1
$\begingroup$

According to the Bohr model, when an electron moves from a ground state to the excited state, it will absorb energy, but what about the distance between electrons and nucleus, will it be changed? Why ? How?

Also, if there is the distance change when from a ground state to the excited state, will there be also any change in distance when electrons move from excited state to ground state ?

$\endgroup$
  • $\begingroup$ What do you want to achieve? The Bohr model is not useful to describe chemical reality. You can use it to calculate numbers, but those cannot be reproduced with other methods. And I don't understand you last sentence. You really ask is the ground state before and after excitation is the same? $\endgroup$ – Karl Oct 1 '17 at 19:49
3
$\begingroup$

Yes.

The Bohr model shows the atom as a small, positively charged nucleus surrounded by orbiting electrons.

orbital

Electrons undergo exothermic or endothermic reactions during ground state or excited state. When the electrons move from ground state to excited state, it means that they gain more energy, resulting in higher energy levels (higher energy shells), and this causes the electrons to orbit further from the nucleus. The same explanation is applicable when electrons move from excited state to ground stage- the electrons lose energy and this results in lower energy levels (lower energy shells), causing the electrons to orbit nearer to the nucleus.

$\endgroup$
  • 1
    $\begingroup$ I am not sure whether the higher excited state always increases distance between electron and nucleus, especially when considering orbitals of higher angular momentum ($d, f \dots$). There might be some exceptions. $\endgroup$ – Feodoran Oct 1 '17 at 8:39
  • $\begingroup$ @Feodoran Bohr knew nothing of orbitals, at least not when he devised the model named after him. And in Schrödingers model of the hydrogen atom, the different orbitals with same quantum number $n$ are degenerate. For anything else than hydrogen atoms, the original Schrödinger is quantitativly nearly as bad as Bohr (although the concept is much better of course). $\endgroup$ – Karl Oct 1 '17 at 19:59
  • $\begingroup$ For Hydrogen yes, but I was thinking about atoms and molecules in general. Why do you say "Schrödinger is quantitatively nearly as bad as Bohr"? The Schrödinger equation is in principle exact (at least in terms of chemical accuracy). Whether you need to make certain approximations to make solving it feasible, is another story. $\endgroup$ – Feodoran Oct 1 '17 at 20:42
0
$\begingroup$

I may be wrong here as this was long ago, but if I remember correctly absorption of light and the excitation process of an electron (for example) are faster than the electronic movement and therefor appear as a straight vertical line on the energy to distance plot (Jablonski-diagram). So you start from an optimized geometry for the ground state which is not necessarily the optimized geometry for the excited that and it will relax in the time while it is being excited before it falls back to the ground state. I remember we used this electronic change to calculate pKs values of acids for example. So yes, if I remember correctly the x-axis on this diagram was the distance of the electron to the nucleus (but I could be wrong there).

There was a rule for the excitation process, I think by Fermi (?). Historically seen, when they first analyzed the emission spectrum of for example hydrogen, they found that not every possible transition is equally likely, so there has to be quantum-mechanical process which descides what transition is more likely. And as the excitation process is much faster it needs to be almost straight up, so without any change in the geometry or any distances. And essentially you will need a good overlap of those amplitudes.

If you look here for example: http://www.geoset.info/wp-content/uploads/2012/05/Electronic-spec.png

You can see that both of these curves represent the different vibrational states and those have amplitudes, which tend to be bigger and the sides and not in the middle. I think in classical mechanics you'd expect that to be somewhere around the point where the oscillating spring is relaxed. And you can show, that the absorption process is better if the overlap of those amplitudes is at it's biggest, so you start from the ground state and then you go straight upwards till you find a higher vibrational mode of the excited state where you end up in a high amplitude. This is why you end up in v'=2 in the picture. Then you still have the same geometry but it is a vibrationally exicted state and will first relax to the excited vibrational ground state v'= 0. But that curve is shifted to the right and this is how the geometry changes and I believe in you example this would be the orientation of the electron to the nucleus.

At least I hope so. If not I'm truly sorry but this was the only example I came up with right now.

$\endgroup$
  • $\begingroup$ In principle you are right, but you are answering a different thing. The question was about change of electron density and you are talking about change in the nuclear geometry. $\endgroup$ – Feodoran Oct 1 '17 at 8:33
  • $\begingroup$ And Fermi and Jablonski are post-Bohr. $\endgroup$ – Karl Oct 1 '17 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.