X-Ray Fluorescence Analytical Techniques - CNSTN : Centre ...

E is the photon energy, ε is the average energy required to produce a free electron-hole pair, F
is the FANO factor and 2.35 converts the root means square deviation to FWHM. For an
equivalent energy, the detector contribution to the resolution is 28 % less for the case of Ge
compared to Si.
The contribution to resolution associated with electronic noise (∆Eelec) is the result of
random fluctuations in thermally generated leakage currents within the detector and in the
early stages of the amplifier components.
III. Spectrum Evaluation
Spectrum evaluation in energy dispersive XRF is certainly more critical than in WD-
XRF, because of the relatively low resolution of the solid-state detectors employed. The aim
is the extraction of the analytically relevant information (net number of counts under a peak)
from experimental spectra.
In EDXRF, the characteristic radiation of a particular line can be described in an
adequate first-order approximation by a Gaussian function (detector response function). The
spectral background results a variety of processes: for photon excitation, the main
contribution is the incoherently scattered primary radiation and therefore depends on the
shape of the excitation spectrum and on the sample composition. For particle-induced X-ray
emission and electron excitation, the background observed is mainly due to Bremsstrahlung.
The most straightforward method to obtain the net data area under a line of interest
consists of interpolating the background under the peak and summing the backgroundcorrected
channel contents in a window over the peak. In practice, this approach is limited by
the curvature of the background and by the presence of other peaks and can therefore not be
used as a general tool for spectrum processing in EDXRF. An example of overlapping peaks
is the analysis of lead and arsenic simultaneously present in a sample (Figure II.7).
Figure II.7: A spectrum of As K, overlapped with a Pb L line spectrum, both excited by
a Mo X-ray tube, under identical conditions. The energy of AsKα1,2 (10.53
keV) and Pb Lα1,2 (10.55 keV) cannot be separated by an Si(Li) detector.