can anyone tell me the difference between absorption spectroscopy and extinction spectroscopy in terms of experiment? and how to get extinction spectroscopy, how to get absorption spectroscopy? Thank you so much.
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https://physicstoday.scitation.org/doi/abs/10.1063/PT.3.4547?journalCode=pto reports experimental distinctions between the two, and credits reference 14 with doing that work: Read https://iopscience.iop.org/article/10.1088/2053-1583/aab670 and https://aip.scitation.org/doi/10.1063/1.5047792 for experimental procedures. My guess is that you'd measure scattered light as well as transmitted. The scattered light would be the difference
- – Sarah Messer Aug 24 '20 at 19:24.
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There should not be any difference in experiment, because the extinction is something you have to calculate by measuring the transmittance/absorbance of a given frequency.
The transmittance $T$ of a material is the ratio between the received intensity of a certain frequency to the transmitted intensity of the same frequency. $$T=I/I_0$$ The absorbance $A$ is defined as the decadic logarithm of the inverse transmittance. $$A=\log_{10}\left(I_0/I\right)=\log_{10}\left(1/T\right)$$
By using Beer-Lambert’s law $$I=I_0\operatorname{e}^{-\varepsilon~c~d}$$ one can calculate the extinction $\varepsilon$ from the measured transmittance/absorbance by $$\varepsilon=\frac{A}{c~d}$$ where $c$ is the concentration of the absorbing material and $d$ is the distance that the light has to travel in that given material.
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