# HCl Bohr Quantum Oscillator

1. Feb 7, 2009

### Dahaka14

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

$$E=n h \nu,~\nu=\frac{\sqrt{\frac{k}{m}}}{2\pi},~E=\frac{h c}{\lambda}$$

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. Feb 8, 2009

### Hootenanny

Staff Emeritus
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.

3. Feb 8, 2009

### Dahaka14

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?

4. Feb 9, 2009

### Hootenanny

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

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

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.