HI community!
Actually i was going through the THOR Lab manual for FTIR spectroscopy and got stuck at understanding the relationship between resolution expressed in wavenumbers (R cm^1-) and in nanometer (R(nm). In the Table-2 (link given below) of that manual, can someone explains the conversion formulae used for calculation ?
Best answer by Sven.Stöttinger
Hi,
your link does not seem to work properly, it only directs back to this thread. But I might have an idea what your question is leading to.
Generally using wavelength in nm is somewhat impractical if you want to express a Δenergy-value, since the same Δ-wavelength value in nm relates to a different ΔE in dependence if you for example are looking at the same Δnm value at 300 nm or at 800nm. This is why the wavenumber (cm^-1) or eV as Δenergy units are more commonly used in spectroscopy.
Just an example: at 500 nm a Δ50nm difference corresponds to a ΔE of roughly 225 meV or 1818 cm^-1,
at 900 nm however the same Δ50nm difference now correpsonds to a ΔE of roughly 72.5 meV or 585 cm^-1.
Here is a link to a online calculator where you can put in wavelength and wavelength-difference to get the energy value calculated (sorry it is in german, but looking at the units should help :) ):
https://optikexpertisen.de/interaktiv/spektralkonverter
Hope this helps,
Sven
Hi,
your link does not seem to work properly, it only directs back to this thread. But I might have an idea what your question is leading to.
Generally using wavelength in nm is somewhat impractical if you want to express a Δenergy-value, since the same Δ-wavelength value in nm relates to a different ΔE in dependence if you for example are looking at the same Δnm value at 300 nm or at 800nm. This is why the wavenumber (cm^-1) or eV as Δenergy units are more commonly used in spectroscopy.
Just an example: at 500 nm a Δ50nm difference corresponds to a ΔE of roughly 225 meV or 1818 cm^-1,
at 900 nm however the same Δ50nm difference now correpsonds to a ΔE of roughly 72.5 meV or 585 cm^-1.
Here is a link to a online calculator where you can put in wavelength and wavelength-difference to get the energy value calculated (sorry it is in german, but looking at the units should help :) ):
https://optikexpertisen.de/interaktiv/spektralkonverter
Hope this helps,
Sven
Thanks
The following relation along with your explanation well explains this artifact :
delta_lanmbda = (delta_nu_prime / nu_prime) * lambda
So, let’s say for resoluting a wavelength of 200 nm with resolution of 1cm^-1,
delta_lambda = (200 nm /50000 cm^-1) * 1 cm^-1 = 0.004 nm
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