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HCl Bohr Quantum Oscillator

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data
    What
 is 
the 
wavelength 
of 
the 
emitted
 photon 
when
 HCl
 de‐excites 
from
 the
 first vibrational 
state?

    Well, I had to solve for the energy of the first vibrational state in the question before, assuming that it behaved like a harmonic oscillator using the reduced mass, which would be at n=1 (here we are only using Bohr's quantization rules). However, how can one de-excite from the n=1 state? Is this not the ground state that we cannot drop any further from, like for the quantum harmonic oscillator?

    2. Relevant equations

    [tex]E=n h \nu,~\nu=\frac{\sqrt{\frac{k}{m}}}{2\pi},~E=\frac{h c}{\lambda}[/tex]

    3. The attempt at a solution

    I would think that if there is a solution, it would just be using the energy from the problem before, and use the de Broglie wavelength equation to solve.
     
  2. jcsd
  3. Feb 8, 2009 #2

    Hootenanny

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    Note that the first excited state is by definition, the first state above the ground state. So if the ground state is n=1, then the first excited state is n=2.
     
  4. Feb 8, 2009 #3
    I understand that. I guess the question is: what is the first vibrational state? Is it the ground state, or one above for a diatomic molecule?
     
  5. Feb 9, 2009 #4

    Hootenanny

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    As you should know, the vibrational energy levels for a quantum harmonic oscillator obey the relationship:

    [tex]E_n = \left(n+\frac{1}{2}\right)h\bar\omega\;\;\;n=0,1,2,3,\ldots[/tex]

    So in this case the ground state is the n=0 state and the first vibrational state is n=1. The reason why the question specifically states vibrational energy levels is that a diatomic molecule may also have translational, rotational and electronic energy levels.
     
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