Question

please explain

0.759 -6.0x106 5.0x106 0.50- F4.0x106 3.0x106 Absorbance Fluorescence (au) 0.25- 2.0x106 1.0x106 0.00+ 300 -0 600 350 550 400 450 500 Wavelength (nm) Figure 2. UV-vis and fluorescence spectrum of the final product, 3,6-Bis(dimethylamino)-2,7-difluoro-10. methylacridin-9(101) one (4)

Answer 1

It's important to remember the difference between this two spectrums. In UV-Vis spectroscopy we measure the absorption of a photon when a molecule is excited by UV or visible light. Whereas in fluorescence we measure the emission of a photon when a molecule return to the ground state (after being excited by fluorescence light).

So, this graph shows the absorption (left spectra) and emission spectra (right spectra) to the molecule:  3,6-Bis(dimethylamino)-2,7-difluoro-10-methylacridin-9-(10H)-one.

The molecule structure is:

I F NI

As you can see this molecule can absorb UV light since is a chromophore and consists in an aromatic system.

The height is the same for both spectra, but there's a difference between the position of the maximum (this means difference in energy).

We have this wavelength:

Abs (UV) = 375 nm (approx.)

Emis(F) = 435 nm (approx.)

Specifically, the fluorescence spectra is shifted approximately 60 nm to the right. Therefore, this indicates that there is a loss of energy when the electron returns to its ground state, it's possible that the electron loss energy as heat in the relaxation process. This difference is known as Stokes shift, sometimes is due to the to the solvent properties or other intermolecular properties.