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7 CHAPTER 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 7.2 All types of electromagnetic radiation travel as waves at the same speed. They differ in both their frequency and wavelength. a) Figure 7.3 describes the electromagnetic spectrum by wavelength and frequency. Wavelength increases from left (102 nm) to right (1012 nm). The trend in increasing wavelength is: xray < ultraviolet < visible < infrared < microwave < radio waves. b) Frequency is inversely proportional to wavelength according to the equation c = , so frequency has the opposite trend: radio < microwave < infrared < visible < ultraviolet < xray. c) Energy is directly proportional to frequency according to the equation E = h. Therefore, the trend in increasing energy matches the trend in increasing frequency: radio < microwave < infrared < visible < ultraviolet < xray. Highenergy electromagnetic radiation disrupts cell function. It makes sense that you want to limit exposure to ultraviolet and xray radiation. 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. 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. a) Frequency: A > B > C b) Energy: A > B > C c) Amplitude: A > C > B d) Since wave A has a higher energy and frequency than B, wave A is more likely to cause a current. e) Wave B is more likely to be infrared radiation since Wave C has a longer wavelength than B. 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. Plan: Wavelength is related to frequency through the equation c = . Recall that a Hz is a reciprocal second, or 1/s = s1. Assume that the number "950" has three significant figures. Solution: c = c 3.00 x 108 m/s = = 315.8 = 316 m (m) = 103 Hz s 1 ( 950.kHz ) 1 kHz Hz
7.3
7.4 7.5
7.6
7.7
71
(nm) =
c
=
1 nm 3.00 x 108 m/s 11 11 9 = 3.158 x 10 = 3.16 x 10 nm 3 1 10 Hz s 10 m ( 950. kHz ) 1k Hz Hz
() =
c
=
1 3.00 x 108 m/s = 3.158 x 1012 = 3.16 x 1012 1 10 10 m 3 10 Hz s ( 950. kHz ) 1 kHz Hz
7.8
Wavelength and frequency relate through the equation c = . Recall that a Hz is a reciprocal second, or 1/s = s1.
(m) =
c
=
3.00 x 108 m/s = 3.208556 = 3.21 m 106 Hz s 1 ( 93.5 MHz ) 1 MHz Hz
(nm) =
c
=
3.00 x 108 m/s 1 nm 9 9 = 3.208556 x 10 = 3.21 x 10 nm 106 Hz s 1 109 m ( 93.5 MHz ) 1 MHz Hz
() =
c
=
1 3.00 x108 m / s 10 10 10 = 3.208556 x 10 = 3.21 x 10 6 1 10 Hz s 10 m ( 93.5 MHz ) 1 MHz Hz
7.9
Plan: Frequency is related to energy through the equation E = h. Note that 1 Hz = 1 s1. Solution: E = h E = (6.626 x 1034 Js) (3.8 x 1010 s1) = 2.51788 x 1023 = 2.5 x 1023 J
hc
7.10
E=
=
( 6.626 x 10
34
J s 3.00 x 108 m/ s 1 15 15 10 = 1.529 x 10 = 1.5 x 10 J 10 m 1.3
)(
)
7.11
Since energy is directly proportional to frequency (E = h) and frequency and wavelength are inversely related hc ). As wavelength decreases, energy ( = c/), it follows that energy is inversely related to wavelength ( E =
increases. In terms of increasing energy the order is red < yellow < blue. 7.12 7.13 Since energy is directly proportional to frequency (E = h): UV ( = 8.0 x 1015 s1) > IR ( = 6.5 x 1013 s1) > microwave ( = 9.8 x 1011 s1) or UV > IR > microwave. Plan: Frequency and wavelength can be calculated using the speed of light: c = . Solution: 1 nm 2.9979 x 108 m/s 7 7 (nm) = c/ = 9 = 1.3482797 x 10 = 1.3483 x 10 nm 9 1 10 Hz s 10 m ( 22.235 GHz ) 1 GHz Hz
() = c/ =
1 2.9979 x 108 m/s = 1.34827973 x 108 = 1.3483 x 108 9 1 10 10 m 10 Hz s ( 22.235 GHz ) 1 GHz Hz
72
7.14
Frequency and wavelength can be calculated using the speed of light: c = . 3.00 x 108 m / s 1 m 13 13 1 a) = c/ = 6 = 3.125 x 10 = 3.1 x 10 s 10 m 9.6 m b) (m) = c/ =
(
1 m 2.9979 x 108 m/s = 3.464979 = 3.465 m s 1 106 m 13 8.652 x 10 Hz Hz
)
7.15
Frequency and energy are related by E = h, and wavelength and energy are related by E = hc/. 106 eV 1.602 x 1019 J (1.33 MeV ) 1 MeV Hz 1 eV E 20 20 (Hz) = = 1 = 3.2156 x 10 = 3.22 x 10 Hz h 6.626 x 1034 J s s = 9.32950 x 1013 = 9.33 x 1013 m 106 eV 1.602 x 1019 J (1.33 MeV ) 1 MeV 1 eV The wavelength can also be found using the frequency calculated in the equation c = .
(m) =
hc = E
( 6.626 x 10
34
J s 3.00 x 108 m/s
)(
)
7.16
Plan: a) The least energetic photon has the longest wavelength (242 nm). b) The most energetic photon has the shortest wavelength (2200 ). Solution: c 3.00 x 108 m/s 1 nm 15 15 1 = a) = 9 = 1.239669 x 10 = 1.24 x 10 s 10 m 242 nm
E=
hc
( 6.626 x 10 =
34
J s 3.00 x 108 m/s 1 nm 19 19 9 = 8.2140 x 10 = 8.21 x 10 J 10 m 242 nm
)(
)
b) =
c
=
3.00 x 108 m/s 1 15 15 1 10 = 1.3636 x 10 = 1.4 x 10 s 10 m 2200
E=
hc
=
( 6.626 x 10
34
J s 3.00 x 108 m/s 1 19 19 10 = 9.03545 x 10 = 9.0 x 10 J 10 m 2200
)(
)
7.17 7.18
"n" in the Rydberg equation is equal to a Bohr orbit of quantum number "n" where n = 1, 2, 3, .... 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. 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. Plan: The quantum number n is related to the energy level of the electron. An electron absorbs energy to change from lower energy (lower n) to higher energy (higher n), giving an absorption spectrum. An electron emits energy as it drops from a higher energy level (higher n) to a lower one (lower n), giving an emission spectrum. Solution: a) absorption b) emission c) emission d) absorption
7.19
7.20
73
7.21 7.22
The Bohr model works for only a oneelectron system. The additional attractions and repulsions in manyelectron systems make it impossible to predict accurately the spectral lines. 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 =
 Z2
18
n2 number for the atom or ion. The line spectra 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 R = 1.096776 x 107 m1 n 1 n2 n 1 = 2 n2 = 5 1 1 1 1 1 = R 2  2 = 1.096776 x 107 m 1 2  2 = 2303229.6 m1 (unrounded) n n2 2 5 1
( )( 2.18 x 10 J ) where Z is the atomic
(
)
1 nm 1 = 434.1729544 = 434.17 nm (nm) = 2303229.6 m 1 109 m
7.24
Calculate wavelength by substituting the given values into the Rydberg equation, 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 1 1 = R 2  2 = 1.096776 x 107 m 1 2  2 = 9749120 m1 (unrounded) n n2 1 3 1
(
)
1 1 () = = 1025.7336 = 1025.7 9749120 m 1 1010 m
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 1 1 = R 2  2 = 1.096776 x 107 m 1 2  2 = 533155 m1 (unrounded) n n2 3 4 1
(
)
1 nm 1 (nm) = = 1875.627 = 1875.6 nm 533155 m 1 109 m 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 1 1 = R 2  2 = 1.096776 x 107 m 1 2  2 = 1523300 m1 (unrounded) n 2 3 1 n2
(
)
74
1 nm 1 = 656.4695 = 656.47 nm (nm) = 1523300 m 1 109 m
7.27
Plan: To find the transition energy, use the equation for the energy of an electron transition and multiply by Avogadro's number. Solution: 1 1 E = 2.18 x 1018 J 2  2 n final n initial 1 1 E = 2.18 x 1018 J 2  2 = 4.578 x 1019 J/photon 2 5
(
)
(
)
4.578 x 1019 J 6.022 x 1023 photons 5 5 E = = 2.75687 x 10 = 2.76 x 10 J/mol photon 1 mol The energy has a negative value since this electron transition to a lower n value is an emission of energy.
7.28
To find the transition energy, use the equation for the energy of an electron transition and multiply by Avogadro's number. 1 1 E = 2.18 x 1018 J 2  2 n final n initial 1 1 E = 2.18 x 1018 J 2  2 = 1.93778 x 1018 J/photon 3 1
(
)
(
)
1.93778 x 1019 J 6.022 x 1023 photons 6 6 E = = 1.1669 x 10 = 1.17 x 10 J/mol photon 1 mol
7.29
Looking at an energy chart will help answer this question. n=5 n=4
(d) (a) (c)
n=3
n=2 (b) n=1
Frequency is proportional to energy so the smallest frequency will be d) n = 4 to n = 3; levels 4 and 5 have a smaller E than the levels in the other transitions. The largest frequency is b) n = 2 to n = 1 since levels 1 and 2 have a larger E than the levels in the other transitions. 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 7.31
b>c>a>d
Plan: Use the Rydberg equation. Solution: 109 m 8 = ( 97.20 nm ) = 9.720 x 10 m 1 nm
ground state: n1 = 1;
n2 = ?
75
1 1 = 1.096776 x 107 m 1 2  2 n 1 n2 1
(
)
1 = 1.096776 x 107 m 1 9.72 x 108 m
1 1 0.9380 = 2  2 1 n 2 1 = 1 0.9380 = 0.0620 n2 2
(
) 11  n1
2 2 2
n 2 = 16.1 2 n2 = 4
7.32
= (1281 nm) x (109 m/1 nm) = 1.281 x 106 m 1 1 1/ = 1.096776 x 107 m 1 2  2 n 1 n2
(
)
1 1 1/(1.281 x 106 m) = 1.096776 x 107 m 1 2  2 n 1 5
(
)
1 1 0.07118 = 2  2 n 1 5 1 = 0.07118 + 0.04000 = 0.11118 2 n1
2 n1 = 8.9944 n1 = 3
7.33 7.34
( 6.626 x 10 E = hc/ =
34
J s 3.00 x 108 m/s 1 nm 19 19 9 = 4.55917 x 10 = 4.56 x 10 J 10 m ( 436 nm )
)(
)
a) Absorptions: A, C, D; Emissions: B, E, F b) Energy of emissions: E < F < B c) Wavelength of absorption: D < A < C If an electron occupies a circular orbit, only integral numbers of wavelengths (= 2nr) 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). de Broglie's concept is supported by the diffraction properties of electrons demonstrated in an electron microscope. 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. 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.
7.35
7.36 7.37
7.38
76
7.39
Plan: Use the de Broglie equation. Mass in pounds must be converted to kg and velocity in mi/h must be converted to m/s because a joule is equivalent to kg im 2 / s 2 . Solution: a) =
6.626 x 1034 J is kgim 2 /s 2 2.205 lb 0.62 mi 1 km 3600 s h = 3 mi J mu 1 kg 1 km 10 m 1 h ( 232 lb ) 19.8 h = 7.099063 x 1037 = 7.10 x 1037 m h 4
(
)
b) x mv
kg im 2 /s 2 J 0.1 mi 4 ( 232 lb ) h 1.11855 x 1035 1 x 1035 m
x h/4 mv
( 6.626 x 10
34
J is
)
2.205 lb 0.62 mi 1 km 3600 s 1 kg 1 km 103 m 1 h
7.40
h a) = = mu
( 6.626 x 10 Jis ) x h/4 mv 0.1 x 10 4 ( 6.6 x 10 g ) h
34 24
kg im 2 /s 2 J 24 7 mi h = 6.59057 x 1015 = 6.6 x 1015 m h b) x i mv 4
( 6.626 x 10 Jis ) ( 6.6 x 10 g ) 3.4 x 10
34
103 g 0.62 mi 1 km 3600 s 1 kg 1 km 103 m 1 h
1.783166 x 1014 2 x 1014 m
kgim 2 /s 2 7 J mi
103 g 0.62 mi 1 km 3600 s 1 kg 1 km 103 m 1 h
7.41
=
h mu
6.626 x 1034 J is h = u= m ( 56.5 g ) ( 5400 )
(
) kgim /s
2
2
J
103 g 1 26 26 = 2.1717 x 10 = 2.2 x 10 m/s 1 kg 1010 m
7.42
=
h mu
6.626 x 1034 J is h = u= m (142 g )(100. pm )
(
) kgim /s
2
2
J
103 g 1 pm = 4.666197 x 1023 = 4.67 x 1023 m/s 1 kg 1012 m
7.43
Plan: The de Broglie wavelength equation will give the mass equivalent of a photon with known wavelength and velocity. The term "massequivalent" 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 = mu
77
m=
6.626 x 1034 J is h = u ( 589 nm ) 3.00 x 108 m/s
(
(
)
)
kg im 2 /s 2 J
1 nm 36 36 9 = 3.7498 x 10 = 3.75 x 10 kg/photon 10 m
7.44
= m=
h mu
6.626 x 1034 J is h = u ( 671 nm ) 3.00 x 108 m/s
(
(
)
)
kg im 2 /s 2 J
1 nm 36 9 = 3.2916 x 10 kg/photon 10 m
3.2916 x 1036 kg 6.022 x 1023 photons 12 12 = 1.9822 x 10 = 1.98 x 10 kg/mol photon mol
7.45 7.46 7.47
The quantity 2 expresses the probability of finding an electron within a specified tiny region of space. 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. 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 . Thus, the probability of finding an electron at 0.529 is much greater for the 1s orbital than for the 2s. 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 threedimensional space. Plan: The following letter designations correlate with the following l quantum numbers: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Remember that allowed ml values are l to + l. The number of orbitals of a particular type is given by the ml quantum number. Solution: a) one b) five c) three d) nine a) There is only a single s orbital in any shell. l = 1 and ml = 0: one value of ml. b) All dorbitals consist of sets of five (ml = 2, 1, 0, +1, +2). c) All porbitals consist of sets of three (ml = 1, 0, +1). d) If n = 3, then there is a 3s (1 orbital), a 3p set (3 orbitals), and a 3d set (5 orbitals) giving 1 + 3 + 5 = 9. a) seven b) three c) five d) four a) All forbitals consist of sets of seven (ml = 3, 2, 1, 0, +1, +2, +3). b) All porbitals consist of sets of three (ml = 1, 0, +1). c) All dorbitals consist 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. Magnetic quantum numbers (ml ) can have integer values from l to + l. a) ml: 2, 1, 0, +1, +2 (if n = 1, then l = 0) b) ml: 0 c) ml: 3, 2, 1, 0, +1, +2, +3
7.48
7.49
7.50
7.51
78
7.52
Magnetic quantum numbers can have integer values from l to +l. a) ml: 3, 2, 1, 0, +1, +2, +3 l = l, ml = 1,0,+1 (if n = 2, then l = 0 or 1) b) ml: l = 0, ml = 0; c) ml: 1, 0, +1 (a) s: spherical (b) px: 2 lobes along the xaxis
7.53
z
z
x y y
x
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) pz: 2 lobes along the zaxis (b) dxy: fourlobes
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 Plan: The following letter designations correlate with the following l quantum numbers: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Remember that allowed ml values are l to + l. Solution: allowable ml # of possible orbitals sublevel 2, 1, 0, +1, +2 5 a) d (l = 2) 1, 0, +1 3 b) p (l = 1) 3, 2, 1, 0, +1, +2, +3 7 c) f (l = 3) sublevel a) s (l = 0) b) d (l = 2) c) p (l = 1) allowable ml 0 2, 1, 0, +1, +2 1, 0, +1 no of possible orbitals 1 5 3
7.56
79
7.57
Plan: The integer in front of the letter represents the n value. The letter designates the l value: l = 0 = s orbital; l = 1 = p orbital; l = 2 = d orbital; l = 3 = f orbital. Solution: 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 orbitals. c) For the 4f subshell, n = 4 and l = 3. Since ml = 3, 2, 1, 0, +1, +2, +3, there are seven orbitals. a) n = 6; l = 4; 9 orbitals b) n = 4; l = 0; 1 orbital c) n = 3; l = 2; 5 orbitals Plan: Allowed values of quantum numbers: n = positive integers; l = integers from 0 to n 1; ml = integers from l through 0 to + l. Solution: 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. c) Combination is allowed. d) With l = 2, ml = 2, 1, 0, +1, +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. 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 Determine the max for carotene by measuring its absorbance in the 610640 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). a) The mass of the electron is 9.1094 x 1031 kg. h2 h2 1 E=  2 =  2 2 2 2 2 8 me a 0 n 8 m e a 0 n = 
7.58
7.59
7.60
7.61
7.62
8 2
( 6.626 x 10 Jis ) ( 9.1094 x 10 kg )(52.92 x 10
34 2 31
12
m
)
2
kgim 2 /s 2 J
1 2 n
1 1 = (2.17963 x 1018 J) 2 = (2.180 x 1018 J) 2 n n This is identical with the result from Bohr's theory. For the H atom, Z = 1 and Bohr's constant = 2.18 x 1018 J. For the hydrogen atom, derivation using classical principles or quantummechanical 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.
710
1 1 E = (2.180 x 1018 J) 2  2 = 3.027778 x 1019 = 3.028 x 1019 J 3 2
c) E =
hc
6.626 x 1034 J is 2.9979 x 108 m/s hc (m) = = = 6.56061 x 107 = 6.561 x 107 m E 3.027778 x 1019 J
(
(
)(
)
)
1 nm m 9 = 656.061 = 656.1 nm 10 m This is the wavelength for the observed red line in the hydrogen spectrum.
( 6.56061 x 10
7
)
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
( 6.626 x 10 a) E (1 photon) = hc/ =
( )
( 6.626 x 10 b) E (1 photon) = hc/ =
( )
J is)(3.00 x 108 m/s 1 nm 19 9 = 2.8397 x 10 J (unrounded) 10 m 700.nm This is the value for each photon, that is, J/photon. 1 photon Number of photons = 2.0 x 1017 J = 70.4296 = 70. photons 19 2.8397 x 10 J J is)(3.00 x 108 m/s 1nm 19 9 = 4.18484 x 10 J (unrounded) 475.nm 10 m This is the value for each photon, that is, J/photon. 1 photon Number of photons = 2.0 x 1017 J = 47.7915 = 48 photons 19 4.18484 x 10 J
34
34
)
)
7.65
Determine the wavelength: = 1/(1953 cm1) = (5.1203277 x 104 cm) (unrounded) 102 m 1 nm 3 (nm) = 5.1203277 x 104 cm = 5120.3277 = 5.120 x 10 nm 1 cm 109 m
(
)
102 m 1 4 () = 5.1203277 x 104 cm 10 = 51203.277 = 5.120 x 10 1 cm 10 m
(
)
= c/ =
2.9979 x 108 m/s 1 cm 1 Hz 13 13 = 5.8548987 x 10 = 5.855 x 10 Hz 5.1203277 x 104 cm 102 m 1 s 1
711
7.66
The Bohr model has been successfully applied to predict the spectral lines for oneelectron species other than H. Common oneelectron 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, solve and Bohr's equation to verify if a whole number for Z 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 highestenergy line corresponds to the transition from n = 1 to n = . E = h = (6.626 x 1034 Js) (2.961 x 1016 Hz) (s1/Hz) = 1.9619586 x 1017 J (unrounded) Z2 E = 2.18 x 1018 J 2 Z = charge of the nucleus n En 2 1.9619586 x 1017 (12 ) = = 8.99998 Z2 = 2.18 x 1018 J 2.18 x 1018 J Then Z2 = 9 and Z = 3. Therefore, the ion is Li2+ with an atomic number of 3.
(
)
7.67
a) 59.5 MHz
(m) = c/ =
2.9979 x 108 m/s = 5.038487 = 5.04 m 106 Hz s 1 ( 59.5 MHz ) 1 MHz Hz
215.8 MHz
2.9979 x 108 m/s = 1.38920 = 1.389 m 106 Hz s 1 ( 215.8 MHz ) 1 MHz Hz Therefore, the VHF band overlaps with the 2.783.41 m FM band. 3.00 x 108 m/s = 545.45 = 550 m b) 550 kHz (m) = c/ = 103 Hz s 1 ( 550 kHz ) 1 kHz Hz
(m) = c/ =
1600 kHz
3.00 x 108 m/s = 187.5 = 190 m 103 Hz s 1 (1600 kHz ) 1 kHz Hz FM width from 2.78 to 3.41 m gives 0.63 m, whereas AM width from 190 to 550 m gives 360 m.
(m) = c/ =
7.68
a) Electron: = h/mv =
b) EK = 1/2 mv2 therefore v2 = 2 EK/m 2 EK v= m Electron: v =
kgim /s ( 6.626 x 10 Jis ) = 2.139214 x 10 = 2.1 x 10 m J ( 9.11 x 10 kg ) 3.4 x 10 s ( 6.626 x 10 Jis ) kgim /s = 1.16696 x 10 = 1.2 x 10 m Proton: = h/mv = m J 1.67 x10 kg ) 3.4 x10 ( s
34 2 2
10
10
m
31
6
34
2
2
13
13
27
6
Proton: v =
( 2 ( 2.7 x 10 J ) kgim /s (1.67 x 10 kg ) J
15 2 27
2 2.7 x 1015 J kgim 2 /s 2 J 9.11 x 1031 kg
2
(
) )
7 = 7.6991 x 10 m/s (unrounded)
6 = 1.7982 x 10 m/s (unrounded)
712
Electron: = h/mv =
kg im /s ( 6.626 x 10 Jis ) = 9.44698 x 10 m J ( 9.11 x 10 kg ) 7.6991 x 10 s kgim /s ( 6.626 x 10 Jis ) Proton: = h/mv = = 2.20646 x 10 m J (1.67 x 10 kg ) 1.7982 x 10 s
34 2 2 31 7 34 2 2 27 6
12
= 9.4 x 1012 m
13
= 2.2 x 1013 m
7.69
The electromagnetic spectrum shows that the visible region goes from 400 to 750 nm (4000 to 7500 ). 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. n = ? n = 1; = 1212.7 (shortest corresponds to the largest E) 1 1 1/ = 1.096776 x 107 m 1 2  2 n 1 n2
(
)
1 1 = 1.096776 x 107 m 1 1 x 1010 m 1212.7
(
) 11  n1
2 2 2
1 1 0.7518456 = 2  2 1 n 2 1 2 = 1 0.7518456 n 2 1 2 = 0.2481544 n 2 n 2 = 4.0297 2
n2 = 2 for line (a) (n = 2 n = 1) n = ? n = 3; = 10,938 (longest corresponds to the smallest E) 1 1 1/ = 1.096776 x 107 m 1 2  2 n 1 n2
(
)
1 1 7 1 = 1.096776 x 10 m 10 10938 1 x 10 m
(
) 31  n1
2 2 2
1 1 0.083357396 = 2  2 3 n 2 1 2 n 2 1 2 n 2 = 0.11111111 0.083357396 = 0.027753715111
n 2 =36.03121 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.
713
7.70
E = hc/
a) = hc/E =
( 6.626 x 10
thus = hc/E
34
( 6.626 x 10 b) = hc/E =
( 6.626 x 10 c) = hc/E =
7.71
J is 3.00 x 108 m/s 1 nm 9 = 432.130 = 432 nm 4.60 x 1019 J 10 m J is 3.00 x 108 m/s 1 nm 9 = 286.4265 = 286 nm 10 m 6.94 x 1019 J
J is 3.00 x 108 m/s 1 nm 9 = 450.748 = 451 nm 10 m 4.41 x 1019 J
34
)(
)
34
)(
)
)(
)
a) The energy of visible light is lower than that of UV light. Thus, metal A must be barium, since of the three metals listed, barium has the smallest work function indicating the attraction between barium's nucleus and outer electron is less than the attraction in tantalum or tungsten. The longest wavelength corresponds to the lowest E = hc/ thus = hc/E energy (work function). Ta: =
6.626 x 1034 J is 3.00 x108 m/s hc = E 6.81 x 1019 J
(
)(
) 1 nm
6.626 x 1034 J is 3.00 x 108 m/s hc Ba: = = E 4.30 x 1019 J
(
)(
= 291.894 = 292 nm 109 m
) 1 nm
= 277.6257 = 278 nm 109 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.
W: =
6.626 x 1034 J is 3.00 x 108 m/s hc = E 7.16 x 1019 J
(
)(
) 1 nm
= 462.279 = 462 nm 109 m
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 Extra significant figures are necessary because of the data presented in the problem. HeNe = 632.8 nm Ar = 6.148 x 1014 s1 ArKr E = 3.499 x 1019 J Dye = 663.7 nm Calculating missing values: Ar = c/ = (2.9979 x 108 m/s)/( 6.148 x 1014 s1) = 4.8762199 x 107 = 4.876 x 107 m ArKr = hc/E = (6.626 x 1034 Js) (2.9979 x 108 m/s)/( 3.499 x 1019 J) = 5.67707 x 107 = 5.677 x 107 m Calculating missing values: HeNe = c/ = (2.9979 x 108 m/s)/[632.8 nm (109 m/nm)] = 4.7375 x 1014 = 4.738 x 1014 s1 ArKr = E/h = (3.499 x 1019 J)/(6.626 x 1034 Js) = 5.28071 x 1014 = 5.281 x 1014 s1 Dye = c/ = (2.9979 x 108 m/s)/[663.7 nm (109 m/nm)] = 4.51695 x 1014 = 4.517 x 1014 s1 Calculating missing E values: HeNe E = hc/ = [(6.626 x 1034 Js) (2.9979 x 108 m/s)]/[632.8 nm (109 m/nm)] = 3.13907797 x 1019 = 3.139 x 1019 J E = h = (6.626 x 1034 Js) (6.148 x 1014 s1) = 4.0736648 x 1019 = 4.074 x 1019 J Ar Dye E = hc/ = [(6.626 x 1034 Js) (2.9979 x 108 m/s)]/[663.7 nm (109 m/nm)] = 2.99293 x 1019 = 2.993 x 1019 J
7.73
714
The colors may be predicted from Figure 7.3 and the frequencies. Orange HeNe = 4.738 x 1014 s1 Green Ar = 6.148 x 1014 s1 Yellow ArKr = 5.281 x 1014 s1 Red Dye = 4.517 x 1014 s1 7.74 The speed of light is necessary; however, the frequency is irrelevant. 103 m 1s = 270 = 2.7 x 102 s a) Time = 8.1 x 107 km 1 km 3.00 x 108 m
(
)
3.00 x 108 m 8 b) Distance = (1.2 s ) = 3.6 x 10 m s
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.
1 1 1/ = 1.096776 x 107 m 1 2  2 n 1 n2
7.76
(
)
a)
1 1 nm = 1.096776 x 107 m 1 94.91 nm 109 m
(
) 11  n1
2 2 2
1 1 0.9606608 = 2  2 1 n 2 n2 = 5 1 nm 1 7 1 b) = 1.096776 x 10 m 1281 nm 109 m
(
) n1  51
2 1 2
1 1 0.071175894 = 2  2 n 1 5 n1 = 3
1 1 c) 1/ = 1.096776 x 107 m 1 2  2 1 3 1/ = 9.74912 x 106 m1 (unrounded) = (1/9.74912 x 106 m1) (1 nm/109 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 the equation for the energy of an electron transition to find the quantity of energy needed to remove completely the electron, called the ionization energy (IE). The charge on the 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 oneelectron species. The general expression is: 1 1 2 6.022 x 1023 E = 2.18 x 1018 J 2  2 Z 1 mol n initial 6 6 2 = 1.312796 x 10 = (1.31 x 10 J/mol) Z for n = 1
(
)
715
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). Z = 5 for B4+. E = IE = (1.312796 x 106 J/mol)( 52) = 3.28199 x 107 = 3.28 x 107 J/mol 1 1 1 1 c) E = ( 2.18 x 10 18 J ) 2  2 Z 2 = 2.18 x1018 J 2  2 Z2 n initial 3 = (2.42222 x 1019 J)Z2 for n = 3 Z = 2 for He+ E = IE = (2.42222 x 1019 J)(22) = 9.68889 x 1019 J
(
)
= hc/E =
( 6.626 x 10
J is 3.00 x 108 m/s 1 nm 9 = 205.162 = 205 nm 9.68889 x 1019 J 10 m
34
)(
)
( 6.626 x 10 = hc/E =
7.78
1 1 2 18 d) E = 2.18 x 1018 J 2  2 Z = 2.18 x10 J n initial (Z = 4 for Be3+) = (5.45 x 1019 J)Z2 for n = 2 19 2 18 E = IE = (5.45 x 10 J)(4 ) = 8.72 x 10 J
(
)
(
) 1
2

1 2 Z 22
J is 3.00 x 108 m/s 1 nm 9 = 22.79587 = 22.8 nm 10 m 8.72 x 1018 J
34
)(
)
a) Orbital D has the largest value of n, given that it is the largest orbital. b) l = 1 indicates a p orbital. Orbitals A and C are p orbitals. l = 2 indicates a d orbital. Orbitals B and D are d orbitals. c) In an atom, there would be four other orbitals with the same value of n and the same shape as orbital B. There would be two other orbitals with the same value of n and the same shape as orbital C. d) Orbital D has the highest energy and Orbital C has the lowest energy. 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. n 2 h 2 0 rn = me e 2 Solution: 2 C2 12 6.626 x 1034 J is 8.854 x 1012 J im kg im 2 /s 2 11 11 a) r1 = = 5.2929377 x 10 = 5.293 x 10 m 2 J 9.109 x 1031 kg 1.602 x 1019 C
7.79
(
)
(
)(
)
2 C2 102 6.626 x 1034 J is 8.854 x 1012 J im kg im 2 /s 2 b) r10 = 2 J 31 19 9.109 x 10 kg 1.602 x 10 C
(
)
(
)(
)
9 9 = 5.2929377 x 10 = 5.293 x 10 m
7.80
Plan: See problem 7.79. Solution: n 2 h 2 0 rn = me e 2
2 C2 32 6.626 x 1034 J is 8.854 x 1012 J im kg im 2 /s 2 a) r3 = 2 J 31 19 9.109 x 10 kg 1.602 x 10 C
(
)
(
)(
)
10 10 = 4.76364 x 10 = 4.764 x 10 m
716
b) Z = 1 for an H atom
En = En =
 Z2
 12
( )( 2.18 x 10 J )
18
32 c) Z = 3 for a Li atom
( )( 2.18 x 10 J ) = 2.42222 x 10
18
n2
19
= 2.42 x 1019 J
En = En =
 Z2
 32
( )( 2.18 x 10 J )
18
J 3 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 orbital becomes more negative with increasing atomic number.
2
( )( 2.18 x 10 J ) = 2.18 x 10
18
n2
18
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/gC) (252 g) (98 20.)C = 8.22407 x 104 J (unrounded)
E=
hc
( 6.626 x 10 =
34
J is 3.00 x 108 m/s
2
)(
)
1.55 x 10 m
= 1.28245 x 1023 J/photon (unrounded)
1 photon 27 27 Number of photons = 8.22407 x104 J = 6.41278 x 10 = 6.4 x 10 photons 1.28245 x 1023 J
(
)
7.82
One sample calculation will be done using the equation in the book:
1 1 = a0
3 2
e
r
a0
1 1 = 52.92 pm
3
2
e
r
a0
= 1.46553 x 103 e
r
a0
(unrounded)
r (pm) 0 50 100 200
(pm3/2) 1.47 x 103 0.570 x 103 0.221 x 103 0.0335 x 103
2 (pm3) 2.15 x 106 0.325 x 106 0.0491 x 106 0.00112 x 106
4r22 (pm1) 0 1.02 x 102 0.616 x 102 0.0563 x 102
717
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 R = 1.096776 x 107 m1 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 1 = 1.096776 x 107 m 1 2  2 1 2 7 = 1.215684272 x 10 = 1.215684 x 107 m Shortest wavelength in the n1 = 2 series: 1 1 1 = 1.096776 x 107 m 1 2  2 2 = 3.647052817 x 107 = 3.647053 x 107 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. 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 1 = 1.096776 x 107 m 1 2  2 3 4 6 = 1.875627163 x 10 = 1.875627 x 106 m Shortest wavelength in n1 = 4 series has n2 = . 1 1 1 = 1.096776 x 107 m 1 2  2 4 6 = 1.458821127 x 10 = 1.458821 x 106 m
1 = 1.096776 x 107 m 1
(
) n1  n1
2 1 2 2
(
(
)
)
(
)
(
)
718
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 1 = 1.096776 x 107 m 1 2  2 5 = 2.27940801 x 106 = 2.279408 x 106 m Similarly for the other combinations: For n1 = 4, n2 = 5, = 4.052281 x 106 m For n1 = 4, n2 = 6, = 2.625878 x 106 m For n1 = 4, n2 = 7, = 2.166128 x 106 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 moreorless 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 1019 J) ( 20 x 1019 J) = 5 x 1019 J = E/h = (5 x 1019 J)/(6.626 x 1034 Js) = 7.54603 x 1014 = 8 x 1014 s1 = c/ = (3.00 x 108 m/s)/(7.54603 x 1014 s1) = 3.97560 x 107 = 4 x 107 m b) The ionization energy (IE) is the same as the reverse of E1. Thus, the value of the IE is 20 x 1019 J/atom. IE = (20 x 1019 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). E = E6 E4 = hc/ = ( 2 x 1019 J) ( 11 x 1019 J) = 9 x 1019 J
( 6.626 x 10 = hc/E =
34
(
J s 3.00 x 108 m/s 1 nm 2 9 = 220.867 = 2 x 10 nm 10 m 19 9 x 10 J
)(
)
)
7.85
color (Figure7.3)
c =
=
c
a) b) c) d) 7.86
red blue orangered yelloworange
= = = =
3.00 x 108 m/s 1 nm 14 14 1 9 = 4.4709 x 10 = 4.47 x 10 s 10 m 671 nm 3.00 x 108 m/s 1 nm 14 14 1 9 = 6.5789 x 10 = 6.58 x 10 s 10 m 456 nm 3.00 x 108 m/s 1 nm 14 14 1 9 = 4.622496 x 10 = 4.62 x 10 s 10 m 649 nm 3.00 x 108 m/s 1 nm 14 14 1 9 = 5.0933786 x 10 = 5.09 x 10 s 10 m 589 nm
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 = (4.854 x 1017 J) (4.098 x 1017 J) = 7.56 x 1018 J =
6.626 x 1034 J is 3.00 x 108 m/s hc = = 2.629365 x 108 = 2.63 x 108 m 18 E 7.56 x 10 J
(
(
)(
)
)
719
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 = (1.024 x 1017 J) + (4.098 x 1017 J) = 5.122 x 1017 J =
6.626 x 1034 J is 3.00 x 108 m/s hc = = 3.88091 x 109 = 3.881 x 109 m 17 E 5.122 x 10 J
(
(
)(
)
)
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 = (5.242 x 1017 J) (5.122 x 1017 J) = 1.2 x 1018 J =
6.626 x 1034 J is 3.00 x 108 m/s hc = = 1.6565 x 107 = 1.66 x 107 m 18 E 1.2 x 10 J
(
(
)(
)
)
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 lightabsorbing pigment, would lead to a darker green color (and vice versa). 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 1031 kg. 2 1 J 9.109 x 1031 kg 6.40 x 105 m/s EK = 1/2 mu2 = kg im 2 /s 2 2 = 1.86552 x 1019 = 1.87 x 1019 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.)
7.88
(
)(
)
E = hc/ =
( 6.626 x 10
34
J is 2.9979 x 108 m/s 1 nm 19 9 = 5.447078 x 10 J (unrounded) 10 m ( 358.1 nm )
)(
)
K = h  EK = (5.447078 x 1019 J) (1.86552 x 1019 J) = 3.581558 x 1019 = 3.58 x 1019 J 7.89
( 6.626 x 10 Jis ) a) = h/mv = ( 9.109 x 10 kg ) 5.5 x 10
34 34
kgim 2 /s 2 11 = 2.424708 x 10 m (unrounded) J 31 7 m s Smallest object = /2 = (2.424708 x 1011 m)/2 = 1.212354 x 1011 = 1.2 x 1011 m
( 6.626 x 10 Jis ) b) = h/mv = ( 9.109 x 10 kg ) 3.0 x 10
kg im 2 /s 2 8 = 1.322568 x 10 m (unrounded) m J 31 4 s Smallest object = /2 = (1.322568 x 108 m)/2 = 6.61284 x 109 = 6.6 x 109 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 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.
720
SrCl2
1 mol SrCl2 6.022 x 1023 photons Number of photons = ( 5.00 g SrCl2 ) 1 mol SrCl 2 158.52 g SrCl2 = 1.8994449 x 1022 photons (unrounded)
J is 3.00 x 108 m/s 1 nm 1 kJ 9 3 10 m 10 J 641 nm 22 = 3.10109 x 10 kJ/photon (unrounded) Etotal = (1.8994449 x 1022 photons) (3.10109 x 1022 kJ/photon) = 5.89035 = 5.89 kJ (Sr) BaCl2 1 mol BaCl2 6.022 x 1023 photons Number of photons = ( 5.00 g BaCl2 ) 1 mol BaCl2 208.2 g BaCl2 = 1.44620557 x 1022 photons (unrounded)
Ephoton =
hc
=
( 6.626 x 10
34
)(
)
J is 3.00 x 108 m/s 1 nm 1 kJ 9 3 10 m 10 J 493 nm 22 = 4.0320487 x 10 kJ/photon (unrounded) Etotal = (1.44620557 x 1022 photons) (4.0320487 x 1022 kJ/photon) = 5.83117 = 5.83 kJ (Ba)
Ephoton =
hc
=
( 6.626 x 10
34
)(
)
7.91
a) The highestenergy line corresponds to the shortest wavelength. The shortestwavelength line is given by 1 1 1 R = 1.096776 x 107 m1 = R 2  2 n n2 1 1 = 1.096776 x 107 m 1
(
) n1  n1
2 1 2 2
304692 m1 = (1.096776 x 107 m1) (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 1 = 1.096776 x 107 m 1 2  n1 ( n + 1)2 1
1 nm 1 7 1 = 1.096776 x 10 m 3282 nm 109 m
(
) n1  1
2 1
2
(
)
1 1 1 1 nm 7 1 9 = 1.096776 x 10 m 2  n1 ( n + 1)2 7460 nm 10 m 1 1 1 134048 m1 = 1.096776 x 107 m 1 2  n1 ( n + 1)2 1 1 1 (134048 m1)/(1.096776 x 107 m1) = 0.0122220 = 2  2 n1 ( n + 1) 1 Rearranging and solving this equation for n1 yields n1 = 5. (You and your students may well need to resort to trialanderror solution of this equation!)
(
)
(
)
721
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 fishliver 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 103 g) is dissolved in a known volume of solvent (250. mL) to give a standard concentration with a known response (1.018 units). This can be used to find the unknown quantity of Vitamin A that gives a response of 0.724 units. An equality can be made between the two concentrationtoabsorbance ratios: A1 A = 2 C1 C2
A1C 2 = A2
C1 =
( 0.724 )
1.67 x 103 g 250. mL
(1.018)
= 4.7508 x 106 g/mL (unrounded)
4.7508 x 106 g Vitamin A ( 500. mL ) 1 mL = 1.92808 x 102 = 1.93 x 102 g Vitamin A/g Oil ( 0.1232 g Oil )
7.93
( 6.626 x 10 Jis )(3.00 x 10 = hc/E = ( 7.59 x 10 J )
34 19
8
m/s 1 nm 9 = 261.897 = 262 nm 10 m
)
Silver is not a good choice for a photocell that uses visible light because 262 nm is in the ultraviolet region. 7.94 a) Absorbance vs Concentration:
0.5 Absorbance 0.4 0.3 0.2 0.1 0 0 0.00001 0.00002 0.00003 0.00004 Concentration (M)
This is a linear plot, thus, the graph is of the type y = mx + b, with m = slope and b = intercept. 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 = (y2 y1)/(x2 x1) = = 13250 = 1.3 x 104/M 5 5 3.0 x 10  1.0 x10 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 105 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 105 M (unrounded) = 1.8 x 105 M This value is Mf in a dilution problem (MiVi = MfVf) with Vi = 20.0 mL and Vf = 150. mL.
722
Thus: Mi = MfVf/Vi = (1.81538 x 105 M) (150. mL)/(20.0 mL) = 1.361538 x 104 = 1.4 x 104 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 =
( 6.626 x 10 Jis ) ( 9.109 x 10 kg ) 1.0169 x 10
34 31 34
kg im 2 /s 2 1 J 9.109 x 1031 kg 2 4.71 x 1015 J
(
)
8 = 1.0169 x 10 m/s (unrounded) kg im 2 /s 2 m J s 12 12 = 7.15323 x 10 = 7.15 x 10 m
8
7.97
The amount of energy is calculated from the wavelength of light:
Ephoton =
hc
( 6.626 x 10 =
J is 3.00 x 108 m/s 1 nm 19 9 = 3.61418 x 10 J/photon (unrounded) 10 m 550 nm
)(
)
1 J/ s 10% 5% Amount of E from the bulb = ( 75 W ) = 0.375 J/s W 100% 100%
1 photon 0.375 J 18 18 Number of photons: = 1.0376 x 10 = 1.0 x 10 photons/s 19 s 3.61418 x 10 J 7.98 a) Sodium ions emit yelloworange light, and potassium ions emit violet light. b) The cobalt glass filter absorbs the yelloworange light, while the violet light passes through. c) Sodium salts used as the oxidizing agents would emit intense yelloworange light which would obscure the light emitted by other salts in the fireworks. a) 6 CO2(g) + 6 H2O(l) C6H12O6(s) + 6 O2(g) Hrxn = [1 mol (HfC6H12O6) + 6 mol (HfO2)] [6 mol (HfCO2) + 6 mol (HfH2O)] 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/ =
7.99
( 6.626 x 10
34
J is 3.00 x 108 m/s 1 nm 19 9 = 2.923235 x 10 J/photon (unrounded) 10 m 680. nm
)(
)
103 J 1 photon Number of photons = ( 2838.7 kJ ) 1 kJ 2.923235 x 1019 J = 9.7108 x 1024 = 9.71 x 1024 photons
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
723
7.101
a) A = (b)c b) b = 4.5 x 103 M1 = slope
0.9 0.7 Absorbance 0.5 0.3 0.1
5 10 20 x 10 Concentration, mol/L A/b = c = (0.55)/(4.5 x 103 M1) = 1.2222 x 104 = 1.2 x 104 M
0
0
724
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