In water, the fluorescence quantum yield and observed fluorescence lifetime of tryptophan are Øf =0.20 and 7, =2.6 ns, respectively. the fluorescence rate constant kç is ? !!
In water, the fluorescence quantum yield and observed fluorescence lifetime of tryptophan are Øf =0.20 and 7, =2.6 ns, respectively. the fluorescence rate constant kç is ? !!
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Fluorescence Quantum Yield and Lifetime of Tryptophan in Water**
In an aqueous medium, the fluorescence quantum yield (ϕ_f) and the observed fluorescence lifetime (τ₀) of the amino acid tryptophan are given.
- The fluorescence quantum yield (ϕ_f) is 0.20.
- The observed fluorescence lifetime (τ₀) is 2.6 nanoseconds (ns).
The task is to determine the fluorescence rate constant (k_f).
Understanding these parameters:
1. **Fluorescence Quantum Yield (ϕ_f):**
- This is the ratio of the number of photons emitted to the number of photons absorbed. It indicates the efficiency of the fluorescence process.
2. **Fluorescence Lifetime (τ₀):**
- This is the average time the molecule stays in its excited state before emitting a photon.
To determine the fluorescence rate constant (k_f), the following relationship between the quantum yield, the rate constant, and the lifetime can be used:
\[ ϕ_f = k_f \times τ₀ \]
Rearranging the equation to solve for the fluorescence rate constant (k_f):
\[ k_f = \frac{ϕ_f}{τ₀} \]
Plugging in the provided values:
\[ k_f = \frac{0.20}{2.6 \, \text{ns}} \]
\[ k_f = \frac{0.20}{2.6 \times 10^{-9} \, \text{s}} \]
\[ k_f \approx 7.69 \times 10^7 \, \text{s}^{-1} \]
Thus, the fluorescence rate constant (k_f) is approximately \( 7.69 \times 10^7 \, \text{s}^{-1} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f91f67c-9902-4555-8b1c-c3ab12dce890%2Fe6c55069-e9dd-4062-b6d1-e8645953c5a3%2Fxttqxus_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Fluorescence Quantum Yield and Lifetime of Tryptophan in Water**
In an aqueous medium, the fluorescence quantum yield (ϕ_f) and the observed fluorescence lifetime (τ₀) of the amino acid tryptophan are given.
- The fluorescence quantum yield (ϕ_f) is 0.20.
- The observed fluorescence lifetime (τ₀) is 2.6 nanoseconds (ns).
The task is to determine the fluorescence rate constant (k_f).
Understanding these parameters:
1. **Fluorescence Quantum Yield (ϕ_f):**
- This is the ratio of the number of photons emitted to the number of photons absorbed. It indicates the efficiency of the fluorescence process.
2. **Fluorescence Lifetime (τ₀):**
- This is the average time the molecule stays in its excited state before emitting a photon.
To determine the fluorescence rate constant (k_f), the following relationship between the quantum yield, the rate constant, and the lifetime can be used:
\[ ϕ_f = k_f \times τ₀ \]
Rearranging the equation to solve for the fluorescence rate constant (k_f):
\[ k_f = \frac{ϕ_f}{τ₀} \]
Plugging in the provided values:
\[ k_f = \frac{0.20}{2.6 \, \text{ns}} \]
\[ k_f = \frac{0.20}{2.6 \times 10^{-9} \, \text{s}} \]
\[ k_f \approx 7.69 \times 10^7 \, \text{s}^{-1} \]
Thus, the fluorescence rate constant (k_f) is approximately \( 7.69 \times 10^7 \, \text{s}^{-1} \).
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