Understanding Absorption and Transmission in Spectroscopy

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SUMMARY

This discussion clarifies the relationship between absorption and transmission in spectroscopy, specifically using the equation A = LOG(1/T), where A represents absorption and T represents transmission. A transmission reading of 60% corresponds to an absorption of approximately 22.2%, confirming the mathematical accuracy of the equation. Participants emphasize the importance of understanding the roles of reflection and scattering in these measurements, as well as the need for proper consultation of manuals and resources when conducting experiments.

PREREQUISITES
  • Understanding of the equation A = LOG(1/T) for calculating absorption.
  • Familiarity with the concepts of light transmission, absorption, and reflection in spectroscopy.
  • Basic knowledge of scattering processes, including elastic and inelastic scattering.
  • Experience with spectrometers and their operational manuals.
NEXT STEPS
  • Research the principles of light scattering in spectroscopy.
  • Learn about the impact of reflection on absorption and transmission measurements.
  • Explore the differences between absorbance and absorption in spectroscopic contexts.
  • Consult resources on the proper use of spectrometers and their equations.
USEFUL FOR

Chemistry students, laboratory technicians, and researchers involved in spectroscopy and light measurement techniques.

Pengwuino
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I'm a little confused, we have tested chemical concentrations based on how much light it can absorb. What i don't understand is why something can have a reading of 60% transmission yet also have an absorption of 22.2% via the equation A = LOG(1/T), T being tranmission, A being absorption. Seems like it should be a 40% absorption. Then i realized i might not quite understand what it means when a spectrometer is reading "Absorption and Transmission". Can someone explain?
 
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The rest goes to scattering processes I guess. One can split the mass-energy absorption factor to elastic and inelastic scattering and photoabsorption.
 
You may have got something wrong with your equation. Maybe you should consult the manual again.
 
Ich said:
You may have got something wrong with your equation. Maybe you should consult the manual again.

No that's how the math works. The meter has a transmission and an absorption scale on the same dial going opposite ways but the Absorption is logarithmic.
 
Reflection accounts for some of the difference...

R + T + A = 1 for all processes.
 
Hey, when T<0.1 -> A>1. That surely does NOT work. You must have forgot something crucial.
 
Ich is right, not only does that allow for absorption to be more than 100%, but A+T > 100% whenever the transmission is less than 13.7%.
 
Ich said:
You may have got something wrong with your equation. Maybe you should consult the manual again.

This is the key to the argument... start consulting.
 
It was a lab for our chem course, we didn't have manuals to consult, we were just told "here, use this equation" and did as we were told haha.
 
  • #10
Pengwuino said:
It was a lab for our chem course, we didn't have manuals to consult, we were just told "here, use this equation" and did as we were told haha.

Ich is right that reflection does come into play, but the percent of light absorbed, transmitted, and reflected depends on what compound you are talking about. Are you sure it wasn't a sort of "introductory thought experiment" to get to used to using equations to model light A&R&T?
 
  • #11
It was probably the only time we used the spectrometer this semester haha. It was one of those "hey look, you can determine concentration via light absorption, how bout that" experiments.
 
  • #12
Got it now - You´re measuring http://en.wikipedia.org/wiki/Absorbance" , not absorption.
 
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  • #13
It's less complicated than that...

Your reading of 22.2 is about right. If transmission = T is 60%, or 0.60, then per your formula absorption = A is log(1/T). Well, 1/.60 is 1.66, and the log of that is .2218, or about 22.2%. So your work stands.
 

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