Chapter 7Quantum 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.
7.1All types of electromagnetic radiation travel as waves at the same speed.They differ in both their frequency and wavelength.
7.2a) 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 ultravioletand x-ray radiation.
7.3a)Refraction isthe 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 isthe separation of light into its component colors (wavelengths), as when light passes through a prism.
d)Interferenceis 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.
7.4Evidence 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.
7.5In 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, called quanta, could be emitted or absorbed.The magnitude of these gains and losses were whole number multiples of the frequency:E = nh.
7.6Radiation (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.
7.7Plan: 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/ = = 312.5 = 312 m
(nm) = c/ = = 3.125 x 1011 = 3.12 x 1011 nm
(Å) = c/ = = 3.125 x 1012 = 3.12 x 1012 Å
7.8Wavelength and frequency relate through the equation c = . Recall that a Hz is a reciprocal second, or 1/s = s–1.
(m) = c/ = = 3208556 = 3.21 m
(nm) = c/ = = 3.208556 x 109 = 3.21 x 109 nm
(Å) = c/ = = 3.208556 x 1010 = 3.21 x 1010 Å
7.9Plan:Frequency is related to energy through the equation E = h.Note that 1 Hz = 1 s–1.
Solution:
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
7.10E = hc/ = = 1.529 x 10–15 = 1.5 x 10–15 J
7.11Since 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 isred < yellow < blue.
7.12Since 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 1011s–1) or UV > IR > microwave.
7.13Plan: Frequency and wavelength can be calculated using the speed of light: c = .
Solution:
(nm) = c/ = = 1.3482887 x 107 = 1.3483 x 107 nm
(Å) = c/ = = 1.3482887 x 108 = 1.3483 x 108 Å
7.14Frequency and wavelength can be calculated using the speed of light: c = .
a) = c/= = 3.125 x 1013 = 3.1 x 1013 s–1
b) (m) = c/ = = 3.465002 = 3.465 m
7.15Frequency and energy are related by E = h, and wavelength and energy are related by E = hc/.
(Hz) = E/h = = 3.2156 x 1020 = 3.22 x 1020 Hz
(m) = hc/E = = 9.32950x 10–13 = 9.33 x 10–13 m
7.16Plan: 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/= = 1.239669 x 1015 = 1.24 x 1015 s–1
E = hc/ = = 8.2140 x 10–19 = 8.21 x 10–19 J
b) = c/= = 1.3636 x 1015 = 1.4 x 1015 s–1
E = hc/ = = 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, ...
7.18Bohr’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.
7.19An 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.
7.20The 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
7.21The 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.
7.22The 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 = where Z is the atomic 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.
7.23Plan: 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:
R = 1.096776 x 107 m–1
n1 = 2n2 = 5
= = 2303229.6 m–1 (unrounded)
(nm) = = 434.1729544 = 434.17 nm
7.24Calculate 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.
= = 9749120 m–1 (unrounded)
(Å) = = 1025.7336 = 1025.7 Å
7.25Plan: 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:
= = 533155 m–1 (unrounded)
(nm) = = 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.
7.26The 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:
= = 1523300 m–1 (unrounded)
(nm) = = 656.4695 = 656.47 nm
7.27Plan: To find the transition energy, apply equation 7.4 and multiply by Avogadro’s number.
Solution:
E = (Avogadro’s number)
E = = -2.75687 x 105 = -2.76 x 105 J/mol
Check: The value is negative and so light is emitted.
7.28To find the transition energy, apply equation 7.4 and multiply by Avogadro’s number.
E = (Avogadro’s number)
E = = 1.1669 x 106 = 1.17 J/mol
7.29Looking at an energy chart will help answer this question.
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.
7.30b > c > a > d
7.31Plan: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/(9.720 x 10–8 m) =
0.9380 =
= 1 - 0.9380 = 0.0620
= 16.1
n2 = 4
7.32 = (1281 nm) x (10–9 m/1 nm) = 1.281 x 10–6 m
1/ =
1/(1.281 x 10–6 m) =
0.07118 =
= 0.07118 + 0.04000 = 0.11118
= 8.9944
n1 = 3
7.33E = hc/ = = 4.55917 x 10–19 = 4.56 x 10–19 J
7.34Plan: The energy can be calculated from E = hc/.
Solution:
E = hc/ = = 3.37487 x 10–19 = 3.37 x 10–19 J/photon
E = = 203.23 = 203 kJ/einstein
7.35If 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 (2L).
7.36de Broglie’s concept is supported by the diffraction properties of electrons demonstrated in an electron microscope.
7.37Macroscopic objects have significant mass.A large m in the denominator of = h/muwill result in a very small wavelength.Macroscopic objects do exhibit a wavelike motion, but the wavelength is too small for humans to see it.
7.38The 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.
7.39Plan:Part a) uses the de Broglie equation, and part b) uses Heisenberg’s relationship.
Solution:
a) = h/mv =
= 7.562675 x 10–37 = 7.6 x 10–37 m
b)x • mv
x h/4 mv
1.17956 x 10–351 x 10–35 m
7.40a) = h/mv =
= 6.59057 x 10–15 = 6.6 x 10–15 m
b)
x h/4 mv
1.783166 x 10–142 x 10–14 m
7.41 = h/mv
v = h/m = = 2.1717 x 10–26 = 2.2 x 10–26 m/s
7.42 = h/mv
v = h/m = = 4.666197 x 10–23 = 4.67 x 10–23 m/s
7.43Plan: 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
m = h/v = = 3.7498 x 10–36 = 3.75 x 10–36 kg/photon
7.44m = h/v =
= 1.9822 x 10–12 = 1.98 x 10–12 kg/mol
7.45The quantity 2 expresses the probability of finding an electron within a specified tiny region of space.
7.46Since 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.
7.47A 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.
7.48a) 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 orientationof the orbital in space in three-dimensional space.
7.49a) 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.
7.50a) 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.
7.51Magnetic 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
7.52Magnetic 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
7.53(a)(b)
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)
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.
7.55sublevelallowable mlno of possible orbitals
a) d(l = 2)-2, -1, 0, +1, +25
b) p(l = 1)-1, 0, +13
c) f(l = 3)-3, -2, -1, 0, +1, +2, +37
7.56sublevelallowable mlno of possible orbitals
a) s(l = 0)01
b) d(l = 2)-2, -1, 0, +1, +25
c) p(l = 1)-1, 0, +13
7.57a) 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.
7.58a) n = 6;l = 4;9 orbitals
b) n = 4;l = 0;1 orbital
c) n = 3;l = 2;5 orbitals
7.59a) With l = 0, the only allowable ml value is 0.To correct, either change l or ml value.
Correctn = 2, l = 1, ml = -1; n = 2, l = 0, ml = 0.
b) Combination is allowed.
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.
7.60a) 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
7.61Determine 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).
7.62a) The mass of the electron is 9.1094 x 10–31 kg.
E = =
=
= -(2.17963 x 10–18 J) = -(2.180 x 10–18 J)
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.
E = -(2.180 x 10–18 J) = -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 = = 6.56061 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.
7.63a) 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 orAg, the kinetic energy of the ejected electron will be the same.
7.64a)E (1 photon) = hc/ = = 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/ = = 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
7.65Determine 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 104Å
= c/= = 5.8548987 x 1013 = 5.855 x 1013Hz
7.66The 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 = En2/(-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+.
7.67a) 59.5 MHz(m) = c/ = = 5.038487 = 5.04 m
215.8 MHz(m) = c/ = = 1.38920 = 1.389 m
Therefore, the VHF band overlaps with the 2.78-3.41 m FM band.
b) 550 kHz(m) = c/ = = 545.45 = 550 m
1600kHz(m) = c/ = = 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.
7.68a) Electron: = h/mv = = 2.4244 x 10–10 = 2.4 x 10–10 m
Proton: = h/mv = = 1.32255 x 10–13 = 1.3 x 10–13 m
b) EK = 1/2 mv2therefore v2 = 2 EK/m
v =
Electron: v = = 7.4084 x 107 m/s (unrounded)
Proton:v = = 1.7303 x 106 m/s (unrounded)
Electron: = h/mv = = 9.8182 x 10–12 = 9.8 x 10–10 m
Proton: = h/mv = = 2.29305 x 10–13 = 2.3 x 10–13 m
7.69The 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 hasnfinal = 1. Wavelength e is in the infrared region of the spectrum and the infrared series has nfinal = 3.
n = ? n = 1; = 1212.7 Å (shortest corresponds to the largest E)
1/ =
=
0.7518456 =
n2 = 2 for line (a) (n = 2 n = 1)
n = ? n = 3; = 10,938 Å (longest corresponds to the smallest E)
1/ =
=
0.083357396 =
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
7.70E = hc/thus = hc/E
a) = hc/E = = 432.130 = 432 nm
b) = hc/E = = 286.4265 = 286 nm
c) = hc/E = = 450.748 = 451 nm
7.71E = 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).
Ta: = hc/E= = 291.894 = 292 nm
Ba: = hc/E = = 462.279 = 462 nm
W: = hc/E = = 277.6257 = 278 nm
Metal A must be barium, because barium is the only metal that emits in the visible range (462 nm).
b) A UV range of 278nm to 292nm is necessary to distinguish between tantalum and tungsten.
7.72Index of refraction = c/vthus v = c/(index of refraction)
a) Waterv = c/(index of refraction) = (3.00 x 108 m/s)/(1.33) = 2.2556 x 108 = 2.26 x 108 m/s
b) Diamondv = c/(index of refraction) = (3.00 x 108 m/s)/(2.42) = 1.239669 x 108 = 1.24 x 108 m/s
7.73Extra 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-KrE = 3.499 x 10–19 J
Dye = 663.7 nm
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-NeE = 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
ArE = 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
DyeE = 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–1Orange
Ar = 6.148 x 1014 s–1Green
Ar-Kr = 5.281 x 1014 s–1Yellow
Dye = 4.517 x 1014 s–1Red
7.74The speed of light is necessary; however, the frequency is irrelevant.
a)Time = (8.1 x 107 km) (103 m/1km) (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
7.75a) 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.