1e03-tut4-a

1e03-tut4-a - CHEM 1E03 TUTORIAL #4 Solutions _ 1. The...

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TUTORIAL #4 Solutions ________________________________________________________________________ Page 1 of 4 1. The McMaster student FM radio station CFMU broadcasts at a frequency of 93.3 MHz (MHz = 1 megahertz = 10 6 s 1 ). a) What is the wavelength of this signal in meters? λ = c / ν = (2.9979 × 10 8 m s -1 ) / (93.3 × 10 6 s 1 ) = 3.21 3 m = 3.21 m b) What is the energy of one photon of this frequency? E = h ν = 6.626 × 10 34 J·s × 9.33 × 10 7 s 1 = 6.18 2 × 10 26 J = 6.18 × 10 26 J c) Compare the photon energy of part b with the energy of a photon of red light with wavelength 685 nm. E = h ν , so determine ν and use to solve for E (or use E = hc/ λ). ν = c / λ = (2.9979 × 10 8 m s -1 ) / (685 × 10 9 m) = 4.37 6 × 10 14 s 1 A red photon (its frequency is about 10 14 s 1 ) has an energy of E = h ν = 6.626 × 10 34 J·s × 4.37 6 × 10 14 s 1 = 2.89 9 × 10 19 J = 2.90 × 10 19 J. Its energy is about five million times larger than the FM radio frequency of part a. 2. Sodium vapour lamps used for street lighting emit two yellow lines that are part of the atomic spectrum of sodium. One of the lines has a wavelength of 589.0 nm, and the other has a wavelength of 589.6 nm. a) Express each of these wavelengths in meters. 589.0 nm × 1 m/10 9 nm = 5.890 × 10 7 m 589.6 nm × 1 m/10 9 nm = 5.896 × 10 7 m b) What is the frequency of each of these lines? ν = c / λ = [2.998 × 10 8 m/s]/5.890 × 10 7 m = 5.089 8 × 10 14 s 1 = 5.090 × 10 14 s 1 (or Hz) ν = c / λ = [2.998 × 10 8 m/s]/5.896 × 10 7 m = 5.084 6 × 10 14 s 1 = 5.085 × 10 14 s 1 (or Hz) c) How much more energy, in joules, does a photon of 589.0 nm possess, as compared to a photon of 589.6 nm? E = h ν 1 - h ν 2 = h( ν 1 - ν 2 ) = 6.626 × 10 34 J·s × (5.089 8 – 5.084 6 ) × 10 14 s 1 E = 6.626 × 10 34 J·s × (0.005 2 ) × 10 14 s 1 = 3. 4
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1e03-tut4-a - CHEM 1E03 TUTORIAL #4 Solutions _ 1. The...

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