There will be no energy transfer from Ba2+ (which is the only way that Ba can be introduced into the cubic Bd2O3 lattice. The question is how did you add the Ba(O-?) to the firing mixture? - RC Ropp

@RcRopp:

then what will be the reason for emitting high intense red light?

Moreover, I used porous silica as host., You can also check the EDX data which is attached with this....

The possibility of energy transfer from Ba to Eu is close to nil. The one of the important requirements of spectral overlap between Ba and Eu cannot be observed since Ba as such is non-luminescent. However, the alternate possibility of having higher red emission probably due to removal defects that are acting as quencher otherwise. In our studies, monovalent ions like lithium and silver enhanced the luminescence intensity of emitter by this way. The second possibility is complete dissolution of Eu ions in presence of barium leading to enhancement. Hope it may be useful to u. Have a nice day.

Another question- How much silica? You have a (Gd,Ba)2SiO3:Eu phosphor plus Gd2O3! However, the structure contains defects, due to the incorporation of Ba2+ on a trivalent cation site (or vice-versa).- RC Ropp

The formula you have contains barium as a cation. It is not an activator. I assume you added BaO to a powder and then fired it. Was the Gd in the form of an oxalate- you did not answer my question concerning silica. If silica is present, it reacts with Gd2O3 to form a silicate. How much barium is present? There is possibility that the BaO acted as a flux. Why not get an x-ray diffraction pattern to see how much silicate is present.- RC Ropp

Eu^3 get incorporated into Gd2O3:Ba host. There occurs Ligand Metal Charge Transfer

(LMCT) in it whose energy is transferred to Eu^3; causing f-f transitions which is observed at 615nm. Of course, it was not expected to be present in Gd2O3:Eu3+ spinel.

Please note that the occurrence LMCT in lanthanide compounds and the transfer of this energy in causing excitation of the neighbouring rare earth ion (Eu^3, the dopant here) is a well reported concept.

I REITERATE- by introducing a divalent cation like barium into the trivalent-cation lattice, you are creating a vacancy-containing lattice because of charge compensation. I do not know whether it is a Frenkel-type or a Schottky-type. If the former, charge-compensation is achieved by a positive oxygen-vacancy plus a divalent barium on a trivalent-Gd-site. If the latter, it is by a neutral oxygen-vacancy plus a positive Gd-vacancy and a divalent barium on a trivalent-Gd-site. Note that the vacancies are associated with the Ba-site.

It would be interesting to determine what happens if equivalent amounts of BaO and ZrO2 (+2 & +4 ions) were added to achieve charge compensation without the formation of lattice vacancies. -RC Ropp

I leave it to your self whether this old man deserves your ‘vote up’ or brickbats (--- in a lighter vein).

The main argument under discussion is the existence of Gd2O3:Ba structure and the presence of Eu^3 in the interstitial sites.The structucture is analogous to the compound Gd2O3:Zn:Eu3+ which has already been reported with the help of PL emission analysis.

CLICK:

Synthesis and Characterization of Luminiscent oxide Nanocrystals by Sooyeon Seo

Then look in:

Page-14 (bottom line where in he discusses the structure of Gd2O3:Zn containing Eu^3)

And I reproduce this line as follows:

“Analysis of the PL emission syggests that Eu^3 ions are incorporated in Gd2O3:Zn host”.

Reply to Manohar Segal- you state "The main argument under discussion is the existence of Gd2O3:Ba structure and the presence of Eu^3 in the interstitial sites.The structucture is analogous to the compound Gd2O3:Zn:Eu3+ which has already been reported with the help of PL emission analysis". I beg to differ. All three structures of Y2O3, Gd2O3 & Eu2O3 have the same cubic structure (SpaceGroup = Ia-3, No. 206). Therefore, it is unlikely that Eu^3+ is in an interstitial site. It must be in a cationic site since it is coupled and perturbed by the phonon modes of the lattice.

I suggest that the statement "Analysis of the PL emission suggests that Eu^3 ions are incorporated in Gd2O3:Zn host” is wrong since it does not comply with the thermodynamic rules governing luminescent inorganic materials. The composition, Gd2O3:Zn does not contain Zn^2+ as an activator but is a defect compound, i.e. - (Gd, [Zn^2+,V^+))2O3.

The paper "Synthesis and Characterization of Luminiscent oxide Nanocrystals by Sooyeon Seo" is quoted. I also point out that the properties of nanocrystals differ from those of regular crystals since the lattice-planes and layers in nanocrystals are limited. As early as 1991, I predicted this phenomenon since no lattice defects were likely to appear and that the physical and optical properties would likely differ from the "norm". (see my book : "Luminescence and the Solid State" by R.C. Ropp, 1st Ed., Elsevier Sci. Publ., Amsterdam & New York (April - 1991). This phenomenon has been reported time after time.-R C Ropp

Knowing that no transfer of energy is permitted between the RE ion and Ba^2 and admitting Dr.RC Ropp’s argument( ignoring Sooyeon Seo over which my previous answer was based) that the crystal structure is Gd2O3.Eu^3 and Ba^2 ions are present as dopants, permit me to put forward my humble view point again for this exalted academic discussion.

Considering the presence of doping ions Ba^2 in the host material(Gd2O3.Eu^3) ,it can be concluded that Introducing ions with mismatching charge will cause a number of oxygen vacancies which will greatly increase crystal asymmetry acting as lattice defects and so whould change the luminescence spectrum and the intensity by enhancing the radiative transition from RE ion.

Dr. Sehgal is correct but I add that at least one of the vacancies must harbor a positive charge. Note that the defect equation is: 2Gd^3 = Ba^2 + V^+, where the right-hand ions are on gadolinium-sites in the lattice. The nature of the oxygen vacancies needs to be further defined (Schottky or Frenkel) but it is clear that the introduction of lattice defects is the cause of the change in PL properties. - RC Ropp

to Sakthivel Gandhi- I forgot to add: run a concentration series of Ba^2 in small steps (if you have not); also one with Sr^2. I predict the Sr-series may be better- RC Ropp

Thks.RCRopp; it gives me some consolation.

@Rc Ropp: sure.... tnks... @Rc Ropp & Manohar Sehgal: Thank you very much...