March 28, 2024, 03:45:38 PM
Forum Rules: Read This Before Posting


Topic: Two-line- and continuum background correction method  (Read 1297 times)

0 Members and 1 Guest are viewing this topic.

Offline Valerio

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
Two-line- and continuum background correction method
« on: December 25, 2019, 10:13:25 AM »
Hello

I think that I don't get the whole picture about the two-line correction method.
What I do understand is that another line is chosen ( as close as possible but not equal to the resonance line). The attenuation of this chosen line is due to the background.

What I don't understand is that it can be used to correct the absorbance of the analyte line.
Does this imply that the intensities of both lines had to be the same before attenuation and that the background attentuates both line to the same degree? If this is the case then I understand that it is usable to correct the analyte line absorbance.

============
As with continuum background correction they do something similar it seems. I don't quite understand why it is alright to substract the absorbance of the continuum source from the absorbance of the analyte line. Below you can see how far I think that I understand.

The deuteriumlamp provide continous radiation throughout the UV-region of the spectrum. The machine is set on the resonance wavelength, so the monochromator only lets a small band of wavelengths through with the resonance line in the middle. At the resonance wavelength, both the analyte and the background attenuate the line. At all the other wavelengths in the band that came out of the monochromator, the background attenuates those wavelengths but the analyte does not. The questions I also ask myself are:

1. All these wavelengths hit the detector (with the machine set on the resonance wavelength), can the detector distinguish between the different wavelengths?

2. If not, what is the intensity that is used to calculate the absorbance? The sum of the intensities of the different wavelengths or an average intensty?



Kind regards


Valerio

Offline billnotgatez

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 4399
  • Mole Snacks: +223/-62
  • Gender: Male
Re: Two-line- and continuum background correction method
« Reply #1 on: January 03, 2020, 03:16:22 PM »
Quote
Reported by Valerio on Today at 12:20:51 PM. They left the following message:
It has been a few days since I posted this topic.
Up until now, I haven't received an answer.
Maybe my questions were not clearly written.
Should I rephrase my questions and repost topic?

I am sorry you have not had a response yet.
We are all volunteers here and someone with the answer has not logged on yet during the holiday season.
You can add more info to this post but please do not re-post since it will cause confusion if 2 separate members post to both threads.

Sometimes a little more information about what you understand so far helps form the best answer.
 

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3471
  • Mole Snacks: +526/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Two-line- and continuum background correction method
« Reply #2 on: January 03, 2020, 05:09:43 PM »
It's not even clear what analytical technique the OP is referring to. Some more context is needed.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Valerio

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
Re: Two-line- and continuum background correction method
« Reply #3 on: January 04, 2020, 03:25:11 AM »
Okay, let's rephrase and condense my first post.

About the continuum background correction used in atomic absorption spectroscopy.
I'm looking from a theoretical point of view, as I haven't had the chance to do labs this term.

What I do understand is the following:

°When a hollow cathode lamp is used, we measure the absorbance at a resonance wavelength of for example 230 nm. The absorbance that is measured, is the sum of the absorbance by the analyte and the absorbance of the background ( molecules that did not atomize,...)
°When a deuterium lamp (continuum source) is used, we also measure the absorbance at the same resonance wavelength of 230 nm. The absorbance measured is approximately equal to the absorbance by the background. This is because of the fact that a portion of the continuum is allowed to exit the monochromator and hit the detector. Less than 1% of the measured absorbance is due to the analyte.


What I'm struggling with is the following:

Suppose that the bandwidth of the light hitting the detector is 0,2nm and the linewidth of the resonance line is 0,002nm. This band of light consists of several wavelengths. Is it necessary that the intensity of all these wavelengths are (both initially and after absorption) approximately the same? When all these wavelengths hit the detector we measure an absorbance value.

This absorbance measured is from several wavelengths hitting the detector and not just the resonance line. How is this measured absorbance an estimate of absorbance by the background at the resonance line? Is this measured absorption maybe an average absorbance over the whole bandwidth?


Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3471
  • Mole Snacks: +526/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Two-line- and continuum background correction method
« Reply #4 on: January 05, 2020, 01:37:02 AM »
In any absorption spectroscopy the detector doesnt care about photon energy. The monoschromoeter selects what photons hit the detector and the detector basically adds them up. The amount that hit the detector with and without the sample are compared, and this is the transmission value which is converted to absorption. 

 The measured absorption is an average over the bandpass opening of course, but in line source mode the resolution of the analyte absorption is provided by the narrow excitation source. The background isnt constant necessarily across the bandpass but the detector doesnt know that. So effectively it's an average across the monochrometer bandpass.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Sponsored Links