Dear Serdar Altin ,

there must be something completely wrong with the height position of the sample stage; I assume accidentally much too high. You may perform some scans at different height positions and see what's happening...

What is about the slit position of the one in front of the sample; is it ok?

Good luck and

best regards

G.M.

Dear Gerhard Martens, thank you for your advice, i checked the stage and the first slit, i think it is okay, i added the picture of the goniometer. may be it can help you to suggest me the solution

Dear Serdar Altin ,

thanks for feed back.

From these pictures all things look ok.

So the fault may be at another place...

a) e.g. movement of the focus position...

Did you change (replace) the tube by another one (e.g. Cu to Mo) or alter the tube position within its support a bit?

... or even switch to another filament (if your tube has got two of them) or even replace the filament (if it is possible in your set up)?

b) It seems there is a monochromator behind the sample (in front of the detector)?

If yes, it may be accidentally detuned (not being capable to provide Mo K alpha yet). According to the pictures, Mo is your current tube type.

Best regards

G.M.

Serdar Altin

Why not contact the manufacturer?

Dear gerhard martens

If it is due to monocromator, and i will remove it, in this case the detector will adjusted at same direction with beam, is it?

I contacted with manufacturer but it is produced before 1980 and they didnt give service

Yes Serdar Altin ,

a) when you remove*) the monochromator the 'detector beam direction' has to be the same as the primary beam (at the 'zero position' of the goinometer).

b) For tuning the monochromator you have also to use the 'zero' position of the goniometer.

But you have to do some precautions to avoid damage of the detector due to high photon flux:

c) please operate at very low current, and

d) put a filter into the beam (e.g. some mm of Al or a very few of Cu; have a look at the manual for these details).

*) I myself would not remove the monochromator. The monochromator is used to prevent background due fluorescence of the sample as well as Compton scatter. Especially for Mo you operate at quite high tube Voltages and thus the bremsspectrum has quite high photon energies giving rise to fluorescence and Compton scatter. Mo K radiation (~17,5keV) itself is able to excite K fluoresecence for elements up to Z=39 (Y)).

Good luck and best regards

G.M.

Serdar Altin If the detector is dead then you have a real problem...

Dear Alan F Rawle ,

please see the 'pattern'. The detector is not dead.

I assume that the monochromator is not properly set at Mo K alpha...

Gerhard Martens Thank you for the correction. Much appreciated. Have a great holiday season and New Year!

Dear all, thank you so much for your incredible help and information, Tomorrow i will try to measure without monochromator and inform you

Have a nice holiday and years….

Dear Serdar Altin , next idea;

before you remove the monochromator please check the applied voltage of the tube.

For Mo the K-edge energy is 20keV; so you need significantly more electron energy to have significant Mo K-alpha line(s); you need at least about 30 to 40kV in tube voltage (or even more) here.

Did you use a voltage setting in that range?

Even without the monochromator you need this quite high voltage...

Best regards

G.M.

Merry Christmas and a happy new year to all of you

Dear Gerhard Martens,

I double check the incoming beam by floresans slit and the beam coming strongly the sample holder part and then after the sample, i didnt observe exactly the beam, generally i use 30 kV and will try above, i will also share detail of the expertience whenever i try, thank you so much

Dear Vadim, thank you for useless information

Dear Gerhard Martens and Alan F Rawle,

first i would like to thank to your valuable contrubitions

i removed the monocromator and made zero calibration qnd than started to make ZnO powder and attached below. the voltage and current value were given below. so i say that the removing the monocramotor is the best solution. and i searched from the google and saw that the Zr foil the best filter for Mo tube. the main problem now is high backround on the graph, do you have any suggestion ?

Dear Serdar Altin ,

thanks for sharing your 'success'.

The Zr filter only blocks part of the x-ray spectrum above 18keV and the very low energy part. Thus the background is due to Zn K fluorescence and Compton scatter of the residual bremsspectrum of the tube.

The monochromator is the best choice to remove your background.

So tuning of the monochromator is your goal now...

Another way out is the use of a set of balanced filters*), here Zr and Y.

Their K-edges are just above and below the Mo K-alphas.

So when taking two XRD pattern (one with Y in front of the detector and the second scan with Zr in front of the detector) and

calculate the difference XRDZr - XRDY, you will have the XRD pattern belonging to the energy range of about 17keV (Y K-edge) to about 18keV (Zr K-edge); so quite 'monochromatic' compared to no filtration.

The challenge here is

a) to calculate the thicknesses of these foils and

b) afterwards to gather them.

The thicknesses should be in the precision range of about 0,1µm.

I will try to match these thicknesses for you, but there is unfortunately only a little time just before Christmas...

*) please google for 'balanced x-ray filter' and/or 'Ross-filter'

Best regards

G. M.

Dear Gerhard Martens

Again thank you so much for your valuable detailed information,

i used LiFePO4 powder and measured by Mo tube and fitted by gsas software as attached figure. it show good fit which provide the system working properly.

first i tried to mount the monochromator and setted to 10 degree with beam and the detector was replaced to 20 degree and played around both to find out the best count values. before the mount of the monochromator, i adjusted the 2 theta the maximum intensity value of the Si wafer and then replaced the monochromator and detectors to new positions. so the count before the monochromator around 1000 cps and then it was decreased to 50-60 cps. so i concluded that the monochromator doesn't work in Mo tube. so the second advice is better now, but i need Zr foil and Y foil or powder. but the thickness which you advised around 100 nm and it is not possible to make it buy using foil. so it can be performed by e-beam, sputter or thermal evaporator. the second can be electrochemical coating by using ZrCl2 for Zr coatings. i can find ZrCl2 but the main problem of this what is the substrate for coating. the second is more difficult to find out of Y and what about the thickness if it is again 100 nm level again we should use different techniques

i know now is holiday time and have a nice holiday and years for all

we can continue after christmas,

again thank you so much.....

Dear Serdar Altin ,

congratulations; you are on the right track...

I can appease you; the precision of the thickness will be in the 0,1µm range (may be a bit more relaxed up to a few µm) but not the total thickness. The total thickness will be in the 50µm range for Y and Zr. So you can buy such foils and you can use it as free standing ones.

I will do some calculations/suggestions with repect to some Y&Zr balanced filter thicknesses for you and present the results after christmas...

Best regards

G. M.

Dear Serdar Altin ,

please see attached a suggestion for Y-Zr balanced filter sets.

The Y K-edge energy is at 17,038 keV and the Zr K edge energy is at 17,997keV.

The Mo K-apha line is just in between at around 17,4keV.

As I wrote above, the difference of the XRD pattern filtered by Zr and Y (XRDZr -XRDY) will give an XRD pattern with strongly reduced background components, which are caused by fluorescence and by Compton scatter.

Balancing of the filters is done by chosing the filter tranmission(s) close to the edges to be equal:

T(Y) = T(Zr) e.g. at 17keV or 18keV or even at 20keV.

From this equation one will have access to the filter thicknesses via:

T(Y) = exp(-µy*dY) = exp(-µZr*dZr) = T(Zr) with µY and µZr being the linear attenuation coefficients of Y and Zr, and dy and dZr being the thicknesses of the filters.

One will get: µy*dy = µZr*dZr and thus dY/dZr = µZr/µY

It turns out*), that this ratio is very close to 1,5 for all the x-ry photon energies mentioned above.

An example of the transmission behavior of two of such filter sets is given in the attachment.

The filter set 75µmY & 50µm Zr is a special one, because these filters are commercially available**) as 'free-standing' foils, even by a superposition of some 25µm filters...

In a second attachment you will see an overlay of a measured x-ray spectrum of a Mo target (35kV) with the 75µm Y & 50µm Zr balanced filter transmission(s).

Here you see, that

a) there is an excellent match to the Mo K-alpha line and

b) that already a Zr filter alone will help to reduce the Mo K-beta contribution and the Compton background in the high energy region (you mentioned Zr already).

It should be noted, that here the width of the Mo K-alpha is due to the 'poor' energy resolution of the HPGe detector in that region of about some 100 eV, and that the natural bandwidth of Mo K-alpha(s) will nicely into the Y/Zr edge gap.

A second reason for showing the Mo spectrum, is to provide you an impression onto the ratio of the Mo K-alpha photon flux compared to the flux of the bremsspectrum. So from this spectrum one can see, that you will have a lot of Compton scatter from above the Mo K-alpha energy but also from below the K-alpha energy. The balanced filter will effectively reduce the XRD background arising from both parts of these energy regions...

*) for µ/rho (energy dependent) values please see for example the NIST/XCOM data base or the Veigele tables in the CRC Handbook of spectroscopy....

**) I will send you a suggestion for a company including the reference numbers for the foils in a privat message here via RG.

Best regards

G.M.

Dear Gerhard Martens

the information which you give is very useful for us. i ordered Y foil and Zr foil and then whenever i get them i will share the results with you.

Again thank you so much

The quantity (m/r) is called the mass absorption coefficient. The ( m/r) of Zr is shown as a function of wavelength at the top of Fig. 8. The sharp discontinuity in the curve is called the absorption edge. They correspond to the entrance of a new absorption mechanism. The highest energy edge corresponds to the ionization potential of a K electron and is called the K edge. Other edges correspond to the Land M electron ionization. The exact location of each of the edges corresponds to the energy required for the removal of that electron. Thus, absorption edges yield fundamental information about the structure of the atom, its environment, and the amount of energy required to excite that series of fluoresced radiation. The fine structure of these edges gives Information on the exact energies of the electrons within the atom. These energies are slightly affected by the bonding environment around the atom. To explain the fine structure would require a full molecular orbital description of the atoms in the absorber and this fact dampened the spirits of many early researchers.