In dye degradation photoreaction, can anybody explain more about role of jablonski diagram with band gap diagram. Ex; degradation of methyl orange with nano TiO2 by UV Light.
First a few points regarding the reaction:
REFERENCE:
Journal of Photochemistry and Photobiology A: Chemistry 148 (2002) 161–168.
Photocatalytic degradation of methyl orange as a model compound
[A]The UV–visible for MO in distilled water shows two absorption maxima at 270 and 465nm. The band at 465 nm is used to monitor the effect of the photo catalysis on the degradation of MO.
[B] When the reaction vessel is exposed to solar radiation for 5hr in the absence of TiO2, the MO is quite stable under solar conditions.
[C] When the reaction vessel containing TiO2 is kept in the dark for 10hr,again no decay is observed
[D] It is only when 40g/L of TiO2 is used while irradiating in visible light, the degradation of MO starts.
The absorption of the visible band at 465nm decreases with time and new band starts to form at 320 nm along with 270nm.
After 1hr, the band intensity at 465nm starts to decrease and the band intensity at 270nm increases. After 3hr, the band intensity at 465 and 270 nm starts to decrease. After 5hr, the bands at 465 and 270 nm disappeare and only one strong band at 254nm increased with time, decreasing after 3h .
[E]This trend suggests that the extended aromatic MO absorbs at 465nm and the aromatic ring absorb in the range 200–270nm.In the first period, the polyaromatic rings inMOstart to degrade creating a mono substituted aromatics thus the band at >200 nm increase in intensity and a new band at 320nm appear. After 2hr, both bands at 270 and 465nm start to decrease and the band at >200 decrease also. This indicates that MO starts to degrade and that CO2 and H2O start to form.
Explanation with Jablonski Diagram[JD]:
As is clear, the reaction needs three things simultaneously- TiO2, Solar light and irradiation at 465nm. MO, being PHOTO INACTIVE, itself, cannot absorb any radiation DIRECTLY from solar ligh. But TiO2 being PHOTOSENSITIVE ,absorbs 465nm radiation[TiO2*] It, then, transmits this energy to MO which then starts the reaction and MO would break up into H2O and CO2.
As you know, JD is a complicated diagram having a number of states[ ground state singlet(S0), Excited Singlet(S1) and Triplet(T) etc] and involve a number of terms[ internal conversion, intersystem crossing, fluorescence, phosphorescence Vibrational Relaxation Etc], but I will restrict up to the terms which are essential for the explanation.
The MO molecules are in So state.Unless they reach in S1 state, the reaction cannot start..Here, the TiO2* helps them by transmitting some energy to MO molecules. They reach S1 state, the excited state singlet(S1).The excited states are short lived and they decay to ground state through RADIATIONLESS TRANSITIONS or EMISSION OF RADIATIONS or UNDERGO PHOTOCHEMICAL RACTIONS involving TI, the TRIPLET STATE AS THEY HAVE LONGER LIFE TIME[up to 10s]THAN SINGLET STATES[ up to 1000ns].
The essential part of JD of both MO alone as well as MO containing TiO2 is attached along with for your kind perusal.
based on the explanation, how do we identify the phosphorencence of flourescence by PL spectrrometer?
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