Chapter 28 Assignment Solutions Page 770 #23-26, 29-30, 43-48, 55 23) Complete the following concept map using these terms: energy levels, fixed electron radii, Bohr model, photon emission and absorption, energy-level difference. 24) Describe how Rutherford determined that the positive charge in an atom is concentrated in a tiny region, rather than spread throughout the atom. (28.1) He directed a beam of charged α particles at a thin sheet of gold foil and measured the number of particles deflected at various angles. The small but significant number deflected at wide angles indicates a concentrated nucleus. 25) How does the Bohr model explain why the absorption spectrum of hydrogen contains exactly the same frequencies as its emission spectrum? (28.1) Bohr said the energy of an emitted photon or absorbed photon is equal to the change in energy of the atom, which can only have specific values. 26) Review the planetary model of the atom. What are some problems with a planetary model of the atom? (28.1) As the electrons undergo centripetal acceleration, they would lose energy and spiral into the nucleus. In addition, all atoms should radiate at all wavelengths, not discrete wavelengths. 29) How does the Bohr model account for the spectra emitted by atoms? (28.1) Photon wavelengths are determined by the difference in energies of allowed levels as electrons jump inward to stationary states.
30) Explain why line spectra produced by hydrogen gas-discharge tubes are different from those produced by helium gas-discharge tubes. (28.1) Each element has a different configuration of electrons and energy levels. 43) A calcium atom drops from 5.16 ev above the ground state to 2.93 ev above the ground state. What is the wavelength of the photon emitted? λ hc = (6.626 10 34 m 2 kg/s)(3.00 10 8 m/s)(10 9 nm/m) 1.60 10 19 J/eV λ = hc ΔE = 1240 ev nm = 556 nm 5.16 ev 2.93 ev = 1240 ev nm 44) A calcium atom in an excited state, E 2, has an energy level 2.93 ev above the ground state. A photon of energy 1.20 ev strikes the calcium atom and is absorbed by it. To what energy level is the calcium atom raised? Refer to Figure 28-22. 2.93 ev + 1.20 ev = 4.13 ev = E 3 45) A calcium atom is in an excited state at the E 6 energy level. How much energy is released when the atom drops down to the E 2 energy level? Refer to Figure 28-22. E 6 E 2 = 5.16 ev 2.93 ev = 2.23 ev 46) A photon of orange light with a wavelength of 6.00 10 2 nm enters a calcium atom in the E 6 excited state and ionizes the atom. What kinetic energy will the electron have as it is ejected from the atom? The kinetic energy of the electron will equal the energy of the incoming photon minus the energy required to ionize the atom (the difference between E 6 and ionization). E photon = hc λ = (6.63 10 34 J s)(3.00 10 8 m/s) 6.00 10 7 m E photon = (3.314 10 19 1 ev 1.60 10 19 J ) E photon = 2.07 ev E ionize = 6.08 ev 5.16 ev = 0.92 ev K electron = E photon E ionize K electron = 2.07 ev 0.92 ev K electron = 1.15 ev
47) Calculate the energy associated with the E 7 and the E 2 energy levels of the hydrogen atom. E n = ( 13.6 ev) ( 1 n 2) E 7 = ( 13.6 ev) ( 1 72) = 0.278 ev E 2 = ( 13.6 ev) ( 1 22) = 3.40 ev 48) Calculate the difference in energy levels in the previous problem. E 7 E 2 = ( 0.278 ev) ( 3.40 ev) = 3.12 ev 55) A hydrogen atom emits a photon of wavelength of 94.3 nm when it falls to the ground state. From what energy level did the electron fall? ΔE = hc λ = (6.626 10 34 J s)(3.00 10 8 m/s) 94.3 10 9 m ΔE = (2.11 10 18 1 ev ΔE = 13.16 ev E 1 = ( 13.6 ev) ( 1 12) = 13.6 ev ΔE = E 1 E n E n = E 1 ΔE = ( 13.6 ev) ( 13.16 ev) E n = 0.44 ev E n = ( 13.6 ev) ( 1 n 2) 0.44 ev = ( 13.6 ev) ( 1 n 2) n 2 = 13.6 ev 0.44 n 2 = 30.8 n = 5.5 The energy level must be a whole number, so n = 6. (The actual wavelength of light from E 6 to E 1 is 93.8 nm, while for E 5 to E 1 it is 95.0 nm. Hydrogen would not release light at 94.3 nm.)
Page 770 #31-33, 41-42, 57-60 31) Lasers: A laboratory laser has a power of only 0.8 mw (8 10 4 W). Why does it seem more powerful than the light of a 100-W lamp? (28.2) The laser light is concentrated into a narrow beam, rather than being spread over a wide area. 32) A device similar to a laser that emits microwave radiation is called a maser. What words likely make up this acronym? (28.2) Microwave Amplification by Stimulated Emission of Radiation 33) What properties of laser light led to its use in light shows? (28.2) Lasers are directional, and they are single, pure colors. 41) Compare the quantum mechanical theory of the atom with the Bohr model. The Bohr model has fixed orbital radii. The present model gives a probability of finding an electron at a location. The Bohr model allows for calculation of only hydrogen atoms. The present model can be used for all elements. 42) Given a red, green, and blue laser, which produces photons with the highest energy? Blue light has the highest frequency and shortest wavelength, and therefore has the highest energy. 57) CD Players: Gallium arsenide lasers are commonly used in CD players. If such a laser emits at 840 nm, what is the difference in ev between the two lasing energy levels? 840 10 9 m ΔE = (2.37 10 19 1 ev ΔE = 1.5 ev 58) A GaInNi laser lases between energy levels that are separated by 2.90 ev. a) What wavelength of light does it emit? λ λ = hc 1240 ev nm = ΔE 2.90 ev λ = 428 nm b) In what part of the spectrum is this light? Blue
59) A carbon dioxide laser emits very high-power infrared radiation. What is the energy difference in ev between the two lasing energy levels? Consult Table 28-1. 10600 10 9 m ΔE = (1.88 10 20 1 ev ΔE = 0.117 ev 60) The power in a laser beam is equal to the energy of each photon times the number of photons per second that are emitted. a) If you want a laser at 840 nm to have the same power as one at 427 nm, how many times more photons per second are needed? Since E = hc 427, the ratio of energy in each photon is = 0.508. Therefore, the ratio of number of λ 840 1 photons per second is = 840 = 1.97. 0.508 427 The 840 nm laser would have to have 1.97 times more photons per second as the 427 nm laser. b) Find the number of photons per second in a 5.0-mW 840-nm laser. P = E t = E photon photons s Let n be the number of photons per second. Let E p be the energy per photon. Then, n = P E p E p = hc 840 10 9 m E p = 2.4 10 19 J/photon n = P E p = 5.0 10 3 J/s 2.4 10 19 J/photon n = 2.1 10 16 photons/s