The transition between the quantum numbers J = 0 -> 1 (the quantum number v is unchanged, v = 0 -> 0) for 1H79B takes place at 500.7216 GHz, while the same transition for 1H81Br takes place at 500.5658 GHz. (a) Calculate the binding distance of these two molecules using the appropriate quantum mechanical model. (b) Calculate the frequency of the transition J = 1 -> 2 for 1H79B.. (c) Calculate the frequency of the transition J = 0 ->1 for 2H79BR under appropriate (and necessary) assumptions. Enter the answer to a), b) and c) in the answer box and attach your calculations.

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Chapter1: Chemical Foundations
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The transition between the quantum numbers J = 0 -> 1 (the quantum number v
is unchanged, v = 0 -> 0) for 1H79B takes place at 500.7216 GHz, while the same
transition for 1H81Br takes place at 500.5658 GHz.
(a) Calculate the binding distance of these two molecules using the appropriate
quantum mechanical model.
(b) Calculate the frequency of the transition J = 1 -> 2 for 1H79B1.
(c) Calculate the frequency of the transition J = 0 ->1 for 2H79BR under
appropriate (and necessary) assumptions.
Enter the answer to a), b) and c) in the answer box and attach your calculations.
Transcribed Image Text:The transition between the quantum numbers J = 0 -> 1 (the quantum number v is unchanged, v = 0 -> 0) for 1H79B takes place at 500.7216 GHz, while the same transition for 1H81Br takes place at 500.5658 GHz. (a) Calculate the binding distance of these two molecules using the appropriate quantum mechanical model. (b) Calculate the frequency of the transition J = 1 -> 2 for 1H79B1. (c) Calculate the frequency of the transition J = 0 ->1 for 2H79BR under appropriate (and necessary) assumptions. Enter the answer to a), b) and c) in the answer box and attach your calculations.
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