Question

The absorption cross section for fluorescein, a dye that is used as a fluorescent probe for biological imaging (see the Chapter Opener image), is 9.25×10^-16 cm² molecule^-1 at 500 nm. How much light is transmitted through a 1-cm path length by a standard 10^-6 M solution of fluorescein used to label biological samples?

Short answer

The amount of light transmitted through 1 cm path length is calculated using Beer – Lambert law.

So, the amount of light transmitted is $500nm$.

Step-by-step solution

Definition of Fluorescein:

Fluorescein is a phthalic indicator dye that is normally yellow – green in colour and bright green in a more alkaline medium. Fluorescein is a fluorescent compound that has a maximum absorbance of and an emission maximum of $\mathbf{\text{521nm}}$

Formula –role="math" localid="1663587258066" ${\mathbf{\text{C}}}_{\mathbf{\text{20}}}{\mathbf{\text{H}}}_{\mathbf{\text{12}}}{\mathbf{\text{O}}}_{\mathbf{\text{5}}}$

Fluorescein, a xanthate dye that is highly fluorescent, detectable even when present in small quantities, is an organic compound that is widely used as a synthetic colouring agent. This agent emits an intense green colour fluorescence in the presence of alkaline solution and therefore is used as a label.

Definition of Beer – Lambert Law:

The Lambert- Beer law states that there is a linear relation between the concentration and absorbance of the solution.

$\mathit{A}\mathbf{=}\mathit{\epsilon }\mathit{c}\mathit{l}$

Beer Lambert law states the fact a constant proportion of light is absorbed per unit length and is independent of the intensity of incident radiation.

role="math" localid="1663587542110" $\frac{{\mathbf{I}}_{\mathbf{t}}}{{\mathbf{I}}_{\mathbf{0}}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{N}}{\mathbf{V}}\mathbf{\sigma }\mathbf{l}}$

Where, ${\mathit{I}}_{\mathbf{t}}$intensity of transmitted radiation

${\mathit{I}}_{\mathbf{0}}$: Intensity of incident radiation

$\frac{\mathbf{N}}{\mathbf{V}}$:Number density

$\mathit{\sigma }$: Cross – section area

$\mathit{l}$: Path length

Using the Beer – Lambert law, the intensity of radiation that is absorbed or transmitted in then given by:

$\begin{array}{rcl}{\mathit{I}}_{\mathbf{t}}\mathbf{-}{\mathbf{I}}_{\mathbf{0}}& \mathbf{=}& {\mathbf{I}}_{\mathbf{0}}\mathbf{-}{\mathbf{I}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{N}}{\mathbf{V}}\mathbf{\sigma }\mathbf{l}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\text{As}}\mathbf{ }\mathbf{ }\frac{{\mathbf{I}}_{\mathbf{t}}}{{\mathbf{I}}_{\mathbf{0}}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{N}}{\mathbf{V}}\mathbf{\sigma }\mathbf{l}}\mathbf{)}\mathbf{}\\ & \mathbf{=}& {\mathbf{I}}_{\mathbf{0}}\mathbf{[}\mathbf{1}\mathbf{-}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{N}}{\mathbf{V}}\mathbf{\sigma }\mathbf{l}}\mathbf{]}\end{array}$

Amount of light transmitted through a standard solution of fluorescein:

The intensity of radiation transmitted is given by (using Beer- Lambert law) the relation:

${\mathrm{I}}_t-{\mathrm{I}}_0={\mathrm{I}}_0[1-{\text{e}}^{-\frac{\text{N}}{\text{V}}\sigma l}]$

Calculate the value of N/V using ideal gas equation i.e.

$\text{PV}\text{=}\text{nRT}$

And, $\text{n}=\frac{\text{N}}{{\text{N}}_{\text{A}}}$ (N_A=Avogadro’s number: 6.022 × 10²³)

Therefore, $\frac{\text{N}}{\text{V}}={\text{P.N}}_{\text{A}}\text{RT}$

Now, put the value of P, V, R and T in the above equation, we get:

$\begin{array}{rcl}\frac{\text{N}}{\text{V}}& =& \frac{1\times 6.022\times {10}^{23}}{0.0821\times 300}\\ & =& 0.244\times {10}^{23}\end{array}$

Now, put the value of $N,V,\sigma ,l$ andI₀ in the equation ${\mathrm{I}}_t-{\mathrm{I}}_0={\mathrm{I}}_0[1-{\text{e}}^{-\frac{\text{N}}{\text{V}}\sigma l}]$-

$$\begin{gathered} =500\times {10}^{-7}\times [1-\exp (-0.244\times {10}^{23}\times 9.25\times {10}^{-16}\times 1\left)\right] \\ =500\times {10}^{-7}\times [1-0] \\ =500\times {10}^{-7} \end{gathered}$$

The amount of light transmitted through a 1-cm path length by a standard 10^-6 M solution of fluorescein used to label biological samples is 500nm.