Tutorial 2 a - CHEM 1A03 TUTORIAL#2 September 21-25 2008...

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CHEM 1A03 TUTORIAL #2 September 21-25, 2008 Solutions ________________________________________________________________________ Page 1 of 7 The following questions (1-14) are based on Chapter 8 + Lecture Notes. 1. The McMaster student FM radio station CFMU broadcasts at a frequency of 93.3 MHz (MHz = 1 megahertz = 10 6 sec 1 ). a) What is the wavelength of this signal in meters? = c / = (2.9979 10 8 m s -1 ) / (9.33 10 7 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. 4 10 22 J = 3 10 22 J (Only 1 sig. fig. in frequency difference, so 1 sig. fig. in energy difference)
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CHEM 1A03 TUTORIAL #2 September 21-25, 2008 Solutions ________________________________________________________________________ Page 2 of 7 3. The longest-wavelength (i.e. lowest energy) light that causes an electron to be emitted from a gaseous lithium atom is 520. nm. Gaseous lithium atoms are irradiated with light of wavelength 400. nm. What is the kinetic energy of the emitted electrons, in kJ/mol? h = hc/ = KE + BE, where KE = kinetic energy and BE = binding energy of the electron to the atom (or threshold energy). From the threshold wavelength, BE = hc/520nm. KE(electron) = E[400 nm] – E[520 nm] = hc
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This note was uploaded on 01/16/2010 for the course CHEMISTRY 1A03 taught by Professor Dr.p.lock during the Fall '08 term at McMaster University.

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Tutorial 2 a - CHEM 1A03 TUTORIAL#2 September 21-25 2008...

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