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In this problem you will investigate the parameters involved in a single- molecule fluorescence experiment. Specifically, the incident photon power needed to see a single molecule with a reasonable signal-to-noise ratio will be determined. a. Rhodamine dye molecules are typically employed in such experiments because their fluorescence quantum yields are large. What is the fluorescence quantum yield for Rhodamine B (a specific rhodamine dye) where \(k_{r}=1 \times 10^{9} \mathrm{s}^{-1}\) and \(k_{i c}=1 \times 10^{8} \mathrm{s}^{-1} ?\) You can ignore intersystem crossing and quenching in deriving this answer. b. If care is taken in selecting the collection optics and detector for the experiment, a detection efficiency of \(10 \%\) can be readily achieved. Furthermore, detector dark noise usually limits these experiments, and dark noise on the order of 10 counts \(s^{-1}\) is typical. If we require a signal- tonoise ratio of \(10: 1,\) then we will need to detect 100 counts \(\mathrm{s}^{-1} .\) Given the detection efficiency, a total emission rate of 1000 fluorescence photons \(s^{-1}\) is required. Using the fluorescence quantum yield and a molar extinction coefficient for Rhodamine \(\mathrm{B}\) of \(\sim 40,000 \mathrm{M}^{-1} \mathrm{cm}^{-1},\) what is the intensity of light needed in this experiment in terms of photons \(\mathrm{cm}^{-2} \mathrm{s}^{-1} ?\) c. The smallest diameter focused spot one can obtain in a microscope using conventional refractive optics is approximately one-half the wavelength of incident light. Studies of Rhodamine B generally employ 532 nm light such that the focused-spot diameter is \(\sim 270 \mathrm{nm}\). Using this diameter, what incident power in watts is required for this experiment? Do not be surprised if this value is relatively modest.

Step by step solution

01

a. Calculate the fluorescence quantum yield for Rhodamine B

To calculate the fluorescence quantum yield, use the equation: Quantum Yield (QY) \(= \frac{k_r}{k_r + k_{ic}}\) where, \(k_r=1 \times 10^9 s^{-1}\) (radiative rate constant) and \(k_{ic}=1 \times 10^8 s^{-1}\) (internal conversion rate constant) Plug in the values and calculate the quantum yield: QY \(= \frac{1 \times 10^9}{1 \times 10^9 + 1 \times 10^8} = \frac{1 \times 10^9}{1.1 \times 10^9} = \frac{10}{11}\)

02

b. Determine the intensity of light needed

We are given the following information: 1. Detection efficiency, DE \(= 10\% = 0.1\) 2. Required detected counts \(= 100\) counts \(s^{-1}\). 3. Molar extinction coefficient, \(\epsilon = 40,000\) M\(^{-1}\) cm\(^{-1}\) First, we need to calculate the total fluorescence emission rate for achieving the required detected counts: Total emission rate \(= \frac{\text{required counts}}{\text{detection efficiency}} = \frac{100}{0.1} = 1000\) fluorescence photons \(s^{-1}\). Now, find the number of absorbed photons required, using the quantum yield (QY) from part a: Absorbed photons rate, A = \(\frac{\text{total emission rate}}{\text{QY}} = \frac{1000}{\frac{10}{11}} = 1100\) photons \(s^{-1}\). We will now use the Beer-Lambert law to calculate the optical density (OD): OD \(= \epsilon \times c \times l\) where, \(c\) is the concentration, and \(l\) is the path length. Here, we are looking for the intensity in terms of absorbed photons per area. To replace concentration with a pseudo concentration, \(c'\): \(c' = \frac{\text{absorbed photons rate}}{\epsilon \times A'} = \frac{1100}{40,000 \times A'}\) where \(A'\) is the surface area. Now, rearrange for the intensity: Intensity, I \(= \frac{1100}{\epsilon \times c' \times l} = \frac{1100}{40,000 \times \frac{1100}{40,000 \times A'} \times l} = \frac{1}{l} \times A'\) photons cm\(^{-2}\) s\(^{-1}\).

03

c. Find the incident power in watts

We're given the following information: 1. Wavelength of incident light, \(\lambda = 532\) nm. 2. Focused-spot diameter, \(d \approx 270\) nm. First, find the area (A) of the focused spot: \(A = \pi \times \left( \frac{d}{2} \right)^2 = \pi \times \left( \frac{270 \times 10^{-7}}{2} \right)^2\) cm\(^2\). Now, the incident photon flux rate can be found by multiplying the intensity (from part b) by the area (A): Incident photon flux rate \(= \text{Intensity} \times A = \frac{1}{l} \times A \times A\) photons s\(^{-1}\). Next, find the energy of each photon using the formula: \(E = \frac{hc}{\lambda}\) where, \(h = 6.63 \times 10^{-34}\) Js (Planck's constant), \(c = 3 \times 10^{8}\) m s\(^{-1}\) (speed of light), and \(\lambda = 532 \times 10^{-9}\) m (wavelength). Now, calculate the incident power (in watts): Incident power \(= \text{Incident photon flux rate} \times E = \frac{1}{l} \times A \times A \times E\) W. Plug in the values for A, E, and l (path length) to find the required incident power in watts.

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Key Concepts

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

Fluorescence Quantum Yield

Fluorescence quantum yield is a crucial parameter when it comes to understanding the efficiency of the fluorescence process in molecules. In simple terms, it tells us how effective a molecule is at converting absorbed light into emitted light. It's like comparing how much of the absorbed energy ends up being usable light.To calculate the fluorescence quantum yield (\(QY\)), you can use the equation:\[QY = \frac{k_r}{k_r + k_{ic}}\]Here,

• \(k_r\) is the radiative rate constant, which relates to how quickly the molecule emits photons.
• \(k_{ic}\) is the internal conversion rate constant associated with non-radiative relaxation processes.

For Rhodamine B, substituting the given values:

• \(k_r = 1 \times 10^9 \,\text{s}^{-1}\)
• \(k_{ic} = 1 \times 10^8 \,\text{s}^{-1}\)

This gives us a quantum yield of \(\frac{10}{11}\). A high quantum yield such as this means Rhodamine B is highly efficient at fluorescence, making it ideal for single-molecule fluorescence experiments.

Signal-to-Noise Ratio

Signal-to-noise ratio (SNR) is an important concept in any measurement, including fluorescence experiments. It tells us how much of the signal we detect is reliable compared to the background noise. Having a high SNR is necessary for clear and accurate detection of single molecules. In scenarios where we require an SNR of 10:1, it indicates that the signal should be ten times greater than the noise to ensure reliable detection. Noise in fluorescence detection comes from various sources, including dark noise from the detector. To achieve the desired SNR, care must be taken to enhance the detected signal:

• Improving the detection efficiency — in our exercise, a 10% efficiency is used.
• Optimizing optical components to maximize light collection.

For this problem, achieving an SNR of 10:1 requires us to detect 100 counts per second, necessitating a total emission rate of 1000 photons per second due to the limited efficiency.

Rhodamine B Fluorescence

Rhodamine B is a popular fluorescent dye in scientific experiments due to its impressive fluorescence characteristics, particularly its high quantum yield.This characteristic, along with a strong molar extinction coefficient around 40,000 M\(^{-1}\) cm\(^{-1}\), ensures Rhodamine B excites efficiently and emits a strong fluorescent signal. That makes it suitable for demanding applications such as single-molecule fluorescence, where seeing each molecule's light emission directly is critical.When involved in experiments like our exercise, Rhodamine B's reliable fluorescence properties allow researchers to predict and calculate the emission rates and absorbed photon requirements accurately. This helps in planning the intensity of light required to visualize molecules with desired precision.

Photon Emission Rate

The photon emission rate is the number of photons emitted per second by a fluorescent source, such as a single molecule labeled with Rhodamine B. In fluorescence experiments, knowing the emission rate is crucial because it influences the amount of incident light required and the detection setup. To achieve specific detection outcomes, like an SNR of 10:1 and 100 counts per second, a total emission rate of 1000 photons per second is necessary. The relationship between the emission rate, quantum yield, and detection efficiency is clear:

• Emission rate depends on the balance between the photon absorption and the molecule's fluorescence characteristics.
• Efficient fluorescence means fewer absorbed photons are needed to achieve high emission rates.

In experiments, calculating this rate allows scientists to determine how much light intensity and power are essential to meet the observation goals without overestimating resources.