Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Microwave/Rotational Spectroscopy

  1. Apr 1, 2009 #1
    In rotational (microwave) spectroscopy, the Kth energy level is BK(K + 1), B = constant.

    It turns out the each Kth level is (2K + 1) degenerate.

    Now, as the energy level increases, does this mean that a single molecule will be in all these states at once, or in one of them? Or does it mean that the number of MOLECULES undergoing a transition to a higher energy state increases?

    I am confused about the interpretation of this.

    I know that the spectrum gives a nice curve shape, increasing first, because of ^^^^ and then decreasing because of the Boltzmann Population Law ( the no. of molecules undergoing transitions to higher states decreases as the energy between the states increases).

    I hope someone can clarify what is actually going on here. Thanks!!!:smile:
    Last edited: Apr 1, 2009
  2. jcsd
  3. Apr 1, 2009 #2


    User Avatar
    Science Advisor

    Quantum mechanically, an unmeasured molecule will occupy several states. But a single molecule can only ever be measured to be in a single state. Degeneracy doesn't enter into it.

    When you measure an ensemble of molecules, you get a Boltzmann distribution of the populations of the states, that is the proportion of molecules in each state. Degeneracy does enter into that equation, directly.

    On the other hand, the Boltzmann distribution say nothing at all about transitions. It just says which states a molecule is likely to be in, when at thermal equilibrium. Nothing about the likelihood of transitions, which isn't at all directly related to either the energy or degeneracy of the original or excited state. It's just that with rotational transitions, the transition-probability is more or less the same between every state, so then the spectral line intensities will depend entirely on the number of molecules available to do a transition, hence the Boltzmann-distributed shape.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook