Chemistry 1000 (Fall 2013) Problem Set #2: Light and Atomic Structure Solutions Answers to Questions in Silberberg (only those w/out answers at the back of the book) 6.1 Microwave and ultraviolet radiation are both forms of electromagnetic radiation. As such, both have wave properties and particle properties. Ultraviolet radiation has a shorter wavelength, higher frequency and higher energy per photon than microwave radiation. 6.5 (a) (b) (c) 6.6 (a) (b) (i) frequency: wave C is lowest then wave B then wave A (ii) energy (per photon): wave C is lowest then wave B then wave A (iii) amplitude: wave B is lowest then wave C then wave A If wave B barely fails to cause a current when shining on a metal, wave C will not cause a current either (since it has less energy per photon). Wave A would be more likely to cause a current (since it has the most energy per photon). If wave B represents visible radiation, wave C is more likely to be IR radiation since IR radiation has a lower frequency than visible radiation. from p. 233: “Einstein proposed that light itself is particulate, quantized into tiny “bundles” of energy, later called photons.” The photoelectric effect exhibits a threshold frequency because the energy of a photon is directly proportional to the light’s frequency (E = hν). If the light does not have a high enough frequency, no individual photon will have enough energy to eject an electron from an atom. So, no current will be observed if the frequency of light is lower than the threshold frequency. The photoelectric effect does not exhibit a time lag because there must be enough energy in a single photon to eject an electron. The effects of multiple photons cannot “build up” or “be added together” in order to provide enough energy for the electron to be ejected (which is what would be implied if a time lag were observed). 6.19 In an absorption spectrum, light of known intensity and wavelength(s) is shone on a sample and we observe the light which is absorbed by it. In other words, we observe which light that does *not* pass through the sample. In an emission spectrum, a sample is excited (can be by light, heat or other forms of energy) then we observe the light emitted as the sample relaxes to a lower energy state. The line spectra Bohr worked with were emission spectra. 6.21 Hydrogen only has one electron. As such, the only electrostatic force that must be considered is the attraction between the nucleus and the electron. All other neutral elements have more than one electron. As such, there are multiple electrostatic forces to consider. Even in helium (2 electrons), there are three forces: attraction between nucleus and electron #1, attraction between nucleus and electron #2, and repulsion between electron #1 and electron #2. 6.34 (a) (b) (c) 6.90 absorptions: A, C, D emissions: B, E, F E (lowest energy), F, B D (highest energy therefore lowest wavelength), A, C low-to-high frequency: Scarlet, Mustard, Green, Peacock, Plum short-to-long wavelength: Plum, Peacock, Green, Mustard (580 nm), Scarlet The dining room indicates a wavelength of 520 nm (Mr. Green), and the lounge and study contain suspects with lower frequencies (Ms. Scarlet and Col. Mustard). The library contains a suspect with the shortest wavelength (Prof. Plum). That leaves Ms. Peacock as the only suspect who could have been in the conservatory at the time of the murder, so she must be the murderer. Additional Practice Problems 1. Which features of the photoelectric effect can’t be explained by classical physics? How does Einstein’s photon hypothesis resolve these issues? see lecture notes and/or textbook 2. What is the stopping potential in the photoelectric effect experiment? What property of the ejected electrons does the stopping potential measure? see lecture notes and/or textbook 3. The threshold wavelength for lanthanum (La) is 376 nm. What is the maximum kinetic energy of electrons ejected from the metal when lanthanum is illuminated with light of wavelength 200 nm? Ek Eincidentlight Ethreshold Ek hc incidentlight hc threshold 1 1 J m 1 1 19 Ek hc 6.626 10 34 2.9979 108 4.65 10 J 7 7 Hz s 2 . 00 10 m 3 . 76 10 m threshold incidentlight 3 sig. fig. 4. Describe the emission spectroscopy experiment. What information do we gain from this experiment? see lecture notes and/or textbook 5. Assume your eyes receive a signal consisting of blue light, = 470 nm. The energy of the signal is 2.50 10–14 J. How many photons reach your eyes? Step 1: Calculate the energy of one photon of blue light ( = 470 nm) E = hυ so: c=υλ and therefore υ = c/ λ E = h c = (6.626 × 10-34 J/Hz)(2.9979 × 108 m/s) × 109 nm = 4.23 × 10-19 J λ 470 nm 1m Step 2: If we know the total energy of the signal and the energy per photon, we can calculate the number of photons: # photons = 6. 2.50 × 10-14 J = 5.92 × 104 photons 4.23 × 10-19 J/photon 3 sig. fig. Excited H atoms give off many different wavelengths of light. One series of wavelengths, called the Pfund series, occurs in the infrared region. It results when an atom drops from a higher energy state to the n = 5 state. Calculate the wavelength and frequency of the lowest energy light of this series. Step 1: The lowest energy light will come from the smallest energy gap. The smallest energy gap between n = 5 and a higher energy level is the gap between n = 5 and n = 6. Therefore, n1 = 5 and n2 = 6. Step 2: We are working with hydrogen, so Z = 1 Step 3: Calculate the energy of the light. En5 RH E n 6 Z2 12 18 8.716 1020 J 2 . 179 10 J 2 2 n 5 Z2 12 18 RH 2 2.179 10 J 2 n 6 6.053 10 20 J 4 sig. fig. E photon En6 En5 6.053 10 20 J 8.716 10 20 J 2.663 10 20 J Step 4: Calculate the frequency (υ) and wavelength (λ) of the light. E h E 2.663 10 20 J 4.019 1013 Hz h 6.626 10 34 J Hz c m c s 7.459 10 6 m 7.459m 13 4.019 10 Hz 2.9979 10 8 4 sig. fig. 7. List the successes and failures of the Bohr atomic theory. see lecture notes and/or textbook
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