# I Transition state of electrons in molecules

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1. May 24, 2017

### goldenboy

Hi !

In my class we have an exercise that I technically understand but that I can't get conceptually.

We have trois molecules (naphtalene, anthracene, tétracene). Considering each molecule as infinite quantum well of length 2L(naphtalene), 3L(anthracene) and 4L(tetracene), we would like to predict the wave length absorbed the electron of each molecules. To do so :

- We find the transition energy for each electron state : ΔE = h*(2n+1)/(8m*Lw)
- Then we find the wave lengths absorbed by each molecules for each transition with : λ = h*c/ΔΕ

So, the exercise is resolved. But my question is : when ALL the electrons at state n = 1 go to n = 2, and n = 2 go to n = 3, ... considering that light is always available, why molecules don't stop absorbing the wave length of the first transition states (consecutively, why the sample doesn't change its colour) through time. And why after a certain state, electron doesn't absorbe the wave length corresponding to it's transition (like in naphtalene, the last transition is from 4 to 5).

Thanks a lot for answers and sorry for my bad english. I hope i have been clear enough

Bye !

2. May 24, 2017

### Staff: Mentor

If all molecules are excited, then indeed you would have no more absorption at that wavelength. But as you can see from the simple particle in a box model, the spacing between adjacent levels is always the same (this is not true for a real molecule, but close enough), so you will still get absorption at that wavelength. Also, you have to factor in the relaxation time. In most cases, before you can even reach a significant proportion of excited molecules, some will already decay back to the ground state and be able to absorb light again.

3. May 24, 2017

### goldenboy

However, i can't get that point : in which way the fact that spacing between adjacent levels is always the same influences that we still get absorption at that wavelength ? =)

4. May 24, 2017

### Staff: Mentor

Because the wavelength of the light must correspond to the energy difference. If the energy difference is constant, then the same wavelength is involved in all n → n+1 transitions, whatever n.

5. May 24, 2017

### mjc123

I think you're perhaps misreading the OP. The energy difference is proportional to 2n-1 because the energy is proportional to n2 for a particle in a box. So the levels are not equally spaced.

6. May 24, 2017

### Staff: Mentor

Don't know why I was thinking of the harmonic oscillator instead of the particle in a box

Sorry for the confusion.

7. May 24, 2017

### Jilang

I would say that at the same time as absorbing they are also emitting and dropping down energy levels.

8. May 25, 2017

### blue_leaf77

Due to spontaneous emission, electron in an excited state will have a finite lifetime to stay in that state. Therefore, it cannot consecutively absorbs photon to get excited to a higher and higher levels. After hitting certain upper level, spontaneous emission becomes much more probable and faster that the probability to de-excite exceed that of excitation.

Last edited: May 27, 2017
9. May 27, 2017

### DrDu

An important point to recognize here is the fact that a molecule, which has been excited, will fall back very rapidly to the ground state again, by so called radiationless transitions. Basically, the electronic excitation energy dissipates rapidly into molecular vibrations. So the probability to sequentially excite various electrons within one molecule are vanishingly small with the exception of molecules in highest power laser fields.