May 23, 2019, 03:13:05 AM
Forum Rules: Read This Before Posting

### Topic: Electronic Spectroscopy: Dissociation into atoms  (Read 191 times)

0 Members and 1 Guest are viewing this topic.

#### chmscrb55

• Very New Member
• Posts: 2
• Mole Snacks: +0/-0
##### Electronic Spectroscopy: Dissociation into atoms
« on: May 03, 2019, 02:35:59 PM »
Hi, currently attempting questions related to electronic spectroscopy and am having difficulty.

https://imgur.com/a/AhZD5ZG

parts (a) to (c) are not related.

My attempt so far:
(d): Got the expression for the energy from the bottom of the ground state energy curve to the dissociation limit, and solved for D0''
D0'' = 51221.333 cm-1

(e): This is the part I got lost at. When a molecule dissociates, can I assume that the dissociation energy is split evenly among the dissociated atoms (in this case, at least since its a diatomic)? If not, how would I approach this question?

(f): Once i get part (e), this shouldn't be a problem.

(g): Yea... I dont even know how to start.

Appreciate the effort to read through and *delete me*

Thanks in advance!

#### Corribus

• Chemist
• Sr. Member
• Posts: 2648
• Mole Snacks: +428/-20
• Gender:
• A lover of spectroscopy and chocolate.
##### Re: Electronic Spectroscopy: Dissociation into atoms
« Reply #1 on: May 03, 2019, 03:21:43 PM »
No, you can't assume the energies are equal for the two dissociated atoms in the excited state. In fact the question explicitly tells you that the energies are different, and gives you the energy difference. But you may assume this, I believe, for the two atoms in the ground electronic state. Maybe this diagram helps, attached. Note you are dealing with the B3Σu- state vs the X3Σg- ground state. This transition is called the Schumann-Runge transition, as noted in the question.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### Corribus

• Chemist
• Sr. Member
• Posts: 2648
• Mole Snacks: +428/-20
• Gender:
• A lover of spectroscopy and chocolate.
##### Re: Electronic Spectroscopy: Dissociation into atoms
« Reply #2 on: May 03, 2019, 04:04:36 PM »
For the last part, think about what how the transition intensity relates to the respective position of the two states and the Franck-Condon factor. (The vibrational wavefunctions tend to have higher intensity at the potential boundaries, with the exception of the lowest-lying state; the Franck-Condon factor implies that the highest intensity transition happens where the bottom of the ground electronic state is positioned directly below the potential boundary of the excited electronic state... all things being equal, of course.) The origin state will be the lowest lying vibrational level in the ground electronic state. The final state will be one of the vibrational levels in the excited electronic state. You are told what the energy of this transition is and you know what the difference in energy between the bottom of these potential wells is. You are told to assume parabolic functions. Based on this and some math, you should be able to estimate where along the x-axis (separation axis) the bottom of the electronically excited state potential well is.

That's how I'd approach it, anyway.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### chmscrb55

• Very New Member
• Posts: 2
• Mole Snacks: +0/-0
##### Re: Electronic Spectroscopy: Dissociation into atoms
« Reply #3 on: May 04, 2019, 01:18:10 AM »
Thanks for the input Corribus. Why was the excitation energy for the most intense transition provided? Using the parameters from the previous parts, i could've calculated the energy at the lowest energy in the excited state.

Also, does k change with electronic excitation? I'm assuming i can take it as the change is negligible and so its constant since of the Franck-Condon Principle?

#### Corribus

• Chemist
• Sr. Member
• Posts: 2648
• Mole Snacks: +428/-20
• Gender:
• A lover of spectroscopy and chocolate.
##### Re: Electronic Spectroscopy: Dissociation into atoms
« Reply #4 on: May 04, 2019, 10:50:51 AM »
If I'm not mistaken in the way I'm approaching the question: the transition energy is provided so you can place the excited potential well on a set of cartesian coordinates and then calculate the difference along the x-axis of the two minima. If you know the: (A) functional forms of the potential (provided), (B) the transition energy (provided), which is the difference between the 0-0 state and some high lying vibronic level along the excited electronic surface, and (C) spectroscopic constants (provided), which allow you to determine where the various vibronic levels are. Based on that, I think you should be able to determine what the unknown higher lying vibronic level is, and therefore position everything accordingly in Cartesian coordinates, and then calculate the excited-state equilibrium bond length. I haven't actually tried to do it, I'm just thinking in my head a general strategy. It may not work out the way I see it in my mind .

Because you're not given any other information, I think you must assume the force constants for the two surfaces are identical. Obviously they aren't - that much is obvious from the diagram I showed. For that matter, a harmonic potential is also assumed for both surfaces, so this isn't probably going to give a good estimate of the real equilibrium bond length anyway.

What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### Corribus

• Chemist
• Sr. Member
• Posts: 2648
• Mole Snacks: +428/-20
• Gender:
• A lover of spectroscopy and chocolate.
##### Re: Electronic Spectroscopy: Dissociation into atoms
« Reply #5 on: May 04, 2019, 02:05:58 PM »
Sorry, scratch that, if you know the fundamental vibrational frequency of each electronic state and the reduced mass, you can calculate the force constant for the harmonically approximated potential. The force constant k will therefore be different for the two states, as it should be. Sorry, I haven't thought about this stuff in a long time: bit rusty. lol
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman