159.PHOTOCHEMISTRY(8) – LAWS OF PHOTOCHEMISTRY(5).

In the last few posts we have been discussing the Beer – Lambert’s law and various terms associated with it. In this post we are going to learn another way of expressing the Beer- lambert’s law , using calculus.

As studied in post 8 , calculus is the study of how quantities change over small intervals of time. We can thus use differentiation to study the rate of change of light intensity with thickness of the medium (Lambert’s law) and its concentration(Beer’s law).

The Lambert’s law , which we studied in post 156 ,can also be stated as –

The fraction of incident radiation absorbed by a transparent medium is independent of the intensity of the incident radiation and that each successive layer of the medium absorbs an equal fraction of incident radiation.

In this statement, the thickness of the medium is referred to as ‘each successive layer’ of the medium. The more the successive layers , greater will be the thickness of the medium.

For differentiating we divide the thickness of the medium / path length(L) into several imaginary small parts of thickness dI as shown below(the figure is blown up in scale for simplifying the illustration – the dotted lines should be very close to each other) –

The dotted lines help divide the entire length of the medium into several small parts of equal thickness dI. According to the Lambert’s law , the intensity of light is directly proportional to the pathlength. The equation can be expressed as follows –

The negative sign in the above expression indicates that the intensity of the light is decreasing as we proceed along the pathlength.

Incorporating the Beer’s law ( intensity proportional to concentration ,C ) we get,

Introducing constant of proportionality(αv), we get –

The quantity Cdl indicates the concentration of the solute per unit area of the layer i.e the amount of solute present in every layer of thickness dl. The concentration is moles per volume and volume is area times thickness-

Therefore, cdl = moles / area.

Now we again consider the equation we derived earlier in this post –

As seen in post 11 , we need to collect the small components that we created while differentiating and bring them together to get the final result . We do this by integrating the above equation with boundary conditions. Boundary conditions help define the behaviour of the function at the boundaries , as shown in the figure below –

The two boundary conditions we set here are –

i) at the beginning ,when path length is zero l = l0 , the intensity of incident light is 0 i.e = 0

and

ii) at the end l = L , the intensity of transmitted light is Ɪ = Ɪt

Integrating with the boundary conditions –

As seen above, after integrating the equation with the boundary conditions, we get equation (1) as the solution. We also know that converting logarithm with base e to logarithm to the base 10 requires multiplying by 2.303 as follows –

Thus , the expression in decadic form is expressed as –

ε = 2.303 (αv) and is called as the molar extinction coefficient or molar absorptivity. It is the measure of how strongly a chemical species absorbs light at a particular wavelength i.e probability of electronic transition at that wavelength .It is an intrinsic property that depends on the structure and chemical composition of the molecule. The units of ε are m2/mol .However in practice it is M-1 cm-1.

And from the previous post , we already know , that –

Therefore, substituting value of A in equation 2 , we get =

A = εCL

This is our expression for the Beer – Lamberts law !! (refer post 156 )

So this is how we arrived at this exact same equation using a different approach !! In the next post we shall further talk about the Beer -Lambert’s law. Till then ,

Be a perpetual student of life and keep learning ….

Good day !!

References and further reading –

  1. Fundamentals of Photochemistry by K K Rohatagi – Mukherjee.

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