The Form of the Absorption Bands in Solutions of the Organic Dyes, and a Relation Between Absorption and Fluorescence

Abstract

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 $h\ensuremath{\nu}$. This leads to a general relation between the intensity of fluorescence $F$ for the frequency $\ensuremath{\nu}$ and the coefficient of absorption for the same frequency, viz. $F={K}^{\ensuremath{'}}\ensuremath{\nu}{e}^{\ensuremath{-}\frac{h\ensuremath{\nu}}{\mathrm{kT}}}$ in which ${K}^{\ensuremath{'}}$ is a function of $T$ and probably of $\ensuremath{\nu}$.The assumed conditions are met if the active molecule acts as an electric doublet whose moment changes from ${\ensuremath{\mu}}_{1}$ in the normal state to ${\ensuremath{\mu}}_{2}$ when excited. In this case it is shown that on the red side of the absorption band we have approximately $\ensuremath{\alpha}={\ensuremath{\alpha}}_{0}{e}^{\frac{\mathrm{ph}\ensuremath{\nu}}{\mathrm{kT}}}, \mathrm{where} p=\frac{{\ensuremath{\mu}}_{1}}{{\ensuremath{\mu}}_{1}}\ensuremath{-}{\ensuremath{\mu}}_{2}$Measurements made with Rhodamine-$\ensuremath{\beta}$ and with Uranine show that Eq. (2) holds through a wide range with $p$ nearly equal to unity, i. e. with ${\ensuremath{\mu}}_{2}$ small compared with ${\ensuremath{\mu}}_{1}$.Eq. (1) has been tested by plotting $logF\ensuremath{-}log\ensuremath{\alpha}$ against $\frac{1}{\ensuremath{\lambda}}$. The result is in each case a straight line. The slant of the line is found to be 1.49\ifmmode\times\else\texttimes\fi{}${10}^{\ensuremath{-}13}$ for rhodamine-$B$ and 1.59\ifmmode\times\else\texttimes\fi{}${10}^{\ensuremath{-}13}$ for uranine as compared with 1.64\ifmmode\times\else\texttimes\fi{}${10}^{\ensuremath{-}13}$ the computed value of $\frac{h}{\mathrm{kT}}$.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\ifmmode\times\else\texttimes\fi{}${10}^{\ensuremath{-}19}$ e.s.u. This is about one-sixth of the value found by Jona for the electric moment of water.