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Unformatted text preview: Chemistry 209 Homework #5 answers Chapter 8: 6 This is most easily done by referring to Figure 8-3, which orders the different wavelengths (or frequencies) of the electromagnetic spectrum, from high energy at the left to low energy at the right end. We have 4 sources; one in the visible, at red, which belongs in the “middle” of our chart of energies; a radio wave in the FM band, thus at the far right portion of the spectrum (91.9 MHz corresponding to 0 . 919 × 10 8 Hz, as M stands for Mega, or 10 6 ); light with frequency 3 × 10 14 corresponds to the infrared region, between our red light and our radio waves; and light at λ = 485 ˚ A; the ˚ A unit is popular in the older literature, and corresponds to 10 − 10 m; so 485 ˚ A is equal to 4 . 85 × 10 − 8 m, and looks to be safely in the ultraviolet region. So in order of increasing frequency –that being our scale at the top, frequency increasing to the left –we have at lowest frequency the radio waves at 91.9 MHz, next the infrared light with frequency 3 × 10 14 , then the red traﬃc light, and last the light at λ = 485 ˚ A. 18 The Lyman series, all of which appears at higher energy than the Balmer lines, is not visible to the naked eye–and thus was discovered after the Balmer series. It involves transitions from any of the “outer” shells down to the lowest ( n = 1) shell in the H atom, and the series of frequencies ν observed are given by ν = 3 . 2881 × 10 15 s − 1 ³ 1 1 2 − 1 n 2 ´ as given in the problem, where the n parameter is the principle quantum number of the excited (source) state (and therefore is an integer greater than 1). (a) The largest energy (remember, frequency is proportional to energy!) arises when the transition is from the most excited state, i.e. for n a maximum. There is no oﬃcial maximum n value for the solutions to the H atom, so it might seem that we have a problem; but that’s not a real problem, as all we need to realize is that for n very large the difference between 1 n 2 and 1 ( n + 1) 2 is negligible, as both are effectively zero. So ν m ax = 3 . 2881 × 10 15 s − 1 ³ 1 1 2...
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This homework help was uploaded on 01/31/2008 for the course CHEM 2090 taught by Professor Zax,d during the Fall '07 term at Cornell University (Engineering School).
- Fall '07