Somewhere in papers, it is written as Mn 2p3/2 XPS spectra can be deconvoluted into four peaks at 640.83, 641.92, 642.97 and 644.56 eV assigned to Mn(II), Mn(III), Mn(IV), and the shakeup peak, respectively, while their surface relative contents were calculated to be 36.3%, 34.50% and 29.2%. How can we measure this surface relative content of deconvoluted peaks?
Dear Dr. Adeel Ahmed ,
I send you , here attached, a manual about the XPS technique containing the spectra and binding energies that distinguish different elements / their compounds which was of great help to me.
Good job and best regards.
Pierluigi Traverso
Dear Adeel Ahmed
To find the surface relative content of Mn(II), Mn(III), Mn(IV) you have to know the XPS peak area of Mn(II), Mn(III), and Mn(IV) from the fitting:
Then apply the following formulas:
Mn(II)=(Mn(II)area/ Mn(II)area+ Mn(III)area+Mn(IV) area)x100%
Mn(III)=(Mn(III)area/ Mn(II)area+ Mn(III)area+Mn(IV) area)x100%
Mn(IV)=(Mn(IV)area/ Mn(II)area+ Mn(III)area+Mn(IV) area)x100%
The following article may help you.
M. S. Alkathy et al 2020 J. Phys. D: Appl. Phys. in press https://doi.org/10.1088/1361-6463/abd12b
Liu, Juan, et al. J. Mater. Sci.: Mater. Electron. 31 (2020) 5205-5213.
Regards
Dear Dr. Adeel Ahmed ,
I send you , here attached, a manual about the XPS technique containing the spectra and binding energies that distinguish different elements / their compounds which was of great help to me.
Good job and best regards.
Pierluigi Traverso
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