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tutorial 4 solutions

# tutorial 4 solutions - CHEM 1E03 TUTORIAL#4 Solutions 1 The...

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CHEM 1E03 TUTORIAL #4 Solutions ________________________________________________________________________ Page 1 of 3 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? = c / = (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 = 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 , so determine and use to solve for E (or use E = hc/  = c / = (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 = 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? = c / = [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) = c / = [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 1 - h 2 = h( 1 - 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.

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tutorial 4 solutions - CHEM 1E03 TUTORIAL#4 Solutions 1 The...

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