Dear Abhishek, if you take a high resolution XPS-spectrum, then a typical FWHM is around 1 eV, this is the result of the convolution of the width of the excitation source (i.e. natural line width of ca.0.6 eV to 0.8 eV for Al K-alpha and Mg K-alpha) and the resolution of the spectrometer you are using. Depending on how the spectrometer is controlled (such as type of spectrometer, pass energy, etc) the resolution of the spectrometer may vary substantially, from about 0.1 eV to several eV. Thus, it is important to check the settings you have used, eventually the pass energy was set inappropriate?

Good luck, Dirk

Dear Abhishek, if you take a high resolution XPS-spectrum, then a typical FWHM is around 1 eV, this is the result of the convolution of the width of the excitation source (i.e. natural line width of ca.0.6 eV to 0.8 eV for Al K-alpha and Mg K-alpha) and the resolution of the spectrometer you are using. Depending on how the spectrometer is controlled (such as type of spectrometer, pass energy, etc) the resolution of the spectrometer may vary substantially, from about 0.1 eV to several eV. Thus, it is important to check the settings you have used, eventually the pass energy was set inappropriate?

Good luck, Dirk

The single-component width is an apparative constant (assuming the settings Dirk Luetzenkirchen-Hecht mentioned are kept the same) so if you are not sure you can determine this width using a reference that has only one atomic species. For C1s, I used Buckminsterfullerene C60 back then.

Dear Abhishek ,

There is another question similar to you. You may find some additional info in the following discussion:

https://www.researchgate.net/post/How_to_set_the_FWHM_full_width_at_half_maxima_for_deconvoluting_the_XPS_spectra

The FWHM is dependent upon the instrumentation, where the Gaussian energy contribution from an hemispherical analyzer (DE) depends on pass energy (Ep), slit width (w) and the mean radius of the hemissphere (r) plays a role when estimating the theoretical resolution contribution of the spectrum. See for example equation 2 in G. Beamson and D. Briggs, High Resolution XPS of Organic Polymers, John Wiley & Sons (1992) where DE=Epw/(2r).

It also depends on the light source, which for simplicity could be estimated to have a Gaussian distribution. For non monochromatic sources the energy width is the life time of the transition, e.g. about 0.8 eV for Al K alpha. If using a monochromated source the crystal of the mono is determining the resolution. Investigating Ag 3d (lifetime of 310 meV) can then give you an idea of the resoltuion. The resolution should also be specified by the manufacturer.

A Gaussian temperature contribution should also be taken into account. For room temperature it is about 90 meV and at 77 K it is around 23 meV. See for example Hüfners Photoelectron Spectroscopy, Springer.

For Lorentizian lifetime contributions to the spectra i can recommend:

Papp, J.L.C.a.T., Widths of the atomic K--‐N7 levels. Atomic Data and Nuclear Data  Tables, 2001. 77: p. 1--‐56.

To calculate the Voight contribution for this one can use the equation from this paper:

Olivero, J.J. and R.L. Longbothum, Empirical fits to voigt line--‐width --‐ brief review. Journal of Quantitative Spectroscopy & Radiative Transfer, 1977. 17(2): p. 233--‐236.

where DE_Voight=((DE_Gauss)^3/2+(DE_Lifitime)^3/2)^2/3

This will give you a theoretical estimation of what to expect under ideal conditions not taking for example vibrational broadening into account.

D. Briggs and M.P, Seah, Practical surface analysis also gives an insight in some of the above John Wiley & Sons (1983) 

The above method was used for example in Eriksson, Review of Scientific Instruments, 85(2014)075119 to calculate the contributions from the instrumentation for a Ag 3d peak.

If you are using CasaXPS there are some films within the area of your question. see for example: https://www.youtube.com/watch?v=-nCchGGAxGE

Best regards,

John

Hi Abhishek,

I would fit the C 1s spectrum using 1 peak and let the peak width vary.

I would be curious to know the peak position of C 1s as I am assuming it will be shifted due to presence of multiple Fluorines.

Dhruv