morex’ray%fluorescence%analysisnotes% - …phys352/more_xrf.pdf · · 2011-03-06microsoft word...
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More X-Ray Fluorescence Analysis Notes M.C. Chen PHYS 352 Queen’s University Physics The Amptek ECLIPSE-‐III x-‐ray generator used in the lab allows you to set the high voltage used to accelerate electrons in the x-‐ray tube that strike a silver target. The energy of those electrons determines the energy spectrum of x-‐rays emitted by the tube. When 30 kV is used, the spectrum that is emitted is shown in Figure 1 below (from the ECLIPSE-‐III manual).
Figure 1. X-‐ray spectrum from the ECLIPSE-‐III generator when 30 kV is used to accelerate the electrons inside the tube.
As expected, the x-‐ray tube emits bremsstrahlung x-‐rays (the broad continuum that peaks around 8 keV) and can also emit the characteristic x-‐rays from the target silver atoms. The older supplementary notes for this lab mention a radioactive source that was used previously to produce the x-‐rays for this lab. Those notes (still useful and completely relevant to the analysis apart from the different source of x-‐rays) discuss the photoelectric cross section for x-‐rays to excite Cu and Zn; but the values given are for the x-‐ray energy from the previous source. To complete your x-‐ray fluorescence lab analysis properly, you will need to know the cross section for x-‐rays to excite Cu and Zn, for x-‐ray energies emitted by the tube. This depends on the high voltage used because that determines the energy of the electrons that bombard the Ag target inside the x-‐ray tube and the resultant energy spectrum of the x-‐rays from the tube.
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It is recommended that you operate the x-‐ray tube at 20 kV instead of the 30 kV that it is capable of (and from which the spectrum in Figure 1 was produced). Why is that? Below 21 kV, the electrons cannot excite the Ag atoms (K shell) in the target. The only energetic x-‐rays emitted from the silver target in the tube will be from bremsstrahlung. The bremsstrahlung spectrum has, approximately, the functional form: N(E) = k (Emax – E) / E and what causes the turnover at low energies is absorption inside the x-‐ray tube (in the target itself, through the window, and through deliberate “filters” in the x-‐ray tube designed to eliminate the not-‐so-‐useful low-‐energy x-‐rays). Thus, whether a voltage of 20 kV or 30 kV is used, the x-‐ray spectrum from the ECLIPSE-‐III will start to turnover around 8 keV. Now, below the “K edge” in Cu and Zn, x-‐rays cannot fluoresce the K x-‐rays from Cu and Zn, so we don’t need to know the precise details of the turnover because x-‐rays at energies lower than ~9 keV aren’t relevant. Figure 2 shows the cross section for Cu (left) and Zn (right) as a function of energy (x-‐ray cross section data from the 5th floor lab).
Figure 2. X-‐ray cross sections on Cu (left) and Zn (right) versus x-‐ray energy. The K edge is clearly visible in the plots.
Reading values off the plots in Figure 2, one finds a power law of about E–2.7 for the cross section between 10-‐20 keV. For the XRF lab analysis, it will be a good
approximation to find the average cross section for x-‐rays on Cu and Zn atoms, integrated over the bremsstrahlung spectrum, by taking the integral from the K edge up to 20 keV. Reading values off the table and doing a little math:
(20 − E)E
CE−2.7
Kedge
20
∫ dE
(20 − E)EKedge
20
∫ dE
I can get a value for the average photoelectric cross section on Cu and on Zn, for x-‐ray energies originating from a bremsstrahlung spectrum that starts from the appropriate K edge value for Cu and Zn, and goes to 20 keV. The calculated average cross sections (for the 20 kV bremsstrahlung spectrum): σCu = 1.5945 x 104 barns σZn = 1.5926 x 104 barns These values are really so close that we will take them as being equal. There are approximations used in this calculation that won’t be precise out to the third decimal place. Thus, in the XRF lab analysis, take σCu=σZn. Note: the average cross section values over a spectrum of the form (Emax–E)/E just happened to be equal for Cu and Zn, when Emax = 20 keV. But, you should note that it is not because the cross sections for Cu and Zn are the same. For example, σCu = 2.297 x 104 barns at 10 keV σZn = 2.625 x 104 barns at 10 keV and if the x-‐ray source used was, instead, a monoenergetic x-‐ray beam of 10 keV, you would need to account for the different photoelectric cross sections on Cu and Zn in the analysis (i.e. the probability to excite a Cu atom and Zn atom could differ, and hence that affects the intensity of the fluorescence x-‐rays from Cu and Zn that is observed).