Answer Key Chapter 3 Worksheet 1 PDF
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Kennesaw State University
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This document is a physics worksheet focusing on waves and energy, particularly concerning the behavior of light. The worksheet contains questions and their corresponding answers relating to definitions of amplitude, wavelength and frequency of waves, the energy of photons, and examples on relevant topics. The worksheet covers concepts like the de Broglie wavelength and Bohr model of the atom in the atomic physics section.
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Answer Key: Chapter 3 Worksheet 1 Match the labeled portions of the diagram of a wave to the correct definitions. 1. Amplitude: C 2. Wavelength: A 3. Frequency: D 4. If the wavelength of a photon of visible light is 546 nm, what is its frequency (GHz)?...
Answer Key: Chapter 3 Worksheet 1 Match the labeled portions of the diagram of a wave to the correct definitions. 1. Amplitude: C 2. Wavelength: A 3. Frequency: D 4. If the wavelength of a photon of visible light is 546 nm, what is its frequency (GHz)? 𝑐𝑐 = 𝜈𝜈 × 𝜆𝜆 10−9 𝑚𝑚 (3 × 108 𝑚𝑚⁄𝑠𝑠) = 𝜈𝜈 × (546 𝑛𝑛𝑛𝑛 × ) 1 𝑛𝑛𝑛𝑛 𝟏𝟏 𝑮𝑮𝑮𝑮𝑮𝑮 𝝂𝝂 = 𝟓𝟓. 𝟒𝟒𝟒𝟒 × 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝑯𝑯𝑯𝑯 × = 𝟓𝟓. 𝟒𝟒𝟒𝟒 × 𝟏𝟏𝟏𝟏𝟓𝟓 𝑮𝑮𝑮𝑮𝑮𝑮 𝟏𝟏𝟏𝟏𝟗𝟗 𝑯𝑯𝑯𝑯 5. A microwave gives off a photon with a wavelength of 8.73 cm. What is the energy (J) of that photon? ℎ𝑐𝑐 𝐸𝐸 = ℎ𝜈𝜈 = 𝜆𝜆 (6.626 × 10−34 𝐽𝐽𝐽𝐽)(3 × 108 𝑚𝑚⁄𝑠𝑠) 𝐸𝐸 = 1 𝑚𝑚 8.73 𝑐𝑐𝑐𝑐 × 100 𝑐𝑐𝑐𝑐 𝑬𝑬 = 𝟐𝟐. 𝟐𝟐𝟐𝟐 × 𝟏𝟏𝟏𝟏−𝟐𝟐𝟐𝟐 𝑱𝑱 6. 2 moles of photons have a total energy of 1.13 J. What is the frequency (Hz) of one photon? 1.13 𝐽𝐽 1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 × = 9.38 × 10−25 𝐽𝐽 𝑖𝑖𝑖𝑖 𝑜𝑜𝑜𝑜𝑜𝑜 𝑝𝑝ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 2 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 6.022 × 1023 𝑝𝑝ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 I had intended 2 𝐸𝐸 = ℎ𝜈𝜈 moles to be an exact measurement, but 9.38 × 10−25 = (6.626 × 10−34 𝐽𝐽𝐽𝐽) 𝜈𝜈 that is not clear in the question, so 1 𝝂𝝂 = 𝟏𝟏 × 𝟏𝟏𝟏𝟏𝟗𝟗 𝑯𝑯𝑯𝑯 𝒐𝒐𝒐𝒐 𝟏𝟏. 𝟒𝟒𝟒𝟒 × 𝟏𝟏𝟏𝟏𝟗𝟗 𝑯𝑯𝑯𝑯 significant digit is the correct number! 7. Which early 20th century experiment or phenomenon specifically demonstrated light’s particle properties? a. Blackbody radiation b. The photoelectric effect c. Atomic emission spectra d. Diffraction of light e. All of the above demonstrate light’s particle properties f. None of the above demonstrate light’s particle properties 8. What is true about the Bohr model of the atom? a. Electrons absorb energy when falling to an orbit with a smaller radius b. Electrons orbit the nucleus in a circular orbit with a fixed radius c. Spacing of the energy levels are equal with increasing n values d. Spacing of the energy levels change with increasing n values e. None of the above is true 9. What is the frequency (Hz) of light absorbed when an electron in a hydrogen atom transitions from an orbital of n = 2 to n = 5? 1 1 ∆𝐸𝐸 = −2.179 × 10−18 𝐽𝐽( 2 − 2 ) 𝑛𝑛𝑓𝑓 𝑛𝑛𝑖𝑖 1 1 1 1 ∆𝐸𝐸 = −2.179 × 10−18 𝐽𝐽 2 − 2 = −2.179 × 10−18 𝐽𝐽 − = 4.576 × 10−19 𝐽𝐽 5 2 25 4 𝐸𝐸 = ℎ𝜈𝜈 4.576 × 10−19 𝐽𝐽 = (6.626 × 10−34 𝐽𝐽𝐽𝐽)𝜈𝜈 𝝂𝝂 = 𝟔𝟔. 𝟗𝟗𝟗𝟗𝟗𝟗 × 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝑯𝑯𝑯𝑯 10. Match each line in the atomic emission spectrum below to the most probable transition. n = 1 to n = 2: D n = 1 to n = 3: C n = 1 to n = 4: B n = 1 to n = 5: A n = 3 to n = 4: F n = 2 to n = 3: E 11. At the end of which transition(s) in the below diagram will the electron be closest to an energy of zero? CE 12. All objects have de Broglie wavelengths, but the dual wave-particle nature of matter only starts manifesting on a sufficiently small scale. What is the de Broglie wavelength (m) of a 144 g baseball thrown by a major league pitcher at 95.3 miles per hour? ℎ 𝜆𝜆 = 𝑚𝑚𝑚𝑚 6.626 × 10−34 𝐽𝐽𝐽𝐽 𝜆𝜆 = 1𝑘𝑘𝑘𝑘 95.3 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 1 ℎ𝑜𝑜𝑜𝑜𝑜𝑜 1 𝑚𝑚𝑚𝑚𝑚𝑚 1609.34 𝑚𝑚 (144𝑔𝑔 × )( × 60 𝑚𝑚𝑚𝑚𝑚𝑚 × 60 𝑠𝑠 × ) 1000𝑔𝑔 1 ℎ𝑜𝑜𝑜𝑜𝑜𝑜 1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝝀𝝀 = 𝟏𝟏. 𝟎𝟎𝟎𝟎 × 𝟏𝟏𝟏𝟏−𝟑𝟑𝟑𝟑 𝒎𝒎 13. Photons do not have mass, but they do have momentum. What is the momentum (kg·m/s) of a photon with a wavelength of 638 nm? ℎ 𝜆𝜆 = 𝑝𝑝 1 𝑚𝑚 6.626 × 10−34 𝐽𝐽𝐽𝐽 638 𝑛𝑛𝑛𝑛 × = 109 𝑛𝑛𝑛𝑛 𝑝𝑝 𝒌𝒌𝒌𝒌 ∙ 𝒎𝒎 𝒑𝒑 = 𝟏𝟏. 𝟎𝟎𝟎𝟎 × 𝟏𝟏𝟏𝟏−𝟐𝟐𝟐𝟐 𝒔𝒔 14. What is the energy (J) of the photon given off when an electron in a hydrogen atom falls from n=3 to n=1? 1 1 ∆𝐸𝐸 = −2.179 × 10−18 𝐽𝐽( 2 − 2) 𝑛𝑛𝑓𝑓 𝑛𝑛𝑖𝑖 1 1 1 1 ∆𝐸𝐸 = −2.179 × 10−18 𝐽𝐽 2 − 2 = −2.179 × 10−18 𝐽𝐽( − ) 1 3 1 9 ∆𝐸𝐸 = −1.937 × 10−18 𝐽𝐽 𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬𝑬 𝒐𝒐𝒐𝒐 𝒕𝒕𝒕𝒕𝒕𝒕 𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑 = −𝟏𝟏. 𝟗𝟗𝟗𝟗𝟗𝟗 × 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏 𝑱𝑱 = 𝟏𝟏. 𝟗𝟗𝟗𝟗𝟗𝟗 × 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏 𝑱𝑱 15. What is the energy (kJ) of 6.75 moles of photons with a wavelength of 612 nm? ℎ𝑐𝑐 𝐸𝐸 = ℎ𝜈𝜈 = 𝜆𝜆 (6.626 × 10−34 𝐽𝐽𝐽𝐽)(3 × 108 𝑚𝑚⁄𝑠𝑠) 𝐸𝐸 = = 3.25 × 10−19 𝐽𝐽 1 𝑚𝑚 612 𝑛𝑛𝑛𝑛 × 9 10 𝑛𝑛𝑛𝑛 𝟑𝟑. 𝟐𝟐𝟐𝟐 × 𝟏𝟏𝟏𝟏−𝟏𝟏𝟏𝟏 𝑱𝑱 𝟔𝟔. 𝟎𝟎𝟎𝟎𝟎𝟎 × 𝟏𝟏𝟏𝟏𝟐𝟐𝟐𝟐 𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑 𝟏𝟏 𝒌𝒌𝒌𝒌 × × 𝟔𝟔. 𝟕𝟕𝟕𝟕 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 × = 𝟏𝟏. 𝟑𝟑𝟑𝟑 × 𝟏𝟏𝟏𝟏𝟑𝟑 𝒌𝒌𝒌𝒌 𝟏𝟏 𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑 𝟏𝟏 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝑱𝑱