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Photoluminescence (PL) spectroscopy is used to find the energy levels in a compound but how does PL spectroscopy works? After exciting a material with a particular wavelength whether PL machine looks for photons emitted with in a range of wavelenght?

The below image was taken from here. enter image description here

To get this spectra the $\ce{CeO2}$ was excited at 345 nm and got peaks at 490 and 520 nm(approx) ie 2.53 and 2.38 eV respectively and its band gap is 3 eV. Whats happening here?

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  • $\begingroup$ What is the question? The experimental procedure? The theory in general? The reason for the shift? The analysis of CeO2 in particular? $\endgroup$ – Greg Dec 4 '15 at 10:34
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The explanation that follows is valid for compound dissolved in liquid solution where PL is understood either as fluorescence or phosphorescence.

PL is measured with a fluorimeter. The main optical components of this instrument are: a lamp that provides the light used to excite the sample, two monochromators (M), one for the excitation (Mex) and the other one for the emission (Mem) and a detector, usually a photomultiplier.

To measure PL the Mex is fixed at the excitation wavelength (in your case 345 nm) and the Mem does the scan, i.e. it let pass the luminescence (a narrow band) in a fixed wavelength range. This is then measured with the photomultiplier. At the end you obtain a spectrum, i.e PL vs wavelength.

From the spectrum that you show this range goes from 200 nm to 1000 nm. Usually the range starts after the excitation, this is to avoid to record the excitation itself that is not desired. Because in the spectrum that you show there is not the excitation band i suppose they use a filter (cut-off) to cut it out.

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