The BeerLambert law states that the absorbance of a sample at a wavenumber # is

Chapter 14, Problem 14.20

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The Beer-Lambert law states that the absorbance of a sample at a wavenumber \(\tilde{v}\) is proportional to the molar concentration [J] of the absorbing species J and to the length l of the sample (eqn 13.4). In this problem you will show that the intensity of fluorescence emission from a sample of J is also proportional to [J] and l. Consider a sample of J that is illuminated with a beam of intensity \(I_0(\tilde{v})\) at the wavenumber \(\tilde{v}\). Before fluorescence can occur, a fraction of \(I_0(\tilde{v})\) must be absorbed and an intensity \(I(\tilde{v})\) will be transmitted. However, not all of the absorbed intensity is emitted and the intensity of fluorescence depends on the fluorescence quantum yield, \(\phi_{\mathrm{f}}\), the efficiency of photon emission. The fluorescence quantum yield ranges from 0 to 1 and is proportional to the ratio of the integral of the fluorescence spectrum over the integrated absorption coefficient. Because of a Stokes shift of magnitude \(\Delta \tilde{v}_{\text {Stokes }}\), fluorescence occurs at a wavenumber \(\tilde{v}_{\mathrm{f}}\), with \(\tilde{v}_{\mathrm{f}}+\Delta \tilde{v}_{\text {Stokes }}=\tilde{v}\). It follows that the fluorescence intensity at \(\tilde{v}_{\mathrm{f}}, I_{\mathrm{f}}\left(\tilde{v}_{\mathrm{f}}\right)\), is proportional to \(\phi_{\mathrm{f}}\) and to the intensity of exciting radiation that is absorbed by \(J, I_{\mathrm{abs}}(\tilde{v})=I_0(\tilde{v})-I(\tilde{v})\).

(a) Use the Beer-Lambert law to express \(I_{\mathrm{abs}}(\tilde{v})\) in terms of \(I_0(\tilde{v}),[J], l\), and \(\varepsilon(\tilde{v})\), the molar absorption coefficient of J at \(\tilde{v}\).

(b) Use your result from part (a) to show that \(I_{\mathrm{f}}\left(\tilde{v}_{\mathrm{f}}\right) \propto I_0(\tilde{v}) \varepsilon(\tilde{v}) \phi_{\mathrm{f}}[\mathrm{J}] l\).