# Transition state of electrons in molecules

• I
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 !

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DrClaude
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.

• goldenboy
[...] 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. [...]

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 ? =)

DrClaude
Mentor
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 ? =)
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.

mjc123
Homework Helper
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.

DrClaude
Mentor
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.
Don't know why I was thinking of the harmonic oscillator instead of the particle in a box Sorry for the confusion.

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

blue_leaf77
Homework Helper
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).
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:
DrDu