In XRD, I think there is a significant relationship between the target and the filter here, and the target has one more proton than the filter, is this true or false? What is the reason in both cases?
Dear Nesa Jafarzadeh,
the reason for that is found in the energy distribution of the electron energy levels of the atoms, which depends on the atomic number (i.e. the number of protons in the nucleus).
For the emission energies*) (K-alpha in the XRD case) of the x-ray targets please have a look at:
https://xdb.lbl.gov/Section1/Sec_1-2.html
(please click on "Table 1-2 (pdf format)" there...
For the energy leves (K-edges; K 1s column) in the case of XRD filters, please have a look at :
https://xdb.lbl.gov/Section1/Sec_1-1.html
(please click on "PDF version of this table") there...
Comparing the K-alpha line energies and the filter K-edge energies according to your blue coloured table, you can confirm the relative shift of 1 atomic number between target material and filter material.
The filter is used to reduce the K-beta lines and the high energy part of the bremsspectrum of the tube output.
In the case of a Mo target, you may use Nb or Zr as filter. Both of the K-edge energies are lower than the Mo K-betas here.
For the case of an Ag target, you may use Pd or Rh as filter. Both of the edge energies are lower than the Ag K-betas here.
You see, that the difference of 1 atomic number is not always stringent. For larger Z, the values of applicable Z-differences goes up...
*) we commonly nowadays deal with energies rather than wavelengths
Best regards
G.M.
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