6
$\begingroup$

When I am supposed to calculate an IR-, Raman-, or UV/Vis-spectrum, the calculations give me some x-y-values, e.g., Frequency-Intensity, or Energy-Oscillator Strength. Those values can either be printed as stick spectra or a sum of Gaussian-/Lorentz-distributions is being used to broaden the peaks to generate a more "natural" spectrum.

I've been asked a couple of times which broadening function to choose when, but actually never had a good answer/given explanation. The only thing that I wrote down quite some time ago, was that for IR- or Raman-spectra of solids a Gaussian broadening and for every other type of IR- or Raman-spectra a Lorentz broadening should be used.

Google searches for "gaussian or lorentz broadening" and "line broadening" did yield results that try to explain it, but I didn't find most of them really helpful.

So the question is: When do I choose the Gaussian or the Lorentz-functions for artificial broadening of IR-, Raman-, or UV-Vis stick spectra and why?

Improve this question
$\endgroup$
2
  • $\begingroup$ Shortly after posting I found this, which is at least related, if not the same chemistry.stackexchange.com/questions/324/… ... but the answer is not enough for my taste. $\endgroup$
    – pH13 - Yet another Philipp
    Commented Aug 17, 2016 at 14:33
  • 1
    $\begingroup$ From the little bit I've learned, Lorentz would be the most accurate for a single broadening event but if you sample a population it would looks more like a Gaussian. That's what someone in my lab said at least... $\endgroup$
    – Jordan Epstein
    Commented Sep 17, 2016 at 19:52

1 Answer 1

Reset to default
2
$\begingroup$

I think that any of these functions, Gaussian, Lorenzian, Voight, will do, they do not differ by much but the Lorenzian has a slightly wider base and in overlapping spectra this may increase background too much. I assume that you will convolute your stick spectrum to get your 'real' spectrum, in this case you can zero pad the transform, it makes nicer looking spectra but adds no more information.
Perhaps the most important point to check is the spectral resolution of the instrument you are simulating spectra from, and adjust the width of you broadening function appropriately.

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.