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CHAPTER 1 THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 1.1 1.2 (a) (b) c c 3.00 108 m s1 8.6 1013 s1 3.00 108 m s1 566 10 9 3.5 106 m 3.5 103 nm 5.30 1014 s1 5.30 1014 Hz m (a) Strategy: We are given the wavelength of an electromagnetic wave and asked to calculate the frequency. Rearranging Equation (1.2) of the text and replacing u with c (the speed of light) gives: c Solution: Because the speed of light is given in meters per second, it is convenient to first convert wavelength to units of meters. Recall that 1 nm 1 109 m. We write: 456 nm 1 109 m 456 109 m 4.56 107 m 1 nm Substituting in the wavelength and the speed of light ( 3.00 108 mŹs1 ), the frequency is: c 3.00 108 m s 1 4.56 107 m 6.58 1014 s1 or 6.58 1014 Hz Check: The answer shows that 6.58 1014 waves pass a fixed point every second. This very high frequency is in accordance with the very high speed of light. (b) Strategy: We are given the frequency of an electromagnetic wave and asked to calculate the wavelength. Rearranging Equation (1.2) of the text and replacing u with c (the speed of light) gives: c Solution: Substituting in the frequency and the speed of light ( 3.00 108 mŹs1 ) into the above equation, the wavelength is: c 3.00 108 m s 1 2.45 109 s 1 0.122 m The problem asks for the wavelength in units of nanometers. Recall that 1 nm 1 109 m. CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 0.122 m 1.3 c 1 nm 1 10 9 195 1.22 108 nm m 3.00 108 m s 1 9192631770 s 1 3.26 102 m 3.26 107 nm This radiation falls in the microwave region of the spectrum. (See Figure 1.6 of the text.) 1.4 1.5 1.6 The wavelength is: 1m 6.05780211 107 m 1,650,763.73 wavelengths c = E h 3.00 108 m s 1 6.05780211 10 hc 7 4.95 1014 s 1 or 4.95 1014 Hz m (6.63 1034 J s)(3.00 108 m s1 ) 624 10 9 3.19 1019 J m (a) Strategy: We are given the frequency of an electromagnetic wave and asked to calculate the wavelength. Rearranging Equation (1.2) of the text and replacing u with c (the speed of light) gives: c Solution: Substituting in the frequency and the speed of light (3.00 108 m s-1) into the above equation, the wavelength is: 3.00 108 m s 1 14 1 7.5 10 s 4.0 107 m 4.0 102 nm Check: The wavelength of 400 nm calculated is in the blue region of the visible spectrum as expected. (b) Strategy: We are given the frequency of an electromagnetic wave and asked to calculate its energy. Equation (1.3) of the text relates the energy and frequency of an electromagnetic wave. E h Solution: Substituting in the frequency and Planck's constant (6.63 1034 J s) into the above equation, the energy of a single photon associated with this frequency is: E h (6.63 1034 J s) 7.5 1014 s1 5.0 1019 J Check: We expect the energy of a single photon to be a very small energy as calculated above, 5.0 1019 J. 196 1.7 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD (a) c 3.00 108 m s1 4 1 6.0 10 s 5.0 103 m 5.0 1012 nm The radiation does not fall in the visible region; it is radio radiation. (See Figure 1.6 of the text.) 1.8 (b) E h (6.63 1034 J s)(6.0 104 s-1) 4.0 1029 J (c) Converting to J mol-1: E 4.0 1029 J 6.022 1023 photons 2.4 105 J mol 1 1 photon 1 mol The energy given in this problem is for 1 mole of photons. To apply E h, we must divide the energy by Avogadro’s number. The energy of one photon is: E 1.0 103 kJ 1 mol 1000 J 1.7 1018 J 23 1 mol 1 kJ 6.022 10 photons The wavelength of this photon can be found using the relationship, E (6.63 1034 J s) 3.00 108 m s1 hc E 1.7 1018 J 1.2 107 m hc . 1 nm 1 10 9 1.2 102 nm m The radiation is in the ultraviolet region (see Figure 1.6 of the text). 1.9 1.10 hc E h ( 6.63 1034 J s )( 3.00 108 m s 1 ) ( 0.154 10 9 1.29 1015 J m) c (a) 3.00 108 m s1 3.70 107 m 3.70 102 nm 14 1 8.11 10 s (b) Checking Figure 1.6 of the text, you should find that the visible region of the spectrum runs from 400 to 700 nm. 370 nm is in the ultraviolet region of the spectrum. (c) E h. Substitute the frequency () into this equation to solve for the energy of one quantum associated with this frequency. E h ( 6.63 1034 J s ) 8.11 1014 s1 5.38 1019 J 1.11 First, we need to calculate the energy of one 600 nm photon. Then, we can determine how many photons are needed to provide 4.0 1017 J of energy. CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD The energy of one 600 nm photon is: E hc (6.63 1034 J s)(3.00 108 m s1 ) 600 10 9 3.32 1019 J m The number of photons needed to produce 4.0 1017 J of energy is: (4.0 1017 J) 1.12 1 photon 3.32 10 19 1.2 102 photons J The heat needed to raise the temperature of 150 mL of water from 20C to 100C is: heat (150 ml)(4.184 J ml-1 C-1)(100 20)C 5.0 104 J The microwave will need to supply more energy than this because only 92.0% of microwave energy is converted to thermal energy of water. The energy that needs to be supplied by the microwave is: 5.0 104 J 5.4 104 J 0.920 The energy supplied by one photon with a wavelength of 1.22 108 nm (0.122 m) is: E hc (6.63 1034 J s)(3.00 108 m s1 ) 1.63 1024 J (0.122 m) The number of photons needed to supply 5.4 104 J of energy is: (5.4 104 J) 1.13 1 photon 1.63 1024 J 3.3 1028 photons This problem must be worked to four significant figure accuracy. We use 6.6256 1034 Js for Planck’s constant and 2.998 108 mŹs1 for the speed of light. First calculate the energy of each of the photons. E E hc hc (6.6256 1034 J s)(2.998 108 m s1 ) 589.0 10 9 m (6.6256 1034 J s)(2.998 108 m s1 ) 589.6 10 9 3.372 1019 J 3.369 1019 J m For one photon the energy difference is: E (3.372 1019 J) (3.369 1019 J) 3 1022 J For one mole of photons the energy difference is: 3 1022 J 6.022 1023 photons 2 102 J mol 1 1 photon 1 mol 197 198 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 1.14 The radiation emitted by the star is measured as a function of wavelength (). The wavelength corresponding to the maximum intensity is then plugged in to Wien’s law (Tmax = 1.44 10-2 K m) to determine the surface temperature. 1.15 Wien’s law: Tmax = 1.44 10-2 Km. Solving for temperature (T) yields: T 1.44 10 2 Km 100nm: ultraviolet T 1.44 10 2 K m 1.44 10 5 K 100 10 9 m 300nm: ultraviolet T 1.44 10 2 K m 4.80 10 4 K 300 10 9 m 500nm: visible (green) T 1.44 10 2 K m 2.88 10 4 K 500 10 9 m 800nm: infrared T 1.44 10 2 K m 1.80 10 4 K 800 10 9 m Most night vision goggles work by converting low energy infrared radiation (which our eyes can not detect) to higher energy visible radiation. 1.16 Because the same number of photons are being delivered by both lasers and because both lasers produce photons with enough energy to eject electrons from the metal, each laser will eject the same number of electrons. The electrons ejected by the blue laser will have higher kinetic energy because the photons from the blue laser have higher energy. 1.17 In the photoelectric effect, light of sufficient energy shining on a metal surface causes electrons to be ejected (photoelectrons). Since the electrons are charged particles, the metal surface becomes positively charged as more electrons are lost. After a long enough period of time, the positive surface charge becomes large enough to increase the amount of energy needed to eject electrons from the metal. 1.18 (a) The maximum wavelength is obtained by solving Equation 1.4 when the kinetic energy of the ejected electrons is equal to zero. In this case, Equation 1.4, together with the relation c / , gives E (b) hc 6.626 10 hc J s 3.00 108 m s 1 520 109 m h (kinetic Energy) (kinetic Energy) 34 3.82 1019 J CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 6.626 10 (kinetic Energy) 34 J s 3.00 108 m s 1 9 450 10 m 199 3.8227 10 19 J 5.9 1020 J Please remember to carry extra significant figures during the calculation and then round at the very end to the appropriate number of significant figures. 1.19 The minimum (threshold) frequency is obtained by solving Equation 1.4 when the kinetic energy of the ejected electrons is equal to zero. In this case, Equation 1.4 gives E h 6.626 1034 J s 8.54 1014 s 1 5.66 1019 J 1.20 The emitted light could be analyzed by passing it through a prism. 1.21 Light emitted by fluorescent materials always has lower energy than the light striking the fluorescent substance. Absorption of visible light could not give rise to emitted ultraviolet light because the latter has higher energy. The reverse process, ultraviolet light producing visible light by fluorescence, is very common. Certain brands of laundry detergents contain materials called “optical brighteners” which, for example, can make a white shirt look much whiter and brighter than a similar shirt washed in ordinary detergent. 1.22 Excited atoms of the chemical elements emit the same characteristic frequencies or lines in a terrestrial laboratory, in the sun, or in a star many light-years distant from earth. 1.23 (a) The energy difference between states E1 and E4 is: E4 E1 (1.0 1019 J) (15 1019 J) 14 1019 J (b) hc (6.63 1034 J s)(3.00 108 m s1 ) 1.4 107 m 140 nm 19 E 14 10 J The energy difference between the states E2 and E3 is: E3 E2 (5.0 1019 J) (10.0 1019 J) 5 1019 J (c) The energy difference between the states E1 and E3 is: E1 E3 (15 1019 J) (5.0 1019 J) 10 1019 J Ignoring the negative sign of E, the wavelength is found as in part (a). 1.24 hc (6.63 1034 J s)(3.00 108 m s1 ) 2.0 107 m 200 nm 19 E 10 10 J We use more accurate values of h and c for this problem. 200 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD E 1.25 hc = ( 6.6256 1034 J s )( 2.998 108 m s1 ) 656.3 109 m 3.027 1019 J In this problem ni 3 and nf 5. 1 1 1 1 E RH 1.55 1019 J (2.18 1018 J) 2 2 2 n 2 nf 3 5 i The sign of E means that this is energy associated with an absorption process. hc (6.626 1034 J s)(2.998 108 m s1 ) 1.28 106 m 1.28 103 nm 19 E 1.55 10 J Is the sign of the energy change consistent with the sign conventions for exo- and endothermic processes? 1.26 Strategy: We are given the initial and final states in the emission process. We can calculate the energy of the emitted photon using Equation (1.5) of the text. Then, from this energy, we can solve for the frequency of the photon, and from the frequency we can solve for the wavelength. The value of Rydberg's constant is 2.18 1018 J. Solution: From Equation (1.5) we write: 1 1 E RH n2 nf2 i 1 1 E (2.18 1018 J) 2 4 22 E 4.09 1019 J The negative sign for E indicates that this is energy associated with an emission process. To calculate the frequency, we will omit the minus sign for E because the frequency of the photon must be positive. We know that E h Rearranging this equation and substituting in the known values, E 4.09 1019 J = 6.17 1014 s1 or 6.17 1014 Hz h 6.63 1034 J s We also know that 3.00 108 m s1 14 1 6.17 10 s c . Substituting the frequency calculated above into this equation gives: 4.86 107 m 486 nm CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 201 Check: This wavelength is in the visible region of the electromagnetic region (see Figure 1.6 of the text). This is consistent with the transition from ni 4 to nf 2 gives rise to a spectral line in the Balmer series (see Figure 1.15 of the text). 1.27 The Balmer series corresponds to transitions to the n 2 level. For He: E R He 1 1 2 n2 n i f hc (6.63 1034 J s)(3.00 108 m s1 ) E E For the transition, n 6 2 1 1 E (8.72 1018 J) 1.94 1018 J 2 62 2 1.99 1025 J m 1.94 1018 J 1.03 107 m 103 nm For the transition, n 6 2, E 1.94 1018 J 103 nm For the transition, n 5 2, E 1.83 1018 J 109 nm For the transition, n 4 2, E 1.64 1018 J 121 nm For the transition, n 3 2, E 1.21 1018 J 164 nm For H, the calculations are identical to those above, except the Rydberg constant for H is 2.18 1018 J. For the transition, n 6 2, E 4.84 1019 J 411 nm For the transition, n 5 2, E 4.58 1019 J 434 nm For the transition, n 4 2, E 4.09 1019 J 487 nm For the transition, n 3 2, E 3.03 1019 J 657 nm All the Balmer transitions for He are in the ultraviolet region; whereas, the transitions for H are all in the visible region. Note the negative sign for energy indicating that a photon has been emitted. 1.28 Equation 1.14: rn 0 h2 n2 Z me e 2 for n = 2: r2 (8.8542 1012 C2 J –1 m –1 )(6.626 1034 J s)2 2 2.117x1010 m = 211.7 pm 1 (9.109 1031 kg)(1.602 1019 C)2 for n = 3: r3 (8.8542 1012 C2 J –1 m –1 )(6.626 1034 J s)2 3 4.764x1010 m = 476.4 pm 1 (9.109 1031 kg)(1.602 1019 C)2 2 2 202 1.29 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD ni 236, nf 235 1 1 E (2.18 1018 J) 3.34 1025 J 2 2362 235 hc (6.63 1034 J s)(3.00 108 m s1 ) 0.596 m E 3.34 1025 J This wavelength is in the microwave region. (See Figure 1.6 of the text.) 1.30 Z 2 RH 1 1 n12 n22 1 3.289832496 1015 s –1 2 1.31 1 1 3.289832496x1015 s –1 3.289832496x1015 Hz 1 First convert the parameters to the appropriate units and then use the following equation: (a) 1.50 108 cm m 1.50 106 m 6 1 100 cm 1.50 10 m s s s (b) h mu 6.626 10 –34 J s 4.85 10 –10 m (9.1094 10 –31 kg)(1.50 10 6 m s–1 ) 60.0 g kg 0.0600 kg 6.00 10 2 kg 1 1000 g 1500 cm m 15.00 m 1 1 1 15.00 m s 1.500 10 m s s 100 cm s 6.626 10 –34 J s 7.36 10 -34 m (6.00 10 –2 kg)(1.500 101 m s –1 ) h 6.63 1034 J s 5.65 1010 m 0.565 nm mu (1.675 1027 kg)(7.00 102 m s1 ) 1.32 1.33 Strategy: We are given the mass and the speed of the proton and asked to calculate the wavelength. We need the de Broglie equation, which is Equation 1.20 of the text. Note that because the units of Planck's constant are J s, m must be in kg and u must be in m s1 (1 J 1 kg m2 s2). Solution: Using Equation 1.20 we write: CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 203 h mu 6.63 1034 kg m2 s1 h 1.37 1015 m mu (1.673 1027 kg)(2.90 108 ms1) The problem asks to express the wavelength in nanometers. (1.37 1015 m) 1.34 1 nm 1 10 9 1.37 106 nm m Converting the velocity to units of ms1: 120 mi 1.61 km 1000 m 1 hr 53.7 m s1 1 hr 1 mi 1 km 3600 s 1.35 h 6.63 1034 J s 9.96 1034 m 9.96 1032 cm 1 mu (0.0124 kg)(53.7 m s ) First, we convert mph to m s1. 35 mi 1.61 km 1000 m 1h 16 m s1 1h 1 mi 1 km 3600 s 1.36 6.63 1034 kg m2 s1 h = 1.7 1032 m 1.7 1023 nm 3 1 mu (2.5 10 kg)(16 ms ) From the Heisenberg uncertainty principle (Equation 1.22) we have x p h 2 For the minimum uncertainty we can change the greater than sign to an equal sign. p h 1.055 1034 J s 1 10 –24 kg m s –1 2 x 2 (4 10 –11m) Because p = m u, the uncertainty in the velocity is given by u p 1 10 –24 kg m s –1 1 10 6 m s –1 m 9.1095 10 –31 kg The minimum uncertainty in the velocity is 1 106 m s–1. 1.37 Below are the probability distributions for the first three energy levels for a one-dimensional particle in a box, as shown in Figure 1.24. The average value for x for all of the states for a one-dimensional particle in a box is L/2 (the middle of the box). This is true even for those states (like the second state) that have no probability of being exactly in the middle of the box. 204 1.38 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD The difference between the energy levels n and (n + 1) for a one-dimensional particle in a box is given by Equation 1.31: Enn1 h2 (2n 1) . 8mL2 All we have to do is to plug the numbers into the equation remembering that 1 Å=10-10 m. Enn1 (6.626 10 34 J s)2 [2(2) 1] 1.20 10 18 J 8(9.109 10 31 kg)[5.00 10 10 m]2 Energy is equal to Planck’s constant times frequency. We can rearrange this equation to solve for frequency. 1.39 E 1.20 10 18 J 1.81 1015 s1 1.81 1015 Hz h 6.626 10 34 J s From Equation 1.15 we can derive an equation for the change in energy going from the n=1 to n=2 states for a hydrogen atom: E 3Z 2 e4 me . 32h 2 02 From Equation 1.31 we can derive an equation for the change in energy going from the n=1 to n=2 states for a particle in a box: E 3h 2 . 8mL2 We can combine the two equations and solve for L. 3Z 2 e4 me 3h 2 32h 2 02 8mL2 L 4h 4 02 2h 2 2(6.626 10 34 J s)2 (8.8542 10 12 C2 J –1 m –1 ) 2 0 3.326 10 10 m 2 4 2 Z e me Ze me 1(1.602 1019 C)2 (9.109 1031 kg) CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 205 The radius of the ground state of the hydrogen atom (the Bohr radius) is 5.29 10-11 m (see Example 1.4). This leads to a diameter of 1.06 10-10 m, which is only a third the size of L that we calculated. So, it seems like a crude estimate. 1.40 The average value for x will be greater than L/2. The larger the electric field, the larger the average value of x. 1.41 (a) The number given in front of the designation of the subshell is the principal quantum number, so in this case n 3. For s orbitals, l 0. ml can have integer values from l to l, therefore, ml 0. The electron spin quantum number, ms, can be either 1/2 or −1/2. Following the same reasoning as part (a) (b) 4p: n 4; l 1; ml −1, 0, or 1; ms 1/2, −1/2 (c) 3d: n 3; l 2; ml −2, −1, 0, 1, or 2; ms 1/2, −1/2 An orbital in a subshell can have any of the allowed values of the magnetic quantum number for that subshell. All the orbitals in a subshell have exactly the same energy. 1.42 (a) (b) (c) 2p: n 2, l 1, ml 1, 0, or 1 6s: n 6, l 0, ml 0 (only allowed value) 5d: n 5, l 2, ml 2, 1, 0, 1, or 2 An orbital in a subshell can have any of the allowed values of the magnetic quantum number for that subshell. All the orbitals in a subshell have exactly the same energy. 1.43 The allowed values of l are 0, 1, 2, 3, and 4. These correspond to the 5s, 5p, 5d, 5f, and 5g subshells. These subshells each have one, three, five, seven, and nine orbitals, respectively. 206 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 1.44 For n 6, the allowed values of l are 0, 1, 2, 3, 4, and 5 [l 0 to (n 1), integer values]. These l values correspond to the 6s, 6p, 6d, 6f, 6g, and 6h subshells. These subshells each have 1, 3, 5, 7, 9, and 11 orbitals, respectively (number of orbitals 2l 1). 1.45 There can be a maximum of two electrons occupying one orbital. (a) two (b) six (c) ten (d) fourteen What rule of nature demands a maximum of two electrons per orbital? Do they have the same energy? How are they different? Would five 4d orbitals hold as many electrons as five 3d orbitals? In other words, does the principal quantum number n affect the number of electrons in a given subshell? 1.46 A 2s orbital is larger than a 1s orbital. Both have the same spherical shape. The 1s orbital is lower in energy than the 2s. The 1s orbital does not have a node while the 2s orbital does. 1.47 The two orbitals are very similar in shape, though the 3py is larger and has a higher energy. The 3py orbital will have 1 radial node, in contrast to the 2px orbital, which has no radial nodes. The two orbitals also differ in their orientation: 3py is oriented parallel to the y-axis and 2px is oriented parallel to the x-axis. Can you assign a specific value of the magnetic quantum number to these orbitals? What are the allowed values of the magnetic quantum number for the 2p subshell? 1.48 We can modify Equation 1.15 to determine the ionization energies of single electron atoms. Ionization is the process of removing an electron (exciting it to a infinitely high energy state). Eionization Z 2 e4 me 1 1 Z 2 e4 me 1 8h2 20 n12 8h2 20 n12 He+ has two protons in the nucleus and hence a +2 nuclear charge, making Z=2. Eionization Z 2 e4 me 8h2 02 1 1 2 2 (1.602 1019 C)4 (9.109 1031 kg) 18 n2 8(6.626 10 34 J s)2 (8.8542 10 12 C2 J –1 m –1 )2 2 8.715 10 J 1 1 We need to convert our answer to kJ mol–1. 8.715 1018 J 1 kJ 6.022 1023 3 -1 Eionization 1000 5.248 10 kJ mol 1 mol Li2+ has three protons in the nucleus and hence a +3 nuclear charge, making Z = 3. Eionization Z 2 e4 me 8h2 02 1 1 32 (1.602 1019 C)4 (9.109 1031 kg) 17 n2 8(6.626 1034 J s)2 (8.8542 10 12 C2 J –1 m –1 )2 2 1.961 10 J 1 1 We need to convert our answer to kJ mol–1. 1.961 1017 Eionization 1 J 1 kJ 6.022 1023 4 -1 1000 1.181 10 kJ mol mol 207 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 1.49 Equation 1.15 yields the energy of the photon absorbed for different transitions. For an emission, Equation 1.15 can be written: Ephoton Z 2 e4 me 8h2 02 1 1 n2 n2 final initial Energy equals Planck’s constant times the speed of light divided by the wavelength: E hc . We can use this relationship to modify the above equation. hc Z 2 e4 me 8h 2 02 1 1 n2 n2 final initial We can now solve for one over ninitial. 1 ninitial 1 2 final n 8h 3 02 c Z 2 e4 me 34 3 12 2 –1 –1 2 8 -1 1 8(6.626 10 J s) (8.8542 10 C J m ) 3.0 10 m s 1 0.20 2 2 19 4 31 9 2 5 1 (1.602 10 C) (9.109 10 kg) 434 10 m Hence, ninitial is 5. 1.50 1.51 (a) False. n 2 is the first excited state. (b) False. In the n 4 state, the electron is (on average) further from the nucleus and hence easier to remove. (c) True. (d) False. The n 4 to n 1 transition is a higher energy transition, which corresponds to a shorter wavelength. (e) True. The energy given in the problem is the energy of 1 mole of gamma rays. We need to convert this to the energy of one gamma ray, and then we can calculate the wavelength and frequency of this gamma ray. 1.29 1011 J 1 mol 2.14 1013 J 23 1 mol 6.022 10 gamma rays Now, we can calculate the wavelength and frequency from this energy. E hc hc (6.63 1034 J s)(3.00 108 m s1 ) 9.29 1013 m 0.929 pm 13 E 2.14 10 J and E h 208 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 1.52 We first calculate the wavelength, then we find the color using Figure 1.6 of the text. 1.53 E 2.14 1013 J 3.23 1020 s1 3.23 1020 Hz h 6.63 1034 J s (a) hc (6.63 1034 J s)(3.00 108 m s1 ) 4.63 107 m 463 nm, which is blue. 19 E 4.30 10 J First, we can calculate the energy of a single photon with a wavelength of 633 nm. E hc (6.63 1034 J s)(3.00 108 m s1 ) 633 109 m 3.14 1019 J The number of photons produced in a 0.376 J pulse is: 0.376 J (b) 1 photon 3.14 10 19 1.20 1018 photons J Since a 1 W 1 J s-1, the power delivered per a 1.00 109 s pulse is: 0.376 J 1.00 109 s 3.76 108 J s1 3.76 108 W Compare this with the power delivered by a 100-W light bulb! 1.54 The energy required to heat the water is: mCsT (368 g)(4.184 J g-1 C-1)(5.00C) 7.70 103 J, where m is the mass, Cs the specific heat of water, and T is the change in temperature. The energy of a photon with a wavelength of 1.06 104 nm can be calculated as follows: E h hc (6.63 1034 J s)(3.00 108 m s1 ) 1.88 1020 J 5 1.06 10 m The number of photons required is: (7.70 103 J) 1.55 1 photon 1.88 10 20 4.10 1023 photons J First, let’s find the energy needed to photodissociate one water molecule. 285.8 kJ 1 mol 4.746 1022 kJ 4.746 1019 J 23 1 mol 6.022 10 molecules CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 209 The maximum wavelength of a photon that would provide the above energy is: hc (6.63 1034 J s)(3.00 108 m s1 ) 4.19 107 m 419 nm 19 E 4.746 10 J This wavelength is in the visible region of the electromagnetic spectrum. Since water is continuously being struck by visible radiation without decomposition, it seems unlikely that photodissociation of water by this method is feasible. 1.56 For the Lyman series nf = 1. The longest wavelength emission corresponds to the smallest change in energy (ni 2). By multiplying both sides of Equation 1.5 by Planck’s constant the following equation is obtained: 1 1 1 1 E h hRH 1.64 1018 J (2.18 1018 J) 2 2 2 n 2 nf 12 i hc (6.63 1034 J s)(3.00 108 m s1 ) 1.21 107 m 121 nm 18 E 1.64 10 J For the Balmer series nf = 2, the shortest wavelength emission corresponds to the largest change in energy (ni ). 1 1 1 1 E h hRH 5.45 1019 J (2.18 1018 J) 2 2 2 2 n nf 2 i hc (6.63 1034 J s)(3.00 108 m s1 ) 3.65 107 m 365 nm 19 E 5.45 10 J The longest wavelength emission for the Lyman series (121 nm) is shorter than the shortest wavelength emission for the Balmer series (365 nm). Hence, the two series do not overlap. 1.57 Since 1 W 1 J s-1, the energy output of the 75-W light bulb in 1 second is 75 J. The actual energy converted to visible light is 15 percent of this value (0.15 75 = 11 J). First, we need to calculate the energy of one 550 nm photon. Then, we can determine how many photons are needed to provide 11 J of energy. The energy of one 550 nm photon is: E hc (6.63 1034 J s)(3.00 108 m s1 ) 3.62 1019 J 9 550 10 m The number of photons needed to produce 11 J of energy is: 11 J 1 photon 3.62 1019 J 3.0 1019 photons 210 1.58 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD The energy needed per photon for the process is: 248 103 J 1 mol 4.12 1019 J 1 mol 6.022 1023 photons hc (6.63 1034 J s)(3.00 108 m s1 ) 4.83 107 m 483 nm 19 E (4.12 10 J) Any wavelength shorter than 483 nm will also promote this reaction. Once a person goes indoors, the reverse reaction Ag Cl AgCl takes place. 1.59 Starting with Equation 1.4, h 1 m u2 , 2 e we can set the kinetic energy equal to eV, h eV , divide both sides by h to yield: e V. h h To use the equation above to determine h and we plot as a function of V. The slope will be equal to e/h and the y-intercept will be equal to /h. Before we can plot, we have to calculate from using v=c, where c is the speed of light. nm) 405.0 435.5 480.0 520.0 577.7 650.0 V (volt) 1.475 1.268 1.027 0.886 0.667 0.381 (x1014 sec-1) 7.402 6.884 6.245 5.765 5.189 4.612 h e 1.602 1019 J V 1 6.125x10-34 Js slope 2.6153 1014 s 1 V 1 7.5 10 14 Y = M0 + M1*X M0 3.5314e+14 M1 2.6153e+14 7 1014 R y intercept 6.125 10 34 Js 2.163 10-19 J 1.60 From Equation 1.17: 6.5 10 14 -1 Frequency (sec ) 3.5314 1014 s 1 6.125 10 34 Js 0.99794 6 1014 5.5 10 14 5 1014 4.5 10 14 0.2 0.4 0.6 0.8 1 Voltage (V) 1.2 1.4 1.6 211 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD RH e 4 me (1.60218 1019 C) 4 (9.10939 1031 kg) 3.28986 1015 s –1 . 8h3 02 8(6.62608 1034 J s)3 (8.85419 1012 C2 J –1 m –1 ) 2 For the hydrogen atom the mass of the nucleus is just equal to the mass of a proton. 9.10939 1031 kg 1.67262 1027 kg 9.10443 1031 kg me mN me mN 9.10939 1031 kg 1.67262 1027 kg RH' e4 (1.60218 1019 C) 4 (9.10443 1031 kg) 3.28806 1015 s –1 3 2 8h 0 8(6.62608 1034 J s)3 (8.85419 1012 C2 J –1 m –1 ) 2 Percentage change RH RH' 3.28986 1015 s –1 3.28806 1015 s –1 100 100 0.0547% RH 3.28986 1015 s –1 For the helium-4 isotope the mass of the nucleus is equal to 6.64610-27 kg. 9.10939 1031 kg 6.646 10-27 kg me mN 9.10814 1031 kg me mN 9.10939 1031 kg 6.646 10-27 kg RH' e4 (1.60218 1019 C) 4 (9.10814 1031 kg) 3.28940 1015 s –1 8h3 02 8(6.62608 1034 J s)3 (8.85419 1012 C2 J –1 m –1 ) 2 Percentage change RH RH' 3.28986 1015 s –1 3.28940 1015 s –1 100 100 0.014% RH 3.28986 1015 s –1 That the percent change for a He+ is smaller illustrates that the bigger the difference in the two masses (the mass of the electron and of the nucleus), the closer the reduced mass is to the lighter mass (the mass of the electron). 1.61 (a) For an s orbital, zero. is zero. Hence, the orbital angular momentum for an electron in an s orbital is Bohr’s model only allowed for orbits with angular momentum equal to an integer multiple of h where is defined as Plank’s constant divided by two times pi h 2 (b) 1.62 , The Fe14+ in the corona is presumed to be formed by the ionization of Fe 13+, which takes 3.5104 kJ mol-1 of energy. The energy for the ionization comes from collisions. Hence, for there to be Fe 14+ in the corona, we can assume that the average kinetic energy of the corona gases has to be atleast equal to the ionization energy of Fe13+. 3RT 3.5 104 kJ mol-1 2 212 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD T 2 3.5 104 kJ mol-1 3R 2 3.5 104 kJ mol-1 3 8.31447 103 kJ K -1mol-1 2.8 106 K The actual temperature can be, and most probably is, higher than this. 1.63 Since the energy corresponding to a photon of wavelength 1 equals the energy of photon of wavelength 2 plus the energy of photon of wavelength 3, then the equation must relate the wavelength to energy. energy of photon 1 (energy of photon 2 energy of photon 3) hc Since E hc 1 hc 2 , then: hc 3 Dividing by hc: 1.64 From the Heisenberg uncertainty principle (Equation 1.22) we have x p h 2 For the minimum uncertainty we can change the greater than sign to an equal sign. We can solve for the uncertainty in momentum (p) by dividing both sides by the uncertainty in the position (x). p h 2 x The uncertainty of the oxygen molecules position in any one direction is equal to the diameter of the alveoli (5.0 x 10-5 m). p 2 x 1.055 1034 J s 1.1 1030 kg m s 1 2 (5.0 10 –5 m) Momentum is equal to mass times velocity. If we assume that the there is no uncertainty in the mass, then the uncertainty in the momentum is equal to mass times the uncertainty in the velocity (p = m u). By dividing both sides by the mass, we see that the uncertainty in the velocity is equal to the uncertainty in momentum divided by the mass. u p 1.1 10 –30 kg m s –1 2.0 10 –5 m s –1 m 5.3 10 –26 kg CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD Because we had started by changing the greater than sign in Heisenberg’s equation to an equal sign, the minimum uncertainty in the velocity is 2.0 10-5 m s-1. 1.65 It takes: (5.0 102 g ice) 334 J 1.67 105 J to melt 5.0 102 g of ice. 1 g ice Energy of a photon with a wavelength of 660 nm: E hc (6.63 1034 J s)(3.00 108 m s1 ) 3.01 1019 J 660 109 m Number of photons needed to melt 5.0 102 g of ice: (1.67 105 J ) 1 photon 5.5 1023 photons 19 3.01 10 J The number of water molecules is: 1 mol H 2O 6.022 1023 H2O molecules (5.0 102 g H 2O) 1.7 1025 H 2O molecules 18.02 g H 2O 1 mol H 2O The number of water molecules converted from ice to water by one photon is: 1.7 1025 H 2O molecules 5.5 1023 photons 1.66 31 H 2O molecules per photon Energy of a photon at 360 nm: E h hc (6.63 1034 J s)(3.00 108 m s1 ) 5.53 1019 J 360 109 m Area of exposed body in cm2: 2 1 cm 0.45 m 2 4.5 103 cm 2 1 102 m The number of photons absorbed by the body in 2 hours is: 0.5 2.0 1016 photons 7200 s (4.5 103 cm2 ) 3.2 1023 photons 2 2 hr cm s The factor of 0.5 is used above because only 50% of the radiation is absorbed. 3.2 1023 photons with a wavelength of 360 nm correspond to an energy of: (3.2 1023 photons) 5.53 1019 J 1.8 105 J 1 photon 213 214 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 1.67 The anti-atom of hydrogen should show the same characteristics as a hydrogen atom. Should an anti-atom of hydrogen collide with a hydrogen atom, they would be annihilated and energy would be given off. 1.68 h Looking at the de Broglie equation , the mass of an N2 molecule (in kg) and the velocity of an N2 mu molecule (in m/s) is needed to calculate the de Broglie wavelength of N2. First, calculate the root-mean-square velocity of N2. M(N2) 28.02 g mol1 0.02802 kg mol1 (3) 8.314 J mol1 K 1 300 K urms = 1 0.02802 kg mol 1 516.8 m s1 Second, calculate the mass of one N2 molecule in kilograms. 28.02 g N 2 1 mol N 2 1 kg 4.653 1026 kg 1 mol N 2 6.022 1023 N 2 molecules 1000 g Now, substitute the mass of an N2 molecule and the root-mean-square velocity into the de Broglie equation to solve for the de Broglie wavelength of an N2 molecule. h (6.63 1034 J s-1 ) 2.76 1011 m 26 1 mu (4.653 10 kg)(516.8 m s ) 1.69 Based on the selection rule, which states that l 1, only (b) and (d) are allowed transitions. 1.70 The kinetic energy acquired by the electrons is equal to the voltage times the charge on the electron. After calculating the kinetic energy, we can calculate the velocity of the electrons (KE 1/2mu2). Finally, we can calculate the wavelength associated with the electrons using the de Broglie equation. KE (5.00 103 V) 1.602 1019 J 8.01 1016 J 1V We can now calculate the velocity of the electrons. KE 1 2 mu 2 8.01 1016 J 1 (9.1094 1031 kg)u 2 2 u 4.19 107 m s-1 Finally, we can calculate the wavelength associated with the electrons using the de Broglie equation. CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 1.71 (a) 215 h mu (6.63 1034 J s) 1.74 1011 m 17.4 pm (9.1094 1031 kg)(4.19 107 m s1 ) We use the Heisenberg Uncertainty Principle to determine the uncertainty in knowing the position of the electron. xp h 4 We use the equal sign in the uncertainty equation to calculate the minimum uncertainty values. p mu, which gives: x(mu) x h 4 h 6.63 1034 J s 1 109 m 4 mu 4 (9.1094 1031 kg)[(0.01)(5 106 m s1 )] Note that because of the unit Joule in Planck’s constant, mass must be in kilograms and velocity must be in m s-1. The uncertainly in the position of the electron is much larger than that radius of the atom. Thus, we have no idea where the electron is in the atom. (b) We again start with the Heisenberg Uncertainly Principle to calculate the uncertainty in the baseball’s position. x h 4p x 6.63 1034 J s 7.9 109 m 4 (1.0 107 )(6.7 kg m s1 ) This uncertainty in position of the baseball is such a small number as to be of no consequence. 1.72 (a) The average kinetic energy of 1 mole of an ideal gas is 3/2RT. Converting to the average kinetic energy per atom, we have: 3 1 mol (8.314 J mol1 K 1 )(298 K) 6.171 1021 J atom 1 2 6.022 1023 atoms (b) To calculate the energy difference between the n 1 and n 2 levels in the hydrogen atom, we modify Equation 1.5 of the text. 1 1 1 1 E RH 2 2 (2.180 1018 J) 2 2 1 nf ni 216 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD E 1.635 1018 J (c) For a collision to excite an electron in a hydrogen atom from the n 1 to n 2 level, KE E 3 RT 2 1.635 1018 J NA T 2 (1.635 1018 J)(6.022 1023 mol1 ) 7.90 104 K 3 (8.314 J mol1 K 1 ) This is an extremely high temperature. Other means of exciting H atoms must be used to be practical. 1.73 To calculate the energy to remove at electron from the n 1 state and the n 5 state in the Li2 ion, we use the following equation: 1 1 E RH Z 2 2 2 . nf ni For ni 1, nf , and Z 3, we have: 1 1 E (2.18 1018 J)(3)2 2 2 1.96 1017 J 1 For ni 5, nf , and Z 3, we have: 1 1 E (2.18 1018 J)(3)2 2 2 7.85 1019 J 5 To calculate the wavelength of the emitted photon in the electronic transition from n 5 to n 1, we first calculate E and then calculate the wavelength. 1 1 1 1 E RH Z 2 2 2 (2.18 1018 J)(3)2 2 2 1.88 1017 J 5 nf 1 ni We ignore the minus sign for E in calculating . hc (6.63 1034 J s1 )(3.00 108 m s1 ) E 1.88 1017 J 1.06 108 m 10.6 nm 1.74 The minimum value of the uncertainty is found using xp x(mu) h 4 h and p mu. Solving for u: 4 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 217 h 4 mx u For the electron with a mass of 9.109 × 10 31 kg and taking x as the radius of the nucleus, we find: 6.63 1034 J s 5.8 1010 m s1 31 15 4 (9.109 10 kg)(1.0 10 m) u This value for the uncertainty is impossible, as it far exceeds the speed of light. Consequently, it is impossible to confine an electron within a nucleus. Repeating the calculation for the proton with a mass of 1.673 × 10 27 kg gives: 6.63 1034 J s 3.2 107 m s1 27 15 4 (1.673 10 kg)(1.0 10 m) u While still a large value, the uncertainty is less than the speed of light, and the confinement of a proton to the nucleus does not represent a physical impossibility. The large value does indicate the necessity of using quantum mechanics to describe nucleons in the nucleus, just as quantum mechanics must be used for electrons in atoms and molecules. 1.75 Below are the solutions for the particle in a one dimensional box (Equation 1.35). 2 n (x) L 1/2 n x sin L n 1, 2,3,.... The equation below yields the average value of x, for a particle in a one dimensional box. L L 2 2 n x 2 L n x f x (x) dx x sin2 dx xsin2 dx L L L 0 L 0 0 We can use the following relation obtained from a table of integrals: xsin ax dx 2 x 2 x sin(2ax) cos 2ax 4 4a 8a 2 2n x 2n x 2 x sin cos 2 2 2 x L L 2 n x f x (x) dx xsin dx 2 n L 0 L L 4 n 0 4 8 L L L L We now evaluate this for x equal to L and 0. 218 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 2 2 L 1 1 2 L f 2 2 L 4 n L n 2 8 8 L L The result above shows us that for all states for a particle in a box, the average value of x is L/2. This makes sense, because the squared wavefunctions are all symmetric about x = L/2. The equation below yields the average value of x2, for a particle in a one dimensional box. L L 2 2 n x 2 L 2 2 n x f x 2 (x) dx x 2 sin2 dx L x sin L dx L L 0 0 0 We can use the following relation obtained from a table of integrals: 2 2 x sin ax dx 2 2 x 3 x cos(2ax) 2a x 1 sin 2ax 6 4a2 8a3 2 2n x 2 n x 2 1 sin 2n x x cos L L L L 2 2 L n x 2 x3 f x 2 (x) dx x 2 sin 2 dx L 6 2 3 L 0 L n n 0 4 8 L L We now evaluate this for x equal to L and 0. 3 2 L L 1 2 1 f L 2 L 6 3 2 n n 4 L 2 for n = 1 the average value of x2 is 0.283 L2 for n = 2 the average value of x2 is 0.308 L2 for n = 3 the average value of x2 is 0.328 L2 1.76 The average value of r using Equation 1.41: r rP r dr r 3 R(r) dr 0 2 0 3 2Zr Z 3/2 Zr Z 3 ao ao 3 3 r r R10 (r) dr r 2 e dr 4 r e dr a0 a0 0 0 0 2 From an integral table: 2 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD x e n ax dx 0 n! a n1 To use the above integral to solve our problem, n = 3, x = r, and a Z r 4 a0 219 3 r e 3 2Zr ao 0 Z dr 4 a0 3 2Z , ao 6ao4 3ao 3 ao . 4 16Z 2Z 2 Hence, the average value for r for a 1s orbital is 3 a . Remember that Z is the nuclear charge and is equal to 1 2 o for a hydrogen atom. To calculate the most probable value of r, we set the derivative of the radial probability function to zero and solve for r. 2 2 2Zr Z 3/2 Zr Z 2 ao ao P r r R r r 2 e 4 r e a0 a0 2 2 2 2 Z Z d a P r 4 r 2e o 4 dr a 0 a0 2Zr 2 2 2Zr 2Zr 2Zr Z 2Z 2 ao Z 2 ao 2re ao r e 8 r r e ao ao a0 ao ao the derivative is zero, which corresponds to the most likely value of r. This is consistent Z with ao being the Bohr radius. When r For the 2s orbital The average value of r: r rP r dr r 3 R(r) dr 0 2 0 2 Zr 2 Z Zr Š 2ao r r 3 R10 (r) dr r 3 2 e dr ao 4 a0 0 0 3 Zr Zr Zr Š Š 1 Z 4Z Z 2 5 Š ao ao ao 3 4 r 4 r e dr r e dr a2 r e dr 8 a0 a o o 2 3/2 Using the same integral solution as above, we can solve for the three integrals in the square brackets 4 r e 3 Š Zr ao dr 4 0 4Z ao 4 r e Š Zr ao 6ao4 Z4 dr 24ao4 Z4 5 4 4Z 24ao 96ao 5 4 ao Z Z 220 CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 6 4 Z 2 5 Š ao Z 2 120ao 120ao r e dr ao2 ao2 Z 6 Z4 Zr When we plug in the results for the integrals into the above equation we get that the average distance from the nucleus for a 2s orbital is six times the Bohr radius. Notice that this is four times the average distance from the nucleus for a 1s orbital. 1 Z r 8 a0 3 24ao4 96ao4 120ao4 1 Z 4 4 Z Z 4 8 a0 Z 3 48ao4 ao 4 6 6ao Z Z To calculate the most probable value of r, we can set the derivative of the radial probability function to zero and solve for r. 2 3 Zr Zr 2 Z 3/2 Zr Š 2ao 1 Z 2 4Zr Z 2 r 2 Š ao P r r R r r 2 e r 4 2 e ao 8 a0 ao 4 a0 ao 3 Zr Zr Zr Š 1 Z 4Zr 3 Š ao Z 2 r 4 Š ao a P r 4r 2 e o e 2 e 8 a0 ao ao 2 2 2 2 Zr Zr Zr d 1 Z d 2 Š ao 4Zr 3 Š ao Z 2 r 4 Š ao 4r e P r e 2 e dr 8 a0 dr ao ao Zr d 1 Z Zr 2 4Z 2 Zr 3 Z 2 3 Zr 4 Š ao P r 4 2r 3r 4r e dr 8 a0 ao ao ao ao2 ao 2 2 d 1 Z 16Zr 2 8Z 2 r 3 Z 3r 4 Š ao P r 8r 3 e dr 8 a0 ao ao2 ao Zr Zr 3 2 2 Ša d 1 Z rZ 3 8a 16ao r ao r 3 P r 3 3o r e o dr 8 a0 ao Z Z Z2 2 The above polynomial can be solved computationally to determine three roots. The largest root, which corresponds to the most probable value is r 3 5a 3 5a . Z o o 1.77 1/2 1/2 L n x m x n x m x 2 2 2 sin sin dx sin sin dx 0 L L L L L0 L L L Using a table of integrals, it can be shown that the last integral listed above is zero if n does not equal m, and is L/2 if n equals m. Hence, the wavefunctions for a one-dimensional particle in a box are orthogonal. 1.78 The five lowest energy levels for a two-dimensional particle in a box correspond to the following sets of quantum numbers: (nx=1, ny=1), (nx=1, ny=2), (nx=2, ny=1), (nx=2, ny=2), (nx=3, ny=1) , (nx=1, ny=3) , (nx=3, ny=2) , (nx=2, ny=3). These wavefunctions are plotted below with the lowest energy wavefunction at the bottom. Notice that the states (nx=1, ny=2) and (nx=2, ny=1) have the same energy as do the states (nx=1, ny=3) and (nx=3, ny=1) and the states (nx=3, ny=2) and (nx=2, ny=3). CHAPTER 1: THE QUANTUM THEORY OF THE SUBMICROSCOPIC WORLD 221