Correct me if I am wrong. Amol and Daniele have already given the answer you require, I think.

You measure a spectrum with intensity vs. wavelength, where the wavelength bins have a fixed width. Now you want to plot the spectrum as intensity vs. energy.

For a single wavelength, E = h * c / Lambda

The differential form is: dE = - h * c / Lambda^2 * dLambda

You transform the x-axis from wavelength to energy using the first above formula.

Since the intensity data consist of bins of finite wavelength ranges, and the energy is inversely proportional to the wavelength (not linearly dependent), you should recalculate the intensity values. The intensity in the same bin has to be the same, of course: I_Lambda * dLambda = I_E * dE. With the second above formula you then arrive at the factor you have to multiply your intensity values with in order to obtain the correct intensity values on the energy scale:

I_E = I_Lambda * Lambda^2 / (h * c)

Don't forget to divide E by the elementary charge (1.602 E-19) if you want to have eV units on the energy scale.

Of course the calibration Gazi described should be performed, if you want to obtain the correct pixel (or channel) to wavelength conversion from your spectral measurement.

Dear Albert

The dispersion distance on the detector is essentially linear in wavelength. Thus, wavelength is the natural unit for this high-resolution x-ray spectrometer.

The conversion between energy and wavelength is provided by the relation:

E ×λ = hc = 12.39852 keVÅ.

For more information about dispersion effect go through

http://cxc.harvard.edu/proposer/POG/html/HETG.html

Dear Albert, the conversion you suggest (and explained by Amol) is correct. So if you have a punctual absorption graph as a function of the wavelength, you can present it as a function of the energy by converting the abscissae in that way.

The situation is different if you display a spectrum as a function of the wavelength, where you have on the y-axis the number of collected photons per wavelength bin, i.e. your wavelength resolving power, d(lambda). In this case, to switch to the energy plot you also have to 1) convert lambda to E in the abscissae and 2) multiply the y-axis with a factor of (12.4/E^2), i.e. the d(lambda)/dE ratio in Å/keV.

Good luck!

Thanks Amol and Daniele.

The approach I know is the one Amol suggested, but I think Daniele's approach is what I needed to learn.

Daniele, could you please give a little more details on that approach. The conversion of wavelength to E in step 1) looks similar to what Amol mentioned. Why would one then multiply the y-axis with this factor? is there a physical meaning to it?

I would appreciate some reference materials if available.

Thanks again.

Dear Albert, step 1) is all you need if you have to convert an absorption/reflectance/transmissivity plot from wavelength to energy and vice versa. You don't have to multiply or divide the y-axis by the same factor.

The reason is that a reflectance plot is the ratio of a reflected spectrum over an incident spectrum. If you want to convert the _individual_ spectrum (either incident or reflected) from wavelengths to energy you also _have_ to multiply the y-axis of both spectra by d(lambda)/dE.

But, when you take the ratio of the two spectra to get the reflectance plot, this factor cancels out. So you do not need to include it any longer.

Hope this helps!

Hi

In my opinion it is correct to perform such conversions. Because the instrument's dispersion has nothing to do with the proper conversion of the scales from wavelength to energy. Dispersion will occur in both scales, or course, but the mathematical relation between the two will automatically take care of the dispersion. In particular, 10 micrometer corresponds to 125 meV or 1000 cm-1. I have some experience with these things because I worked on mid-infrared quantum cascade lasers and detectors....

Daniel

Albert, monochromators usually need to be calibrated. Calibration can be done with or without taking into account the dispersion of the monochromator. If monochromator is calibrated using the first approach, then you need only to use the conversion factor given by Amol. This also slightly depends on media in which your sample is immersed, e.g. air or liquid helium. If your monochromator is calibrated using the second approach and the monocromator has dispersion (they always have some), then you need 1)correct the x-axis according to the dispersion, i.e. calibrate your spectrum and 2)convert wavelength to energy using the conversion factor given above.

Thank you Gazi. I would like to learn more on how to do the correction of the x-axis according to the dispersion as you have mentioned.

Correct me if I am wrong. Amol and Daniele have already given the answer you require, I think.

You measure a spectrum with intensity vs. wavelength, where the wavelength bins have a fixed width. Now you want to plot the spectrum as intensity vs. energy.

For a single wavelength, E = h * c / Lambda

The differential form is: dE = - h * c / Lambda^2 * dLambda

You transform the x-axis from wavelength to energy using the first above formula.

Since the intensity data consist of bins of finite wavelength ranges, and the energy is inversely proportional to the wavelength (not linearly dependent), you should recalculate the intensity values. The intensity in the same bin has to be the same, of course: I_Lambda * dLambda = I_E * dE. With the second above formula you then arrive at the factor you have to multiply your intensity values with in order to obtain the correct intensity values on the energy scale:

I_E = I_Lambda * Lambda^2 / (h * c)

Don't forget to divide E by the elementary charge (1.602 E-19) if you want to have eV units on the energy scale.

Of course the calibration Gazi described should be performed, if you want to obtain the correct pixel (or channel) to wavelength conversion from your spectral measurement.

Hi all,

Benjamin Bruhn has answered the question very fully and I agree - just to add to that answer, a diffraction grating based spectrometer with a fixed entrance slit width will have a nearly constant dispersion in wavelength - there's nothing experimentally you can do to change that. This is why when you want to plot your spectrum in energy units, you must scale the intensity values as expained by Benjamin. In order to compare or fit your data with any quantum mechanical theory for a spectral lineshape (Lorentz, Gauss, etc), you need the spectrum to be plotted as if it had been measured with constant energy bandwidth.

I have deleted my answer explaining how to solve dispersion related calibration problem because it was not clearly written.

I agree that Benjamin Bruhn has tackled it very well. Thank you all.

Is there any book or publication that gives this explanation? I would like to read and cite it in my work.

Now I can understand the explanation by Daniele Spiga as well.

There is any book or paper for detail

I think this can help: http://pubs.acs.org/doi/abs/10.1021/jz401508t

I just want to remind that the conversion nicely described by Benjamin applies only to emission spectra (and not to absorption and excitation ones).

In the paper cited by Fabian the authors erroneously apply it also to an absorption spectrum (see Fig. 2a). In that case, however, only the x-axis has to be transformed.

In strict mathematical language, this difference is due to the fact that an emission spectrum is a distribution while an absorption one is a function. In the first case it is the area under the curve that must be preserved in the transformation, whereas in the second one it is the y-value (i.e. the absorption coefficient at a given wavelength).

Thank you for the contributions on the question.

@Mauro_Fasoli: a correction to this article was issued about a year later, stating that it doesn't apply to the absorbance example:-) See here: https://pubs.acs.org/doi/abs/10.1021/jz502066v#

For quantitative analysis the spectra needs to be transformed further (after conversion to wavenumber). This is due the different nature of transitions in absorption (stimulated) and emission (spontaneous).

Article Recalling the appropriate representation of electronic spectra

http://rspa.royalsocietypublishing.org/content/275/1360/135

@Friedrich_Menges: Thanks, that indeed fixes the problem.

Given the topic of this discussion I would like to recommend a paper describing potential problems related to luminescence data processing.

Article Potential problems in collection and data processing of lumi...