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If \(\tau_{f}=1 \times 10^{-10} \mathrm{s}\) and \(k_{i c}=5 \times 10^{8} \mathrm{s}^{-1},\) what is \(\phi_{f} ?\) Assume that the rate constants for intersystem crossing and quenching are sufficiently small that these processes can be neglected.

Short Answer

Expert verified
Answer: The fluorescence quantum yield for the given molecule is approximately 0.0476, which means about 4.76% of the absorbed photons are emitted as fluorescence photons.

Step by step solution

01

Write down the given values

We have the following values given: - Fluorescence lifetime, \(\tau_{f} = 1 \times 10^{-10}\ \mathrm{s}\) - Rate of radiative decay, \(k_{i c} = 5 \times 10^{8}\ \mathrm{s}^{-1}\)
02

Write down the formula for fluorescence quantum yield

We will be using the formula: $$\phi_{f} = \frac{k_{i c} \tau_{f}}{1 + k_{i c} \tau_{f}}$$
03

Plug in the given values and solve for \(\phi_{f}\)

Now, we will substitute the given values for \(\tau_{f}\) and \(k_{i c}\) in the formula: $$\phi_{f} = \frac{(5 \times 10^{8}\ \mathrm{s}^{-1})(1 \times 10^{-10}\ \mathrm{s})}{1 + (5 \times10^{8}\ \mathrm{s}^{-1})(1 \times 10^{-10}\ \mathrm{s})}$$
04

Perform the calculations

Calculate the numerator and denominator separately: Numerator: $$(5 \times 10^{8}\ \mathrm{s}^{-1})(1 \times 10^{-10}\ \mathrm{s}) = 5 \times 10^{-2}$$ Denominator: $$1+ (5 \times 10^{8}\ \mathrm{s}^{-1})(1 \times 10^{-10}\ \mathrm{s}) = 1 + 5 \times 10^{-2} = 1.05$$ Now, calculate \(\phi_{f}\): $$\phi_{f} =\frac{5 \times 10^{-2}}{1.05} \approx 0.0476$$
05

Interpret the result

The fluorescence quantum yield, \(\phi_{f}\), is approximately 0.0476. This value represents the efficiency with which absorbed photons are emitted as fluorescence photons. In this case, about 4.76% of the absorbed photons are emitted as fluorescence.

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Key Concepts

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

Fluorescence Lifetime
Fluorescence lifetime, denoted by \(\tau_f\), is a parameter that represents the average time a molecule remains in its excited state before returning to the ground state via fluorescence. This concept is crucial in understanding how quickly fluorescence occurs. A longer \(\tau_f\) indicates that the excited state is relatively stable and that emitted photons are released over a longer period.

In practical terms, fluorescence lifetime is measured in nanoseconds (ns), and it's affected by various factors, such as the molecular structure of the fluorophore, the solvent, and temperature. In the given exercise, we're working with a fluorescence lifetime of \(1 \times 10^{-10}\) seconds, which suggests a very fast fluorescence process.

It's important to note that fluorescence lifetime doesn't give us a direct measure of fluorescence efficiency; it only tells us about the duration of fluorescence. However, it is directly related to both non-radiative and radiative decay rates, which leads us to our next concept.
Radiative Decay Rate
The radiative decay rate, represented as \(k_{ic}\), is the probability per unit time of the excited state returning to the ground state by emitting a photon, which in effect contributes to fluorescence. Therefore, it provides an indication of how likely it is for fluorescence to occur compared to other non-radiative processes such as heat loss.

The rate is typically measured in reciprocal seconds (\(s^{-1}\)) and can vary widely depending on the nature of the fluorescent molecule and its environment. A higher radiative decay rate suggests a greater propensity for a molecule to fluoresce.

In our exercise, the radiative decay rate is \(5 \times 10^{8}\ \mathrm{s}^{-1}\). This signifies a strong likelihood of fluorescence occurring rapidly once the molecule is excited. The radiative decay rate plays a pivotal role in the quantum yield calculation as illustrated in our exercise solution.
Quantum Yield Calculation
The quantum yield of fluorescence, symbolized by \(\phi_f\), is the ratio of the number of photons emitted to the number of photons absorbed. This yield is a dimensionless number ranging from 0 to 1, representing the fluorescence efficiency of a molecule.

For the calculation of the quantum yield, we employ the formula:
\[\phi_{f} = \frac{k_{i c} \tau_{f}}{1 + k_{i c} \tau_{f}}\]
This shows us that the quantum yield is dependent on both the fluorescence lifetime \(\tau_{f}\) and the radiative decay rate \(k_{i c}\). The operation is as follows: calculate the product of \(k_{i c}\) and \(\tau_{f}\), and then divide this product by the sum of 1 and the same product.

Within the context of our exercise, we performed these calculations and found that the quantum yield is approximately 0.0476, indicating that under the given conditions, around 4.76% of the absorbed photons result in fluorescence. This value helps assess the effectiveness of fluorescence and is particularly important in fields like photochemistry and photobiology, where the understanding of light-matter interactions is essential.
When considering improvements to the exercise, it's crucial to explain the physical meaning of each term and the relevance of quantum yield in practical applications, such as in the design of fluorescent probes for biological imaging.

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Most popular questions from this chapter

The quantum yield for \(\mathrm{CO}(g)\) production in the photolysis of gaseous acetone is unity for wavelengths between 250 and \(320 \mathrm{nm} .\) After 20.0 min of irradiation at \(313 \mathrm{nm}\) \(18.4 \mathrm{cm}^{3}\) of \(\mathrm{CO}(g)\) (measured at \(1008 \mathrm{Pa}\) and \(22^{\circ} \mathrm{C}\) ) is produced. Calculate the number of photons absorbed and the absorbed intensity in \(\mathrm{J} \mathrm{s}^{-1}\)

The reaction of nitric oxide \((\mathrm{NO}(g))\) with molecular hydrogen \(\left(\mathrm{H}_{2}(g)\right)\) results in the production of molecular nitrogen and water as follows: \\[ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \\] The experimentally-determined rate-law expression for this reaction is first order in \(\mathrm{H}_{2}(g)\) and second order in \(\mathrm{NO}(g)\) a. Is the reaction as written consistent with the experimental order dependence for this reaction? b. One potential mechanism for this reaction is as follows: \\[ \begin{array}{l} \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} \\] Is this mechanism consistent with the experimental rate law? c. An alternative mechanism for the reaction is \\[ \begin{array}{l} 2 \mathrm{NO}(g) \frac{k_{1}}{\sum_{k-1}} \mathrm{N}_{2} \mathrm{O}_{2}(g) \text { (fast) } \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{2}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} \\] Show that this mechanism is consistent with the experimental rate law.

For the reaction \(\mathrm{I}^{-}(a q)+\mathrm{OCl}^{-}(a q) \rightleftharpoons\) \(\mathrm{OI}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) occurring in aqueous solution, the following mechanism has been proposed: \\[ \begin{array}{l} \mathrm{OCl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \quad \frac{k_{1}}{\overrightarrow{k_{-1}}} \quad \mathrm{HOCl}(a q)+\mathrm{OH}^{-}(a q) \\ \mathrm{I}(a q)+\mathrm{HOCl}(a q) \stackrel{k_{2}}{\longrightarrow} \mathrm{HOI}(a q)+\mathrm{Cl}^{-}(a q) \\ \mathrm{HOI}(a q)+\mathrm{OH}^{-}(a q) \stackrel{k_{3}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{OI}^{-}(a q) \end{array} \\] a. Derive the rate law expression for this reaction based on this mechanism. (Hint: \(\left[\mathrm{OH}^{-}\right]\) should appear in the rate law. b. The initial rate of reaction was studied as a function of concentration by Chia and Connick [J. Physical Chemistry \(63(1959): 1518]\), and the following data were obtained: $$\begin{array}{lccc} & & & \text { Initial Rate } \\ {\left[\mathbf{I}^{-}\right]_{0}(\mathbf{M})} & {\left[\mathbf{O C l}^{-}\right]_{0}(\mathbf{M})} & {\left[\mathbf{O H}^{-}\right]_{0}(\mathbf{M})} & \left(\mathbf{M} \mathrm{s}^{-1}\right) \\ \hline 2.0 \times 10^{-3} & 1.5 \times 10^{-3} & 1.00 & 1.8 \times 10^{-4} \\ 4.0 \times 10^{-3} & 1.5 \times 10^{-3} & 1.00 & 3.6 \times 10^{-4} \\ 2.0 \times 10^{-3} & 3.0 \times 10^{-3} & 2.00 & 1.8 \times 10^{-4} \\ 4.0 \times 10^{-3} & 3.0 \times 10^{-3} & 1.00 & 7.2 \times 10^{-4} \end{array}$$ Is the predicted rate law expression derived from the mechanism consistent with these data?

Consider the following mechanism for ozone thermal decomposition: \\[ \begin{array}{l} \mathrm{O}_{3}(g) \stackrel{k_{1}}{\rightleftharpoons} \mathrm{O}_{2}(g)+\mathrm{O}(g) \\ \mathrm{O}_{3}(g)+\mathrm{O}(g) \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{O}_{2}(g) \end{array} \\] a. Derive the rate law expression for the loss of \(\mathrm{O}_{3}(g)\) b. Under what conditions will the rate law expression for \(\mathrm{O}_{3}(g)\) decomposition be first order with respect to \(\mathrm{O}_{3}(g) ?\)

Consider the following mechanism, which results in the formation of product \(P:\) \\[ \begin{array}{l} \mathrm{A} \stackrel{k_{1}}{\rightleftharpoons_{k-1}} \mathrm{B} \frac{k_{2}}{\rightleftharpoons_{-2}} \mathrm{C} \\ \mathrm{B} \stackrel{k_{3}}{\rightarrow} \mathrm{P} \end{array} \\] If only the species \(A\) is present at \(t=0,\) what is the expression for the concentration of \(\mathrm{P}\) as a function of time? You can apply the preequilibrium approximation in deriving your answer.

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