I noticed that in some X-ray based analysis, such EDS, the spectra of elements are given in cps/eV. Why is this the case? why not only cps is provided on the y-axis?
Thanks
Dear Faisal Mohammed ,
the advantage of such an representation of the 'intensity' scale in cps/energy is, that an integral over a peak or an energy section is
a) independent on the energy resolution of the detector in use , and
b) independent on the energy step width deltaE of the spectrum points.
So you can easily compare spectra (based on different energy resolutions and different step widths) with respect to spectrum integrals without figuring out the individual energy resolutions, which in most cases are not mentioned.
Spectrum integral [cts] = sum (intensity[cps/energy]*energy step width)=
energy step width *sum( intensity [cps/energy]).
Conversion from cps to cps/energy (e.g. cps/eV) is done by dividing the measured cps by the energy resolution of the detector.
Best regards
G.M.
The number of counts per second is proportional to the reading time of a X-Ray fluorescence band. If we vary the time we could have the same emission band of interest, but of different intensities. So, to my understanding, if cps / KeV is divided, it would be possible to compare the same band, but with different instrumental conditions.
Dear Claudio Hernán González-Rojas ,
sorry, for correcting you;
but 'the number of counts per second' is not proportional to the reading time.
The 'number of counts' is proportional to the reading time and thus, if varying the reading/acquisition time, the 'number of counts per second (cps)' is constant.
Your statement with respect to cps/keV is right; it allows comparison of bands taken under different instrumental conditions such as energy resolution and/or step widths as I have described above.
However, we have to take care, when different detector distances and even different (net) detector areas are involved. Here we have to replace the cps/keV by cps/(sr*keV).
In other words: here we have to divide the cps/keV additionally by the solid angle omega of detection; omega= area/distance².
Dear Professor Faisal Mohammed: Thanks for your correction, I will revisit the fundamentals of X-ray fluorescence, to continue advancing
Sincerely,
Claudio Gonzalez-Rojas
Claudio Hernán González-Rojas Gerhard Martens
Thank you for your responses.
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