0
$\begingroup$

In XPS analysis, the FAT (Fixed Analyzer Transmission) parameter controls the electrostatic field of an emispheric analyzer, called Pass Energy $E_P$.

The FWHM of a peak of signal is given by $$\Delta E = E_P \cdot \text{const.} $$ and the Pass energy is proportional to the FAT number (e.g. a peak of signal acquired at FAT10 is thinner than a peak acquired at FAT90).

Now the question is: as one raises the FAT parameter, the peak becomes larger, but why does the total signal increase?

a measure my team and me made varying FAT parameter

This plot is a measure made by my team and me varying the FAT parameter on peaks of a sample of Silver.

$\endgroup$
0
$\begingroup$

That is because increasing the pass energy not only increases the FWHM of the peaks but also the number of electrons that can reach the detector. In other words by increasing the pass energy one lets more electrons in total to enter the analyser.

Think in this way; if the pass energy is 10 eV then electrons with energy of 9-11 eV (±10% of the pass energy) will enter the analyser but if the pass energy is 100 eV then electrons with energy of 90-110 eV will enter the analyser. That will result an increase in the signal. This 10% thing is a rough and analyser dependent number. As far as I remember it, among some other things, depends on the radius of the analyser (assuming hemispherical).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.