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CHEMISTRY 161 HW#7A 32, 34, 38, 40, 44, 46, 50, 52, 54, 58, 60, 66. 7-32 A photon is the smallest indivisible packet or particle of light energy 7-34 If we simply apply the equation E = hc/, we definitely would think that the photon moving away from us has lost energy. However, we have to consider here the frame of reference. It only appears to us that the photon lost energy due to its motion relative to us. If we were traveling right along with the photon the energy would not change. From the point of view of the photon, its energy has not changed. 7-38 6.24 10 –19 J 9.417 1014 s –1 6.626 10 –34 J s 3.00 108 m/s 3.19 10 –7 m or 319 nm 9.417 1014 s –1 7-40 The energy of light of 500 nm wavelength can be found as 6.626 10 –34 J s 3.00 108 m/s E 3.98 10 –19 J 5.00 10 –7 m This energy is less than the work function for tantalum ( = 6.81 10–19 J), so tantalum could not be used to convert solar energy at 500 nm to electricity. 7-44 Energy of the photons of light at 500 nm: 6.626 10 –34 J s 3.00 108 m/s E = 3.975 10 –19 J/photon 5.00 10 –7 m Number of photons in 1 m3: 1 10 –15 J 1 photon 2520 photons/m3 m3 3.975 10 –19 J 7-46 The principal quantum number, n, describes, in part, the energy of an electron and can be seen as a description of how far away from the nucleus the electron resides (in an orbit). 7-50 Because energy levels in hydrogen become more closely spaced as n increases, ∆E for the energy levels at lower n values must be greater and therefore have higher frequencies associated with them. In order of increasing frequencies: (d) n = 11 to n = 13 < (c) n = 9 to n = 11 < (b) n = 6 to n = 8 < (a) n = 4 to n = 6 7-52 These emissions are in the Lyman series which all fall into the ultraviolet (UV) region of the spectrum. 7-54 Because both atoms have only one electron, there is no e –– e– repulsion in the atoms. Their spectra, therefore, should both be fairly simple, as should emissions based on the transitions from higher to lower n orbits. The He+ ion, however, pulls the electron in closer to the nucleus because of its increased nuclear charge. This shifts all the n orbits to lower energy, and the ∆n energies are observed at shorter wavelengths. 7-58 1 1 E (2.18 10–18 J)(2) 2 2 – 2 1.21 10–18 J 2 3 6.626 10 –34 J s 3.00 108 m/s 1.64 10 –7 m or 164 nm 1.21 10 –18 J This is not in the visible range. Therefore, the transition from higher energy states to n = 2 in He+ will never produce visible light. 7-60 (a) Because this line is probably in the Lyman series, the transition is associated with the emission of light from an electron in an excited state relaxing to the ground state (n = 1). (b) Here, ni = n2. 1 1 1 [1.097 10 –2 (nm) –1 ] 2 2 92.3 nm 1 n 2 1 1 0.988 12 n22 1 0.012 n22 n2 9 (c) The longest wavelength (lowest energy) transition from the ground state (n = 1) is n = 1 to n = 2. 1 1 E 2.18 10–18 J 2 – 2 1.64 10–18 J 2 1 7-66 (a) False. Heavier particles have a shorter wavelength than lighter particles. (b) True. When m2 = 2m1, then 2 h 2m1u 1 1 h m1u 2 (c) True. Doubling the speed gives u2 = 2u1: 2 h m2u1 1 1 h mu1 2 This is the same as doubling the mass in b.