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Unformatted text preview: Chemistry 350 Assignment 2 Solutions October 13, 2004 1. The energies and degeneracies of the two lowest electronic states of atomic iodine are: e 1 = 0 cm 1 , g e 1 = 4; e 2 = 7603.2 cm 1 , g e 2 = 2. What temperature is required so that 10% of the atoms are in the excited state? Disregard any contribution from higher electronic states. Answer: As in any question like this, this requires a simple calculation of the Boltzmann factor, p j , for the desired state. We can assume we can state the percentage to as many significant digits as we desire, so we’ll choose six here. p j = e e j /kT ∑ i e e i /kT . 100000 = 2 e 7603 . 2 cm 1 / (0 . 69509 cm 1 K 1 ) T 4 + 2 e 7603 . 2 cm 1 / (0 . 69509 cm 1 K 1 ) T ln . 400000 1 . 80000 = 7603 . 2 cm 1 (0 . 69509 cm 1 K 1 ) T T = 7272 . 5 K 2. Show that the fraction of harmonic oscillators in the ground vibrational state is given by f = 1 e hν e /kT Calculate f for 15 N 2 (g) ( k f = 2243 N m 1 ) at 300, 500 and 1500 K....
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This note was uploaded on 04/11/2010 for the course CHEM 1101 taught by Professor Leroy during the Spring '10 term at University of Toronto.
 Spring '10
 Leroy
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