Why in some cases excitation spectra of a dye solution monitored at its emission maximum shows higher emission intensity than that of the emission spectra measured by exciting at its absorption maximum under the similar instrumental conditions?
Thank you in advance
Iam not able to get your question. Usually the excitation peak at its lambda max will have higher intensity than its emission
It may be that your fluorophore has undergone photobleaching during the course of making all the measurements. You should reduce the intensity of the exciting light to reduce photobleaching. You can also replace the sample between measurements with more of the same solution that has not been exposed to illumination.
What is the "higher emission intensity?" You should count the number of photons and the energy of each photon
if the sample absorbs over a wide wavelength range and at a longer wavelength than the emission then the narrow emission peak contains less energy than was gained by absorption even if the peak height is greater so conservation of energy is maintained
Excitation of dye molecules with light having lower wavelength (higher energy photons) will be more efficient and molecule will reach in some vibrational level of S1 electronic state. Then excited molecules will loose energy by collision and come to lower level of S1 state. From there fluorescence will take place efficiently. But using a light having wavelength equal to the absorption wavelength (absorption peak wavelength), number of excited molecules will be less (only half of the band are excited) resulting in less fluorescence.
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