To account for the breadth of the absorption and fluorescence bands in solutions and for the fact that the absorption and emission bands are not coincident it is assumed that for a given position relative to the solvent molecules the potential energy of an excited molecule is different from that of a normal molecule, and that the change in the potential energy that occurs during excitation or emission is to be counted as a part of the energy that determines hν. This leads to a general relation between the intensity of fluorescence F for the frequency ν and the coefficient of absorption for the same frequency, viz. F=K^'νe^-hνkT in which K^' is a function of T and probably of ν.The assumed conditions are met if the active molecule acts as an electric doublet whose moment changes from μ₁ in the normal state to μ₂ when excited. In this case it is shown that on the red side of the absorption band we have approximately ${α}={{α}}₀{e}^{{ph{ν}}{kT}}, where p={{{μ}}₁}{{{μ}}₁}{-}{{μ}}₂$Measurements made with Rhodamine-${β}$ and with Uranine show that Eq. (2) holds through a wide range with $p$ nearly equal to unity, i. e. with ${{μ}}₂$ small compared with ${{μ}}₁$.Eq. (1) has been tested by plotting $logF{-}log{α}$ against ${1}{{λ}}. The result is in each case a straight line. The slant of the line is found to be 1.49×{10}^{{-}13}$ for rhodamine-$Band 1.59×{10}^{{-}13}for uranine as compared with 1.64×{10}^{{-}13}$ the computed value of $h/kT.To account for the observed breadth of the fluorescence band of uranine it is sufficient to assume that the electric moment of the active molecule is of the order 3×{10}^{{-}19}$ e.s.u. This is about one-sixth of the value found by Jona for the electric moment of water.
No takes yet. Share an insight, caveat, or question.
Ernest Merritt (1926) studied this question.