Isotope shift in molecules

[kelly0303]

Hello! Is the isotope shift between 2 (low lying) vibrational levels of the lowest 2 electronic levels (of a diatomic molecule) the same, no matter what the 2 levels are? For example, is the isotope shift associated to the $0 \to 0$ vibrational transition of 2 molecular isotopes (here I mean the lowest vibrational level of the electronic ground state and the lowest vibrational level of the first excited electronic state) the same as the $1 \to 1$ isotope shift (here I mean the first excited vibrational level of the electronic ground state and the first excited vibrational level of the first excited electronic state)? And if they are not the same, how big (order of magnitude) is the difference? Any insight or suggested reading would be greatly appreciated. Thank you!

 

[mfb]

As far as I remember the two energy spectra should be roughly independent. In other words:
E_total = E_electronic + E_vibration where both components have their own isotope shift. If you consider the isotope effect on a transition only then E_transition_isotope = E_electronic_transition_isotope + E_vibration_transition_isotope. As long as you don't change the vibrational state I don't expect the isotope shift to depend on that state.

 

[mjc123]

I think the principle is that the energies are separable, i.e. E_total = E_electronic + (n+1/2)hν (ignoring anharmonicity). But ν may, and generally does, vary from one electronic state to another, as the different electronic distribution affects the stiffness of bonds. Then
E(0→0) = ΔE_e + 1/2 hΔν
E(1→1) = ΔE_e + 3/2 hΔν
So E(0→0) is different from E(1→1) for the same isotopomer.
The isotope shift for E(0→0) will be 1/2 h(Δν₁ - Δν₂) = 1/2 hΔν₁(1-μ₁/μ₂)
where the subscript 1 denotes the lighter isotopomer, and μ is the reduced mass, and for E(1→1) will be
3/2 hΔν₁(1-μ₁/μ₂)

 

[mfb]

as the different electronic distribution affects the stiffness of bonds

Then there is a difference, but that sounds like a tiny higher order effect.

 

[DrClaude]

Then there is a difference, but that sounds like a tiny higher order effect.

Certainly not!

Here is an example for Cs₂

and this one for O₂

These curves definitely correspond to very different bond stiffness, and thus different $\nu$.

 

[kelly0303]

I think the principle is that the energies are separable, i.e. E_total = E_electronic + (n+1/2)hν (ignoring anharmonicity). But ν may, and generally does, vary from one electronic state to another, as the different electronic distribution affects the stiffness of bonds. Then
E(0→0) = ΔE_e + 1/2 hΔν
E(1→1) = ΔE_e + 3/2 hΔν
So E(0→0) is different from E(1→1) for the same isotopomer.
The isotope shift for E(0→0) will be 1/2 h(Δν₁ - Δν₂) = 1/2 hΔν₁(1-μ₁/μ₂)
where the subscript 1 denotes the lighter isotopomer, and μ is the reduced mass, and for E(1→1) will be
3/2 hΔν₁(1-μ₁/μ₂)

Thank you for this! Actually I should have been more clear in my question (sorry for that). I am interested in high mass isotopes, where the mass shift is basically negligible i.e. $\mu1/\mu2$ is basically 1. What I am interested in is the field (volume) effect of isotope shift. I didn't find much online so any suggested reading would be really welcome.

 

[mjc123]

I haven't heard of the field (volume) effect of isotope shift. Can you enlighten me?

 

[kelly0303]

I haven't heard of the field (volume) effect of isotope shift. Can you enlighten me?

The field shift is just the change in the volume of the nucleus for different isotopes. In short, adding more neutrons has an influence on the distribution of the protons in the nucleus, and the electrons whose orbits have a non-zero probability of being inside the nucleus get affected by this (this change in volume effect is actually significantly higher than the change in mass for higher A nuclei). Here is a nice introduction to that (page 236). But as I said before, I would like some readings where this effect is analyzed in the molecules, not just atoms i.e. how does the field effect of isotope shift affects the vibrational and rotational levels of a molecule. Thank you!