p40_029

# p40_029 - f = 11 . 00 3 . 00 = 8 . 00. or a transition...

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29. The ratios computed in problem 28 can be related to the frequencies emitted using f =∆ E/h ,where each level E is equal to one of those ratios multiplied by h 2 / 8 mL 2 . This eFectively involves no more than a cancellation of one of the factors of h . Thus, for a transition from the second excited state (see part (b) of problem 28) to the ground state (treated in part (a) of that problem), we ±nd f =(9 . 00 3 . 00) µ h 8 mL 2 =(6 . 00) µ h 8 mL 2 . In the following, we omit the h/ 8 mL 2 factors. ²or a transition between the fourth excited state and the ground state, we have f =12 . 00 3 . 00 = 9 . 00. ²or a transition between the third excited state and the ground state, we have
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Unformatted text preview: f = 11 . 00 3 . 00 = 8 . 00. or a transition between the third excited state and the rst excited state, we have f = 11 . 00 6 . 00 = 5 . 00. or a transition between the fourth excited state and the third excited state, we have f = 12 . 00 11 . 00 = 1 . 00. or a transition between the third excited state and the second excited state, we have f = 11 . 00 9 . 00 = 2 . 00. or a transition between the second excited state and the rst excited state, we have f = 9 . 00 6 . 00 = 3 . 00, which also results from some other transitions....
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