The Beer-Lambert law states that the absorbance of a sample at a wavenumber \(\$ is proportional to the molar concentration \)[1]\( of the absorbing species \)\mathrm{J}\( and to the length 1 of the sample (eqn \)13.4)\(. In this problem you will show that the intensity of fluorescence emission from a sample of \)\mathrm{J}\( is also proportional to \)[\mathrm{J}]\( and \)l\(. Consider a sample of \)\mathrm{J}\( that is illuminated with a beam of intensity \){ }^{I_{0}(0)}\( at the wavenumber \){ }^{0 .}\( Before fluorescence can occur, a fraction of \)\mathrm{I}_{0}\left({ }^{0)}\right.\( must be absorbed and an intensity \){ }^{\prime(0)}\( 'will be transmitted. However, not all of the absorbed intensity is emitted and the intensity of fluorescence depends on the fluorescence quantum yield, \)\varphi_{\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 \mathrm{v}_{\text {Slokes }}\(, fluorescence occurs at a wavenumber \)\mathrm{v}_{\mathrm{f}}\(, with \)\mathrm{v}_{\mathrm{f}}+\Delta \mathrm{v}_{\text {stokes }}=\mathrm{v} .\( It follows that the fluorescence intensity at \)\mathrm{v}_{\mathrm{f}}, \mathrm{I}_{\mathrm{f}}\left(\mathrm{v}_{\mathrm{f}}\right)\(, is proportional to \)\varphi_{\mathrm{f}}\( and to the intensity of exciting radiation that is absorbed by \)\mathrm{J}, \mathrm{I}_{\mathrm{abs}}(\mathrm{v})=\mathrm{I}_{0}(\mathrm{v})-\mathrm{I}(\mathrm{v})\(. (a) Use the Beer-Lambert law to express \)\mathrm{I}_{\mathrm{abs}}(\mathrm{v})\( in terms of \)\mathrm{I}_{0}(\mathrm{v}),[\mathrm{J}], \mathrm{I}\(, and \)\varepsilon(\mathrm{v})\(, the molar absorption coefficient of \)\mathrm{J}\( at \)\mathrm{V}\(. (b) Use your result from part (a) to show that \)\left.\mathrm{I}_{\mathrm{f}}\left(\mathrm{v}_{\mathrm{f}}\right) \infty \mathrm{I}_{0}(\mathrm{v}) \varepsilon(\mathrm{v}) \varphi_{\mathrm{f}} \mathrm{f}\right] \mathrm{I}$.

Step by step solution

01

Review the Beer-Lambert Law

The Beer-Lambert law relates the absorbance (A) of a sample to the molar concentration ([J]), the path length (l), and the molar absorption coefficient (\(\varepsilon(u)\)). The law is expressed by the equation \(A = \varepsilon(u) [J] l\). The absorbance is also related to the incident intensity (\(I_0(u)\)) and the transmitted intensity (\(I(u)\)) through \(A = -\log_{10}(I(u)/I_0(u))\).

02

Express Transmitted Intensity

Solve the Beer-Lambert Law equation for the transmitted intensity \(I(u)\), using \(A = -\log_{10}(I(u)/I_0(u))\). This gives us \(I(u) = I_0(u) 10^{-A} = I_0(u) 10^{-\varepsilon(u) [J] l}\).

03

Determine Absorbed Intensity

The absorbed intensity \(I_{\text{abs}}(u)\) is the difference between the incident intensity and the transmitted intensity: \(I_{\text{abs}}(u) = I_0(u) - I(u)\). Substitute the expression for transmitted intensity from Step 2 to find \(I_{\text{abs}}(u)\) in terms of \([J]\), \(l\), and \(\varepsilon(u)\).

04

Relate Fluorescence Intensity to Absorbed Intensity

The fluorescence intensity at the wavenumber \(u_f\), \(I_f(u_f)\), is proportional to the fluorescence quantum yield (\(\varphi_f\)) and the absorbed intensity. We can then write \(I_f(u_f) \propto \varphi_f I_{\text{abs}}(u)\).

05

Combine Expressions

Combine the expression for \(I_{\text{abs}}(u)\) from Step 3 with the proportionality from Step 4 to express \(I_f(u_f)\) in terms of \(I_0(u)\), \(\varepsilon(u)\), \(\varphi_f\), \([J]\), and \(l\). This shows that \(I_f(u_f)\) is indeed proportional to the molar concentration and path length, as well as the quantum yield. The final proportionality is: \(I_f(u_f) \propto I_0(u) \varepsilon(u) \varphi_f [J] l\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Concentration

When studying the interaction of light with substances, understanding molar concentration is crucial. Molar concentration, denoted by \[\text{[J]}\], refers to the amount of a substance, expressed in moles, present in a specified volume of solution. To comprehend this, imagine measuring out a certain number of moles of a substance and dissolving them into a set volume of liquid; the resulting mixture's molar concentration tells you how 'strong' or 'concentrated' that solution is.

In the context of the Beer-Lambert law, the absorbance of light by a sample is directly proportional to the molar concentration of the absorbing species. This implies that the more concentrated the solution, the more it will absorb light of a particular wavelength, given the path length and molar absorption coefficient remain constant. Students should remember that molar concentration is not just a number; it embodies how densely packed the absorbing species are in the solution, influencing the light-absorption process.

For clarification, let's consider a practical scenario: if you have two solutions of dye, one more concentrated than the other, and shine a light through each, you'll observe that the more concentrated solution appears darker. This observation is a direct consequence of the higher molar concentration absorbing more light, as described by the Beer-Lambert law.

Fluorescence Quantum Yield

Fluorescence quantum yield, represented by \(\varphi_{f}\), is a critical concept in the study of light and matter interactions, especially when exploring the phenomenon of fluorescence. Imagine a substance that, upon absorbing light, re-emits a portion of that light at different wavelengths — this re-emission is fluorescence. The fluorescence quantum yield quantifies the efficiency of this process, essentially answering the question: What fraction of absorbed photons are re-emitted as fluorescent light? This value ranges from 0 to 1, where 0 signifies no fluorescence and 1 implies that each absorbed photon results in a fluorescent photon.

The quantum yield is influenced by various factors, including the molecular structure of the fluorescing substance and the surrounding environment. It's important for students to grasp that while a higher quantum yield indicates more efficient fluorescence, this doesn't necessarily mean brighter fluorescence, as it also depends on the amount of light initially absorbed — a reflection of the molar concentration and molar absorption coefficient.

Molar Absorption Coefficient

To fully grasp how substances interact with light, we must understand the molar absorption coefficient, symbolized as \(\varepsilon(v)\). This coefficient measures a substance's propensity to absorb light at a specific wavelength. A higher molar absorption coefficient means the substance is very effective at absorbing light, and a smaller amount of this substance is needed to show significant absorbance. It is measured in units typically reported as liters per mole-centimeter (L·mol⁻¹·cm⁻¹).

When a beam of light encounters a sample, the molar absorption coefficient helps us predict how much of that light will be absorbed, based on the substance's concentration and the light's path length through it. In the context of the Beer-Lambert law, \(\varepsilon(v)\) is pivotal; along with molar concentration and path length, it defines the sample's absorbance at a particular wavelength.

It's useful for students to note that the molar absorption coefficient isn't an arbitrary constant — it reflects intrinsic properties of the substance, including how it interacts with light in terms of its electronic structure. In educational demonstrations, substances with high \(\varepsilon(v)\) values are often chosen to clearly show the effects of concentration changes on absorbance.