the mass equivalent of a
photon with known wavelength and velocity. The term “mass-equivalent” is used
instead of “mass of photon” because photons are quanta of
electromagnetic energy
that have no mass. A light photon’s velocity is the speed of light, 3.00 x 108 m/s.
Solution:
 = h/mv
 6.626 x 10 J • s   kg • m /s
J
 589 nm   3.00 x 10 m/s  
34
m = h/v =
2
2
 1 nm 

  109 m 


2
  1 nm   6.022 x 1023 photons 
  9  

mol
  10 m  

8
= 3.7498 x 10–36 = 3.75 x 10–36
kg/photon
 6.626 x 10 J • s   kg • m /s
J
 671 nm  3.00 x 10 m/s  
34
7.44 m = h/v =
2
8
= 1.9822 x 10–12 = 1.98 x 10–12 kg/mol
7.45 The quantity 2 expresses the probability of finding an electron within a specified
tiny region of space.
7.46 Since 2 is the probability of finding an electron within a small region or volume,
electron density would represent
a probability per unit volume and would more
accurately be called electron probability density.
7.47 A peak in the radial probability distribution at a certain distance means that the
total probability of finding the electron is greatest within a thin spherical volume
having a radius very close to that distance. Since principal
quantum number (n)
correlates with distance from the nucleus, the peak for n = 2 would occur at a greater
distance
from the nucleus than 0.529 Å. Thus, the probability of finding an electron
at 0.529 Å is much greater for the 1s orbital than for the 2s.
7.48 a) Principal quantum number, n, relates to the size of the orbital. More
specifically, it relates to the distance from the nucleus at which the probability of
finding an electron is greatest. This distance is determined by the energy of the
electron.
b) Angular momentum quantum number, l, relates to the shape of the orbital. It is
also called the azimuthal quantum number.
c) Magnetic quantum number, ml, relates to the orientation of the orbital in space
in three-dimensional space.
7-10
7.49 a) one b) five c) three d) nine
a) There is only a single s orbital in any shell.
b) All d-orbitals consists of sets of five (ml = -2, -1, 0, +1, +2).
c) All p-orbitals consists of sets of three (ml = -1, 0, +1).
d) If n = 3, then there is a 3s (1 orbital), a 3p (3 orbitals), and a 3d (5 orbitals)
giving 1 + 3 + 5 = 9.
7.50 a) seven b) three c) five d) four
a) All f-orbitals consists of sets of seven (ml = -3, -2, -1, 0, +1, +2, +3).
b) All p-orbitals consists of sets of three (ml = -1, 0, +1).
c) All d-orbitals consists of sets of five (ml = -2, -1, 0, +1, +2).
d) If n = 2, then there is a 2s (1 orbital) and a 2p (3 orbitals) giving 1 + 3 = 4.
7.51 Magnetic quantum numbers can have integer values from -l to + l.
a) ml: -2, -1, 0, +1, +2
b) ml: 0
(if n = 1, then l = 0)
c) ml: -3, -2, -1, 0, +1, +2, +3
7.52 Magnetic quantum numbers can have integer values from -l to +l.
a) ml: -3, -2, -1, 0, +1, +2, +3
b) ml: l = 0, ml = 0; l = l, ml = -1,0,+1 (if n = 2, then l = 0 or 1)
c) ml: -1, 0, +1
7-11
7.53 (a)
(b)
z
z
x
x
y
y
The variations in coloring of the p orbital are a consequence of the quantum
mechanical derivation of atomic
orbitals that are beyond the scope of this course.
7.54
(a)
(b)
z
y
y
x
x
The variations in coloring of the p and d orbitals are a consequence of the
quantum mechanical derivation of
atomic orbitals that are beyond the scope of this
course.
7.55
sublevel
allowable ml
a) d (l = 2) -2, -1, 0, +1, +2
b) p (l = 1) -1, 0, +1
c) f (l = 3) -3, -2, -1, 0, +1, +2, +3
no of possible orbitals
5
3
7
7.56
sublevel
allowable ml
a) s (l = 0) 0
b) d (l = 2) -2, -1, 0, +1, +2
c) p (l = 1) -1, 0, +1
no of possible orbitals
1
5
3
7.57 a) For the 5s subshell, n = 5 and l = 0. Since ml = 0, there is one orbital.
b) For the 3p subshell, n = 3 and l = 1. Since ml = -1, 0, +1, there are three
7-12
orbitals.
c) For the 4f subshell, n = 4 and l = 3. Since ml = -3, -2, -1, 0, +1, +2, +3, there are
seven orbitals.
7.58 a) n = 6; l = 4; 9 orbitals
b) n = 4; l = 0; 1 orbital
c) n = 3; l = 2; 5 orbitals
7.59 a) With l = 0, the only allowable ml value is 0. To correct, either change l or ml
value.
Correct n = 2, l = 1, ml = -1; n = 2, l = 0, ml = 0.
b) Combination is allowed.
7-13
c) Combination is allowed.
d) With l = 2, +3 is not an allowable ml value. To correct, either change l or ml
value.
Correct: n = 5, l = 3, ml = +3; n = 5, l = 2, ml = 0.
7.60 a) Combination is allowed.
b) No;
n = 2, l = 1; ml = +1
n = 2, l = 1; ml = 0
c) No; n = 7, l = 1; ml = +1
n = 7, l = 3; ml = 0
d) No;
n = 3, l = 1; ml = -1
n = 3, l = 2; ml = -2
7.61 Determine the max for -carotene by measuring its absorbance in the 610-640 nm
region of the visible spectrum. Prepare a series of solutions of -carotene of accurately
known concentration (using benzene or chloroform as
solvent), and measure the
absorbance for each solution. Prepare a graph of absorbance versus concentration for
these solutions and determine its slope (assuming that this material obeys Beer’s Law).
Measure the absorbance of the oil expressed from orange peel (diluting with solvent if
necessary). The -carotene concentration can then either be read directly from the
calibration curve or calculated from the slope (A = kC, where k = slope of the line and
C = concentration).
7.62 a) The mass of the electron is 9.1094 x 10–31 kg.
E=

h2
8
2
=
me a 02 n 2


=

h2
8
2
me a 02
 1 
 2
n 
 6.626 x 10
34
J•s


2
82 9.1094 x 1031 kg 52.92 x 1012 m
= -(2.17963 x 10–18 J) 
1 

2
n
 

2
 kg • m2 /s2

J

 1 
  2 
 n 
= -(2.180 x 10–18 J) 
1 

2
n
 
This is identical with the result from Bohr’s theory. For the H atom, Z = 1 and
Bohr’s constant = -2.18 x 10–18 J.
For the hydrogen atom, derivation using classical
principles or quantum-mechanical principles yields the same
constant.
b) The n = 3 energy level is higher in energy than the n = 2 level. Because the zero
point of the atom’s energy is
defined as an electron’s infinite distance from the
nucleus, a larger negative number describes a lower energy
level. Although this
may be confusing, it makes sense that an energy change would be a positive number.
7-14
E = -(2.180 x 10–18 J) 
1
2
2

1 

32 
= -3.027778 x 10–19 = 3.028 x 10–19 J
c) Find the frequency that corresponds to this energy using E/h, and then
calculate wavelength using = c/
 (m) = hc/E =
 6.626 x 10 J • s  2.9979 x 10 m/s 
3.027778 x 10 J 
34
8
19
= 6.56061 x 10–7 = 6.561 x 10–13 m
= 6.562 x 10–7 m = 656.2 nm
This is the wavelength for the observed red line in the hydrogen spectrum.
7.63 a) The lines do not begin at the origin because an electron must absorb a minimum
amount of energy before it has enough energy to overcome the attraction of the nucleus
and leave the atom. This minimum energy is the energy of photons of light at the
threshold frequency.
b) The lines for K and Ag do not begin at the same point. The amount of energy
that an electron must absorb to leave the K atom is less than the amount of energy that
an electron must absorb to leave the Ag atom, where the attraction between the nucleus
and outer electron is stronger than in a K atom.
c) Wavelength is inversely proportional to energy. Thus, the metal that requires a
larger amount of energy to be absorbed before electrons are emitted will require a
shorter wavelength of light. Electrons in Ag atoms require
more energy to leave,
so Ag requires a shorter wavelength of light than K to eject an electron.
d) The slopes of the line show an increase in kinetic energy as the frequency (or
energy) of light is increased.
Since the slopes are the same, this means that for an
increase of one unit of frequency (or energy) of light, the increase in kinetic energy of
an electron ejected from K is the same as the increase in the kinetic energy of an
electron ejected from Ag. After an electron is ejected, the energy that it absorbs
above the threshold energy
becomes kinetic energy of the electron. For the same
increase in energy above the threshold energy, for either K
or Ag, the kinetic
energy of the ejected electron will be the same.
7.64 a) E (1 photon) = hc/ =
 6.626 x 10

34
J • s)(3.00 x 10 8 m/s  1 nm 
 9 
700.nm
 10 m 
= 2.8397 x 10–19 J
(unrounded)
This is the value for each photon, that is, J/photon.
Number of photons = (2.0 x 10–17 J)/(2.8397 x 10–19 J/photon) = 70.4296 = 70.
photons
b) E (1 photon) = hc/ =
 6.626 x 10

34
J • s)(3.00 x 10 8 m/s  1nm 
 9 
475.nm
 10 m 
7-15
= 4.18484 x 10–19 J
(unrounded)
This is the value for each photon, that is, J/photon.
Number of photons = (2.0 x 10–17 J)/(4.18484 x 10–19 J/photon) = 47.7915 = 48
photons
7.65 Determine the wavelength:
 = 1/(1953 cm–1) = (5.1203277 x 10–4 cm) (unrounded)
 (nm) = (5.1203277 x 10–4 cm) (10–2 m/1 cm) (1 nm/10–9 m) = 5120.3277 =
5.120 x 103 nm
 (Å) = (5.1203277 x 10–4 cm) (10–2 m/1 cm) (0.01 Å/10–12 m) = 51203.277 =
5.120 x 104 Å
= c/=
 1 cm  1 Hz 

 1 
5.1203277 x 10 cm  10 2 m 
 1 s 
2.9979 x 108 m/s
4
= 5.8548987 x 1013 = 5.855 x 1013 Hz
7.66 The Bohr model has been successfully applied to predict the spectral lines for
one-electron species other than H.
Common one-electron species are small cations
with all but one electron removed. Since the problem specifies a metal ion, assume that
the possible choices are Li+2 or Be+3, and solve Bohr’s equation to verify if a whole
number
for n can be calculated. Recall that the negative sign is a convention based
on the zero point of the atom’s energy;
it is deleted in this calculation to avoid
taking the square root of a negative number.
The highest-energy line corresponds to the transition from n = 1 to n = .
E = h = (6.626 x 10–34 Js) (2.961 x 1016 Hz) (s–1/Hz) = -1.9619586 x 10–17 J
(unrounded)
E = -Z2 (2.18 x 10–18 J)/n2
Z2 = E n2/(-2.18 x 10–18 J) = (-1.9619586 x 10–17 J) 12/(-2.18 x 10–18 J) = 8.99998
Then Z2 = 9 and Z = 3.
Therefore, the ion is Li2+.
7.67 a) 59.5 MHz (m) = c/ =
215.8 MHz (m) = c/ =
2.9979 x 108 m/s
 106 Hz  s 1 
 59.5 MHz  



 1 MHz  Hz 
2.9979 x 108 m/s
 106 Hz  s 1 
 215.8 MHz  



 1 MHz  Hz 
= 5.038487 = 5.04 m
= 1.38920 = 1.389 m
Therefore, the VHF band overlaps with the 2.78-3.41 m FM band.
7-16
b) 550 kHz (m) = c/ =
1600kHz
(m) = c/ =
3.00 x 108 m/s
 103 Hz  s 1 
 550 kHz  



 1 kHz  Hz 
3.00 x 108 m/s
 103 Hz  s 1 
1600 kHz  



 1 kHz  Hz 
= 545.45 = 550 m
= 187.5 = 190 m
FM width from 2.78 to 3.41 m gives 0.63 m, whereas AM width from 190 to 550
m gives 360 m.
7.68 a) Electron:  = h/mv =
x 10–10 m
Proton:  = h/mv =
10–13 m
b) EK = 1/2 mv2
v=
2 EK
m
Electron: v =
 6.626 x 10 J • s 
9.11 x 10 kg   3.0 x 10
34
31
 6.626 x 10 J • s 
1.67 x10 kg   3.0 x10
34
27
6
6
 kg • m 2 /s 2

m  
J

s 
 kg • m 2 /s 2

m  
J
s 






= 2.4244 x 10–10 = 2.4
= 1.32255 x 10–13 = 1.3 x
therefore v2 = 2 EK/m



2 2.5 x 1015 J  kg • m2 /s 2

J
9.11 x 1031 kg 



= 7.4084 x 107 m/s (unrounded)

2  2.5 x 10 J   kg • m /s 
6
Proton: v =

 = 1.7303 x 10 m/s (unrounded)
J

1.67 x 10 kg  
 kg • m /s 
 6.626 x 10 J • s 
–12
Electron:  = h/mv =
= 9.8

 = 9.8182 x 10
m
J


9.11
x
10
kg
7.4084
x
10

 
s 
15
2
2
27
34
2
31
x 10–10 m
Proton:  = h/mv =
2
7
 6.626 x 10 J • s 
1.67 x 10 kg  1.7303 x 10
34
27
6
 kg • m 2 /s 2

m  
J
s 



= 2.29305 x 10–13 = 2.3 x
10–13 m
7.69 The electromagnetic spectrum shows that the visible region goes from 400 to 750
nm. Thus, wavelengths b, c, and
d are for the three transitions in the visible series
with nfinal = 2. Wavelength a is in the ultraviolet region of the spectrum and the
ultraviolet series has nfinal = 1. Wavelength e is in the infrared region of the spectrum and
the infrared series has nfinal = 3.
7-17
n = ?  n = 1;  = 1212.7 Å (shortest  corresponds to the largest E)
1/ =


1.096776 x 10 m   n1  n1 
1
7
2
1

 0.01 Å 
1


1212.7 Å  1 x 1012 m 
=
1 1
 2  2
 1 n2
0.7518456 =
2
2



1.096776 x 10 m   11  n1 
7
1

2
2
2




n2 = 2 for line (a) (n = 2  n = 1)
n = ?  n = 3;  = 10,938 Å (longest  corresponds to the smallest E)
1/ =


1.096776 x 10 m   n1  n1 
1
7
2
1

 0.01 Å 
1


10938 Å  1 x 1012 m 
0.083357396 =
=
2
2



1.096776 x 10 m   31  n1 
7
1

 1 1
 2  2
 3 n2
2
2
2




n2 = 6 for line (e) (n = 6  n = 3)
For the other three lines, n1 = 2.
For line (d), n2 = 3 (largest   smallest E).
For line (b), n2 = 5 (smallest   largest E).
For line (c), n2 = 4
7.70 E = hc/
thus  = hc/E
a)  = hc/E =
 6.626 x 10
34




= 286.4265 = 286 nm


= 450.748 = 451 nm
J • s 3.00 x 10 8 m/s  1 nm 
 9 
4.60 x 10 19 J
 10 m 
b)  = hc/E =
 6.626 x 10
c)  = hc/E =
 6.626 x 10
34
J • s 3.00 x 10 8m/s  1 nm 
 9 
6.94 x 10 19 J
 10 m 
34
J • s 3.00 x 10 8 m/s  1 nm 
 9 
4.41 x 10 19 J
 10 m 
= 432.130 = 432 nm
7.71 E = hc/
thus  = hc/E
a) The energy of visible light is lower than that of UV light. Thus, metal A must
be barium because the attraction
between barium’s nucleus and outer electron is
less than the attraction in tantalum or tungsten. In all three
elements, the outer
electron is in the same energy level (6s) but the nuclear charge is less in barium than
tantalum
or tungsten. The longest wavelength corresponds to the lowest energy
(work function).
7-18
Ta:  = hc/E =
 6.626 x 10
Ba:  = hc/E =
 6.626 x 10
W:  = hc/E =
 6.626 x 10
34


J • s 3.00 x10 8 m/s  1 nm 
 9 
6.81 x 10 19 J
 10 m 
34
= 291.894 = 292 nm


= 462.279 = 462 nm


= 277.6257 = 278 nm
J • s 3.00 x 10 8m/s  1 nm 
 9 
4.30 x 10 19 J
 10 m 
34
J • s 3.00 x 10 8 m/s  1 nm 
 9 
7.16 x 10 19 J
 10 m 
Metal A must be barium, because barium is the only metal that emits in the visible
range (462 nm).
b) A UV range of 278 nm to 292 nm is necessary to distinguish between tantalum
and tungsten.
7.72 Index of refraction = c/v thus v = c/(index of refraction)
a) Water
v = c/(index of refraction) = (3.00 x 108 m/s)/(1.33) = 2.2556 x
108 = 2.26 x 108 m/s
b) Diamond v = c/(index of refraction) = (3.00 x 108 m/s)/(2.42) = 1.239669 x 108
= 1.24 x 108 m/s
7.73 Extra significant figures are necessary because of the data presented in the
problem.
He-Ne
 = 632.8 nm
Ar
 = 6.148 x 1014 s–1
Ar-Kr
E = 3.499 x 10–19 J
Dye  = 663.7 nm
7-19
Calculating missing  values:
Ar
 = c/ = (2.9979 x 108 m/s)/( 6.148 x 1014 s–1) = 4.8762199 x 10–7 = 4.876
x 10–7 m
Ar-Kr  = hc/E = (2.9979 x 108 m/s) (6.626 x 10–34 J•s)/( 3.499 x 10–19 J)
= 5.67707 x 10–7 = 5.677 x 10–7 m
Calculating missing  values:
He-Ne
 = c/ = (2.9979 x 108 m/s)/[632.8 nm (10–9 m/nm)]
= 4.7375 x 1014 = 4.738 x 1014 s–1
Ar-Kr  = E/h = (3.499 x 10–19 J)/(6.626 x 10–34 J•s) = 5.28071 x 1014 = 5.281 x
1014 s–1
Dye  = c/ = (2.9979 x 108 m/s)/[663.7 nm (10–9 m/nm)]
= 4.51695 x 1014 = 4.517 x 1014 s–1
Calculating missing E values:
He-Ne
E = hc/ = [(6.626 x 10–34 J•s) (2.9979 x 108 m/s)]/[632.8 nm (10–9
m/nm)]
= 3.13907797 x 10–19 = 3.139 x 10–19 J
Ar
E = h = (6.626 x 10–34 J•s) (6.148 x 1014 s–1) = 4.0736648 x 10–19 = 4.074
x 10–19 J
Dye E = hc/ = [(6.626 x 10–34 J•s) (2.9979 x 108 m/s)]/[663.7 nm (10–9 m/nm)]
= 2.99293 x 10–19 = 2.993 x 10–19 J
The colors may be predicted from Figure 7.3 and the frequencies.
He-Ne
 = 4.738 x 1014 s–1
Orange
14 –1
Ar
 = 6.148 x 10 s
Green
14 –1
Ar-Kr  = 5.281 x 10 s
Yellow
14 –1
Dye  = 4.517 x 10 s
Red
7.74 The speed of light is necessary; however, the frequency is irrelevant.
a) Time = (8.1 x 107 km) (103 m/1 km) (1 s/3.00 x 108 m) = 270 = 2.7 x 102 s
b) Distance = (1.2 s) (3.00 x 108 m/s) = 3.6 x 108 m
7.75 a) The l value must be at least 1 for ml to be -1, but cannot be greater than n - 1 =
2. Increase the l value to 1 or 2 to create an allowable combination.
b) The l value must be at least 1 for ml to be +1, but cannot be greater than n - 1 =
2. Decrease the l value to 1 or 2 to create an allowable combination.
c) The l value must be at least 3 for ml to be +3, but cannot be greater than n - 1 =
6. Increase the l value to 3, 4, 5,
or 6 to create an allowable combination.
d) The l value must be at least 2 for ml to be -2, but cannot be greater than n - 1 =
3. Increase the l value to 2 or 3 to create an allowable combination.
7-20
7.76 1/ =
a)
1.096776 x 10
7
1
 1 nm 
94.91 nm  109 m 
0.9606608 =
 1
1 
m1  2  2 
n

 1 n2 
=



1.096776 x 10 m   11  n1 
7
1

2
2
2

1 1 
 2  2 
 1 n2 
n2 = 5
b)
 1 nm 
1


1281 nm  109 m 
0.071175894 =
=


1.096776 x 10 m   n1  51 
7
1

2
1
 1 1 
 2  2 
 n1 5 
n1 = 3
7-21
2

c) 1/ =
1.096776 x 10 m   11  31 
1
7
2
2
1/ = 9.74912 x 106 m–1 (unrounded)
 = (1/9.74912 x 106 m–1) (1 nm/10–9 m) = 102.573 = 102.6 nm
7.77 a) Ionization occurs when the electron is completely removed from the atom, or
when n = . We can use
equation 7.4 to find the quantity of energy needed to remove completely the
electron, called the ionization energy (IE). The charge on nucleus must affect the IE
because a larger nucleus would exert a greater pull on the escaping
electron. The
Bohr equation applies to H and other one-electron species. The general expression is:
E =

 2.18 x 10 J   1
18
2


1
2
n initial
 2  6.022 x 1023 
 Z 

1 mol

 
= 1.312796 x 106 = (1.31 x 106 J/mol) Z2 for n = 1
b) In the ground state n = 1, the initial energy level for the single electron in B4+.
Once ionized, n =  is the final energy level. The ionization energy is calculated as
follows (remember that 1 /  = 0).
E = IE = (1.312796 x 106 J/mol) 52 = 3.28199 x 107 = 3.28 x 107 J/mol
c) E =  2.18 x 10
18
 1
1 
J  2  2  Z 2

n
initial 

= (2.42222 x 10–19 J)Z2 for n = 3
E = IE = (2.42222 x 10–19 J)22 = 9.68889 x 10–19 J
 = hc/E =
d) E =
 6.626 x 10

34

J • s 3.00 x 10 8 m/s  1 nm 
 9 
9.68889 x 10 19 J
 10 m 

 2.18 x 10 J   1
18

2

1
2
n initial
= 205.162 = 205 nm
 2
 Z

= (5.45 x 10–19 J)Z2 for n = 2
E = IE = (5.45 x 10–19 J)42 = 8.72 x 10–18 J
 = hc/E =
 6.626 x 10
34


J • s 3.00 x 10 8 m/s  1 nm 
 9 
8.72 x 10 18 J
 10 m 
= 22.79587 = 22.8 nm
7.78 The wavelengths of light responsible for the spectral lines (for the different series)
for hydrogen are related by the Rydberg equation:
1

=
 1
1
R 2  2
n
 1 n2



where n = 1, 2, 3.... and n2 > n1
As the values of n increase, the energies associated with n move closer. The result
is that E values within a series
increase by continually smaller values and thus,
the smaller wavelength associated with these E values moves closer together. The
7-22
spectral lines become closer and closer together in the short wavelength region of each
series because the difference in energy associated with the transition from ni to nf
becomes smaller and smaller with
increasing distance from the nucleus.
7.79 Plan: Use the values and the equation given in the problem to calculate the
appropriate values. Additional information is given in the back of the textbook.
rn =
n 2 h 2 0
me e 2
Solution:
2
C2 
12 6.626 x 1034 J • s  8.854 x 1012

J • m   kg • m 2 /s 2


2
J

 9.109 x 1031 kg 1.602 x 1019 C

a) r1 =







= 5.2929377 x 10–11 = 5.293 x
10–11 m
2
C2 
102 6.626 x 1034 J • s  8.854 x 1012

J • m   kg • m 2 /s 2


2
J

 9.109 x 1031 kg 1.602 x 1019 C

b) r10 =







= 5.2929377 x 10–9 = 5.293
x 10–9 m
7.80 Plan: See problems 7.79 and 7.22.
Solution:
2
C2 
32 6.626 x 1034 J • s  8.854 x 10 12

J • m   kg • m 2 /s 2


2
J

 9.109 x 1031 kg 1.602 x 1019 C

a) r3 =







= 4.76364 x 10–10 = 4.764 x
10–10 m
b) H atom so Z = 1
En =
En =
  2.18 x 10
 Z2
 
 12
n
18
J
2.18 x 10 18 J

2
3
c) Li ion so Z = 3
En =
En =

2
  2.18 x 10
 Z2
 
 32
n
18
J

2
2.18 x 10 18 J
2
3
= -2.42222 x 10–19 = -2.42 x 10–19 J

= -2.18 x 10–18 J
d) The greater number of protons in the Li nucleus results in a greater interaction
between the Li nucleus and its electrons. Thus, the energy of an electron in a particular
7-23
orbital becomes more negative with increasing atomic
number.
7.81 Plan: Refer to Chapter 6 for the calculation of the amount of heat energy absorbed
by a substance from its heat
capacity and temperature change (q = C x mass x T).
Using this equation, calculate the energy absorbed by the water. This energy equals the
energy from the microwave photons. The energy of each photon can be calculated
from its wavelength: E = hc/. Dividing the total energy by the energy of each
photon gives the number of
photons absorbed by the water.
Solution:
q = C x mass x T
q = (4.184 J/g°C) (252 g) (98 - 20.)°C = 8.22407 x 104 J (unrounded)
E = hc/ = [(6.626 x 10–34 J•s) (3.00 x 108 m/s)]/(1.55 x 10–2 m)
E = 1.28245 x 10–23 J/photon (unrounded)
Number of photons = (8.22407 x 104 J) (1 photon/1.28245 x 10–23 J)
= 6.41278 x 1027 = 6.4 x 1027 photons
Check: The order of magnitude appears to be correct for the calculation of total
energy absorbed: 101 x 102 x 101 =
104. The order of magnitude of the energy for one
photon also appears correct: 10–34 x 108/10–2 = 10–24. In addition,
the order of
4
–24
28
magnitude of the number of photons is estimated as 10 /10 = 10 , which agrees with the
calculated 6x1027.
7.82 One sample calculation will be done using the equation in the book:
=
r
e
a0
 1  1 

 
    a0 
3
r
2
e
a0
=
 1 pm  
 1 
1
 12  

 
m  
    52.92 pm  10
(unrounded)
7-24
3
r
2
e
a0
= 1.46553 x 10–21
r
(pm)
0
50
100
200
 (pm–3/2)
2 (pm–3)
1.47 x 10–3
0.570 x 10–3
0.221 x 10–3
0.0335 x 10–3
2.15 x 10–6
0.325 x 10–6
0.0491 x 10–6
0.00112 x 10–6
4r22 (pm–
1
)
0
1.02 x 10–2
0.616 x 10–2
0.0563 x 10–
2
The plots are similar to Figure 7.17A in the text.
7.83 In general, to test for overlap of the two series, compare the longest wavelength in
the “n” series with the shortest wavelength in the “n+1” series. The longest wavelength
in any series corresponds to the transition between the n1 level and the next level above
it; the shortest wavelength corresponds to the transition between the n1 level and the
n =  level. Use:
1

=
 1
1 
R 2  2 
n

 1 n2 
1

=
1.096776 x 10
R = 1.096776 x 107 m–1
7
 1
1
m1  2  2
n
 1 n2




a) The overlap between the n1 = 1 series and the n1 = 2 series would occur between
the longest wavelengths for
n1 = 1 and the shortest wavelengths for n1 = 2.
Longest wavelength in n1 = 1 series has n2 equal to 2.
1

=
1.096776 x 10
7

1 1 
m 1  2  2 
1 2 
 = 1.215684272 x 10–7 = 1.215684 x 10–7 m
7-25
Shortest wavelength in the n1 = 2 series:
1

=
1.096776 x 10 m   21  1
7
1
2
2



 = 3.647052817 x 10–7 = 3.647053 x 10–7 m
Since the longest wavelength for n1 = 1 series is shorter than shortest wavelength
for n1 = 2 series, there
is no overlap between the two series.
7-26
b) The overlap between the n1 = 3 series and the n1 = 4 series would occur
between the longest wavelengths for
n1 = 3 and the shortest wavelengths for n1 = 4.
Longest wavelength in n1 = 3 series has n2 equal to 4.
1

=
1.096776 x 10
7

 1 1 
m 1  2  2 
3 4 
 = 1.875627163 x 10–6 = 1.875627 x 10–6 m
Shortest wavelength in n1 = 4 series has n2 = .
1

=
1.096776 x 10
7

1 
 1
m 1  2  2 
 
4
 = 1.458821127 x 10–6 = 1.458821 x 10–6 m
Since the n1 = 4 series shortest wavelength is shorter than the n1 = 3 series longest
wavelength, the series
do overlap.
c) Shortest wavelength in n1 = 5 series:
1

=
1.096776 x 10
7

1 
 1
m 1  2  2 
5  
 = 2.27940801 x 10–6 = 2.279408 x 10–6 m
Similarly for the other combinations:
For n1 = 4, n2 = 5,  = 4.052281 x 10–6 m
For n1 = 4, n2 = 6,  = 2.625878 x 10–6 m
For n1 = 4, n2 = 7,  = 2.166128 x 10–6 m
Therefore, only the first two lines of the n1 = 4 series overlap with the n1 = 5
series.
d) At longer wavelengths (i.e., lower energies), there is increasing overlap between
the lines from different series (i.e.,
with different n1 values). The hydrogen spectrum
becomes more complex, since the lines begin to merge into a moreor-less
continuous band, and much more care is needed to interpret the information.
7.84 a) The highest frequency would correspond to the greatest energy difference. In
this case, the greatest energy
difference would be between E3 and E1.
E = E3 - E1 = h = (-15 x 10–19 J) - (-20 x 10–19 J) = 5 x 10–19 J
 = E/h = (5 x 10–19 J)/(6.626 x 10–34 J•s) = 7.54603 x 1014 = 8 x 1014 s–1
 = c/ = (3.00 x 108 m/s)/(7.54603 x 1014 s–1) = 3.97560 x 10–7 = 4 x 10–7 m
b) The ionization energy (IE) is the same as the reverse of E1. Thus, the value of
the IE is 20 x 10–19 J/atom.
IE = (20 x 10–19 J/atom) (1kJ/103 J) (6.022 x 1023 atoms/mol) = 1204.4 = 1.2 x 103
kJ/mol
c) The shortest wavelength would correspond to an electron moving from the n =
4 level to the highest level
available in the problem (n = 6).
7-27
E = E6 - E4 = hc/ = (-2 x 10–19 J) - (-11 x 10–19 J) = 9 x 10–19 J
 = hc/E =

7.85
 6.626 x 10

34


J • s 3.00 x 108 m/s  1 nm 
 9 
9 x 1019 J
 10 m 

= 220.867 = 2 x 102 nm
color
 = c/
a)
red
=
3.00 x 108 m/s  1 nm 
 9 
671 nm
 10 m 
= 4.4709 x 1014 = 4.47 x 1014 s–1
b)
blue
=
3.00 x 108 m/s  1 nm 
 9 
456 nm
 10 m 
= 6.5789 x 1014 = 6.58 x 1014 s–1
c)
orange-red  =
3.00 x 108 m/s  1 nm 
 9 
649 nm
 10 m 
= 4.622496 x 1014 = 4.62 x 1014 s–1
d)
yellow-orange
=
3.00 x 108 m/s  1 nm 
 9 
589 nm
 10 m 
= 5.0933786 x 1014 = 5.09 x 1014
s–1
7.86 Plan: The energy differences sought may be determined by looking at the energy
changes in steps.
Solution:
a) The difference between levels 3 and 2 (E32) may be found by taking the
difference in the energies for the 3  1
transition (E31) and the 2  1 transition
(E21).
E32 = E31 - E21 = (4.844 x 10–17 J) - (4.088 x 10–17 J) = 7.56 x 10–18 J
 = hc/E =
 6.626 x 10 J • s 3.00 x 10
 7.56 x 10 J 
34
8
m/s

18
= 2.629365 x 10–8 = 2.63 x 10–8 m
b) The difference between levels 4 and 1 (E41) may be found by adding the
energies for the 4  2 transition (E42) and the 2  1 transition (E21).
E41 = E42 + E21 = (1.022 x 10–17 J) + (4.088 x 10–17 J) = 5.110 x 10–17 J
 = hc/E =
 6.626 x 10 J • s 3.00 x 10
5.110 x 10 J 
34
8
m/s

17
= 3.8900 x 10–9 = 3.890 x 10–9 m
c) The difference between levels 5 and 4 (E54) may be found by taking the
difference in the energies for the 5  1
transition (E51) and the 4  1 transition
(see part b).
E54 = E51 - E41 = (5.232 x 10–17 J) - (5.110 x 10–17 J) = 1.22 x 10–18 J
 = hc/E =
 6.626 x 10 J • s 3.00 x 10
1.22 x 10 J 
34
8
m/s

18
7-28
= 1.629344 x 10–7 = 1.63 x 10–7 m
7.87 a) A dark green color implies that relatively few photons are being reflected from
the leaf. A large fraction of the photons is being absorbed, particularly in the red region
of the spectrum. A plant might adapt in this way when
photons are in short supply —
i.e., in conditions of low light intensity.
b) An increase in the concentration of chlorophyll, the light-absorbing pigment,
would lead to a darker green
color (and vice versa).
7.88 a) The energy of the electron is a function of its speed leaving the surface of the
metal. The mass of the electron is
9.109 x 10–31 kg.
EK = 1/2 mu2 =


1
9.109 x 1031 kg 6.40 x 105 m/s
2

2
J

2 2
 kg • m /s



= 1.86552 x 10–19 = 1.87 x 10–19 J
b) The minimum energy required to dislodge the electron () is a function of the
incident light. 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, h, to yield the work function, . (The number of significant
figures given in the wavelength requires more significant figures in the speed of
light.)
E = hc/ =
 6.626 x 10

34

J • s 2.9979 x 10 8 m/s  1 nm 
 9 
 358.1 nm 
 10 m 
= 5.447078 x 10–19 J (unrounded)
K = h - EK = (5.447078 x 10–19 J) - (1.86552 x 10–19 J) = 3.581558 x 10–19 =
3.58 x 10–19 J
7.89 a)  = h/mv =
 6.626 x 10 J • s 
9.109 x 10 kg   5.5 x 10
34
31
4
 kg • m 2 /s 2

m  
J
s 



= 1.322568 x 10–8 m
(unrounded)
Smallest object = /2 = (1.322568 x 10–8 m)/2 = 6.61284 x 10–9 = 6.6 x 10–9 m
b)  = h/mv =
 6.626 x 10 J • s 
9.109 x 10 kg   3.0 x 10
34
31
7
 kg • m 2 /s 2

m  
J

s 



= 2.424708 x 10–11 m
(unrounded)
Smallest object = /2 = (2.424708 x 10–11 m)/2 = 1.212354 x 10–11 = 1.2 x 10–11 m
7.90 a) Figure 7.3 indicates that the 641 nm wavelength of Sr falls in the red region
and the 493 nm wavelength of Ba
falls in the green region.
b) Use the formula E = hc/ to calculate kJ/photon. Convert the 1.00 g amounts of
7-29
BaCl2 (M = 208.2 g/mol) and SrCl2 (M = 158.52 g/mol) to moles, then to atoms, and
assume each atom undergoes one electron transition
(which produces the colored
light) to find number of photons. Multiply kJ/photon by number of photons to find total
energy.
SrCl2
Number of photons = (1.00 g SrCl2) (1 mol SrCl2/158.52 g SrCl2) (6.022 x 1023
photons/1 mol SrCl2)
= 3.7988897 x 1021 photons (unrounded)
 6.626 x 10
Ephoton = hc/=
34


J • s 3.00 x 108 m/s  1 nm  1 kJ 
 9 
 3 
641 nm
 10 m  10 J 
= 3.10109 x 10–22 kJ/photon (unrounded)
Etotal = (3.7988897 x 1021 photons) (3.10109 x 10–22 kJ/photon) = 1.17807 = 1.18
kJ (Sr)
BaCl2
Number of photons = (1.00 g BaCl2) (1 mol BaCl2/208.2 g BaCl2) (6.022 x 1023
photons/1 mol BaCl2)
= 2.892411 x 1021 photons (unrounded)
 6.626 x 10
Ephoton = hc/=
34


J • s 3.00 x 108 m/s  1 nm  1 kJ 
 9 
 3 
493 nm
 10 m  10 J 
= 4.0320487 x 10–22 kJ/photon (unrounded)
Etotal = (2.892411 x 1021 photons) (4.0320487 x 10–22 kJ/photon) = 1.16623 = 1.17
kJ (Ba)
7.91 a) The highest-energy line corresponds to the shortest wavelength. The shortestwavelength line is given by
1

=
 1
1 
R 2  2 
n

 1 n2 
1

=
1.096776 x 10
R = 1.096776 x 107 m–1
7
 1
1
m1  2  2
n
 1 n2




 1 nm 
1
7
1

 = 1.096776 x 10 m
3282 nm  109 m 

–1

  n1  1

2
1
2
–1
304692 m = (1.096776 x 10 m ) (1/n2)
1/n2 = 0.0277807
n=6
b) The line with the smallest energy (corresponding to the line with the longest
wavelength),
1

=
1.096776 x 10
7



7
 1
1
m 1  2 
 n1  n  12
1






7-30

1
1
 1 nm 
7
1  1
=
1.096776
x
10
m



9
2
7460 nm  10 m 
 n1  n  12
1



134048 m–1 = 1.096776 x 107 m1  12  1 2 
 n1  n  1 
1










(134048 m–1)/(1.096776 x 107 m–1) = 0.0122220 =
 1
1
 
 n12  n  12
1





Rearranging and solving this equation for n1 yields n1 = 5. (You and your students
may well need to resort
to trial-and-error solution of this equation!)
7.92 a) At this wavelength the sensitivity to absorbance of light by Vitamin A is
maximized while minimizing interference due to the absorbance of light by other
substances in the fish-liver oil.
b) The wavelength 329 nm lies in the ultraviolet region of the electromagnetic
spectrum.
c) A known quantity of vitamin A (1.67 x 10–3 g) is dissolved in a known volume
of solvent (250. mL) to give a standard concentration with a known response (1.018
units). An equality can be made between the two concentration-to-absorbance ratios:
A1/C1 = A2/C2
 1.67 x 10 3 g 

 250. mL 
 0.724  
= 4.7508 x 10–6 g/mL (unrounded)
1.018 
 4.7508 x 106 g Vitamin A/mL  500. mL  = 1.92808 x 10–2 = 1.93 x 10–2 g Vitamin A/g
 0.1232 g Oil 
C1 = A1C2/A2 =
Oil
7.93  = hc/E =
 6.626 x 10 J • s 3.00 x 10
 7.59 x 10 J 
34
8
19

m/s  1 nm 
 9 
 10 m 
= 2.61897 = 262 nm
Silver is not a good choice for a photocell that uses visible light. 262 nm is in the
ultraviolet region.
7.94 a) All concentrations are multiplied by 105.
7-31
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
1
1.5
2
2.5
3
Concentration (M x 105)
This is a linear plot, thus, the graph is of the type y = mx + b, with m = slope and
b = intercept.
7-32
Since there appears to be no scatter in the data plot, any two points may be used to
find the slope. Using the first and last points given:
 0.396  0.131
m = (y - y )/(x - x ) =
= 13250 = 1.3 x 104/M
2
1
2
1
3.0 x 10
5

 1.0 x105 M
Using the slope just calculated and any of the data points, the value of the
intercept may be found.
b = y - mx = 0.396 Abs - (13250 Abs/M) (3.0 x 10–5 M) = -0.0015 = 0.00
(absorbance has no units)
b) Using the equation just determined (y = (1.3 x 104 Abs/M) x + 0.00 Abs)
x = (y - 0.00)/(1.3 x 104/M) = (0.236/1.3 x 104) M = 1.81538 x 10–5 M
(unrounded)
This value is Mf in a dilution problem (MiVi = MfVf) with Vi = 20.0 mL and Vf =
150. mL
Thus:
Mi = MfVf/Vi = (1.81538 x 10–5 M) (150. mL)/(20.0 mL) = 1.361538 x 10–4 = 1.4
x 10–4 M
7.95 Mr. Green must be in the dining room where green light (520 nm) is reflected.
Lower frequency, longer wavelength light is reflected in the lounge and study. Both
yellow and red light have longer wavelengths than green light. Therefore, Col. Mustard
and Ms. Scarlet must be in either the lounge or study. The shortest
wavelengths are
violet. Prof. Plum must be in the library. Ms. Peacock must be the murderer.
7.96 EK = 1/2 mv2
v=
EK
1
m
2
=
 = h/mv =
12
 kg • m 2 /s 2

1
J
9.109 x 1031 kg 
2
4.71 x 1015 J


 6.626 x 10 J • s 
9.109 x 10 kg  1.0169 x 10
34
31
8



= 1.0169 x 108 m/s (unrounded)
 kg • m 2 /s 2

m  
J
s 



= 7.15323 x 10–12 = 7.15 x 10–
m
7.97 The amount of energy is calculated from the wavelength of light:
Ephoton = hc/=
 6.626 x 10
34


J • s 3.00 x 10 8 m/s  1 nm 
 9 
550 nm
 10 m 
= 3.61418 x 10–19 J/photon
(unrounded)

1 photon
 5%  



19 
 W  100%  100%   3.61418 x 10 J 
10%
Number of photons =  75 W   1 J/ s 

= 1.0376 x 1018 = 1.0 x 1018 photons/s
7-33
7.98 a) Sodium ions emit yellow-orange light, and potassium ions emit violet light.
b) The cobalt glass filter absorbs the yellow-orange light, while the violet light
passes through.
7.99 a) 6 CO2(g) + 6 H2O(l)  C6H12O6(s) + 6 O2(g)
Hrxn = [1 mol Hf(C6H12O6) + 6 mol Hf(O2)] - [6 mol Hf(CO2) + 6 mol
Hf(H2O)]
Hrxn = [-1237.3 kJ + 6(0.0 kJ)] - [6(-393.5 kJ) + 6(-285.840 kJ)] = 2838.74 =
2838.7 kJ
6 CO2(g) + 6 H2O(l)  C6H12O6(s) + 6 O2(g)
Hrxn = 2838.7 kJ (for 1.00
mol C6H12O6)
b) Ephoton = hc/=
 6.626 x 10
34


J • s 3.00 x 10 8 m/s  1 nm 
 9 
680. nm
 10 m 
J/photon (unrounded)


J 
1 photon



19

 1 kJ   2.923235 x 10 J 
3
Number of photons =  2838.7 kJ   10
= 9.7108 x 1024 = 9.71 x 1024 photons
7-34
= 2.923235 x 10–19
7.100 In the visible series with nfinal = 2, the transitions will end in either the 2s or 2p
orbitals. With the restriction that
the angular momentum quantum number can
change by only ±1, the allowable transitions are from a p orbital to
2s, from an s
orbital to 2p, and from a d orbital to 2p. The problem specifies a change in energy level,
so ninit must be 3, 4, 5, etc. (Although a change from 2p to 2s would result in a +1
change in l, this is not a change in energy
level.)
The first four transitions are as follows:
3s → 2p
3d → 2p
4s → 2p
3p → 2s
7.101 a) A = (b)c
0.9
Absorbance
0.7
0.5
0.3
0.1
0
0
10
20
Concentration, mol/L
x 105
b) b = 4.5 x 103 M–1 = slope
A/b = c = (0.55)/(4.5 x 103 M–1) = 1.2222 x 10–4 = 1.2 x 10–4 M
7-35