Determine energy state difference using absorption-emission spectra

In summary: The first peak is S1,v=1 -> S0,v=1 at λ = 400 nm and the second peak is S1,v=1 -> S0,v=1 at λ = 380 nm.
  • #1
JoJoQuinoa
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Hello,

I was wondering if someone could help clarifying this question.

The question asks to estimate the energy state difference between the vibrational ground state of S0,v=0 and the first excited vibrational ground state S0,v=1 of the spectra below.

The given solution: S1,v=1 -> S0,v=1 at [itex]\lambda [/itex] = 400 nm and S1,v=2 -> S0,v=1 at [itex]\lambda [/itex] = 380 nm.

There are two things I'm confused about the solution:
1) From Figure B, I would assume that the first Fluorescence peak at 380 nm in Figure A corresponds to S1,v=0 -> S0,v=1 and the second peak at 400 nm corresponds to S1,v=0 -> S0,v=2. Larger transition results in higher emission energy or smaller wavelength.

2) Why is S0,v=1 being used as the final state for both peaks?
The S1,v=1 -> S0,v=1 would occur as S1,v=1 ->S1,v=0 ->S0,v=1 and
S1,v=2 ->S1,v=0 ->S0,v=1.
Since S1,v=2 or 1 ->S1,v=0 is internal conversion, wouldn't S1,v=0 ->S0,v=1 for the two transitions give off the same energy?

Thanks in advance!
anthraspec.PNG
peakid.PNG
 
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  • #2
I don't think the solution matches what the question asks for. Could you write the fully replicated question and the solution? Perhaps, a screenshot/photo of the question and solution?

For example, the question asks for the energy state difference, which means that the solution should be in the units of eV or cm-1 (or Hartree). It doesn't make any sense that the solution is in wavelength (nm) because wavelength is inversely proportional to energy and is not linear. Thus, you can't just add/subtract wavelength differences (unless you know what you are doing).

Also,
1) No, the fluorescence peak at 380 is not S1,v=0 → S0,v=1, and the second peak at 400 nm is not S1,v=0 → S0,v=2. What is your reasoning that you are not seeing any v=0→0 transition?
 
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  • #3
Hi Hayao,
Thank you for your response!

The question was: Use the anthracene absorption and emission spectrum in Figure 8.5 (Figure 1 in this post) to construct a combined Jablonski diagram/absorption/emission spectrum sketch. Label all states and calculate transition energies in cm-1.

The solution provided by the Professor and the one I found on Google were similar and as provided in the original post: S1,v=1 -> S0,v=1 at λ = 400 nm and S1,v=2 -> S0,v=1 at λ = 380 nm. I believe both were mislabeled. I found the figure below from ScienceDirect and I think it is correct.

"No, the fluorescence peak at 380 is not S1,v=0 → S0,v=1, and the second peak at 400 nm is not S1,v=0 → S0,v=2. What is your reasoning that you are not seeing any v=0→0 transition?"

Initially, I labeled the first peak as v=0->0 (overlapping peaks in Absorption and Emission spectrum are v=0->0 transition) and second peak to be v=0->1. I got confused by the solution and modified my answer.

anthraspeccorrect.PNG
 
  • #4
Yes, that figure is correct.
 
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Related to Determine energy state difference using absorption-emission spectra

1. What is an absorption-emission spectrum?

An absorption-emission spectrum is a graphical representation of the energy absorbed or emitted by a substance at different wavelengths of light. It shows the specific wavelengths of light that are absorbed or emitted by the substance, which can provide information about its energy state.

2. How is an absorption-emission spectrum used to determine energy state differences?

By analyzing the peaks and valleys on an absorption-emission spectrum, scientists can determine the specific wavelengths of light that are absorbed or emitted by a substance. These wavelengths correspond to the energy differences between the different energy states of the substance.

3. What is the relationship between absorption and emission spectra?

The absorption spectrum shows the wavelengths of light that are absorbed by a substance, while the emission spectrum shows the wavelengths of light that are emitted by a substance. They are related because the wavelengths of light that are absorbed are the same as the wavelengths of light that are emitted when the substance returns to a lower energy state.

4. What type of information can be obtained from an absorption-emission spectrum?

An absorption-emission spectrum can provide information about the energy levels and transitions of a substance, as well as its chemical composition and structure. It can also be used to identify unknown substances and monitor changes in a substance over time.

5. How is an absorption-emission spectrum experimentally obtained?

An absorption-emission spectrum is obtained by passing light of different wavelengths through a sample of the substance and measuring the amount of light absorbed or emitted at each wavelength. This can be done using a spectrophotometer or other specialized equipment.

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