Solutions for Chapter 4 End-of-Chapter Problems Assignment

Solutions for Chapter 4 End-of-Chapter Problems Problem 4.1. (a) Mendeleev could not have based his periodic table on the structure of atoms, because the structure of atoms was not understood until considerably after the time of Mendeleev. The basis for understanding the present periodic arrangement of atoms, the number of protons in the nucleus, was not experimental evidence available to Mendeleev in 1869. (b) Mendeleev’s periodic table was based on empirical correlations among the physical and chemical properties of the elements and their compounds. The experimental properties of the 60 elements known at that time showed repeated patterns based on their relative atomic masses. (c) To align known elements with similar chemical properties in families, Mendeleev had to leave gaps in the first periodic table. Mendeleev assumed that the gaps would be filled by elements that had not yet been discovered and he made predictions about the properties of these yet-to-be-discovered elements. Problem 4.2. The data for this problem are in this table. Atomic Radii (pm) Element Radius Element Radius Element Radius H He Li Be B C 37 32 134 90 82 77 N O F Ne Na Mg 75 73 71 69 154 130 Al Si P S Cl Ar 118 111 105 102 99 97 (a) We see that there is a large jump in the atomic radius going from He to Li and then eight elements later another large jump going from Ne to Na. In each case, the atomic radii of the following elements decreases as the atomic number increases until the next jump in size occurs. On the basis of the data here, we predict that there will be another jump in radius going from Ar to the next element, K, since these are another eight elements after the second jump. (b) An obvious way to plot these data is a graph similar to Figure 4.1 or 4.2 in the text, but with the atomic number (or equal spacing between elements) as the independent variable, as in Figure 4.3. Here is such a plot on which the periodic pattern is quite obvious. (The points are connected only to help visualize the trends; this is not a continuous plot but a series of discrete points with no values between them.) 160 Atomic radius, pm 140 120 100 80 60 40 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 He Li Ne Na Ar 0 Atomic number Another possible way to display the data graphically is to show the relative sizes of the atoms as circles of radii proportional to the atomic radii. Such a graphical array in a pattern that is like a conventional periodic table is shown here: Period 1 2 I 3 II III IV V Family VI VII VIII (c) It is almost always easier to visualize trends from a well-made graphical representation than trying to pick out the trends from a table of data. Exact numerical values are more easily found from a table. (d) The atomic radius of K will be larger than that of Na, just as Na is larger than Li, since the size of the atoms appears to increase down a family of the periodic table. If we look at the differences between the sizes of the atoms in the second and third periods on our plots (or from the table), we see that third period radii are 30-40 pm larger than the second period radii. noble gases, He, Ne, and Ar in these data, we see that the difference in radii between Ne and H is 37 pm and between Ar and Ne is 28 pm. We might conclude that the difference between the radii of K (fourth period) and Na (third period) will be somewhere in this range and would then predict that the radius of K would be about 35 pm larger than that for Na, or about 189 pm. A similar analysis for Be, Mg, and Ca leads to a prediction of approximately 165 pm for the radius of Ca. (e) The data in the table come from the WebElements web site, www.webelements.com, where the values for the covalent atomic radii of K and Ca are 196 pm and 174 pm, respectively. Our predictions in part (d) are in the ballpark, although about 5% low. This relatively good agreement shows the usefulness of periodic trends to make estimates of elemental atomic properties. The data for most elements are on this plot (a screen capture from the web site): Problem 4.3. (a) The energy required to remove a second electron from helium should be considerably greater than that required to remove the first electron. Removing the second electron leaves behind a doubly positively charged ion, He2+. The attraction the electron feels as it is removed is at least twice that for the removal of the first electron, which leaves behind only a singly charged ion, He+. Also, when the first electron is close to the nucleus, it is repelled by the second, and this makes it easier to remove the first electron. The experimental data are: 2372 kJ.mol–1 First ionization energy: He(g) He+(g) + e– Second ionization energy: He+(g) He2+(g) + e– 7300 kJ.mol–1 (b) The second ionization energy for beryllium should be greater than the first because the electron is being removed from the positively charged ion, leaving behind a doubly positively charged ion, as in part (a) for helium. The third ionization energy should be much greater, because this electron is coming from the electron core, not the outer valence electrons. First ionization energy: Be(g) Be+(g) + e– 900 kJ·mol–1 Second ionization energy: Be+(g) Be2+(g) + e– 1760 kJ·mol–1 Third ionization energy: Be2+(g) Be3+(g) + e– 14,850 kJ·mol–1 Problem 4.4. This problem asks you to compare Figure 4.4 with this representation of first ionization energies for the representative (main group) elements (families labeled I through VIII). 250 250 0 0 200 0 20 00 15 15 00 10 10 00 50 00 00 50 0 0 0 0 Pe riod 5 I Pe pI pI II riod 4 Pe pV riod 3 Pe VII I riod 1 Gro up pV Gr 2 Pe II pV I riod ou ou Gr Gr ou pI V Gr ou Gr ou ou ou pI riod Gr Gr 6 Pe Although we may be more familiar with the sort of graphical representation in Figure 4.4, this three-dimensional representation has some added advantages. One of the main advantages is the positioning of the families (groups) that makes it easier to see the relationships within families as well as within periods. This is particularly important in this representation that shows main group elements for six periods, not just the first three, as in Figure 4.4. Both representations show the change in ionization energy within each period and group. It is easier to read numeric values from the two-dimensional graph in Figure 4.4, but it may be easier to observe some apparent irregularities in the patterns with the three-dimensional graph. For example, why is the value for nitrogen so much higher than its neighbors within the second period and within group V? Both representations show the shell structure of atoms and this structure falls naturally out of the two-dimensional plot when the ionization energies are plotted as a function of atomic number. However, in the three-dimensional representation, the repetitive structure (the substantial drops in ionization energies after each noble gas) is imposed on the data in order to position the third dimension (the families) appropriately. Thus, the repeating (shell) structure is used in the construction, rather than appearing naturally. Problem 4.5. The data for this problem are in this table. Second Ionization Energy (IE2), kJ·mol–1 Element H He Li Be B C N IE2 Element IE2 Element IE2 5248 7300 1757 2430 2350 2860 O F Ne Na Mg Al Si 3390 3370 3950 4560 1450 1820 1580 P S Cl Ar K Ca 1904 2250 2297 2666 3052 1145 (a) There is no value for the second ionization energy of the hydrogen atom, because the atom has only one electron; there is not a second one to lose. (b) A plot of the second ionization energy data in the table (together with the data for the first ionization energy from Figure 4.4 and Table 4.2) as a function of atomic number gives: 2nd ionization energy, kJ/mol 8000 7000 1st IE 6000 2nd IE 5000 4000 3000 2000 1000 0 0 5 10 atomic number 15 20 Comparison of the plots of the first and second ionization energies shows that the shape of the two curves is essentially identical, except that the second ionization energy plot is moved over one atomic number higher and the energies on this plot are a good deal higher. It looks like the ionization energies of the mononegative ions from He+ through Ca+, follow exactly the same pattern as the first ionization energies for H through K. Each ion in this series of ions has the same number electrons as the corresponding atom in the other series. That is, He+ has one electron and H has one electron and so on to Ca+ and K, each of which has 19 electrons. The shell structure applies to ions as well as atoms and depends on the number of electrons surrounding the nucleus. The second ionization energies for an elemental atom are all higher than the first, because the electron is being removed from a positive ion and is held more tightly. (Compare with the solution to Problem 4.3.) Problem 4.6. (a) Three adjacent pairs of atomic masses that are “out of order” and one pair that are almost the same are shown on this screen capture of a plot of molar mass as a function of atomic number from the interactive periodic table in the Web Companion. The modern relative atomic masses are shown for each element. (Other pairs at higher atomic number are also “out of order,” but in these cases, one or both of the elements does not exist naturally, so only the isotopes that have been prepared synthetically are available. We do not know what a “natural” distribution of isotopes would be, nor, of course, did Mendeleev know about any of these.) (b) Since Mendeleev ordered the elements by relative atomic mass (or molar mass), he would have problems with the three pairs that are “out of order,” because their chemical properties would not “fit” in the family in which their mass would have put them. This was not a problem for the Ar-K pair, since Ar (and the other noble gases) had not been discovered when Mendeleev created his first table. Note that Mendeleev assigned both of the Co-Ni pair the same relative atomic mass; these were probably the best data available in 1869. For the Te-I pair, Mendeleev gives the correct relative atomic masses (he uses “J,” the German symbol for iodine, since he published in a German journal), but reverses the order given by the atomic mass, in order to put the elements in the correct chemical families (iodine with the other halogens). He indicates his concern about this by placing a question mark after the value of Te, presumably to indicate that this value might be suspect. Problem 4.7. Sodium metal is a solid at room temperature. The metal melts at 97.8 C and boils at 882.9 C. It is not necessary to heat the sample to its boiling point to do the experiment represented in Figure 4.6(b); a vapor pressure less than one atmosphere provides sufficient gaseous atoms to observe the absorption spectrum shown in the figure, so the sample only needs to be heated to a few hundred degrees. Problem 4.8. Atoms strongly heated at the sun’s surface, like the atoms in an electrical discharge tube in Figure 4.6(a), are given a great deal of energy that they lose by emitting light. Atoms further from the sun are not heated strongly enough to make them emit, but they can absorb light energy from the sun, just as the sample of atoms in Figure 4.6(b) does. Problem 4.9. (a), (b), (c) See Figure 4.8 for a representation of a wave with the wavelength, amplitude, and node identified. (d) A cycle is one complete traversal of the wave plot from some point to the first identical point further along the plot. In Figure 4.8, for example, the adjacent maximum amplitudes that denote the wavelength also denote one full cycle of the wave. The entire wave is composed of repeated cycles. Problem 4.10. (a) Because, even at its closest approach to Earth, Mars is 56 million km from Earth, it takes time for a radio message, traveling at the speed of light, 3.00 108 m·s–1, from a space probe on Mars to reach the Earth. The time required is: time = (56 106 km) 1000 m 1 km 1s 3.00 10 8 m = 187 s (a little more than 3 minutes) (b) Controllers on Earth have problems maneuvering a remote-controlled vehicle on the Martian surface because it takes a minimum of about three minutes to get a signal from the vehicle and another three minutes for a command to reach the vehicle. This is one of the reasons that the vehicles are designed to move so slowly. The more recent rovers, Spirit and Opportunity that landed in early 2004, are more sophisticated than the 1997 Sojourner Rover, Figure 1.39, and do some of their own computer-controlled navigation (with less immediate guidance from the Earth-based human controllers), Problem 4.11. c , For each of the frequencies, , calculate the wavelength, , using the relationship where c is the speed of light, 3.00 108 m·s–1, and Figure 4.13 to identify the type of radiation (region of the electromagnetic spectrum). (a) 101 MHz: = 3.00 10 8 m s–1 101 106 s–1 = 297 m (AM radio) 15 3.00 10 8 m s–1 = 25 1015 s –1 = 1.2 10–8 m = 12 nm (ultraviolet) (c) 200 GHz 3.00 10 8 m s–1 = 200 109 s–1 = 1.5 10–3 m = 1.5 mm (microwave) (b) 25 10 Hz Problem 4.12. c , where For each of the wavelengths, , calculate the frequency, , using the relationship c is the speed of light, 3.00 108 m·s–1, and Figure 4.13 to identify the type of radiation (region of the electromagnetic spectrum). (a) 400 nm = 3.00 10 8 m s –1 400 10–9 m = 7.5 10–14 s–1 (visible) (b) 25 km = 3.00 10 8 m s –1 25 10 3 m = 1.2 104 s–1 (radio, very low frequency) = 3.00 10 8 m s –1 4.5 10 –4 m = 6.7 1011 s–1 (radar, UHF-TV) (c) 4.5 10–4 m Problem 4.13. (a) A wave with a wavelength of 0.34 m and a frequency of 0.75 s-1 has a speed, u, of: u = · = (0.75 s-1)(0.34 m) = 0.26 m·s–1 (b) This cannot be a wave of electromagnetic radiation, because electromagnetic waves travel at the speed of light, 3.00 108 m·s–1. Problem 4.14. The wavelength of microwave radiation with a frequency of 1.145 1010 s-1 is: c 2.998 10 8 m s–1 = 1.145 1010 s–1 = 0.02598 m = 25.98 mm Problem 4.15. The frequency corresponding to an ultraviolet wavelength of 280 nm is: c 3.00 10 8 m s –1 = 280 10–9 m = 1.07 1015 s–1 Problem 4.16. The length of a crossbar in an antenna that receives a signal with a frequency of 3 102 MHz should have a length that is approximately the same as the wavelength of this signal. The wavelength is: c 3.00 10 8 m s –1 = 3 108 s –1 =1m The crossbars on TV antennas are 1 to 2 m long, so this answer is in the right ballpark for a VHF-TV antenna. Problem 4.17. (a) A screen capture from the Web Companion, Chapter 4, Section 4.2, page 4 (with the correct set of oscillating arrows representing the electromagnetic waves chosen) is shown here. [NOTE: The oscillating arrows in the Companion are slightly misleading because the blue arrow is 180 out of phase with the representation of the wave. That is, when the blue electric field wave is moving up, as in this screen capture, the blue arrow in the box is pointing down and vice versa. In this picture, the screen capture has been manipulated to flip the blue arrow, so that it corresponds to the position of the motion of the blue wave in the electromagnetic wave representation.] The feedback box in the Companion tells you how the correct set of oscillating arrows is related to the representation of the electromagnetic wave. The electric and magnetic waves oscillate perpendicular to one another, as shown by the arrows. At any instant (as captured here) the amplitudes of the two waves are the same, as shown by the equal lengths of the arrows. (b) The description of electromagnetic radiation given in this chapter of the text is essentially identical to that outlined in part (a). Thus, the description of the relationship between the oscillating arrows in the Companion and the text description, including Figure 4.12 (which is essentially identical to this one in the Companion), is the same as in part (a). Problem 4.18. Figure 4.14 shows that the wave theory (which associates wave amplitude with energy) predicts emission from glowing bodies that is more intense than is observed and does not drop off at shorter wavelengths. Problem 4.19. (a) The results of photoelectric effect experiments showed that the intensity of light did not affect the maximum energy of the electrons being ejected. The maximum energy is a function of the wavelength of the incident light. The number of electrons ejected is a function of the intensity of the incident light. These effects are consistent with a particle description of light in which each particle carries an amount of energy that is determined by the wavelength and the intensity of the light is associated with the number of particles. Each particle (with sufficient energy, that is, short enough wavelength) can eject an electron whose energy is determined by the energy of the particle. The more particles there are (higher intensity), the more electrons can be ejected. (b) The wave theory predicted that, if a light beam shining on a metal surface were bright enough (had a large enough amplitude or intensity), any frequency (wavelength) of light could eject an electron from the metal surface. The experimental results however do not support this prediction, because electrons are ejected only by light of a frequency higher than some minimum that is dependent on the metal being investigated. Problem 4.20. When Einstein analyzed the photoelectric experimental data quantitatively, he found that the maximum energy of the ejected electrons could be expressed in terms of the frequency of the incident light, , and a threshold frequency, 0, as Emax = (constant)( – 0). The numeric value of the constant in this relationship was identical to the numeric value of the constant, h, that Planck had had to use to relate the frequency and energy of light to explain the emission from glowing hot objects. This finding reinforced the Planck hypothesis that light is quantized and comes in discrete energy packets, E = h . Einstein used the term photon to describe the energy packets of light. Photons of higher-frequency radiation have higher energies, and photons of lower-frequency radiation have lower energy. If the photons of the radiation in a photoelectric experiment have less energy than the energy required for an electron to overcome the attractive forces that hold it within the metal, the electron cannot escape from the metal. If the incident photon energy is higher than the threshold energy for ejection, h 0, the maximum energy of the ejected electron is the difference between the incident and threshold energy, h – h 0. Problem 4.21. Two examples of quantization in everyday life are: monetary systems such as that used in the United States where all values are based on the “quantum” of one cent (a penny). [NOTE; For some financial transactions and computations, the system is further subdivided into mils, 1 mil = 0.1 cent, but there is no object that represents a mil, so for actual monetary transactions, the cent is the smallest quantum unit.] the notes of musical instruments constrained to only certain values that depend upon the length of strings (pianos and guitars) or a resonant pipe (flutes and trombones). What others did you think of? Problem 4.22. The energy of a photon with a frequency of 1.255 106 s-1 is: E = h = (6.626 10–34 J·s)( 1.255 106 s-1) = 8.316 10–28 J Problem 4.23. (a) The frequency of the radiation associated with one photon of microwave radiation that has an energy of 1.64 10-24 J is: 1.64 10 –24 J = 2.48 109 s–1 h 6.626 10 –34 J s (b) To compare the frequency of the microwave photon in part (a) with the frequency of green light with a wavelength of 515 nm, we have to calculate the frequency of the green light: = E = = 3.00 10 8 m s –1 515 10 –9 m = 5.83 10–14 s–1 This frequency is approximately 105 or 100,000 times higher than the frequency of the microwave radiation. Note: This frequency could also be found from the energy of green light calculated in Check This 4.33(a) from the relationship E hv . (c) Frequency is directly proportional to energy, so the photons of green light have about 105 or 100,000 times more energy than photons of microwaves. Problem 4.24. (a) The wavelength (in meters) of the radiation used in a 500-MHz nuclear magnetic resonance (NMR) spectrometer is: c 3.00 10 8 m s –1 = = 0.600 m = 600 mm 500 10 6 s –1 (b) This NMR instrument operates in approximately the cellular phone wavelength region of the electromagnetic spectrum. (c) The energy (in kJ·mol–1) corresponding to a frequency of 500 MHz is: E = h NA = (6.626 = 2.00 10–34 J·s)(500 106 s-1)(6.022 1023 mol–1) 1 kJ 1000 J 10–4 kJ Problem 4.25. The energy of one photon and one mole of photons with (a) a frequency of 101 MHz is: E (per photon) = h = (6.626 10–34 J·s)(101 106 s-1) = 6.69 10–26 J E (per mole) = h NA = (6.69 10–26 J) (6.022 1023 mol–1) = 4.03 10–2 J (b) a wavelength of 400 nm is: E (per photon) = hc = 4.97 E (per mole) = hc –34 8 –1 = (6.626 10 J·s)(3.00 10 m s ) 10–19 J NA = (4.97 10–19 J) (6.022 (400 10 –9 m) 1023 mol–1) = 2.99 105 J = 299 kJ Problem 4.26. The energy of a quantum of light with a wavelength of 589 nm (yellow light) is: E = hc –34 8 –1 = (6.626 10 J·s)(3.00 10 m s ) –9 = 3.37 10-19 J (589 10 m) The student’s calculation that gave 4.2 10 J was not correct. Although different amounts of energy can be emitted by the light, depending upon the number of photons emitted, these energies must all be multiples of 3.37 10-19 J. The light source cannot lose energy except in multiples of 3.37 10-19J and the student's answer is not a multiple of 3.37 10-19J. –19 Problem 4.27. The lenses in continuous “photo-gray” or “photo-sun” eyeglasses contain silver chloride crystals evenly distributed in a glass or highly crystalline plastic matrix. As light hits the crystal, it provides the energy to transfer an electron from chloride ion to silver ion, forming tiny specks of silver metal that cause the lens to darken. The chlorine atom is not free to join other chlorine atoms and escape as gaseous chlorine because it is still locked within the crystalline matrix. This reaction is reversible, so in the absence of light, silver metal and chlorine atoms rejoin to form silver chloride crystals, which do not block light. The lens regains its former transparent appearance. Ag Cl light energy (s) Ag (s) silver metal Cl (aq) chlorine atom Problem 4.28. In Investigate This 4.29, we found that red light does not cause the formation of Ag metal in a crystal of silver chloride, whereas blue and green light do cause this transformation. A red safelight is used in a photographic darkroom to prevent exposure of the film to light that will cause further reaction in the film being developed. Problem 4.29. In a Science article, scientists reported variations in the concentrations of thorium and potassium on the moon’s surface. They used gamma ray spectrometry that recorded the gamma ray emission from potassium at approximately 1.4 MeV and from thorium at approximately 2.6 MeV. Calculate the frequencies of these two gamma ray emissions. Calculate the wavelengths 6 of these two gamma ray emissions. Note: 1 MeV = 1 10 eV (electron volt) and -19 1 eV = 1.602 10 J. Answer to 4.29: To find the frequencies and wavelengths of gamma ray emissions from potassium (emission energy of approximately 1.4 MeV) and thorium (emission energy of approximately 2.6 MeV), we need to convert the energies to joules and use the Planck relationship to get the frequency and the relationship among frequency, wavelength, and speed of light to calculate the wavelength. For potassium, the energy conversion and calculations are: E = 1.4 MeV = (1.4 MeV) = E h = 1 10 6 eV 1 MeV –13 2.2 10 J = 3.4 6.626 10 –34 J s 1.602 10 –19 J 1 eV 1020 s–1 = 2.2 -13 10 J c = 8 –1 3.00 10 m s 3.4 1020 s –1 = 8.8 10–13 m For thorium, the energy conversion and calculations are: 1 10 6 eV E = 2.6 MeV = (2.6 MeV) 1 MeV = E h c = 4.2 10 –13 J = 6.3 6.626 10 –34 J s = 3.00 10 8 m s –1 6.3 10 20 s –1 1.602 10 –19 J 1 eV = 4.2 -13 10 J 1020 s–1 = 4.8 10–13 m Problem 4.30. (a) The spectrum of radiation emitted by a substance is called an emission spectrum. When the emitted light is from atoms that have been given energy (excited) by heating or by an electric discharge, only certain wavelengths are present in the spectrum. The emission spectrum consists of several sharp lines. In an excited atom some of the electrons are in higher energy levels of the atom and when they lose energy to go to lower energy levels, the difference in energy between the two levels is emitted in the form of a photon of light with a frequency/wavelength corresponding to this energy. Since only certain energy levels are allowed in the atom, the atomic emissions occur only at discrete energies (frequencies/wavelengths) corresponding to the differences in energy between the energy levels in the atom. (b) When continuous electromagnetic radiation, such as that from a light bulb, passes through a substance, some wavelengths may be absorbed. The spectrum of the radiation that passes through is called an absorption spectrum. The absorption spectrum from gaseous atoms consists of what looks like a continuous spectrum with several black lines where light of these wavelengths has been absorbed by the atoms in the sample. The electrons in the atoms absorb energy and go from lower to higher energy levels. The wavelengths of the absorbed light corresponds to the photons with energies equal to the energy spacing between the lower and higher energy levels. Since only certain energy levels are allowed in the atom, the atomic absorptions occur only at certain energies (frequencies/wavelengths). (c) For a given element, the atomic absorption and emission spectra are complementary. The wavelengths present in the emission spectrum are the same as those missing in the absorption spectrum. In the emission and absorption processes, light of exactly the same wavelength is emitted (when an electron changes from a higher to a lower energy level) or absorbed (when an electron changes from the lower to the higher energy level). Problem 4.31. The intensity of a spectral line in an atomic emission spectrum can be directly related to (iii) the number of electrons undergoing the transition that gives rise to the line. If more electrons undergo a transition, more photons will be emitted and a more intense line will be observed. Only two energy levels are involved in the transition that gives rise to an emission, so phrase (i) cannot be appropriate. The speed of transition between, phrase (ii), has no effect on the intensity of the emission. Indeed, the concept of the speed of the transition may be meaningless. If the electron took a finite amount of time to go from the upper to lower state, it would presumably have different amounts of energy, depending on how far along it was in the transition, and we might expect to observe emissions of different amounts of energy (more than one wavelength). As atoms are probed experimentally at shorter and shorter times, now approaching attoseconds, 1attosecond = 10–18 seconds, we may discover new phenomena associated with atomic energy level transitions, but they do not affect the phenomena observed on the long time scales of atomic emission spectroscopy. Problem 4.32. When electrons are excited from a ground state to an excited state, (iii) energy is absorbed. The excited state is a state with higher energy. An electron needs to absorb energy in order to reach this state. Phrases (i) and (iv) both have to do with emission of energy, which involves transitions from an excited state to a lower energy state (sometimes the ground state), so these are not appropriate to complete the sentence. Energy in the form of heat might be responsible for the excitation of an electron from the ground to the excited state, but the heat energy would have to be absorbed, not released, as phrase (ii) suggests. Problem 4.33. Excited atoms of neon gas inside glass tubing produce the familiar red-orange glow of many “neon” lights. The electrons in the neon atoms are excited by an electrical discharge. The atoms emit light when the high-energy, excited electrons return to lower energy levels. When the emitted light is dispersed by a prism, only certain wavelength are observed. These wavelengths correspond to the differences in energy between the atomic energy levels. Since only certain energy levels are allowed in the atoms, only certain frequencies of light are emitted and the spectrum is not continuous but consists of several sharp lines. Problem 4.34. When you add so much energy to an atom that an electron completely leaves the atom, it becomes an ion (a cation), since the electron is completely removed from the atom. Problem 4.35. (a) In a flame emission test, the yellow flame from a solution containing Na+ ions corresponds to a wavelength of about 600 nm while the violet flame from a solution containing K+ ions corresponds to a wavelength of about 400 nm. (See the visible electromagnetic spectrum in Check This 4.9.) (b) Figure 4.6(a) represents the emission of light from an electrically excited sample of sodium gas. In a flame emission test the heat of the flame excites the atoms/ions in the sample. The energy levels of the atoms/ions are the same, no matter how they are excited, so emissions from the two sources are essentially the same. [More energy can usually be given the excited atoms in an electrical discharge, so, often, more lines are observed in the emission from a discharge than from a flame, but all the lines in the flame emission are also present in the discharge emission.] (c) The emission from the K+ ion solution has the higher energy. Since energy is inversely proportional to wavelength, a lower wavelength corresponds to a higher energy. Problem 4.36. The colors to which the six wavelengths in the emission spectrum of mercury in the visible range correspond are: blue-violet to blue for the 405, 408, and 436 nm lines; green for the 546 nm line; and yellow-green for the 577 and 579 nm lines. (b) Since the energy of an emission is inversely proportional to the wavelength of the emission, the 405 nm wavelength, the shortest wavelength, corresponds to the largest difference in energy between the atomic states of mercury responsible for the emission. (c) The 579 nm wavelength, the longest wavelength, corresponds to the smallest difference in energy between the atomic states of mercury responsible for the emission. Problem 4.37. The line spectra of hydrogen, helium, mercury, and neon support the idea that the energy of electrons in atoms is quantized, because these discrete emission (or absorption) wavelengths represent energy changes of the electrons in the atoms. If the energies in the atoms could take any value, then the emissions would be at all wavelengths (that is, continuous). The fact that the observed emissions are at only certain wavelengths is consistent with an atomic model in which electrons can have only certain energies, that is, that their energies are quantized. Problem 4.38. The electronic energy transitions in this problem refer to the energy levels on this energy level diagram (a) The transition E1 E2 requires absorption of a photon to increase the energy of the electron from the ground state to the first excited state. (b) The transition E2 E1 results in emission of a photon as the electron energy decreases going from the first excited state to the ground state. The wavelength of the emitted photon is the same as the wavelength of the photon absorbed in part (a). (c) The transition E4 E3 results in emission of a photon as the electron energy decreases going from the third to the second excited state. (d) The transition E4 E5 requires absorption of a photon to increase the energy of the electron from the third to the fourth excited state. (e) The transition E6 E2 results in emission of a photon as the electron energy decreases going from the fifth to the first excited state. (f) The transition E6 E1 results in emission of a photon as the electron energy decreases going from the fifth excited state to the ground state. To rank the relative wavelengths of emission and absorption (shortest to longest) we need to rank the energy changes from largest to smallest. We can do this by comparing the distances between levels on the energy level diagram. Going from the highest level, E6, to the lowest level, E1, as in (f) above, is obviously the largest energy change. The changes between energy levels E1 and E2, (a) and (b), are the same; this is the second largest energy change. The change between E6 and E2, (e), brackets the remaining two changes, so it must be the third largest energy change. The remaining two changes both involve energy level E4. It is obvious from the diagram that the energy difference between E4 and E3 is larger than that between E4 and E5, so (c) represents the fourth largest energy change and (d) the fifth largest (or smallest) energy change. Thus, the relative wavelengths of emission and absorption (shortest to longest) are: (f) < (a) = (b) < (e) < (c) < (d) Problem 4.39. In theory, any moving object gives rise to a characteristic matter wave and de Broglie’s equation can be applied to both macroscopic and submicroscopic (atomic/molecular level) objects. However, the wave properties become observable only for submicroscopic objects. The wavelengths of large moving objects, (like vehicles or balls), are so small that they can not be detected by any existing measuring device. Problem 4.40. We use the de Broglie wavelength equation to calculate the wavelength of a baseball of mass 0.5 kg traveling at 30 m·s–1: –34 6.626 10 J s = hm u = =4 0.5 kg 30 m s–1 10–35 m The wavelength of the moving baseball is so short that it is undetectable by any measuring device, so any affect on its behavior is undetectable.