HCl Bohr Quantum Oscillator

[Dahaka14]

Homework Statement

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?

Homework Equations

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

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.

 

[Hootenanny]

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?

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.

 

[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?

 

[Hootenanny]

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?

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.