Quantum Chemistry 6.6 - Diatomic Rovibrational Spectra

[@amyliu2394]

Thank you so much! When I watch your videos before class, I get so much more out of lecture because I'm not spending 3/4 of it confused.

[@dionsilverman4195]

Are the height of the intensities on the P and R branches the same for a given initial J? I am assuming that the starting J increases as you go either direction from the central frequency (wavenumber). They look the same, that is symmetrical, but if for a given rotational number J, the probability of increasing or decreasing J were equal, then wouldn't the resultant distribution be flat? That is, isn't the Boltzmann distribution based off that at higher energies there are more ways a particle can lose energy to a lower energy particle than gain energy?

[@tag_of_frank]

So selection rule means if quantum number N increases, quantum number J must increase or decrease by +- 1? J cannot stay the same when N changes?

[@ZIP854]

Why does the rotational energy levels being much bigger than the vibrational energy levels mean that the rotational selection rule is always positive? I don’t follow the logic there.

[@dionsilverman4195]

Because vibrational levels are bigger than rotational, if a photon is being absorbed (∆E system > 0) then ∆n = +1, regardless of whether ∆J = +1 or -1. If a photon were being emitted, the reverse would be true, ∆n would be guaranteed to be negative as energy leaves the system.

[@nkyu3035]

let's pretend w-bar=W. If a quantum system(n=0,J=0) is hit by a photon of wavenumber=W_e +2B_bar (1+1), does the quantum number of system become n=1 and J=2? this interpretation seems wrong considering Selection Rule, but not sure how to see it another way.

[@TMPChem]

The lowest energy peak in the R branch is where we jump from n = 0, J = 0 to n = 1, J = 1. The selection rule in the R branch is delta_n = +1, delta_J = +1. Both n and J have increased by 1.

[@nkyu3035]

but the lowest energy peak in the R branch is not at wavenumber=W_e +2B_bar (1+1). it is at wavenumber=W_e +2B_bar (0+1). I see when the system is hit by a photon of wavenumber=W_e +2B_bar (0+1), the system of n=0 and J=0 becomes the system of n=1 and J=1. I'm more interested in for the case of wavenumber=W_e +2B_bar (1+1). does the system of n=0 and J=0 become the system of n=1 and J=2 after it's hit by a photon of that wavenumber? If not, is there no change in quantum number(Q.N)? then for the Q.N of system of n=0 and J=0 to become n=1 and J=2, should the system hit by 2 photon? I mean, like (n=0,J=0)->(n=1,J=1)->(n=1,J=2).

[@TMPChem]

The second peak in the R branch is the jump between states n=0, J=1 -> n=1, J=2. J=1 has an energy of 2hcBbar. J=2 has an energy of 6hcBbar. The difference is 4hcBbar. The vibrational energy difference is the same. The result is a peak of frequency [ omega_bar + 4Bbar ].

[@nkyu3035]

What was I thinking o_o;;; Thanks

[@DeadAnenome]

Hello again, How come J can be plus or minus 1 while n can only be plus 1? Do vibration energies not release photons?

[@TMPChem]

To simplify the analysis we are only considering absorption spectra, where the photon is absorbed and the system energy increases. Since vibrational energy level separations are typically several orders of magnitude larger than rotational energy level separations, in order for this to be an absorption we must have delta_n = +1. We would observe a fairly symmetric spectrum for rovibrational emission spectra, except the R branch would then be where delta_J = -1 instead of +1 (and vice-versa for the P branch).

[@DeadAnenome]

Got it! Thank you!

[@pollysplash]

@@TMPChem Hello! I didn't quiet get why your answer explains ΔJ=-1. I thought it had to do with the values of the quantum number m... Am I completely wrong?

[@dionsilverman4195]

I think, and someone can correct me if I'm wrong, in a given spectrum you're either detecting absorptions or emissions, not doing both at once. Since the energy difference between vibrational levels is larger than that between rotational levels, if ∆n = +1 it's absorbing a photon, and if ∆n = -1, it's emitting one. So it would make sense that a spectrum would show only one of ∆n = ±1.

[@nkyu3035]

what decides the intensity of peaks?

[@TMPChem]

The intensity of peaks in the rovibrational spectrum is determined by the population (number / fraction of molecules) in each initial rotational energy level of the molecule. This is determined by the spacing of the energy levels and their degeneracies. In statistical mechanics, the probability a state is occupied increases linearly with increasing degeneracy and decreases exponentially with increasing energy. The rigid rotor has a degeneracy of 2J+1, and an energy of hBJ(J+1), which gives rise to the gull-wing shape of peak occupations we see here. That function can be roughly approximated by the function x^2 * exp(-kx).

[@solsticetwo3476]

Man, don't take it wrong, but you spend too much time reading out the equations. It would be more valuable to explains more, to offer insights, to give your thoughts about it.

[@TMPChem]

I typically do that, once I've established the baseline of what the equations say. People have different backgrounds. Some have seen it before, some haven't. Some are reviewing the concept before they go give a lecture, and some have no chemical background outside of this series. Some people have never seen Greek letters before, and wouldn't know how to pronounce it if I didn't say it. Some people are watching in 144p on a phone in India and can't read fine details on the screen. Some people don't know English, and without KZbin's attempt at a transcript they have nothing to throw into Google Translate and have a chance at learning something for which there is no content in they're native language. All of these things show up in my Analytics, most of them in double-digit percentages. I believe this is a happy medium between all the use cases the video could serve. All the notes are self-contained in one slide, so anyone who just wants the notes, or wants to skim quickly can do so. Also, every element of the screen is vocalized, so that on the other end of the spectrum everyone at least knows what's on the screen before we go into further detail about what that means and why it's important.