Crystallite Size from XRD (comments on method required)

[Model 500]

Hi, the XRD in question is a Siemens (Bruker-Axs) D5000 and apologies as this may be a somewhat long post.

A member of staff tried to change password permissions on our XRD Crystallite Size software (Win-Crysize) and the copyright dongle for the software didn’t like this, so we now have a not-altogether inexpensive brick. Unfortunately our budget is already maxed out, so we cannot afford replacement software for the foreseeable future (this would be TOPAS P from Bruker-Axs).

However, luckily the XRD Control software (XRD Commander) and the XRD plot viewing software (Eva) are unaffected, and as such we can still perform XRD analysis (admittedly of a somewhat limited fashion).

This was fine until now, where I have had a request to obtain crystallite size for a sample. I now need a critical eye and input/suggestions on my manual calculation method.

I have obtained a 30-day trial of software called PeakFit, with this I have managed to use the extracted raw data (see attached .txt files), which when converted it into an Excel format can be imported into PeakFit.

View attachment sample raw data.txt
View attachment NaCl raw data.txt

Now PeakFit isn’t strictly speaking XRD software, however, it has allowed me to do a baseline subtraction, FFT Filter smoothing of the raw data and do a Voigt(amplitude) peak fit.

All this has enabled me to obtain a FWHM value for the Reflections of interest i.e. those at 2-Theta 7.0, 10.5 and 18.5, the FWHM values (from PeakFit not raw data):

7.0 = 0.202
10.5 = 0.230
18.5 = 0.265

Using the same approach above on an NaCl Std, to take account of instrumental broadening etc, the NaCl FWHM = 0.153.

Other parameters are:
x-axis is 2-Theta scanned from 3 to 35
y-axis is counts (scintillation detector)
λ = 1.54
k = 0.9

Now, as the crystallite size isn’t in the μm range I have went with the following version of the Sherrer Equation:

d = 0.9λ
...BcosΘ

Where B = bsample-bstandard


Please see attached Excel spreadsheet for details. (please rename XRD Crystallite Size - Sample Calc.txt to XRD Crystallite Size - Sample Calc.xls)

View attachment XRD Crystallite Size - Sample Calc.txt

We expect the sample to have an average primary crystal diameter of between 40 and 120nm.
Crystallite diameter perpendicular to β (001) plane of between 22 and 33nm
Crystallite diameter perpendicular to monoclinic β (200) plane of between 17 and 27nm.
Crystallite diameter perpendicular to β (010) plane of between 10 and 20nm.

Although it looks like I have obtained an answer within the ranges above I really need to know; Am I fudging it or does my method stand up to scrutiny?

All suggestions, comments and questions welcome.

 

[kobela0312]

the lamda should be A, right?

Hi, the XRD in question is a Siemens (Bruker-Axs) D5000 and apologies as this may be a somewhat long post.

A member of staff tried to change password permissions on our XRD Crystallite Size software (Win-Crysize) and the copyright dongle for the software didn’t like this, so we now have a not-altogether inexpensive brick. Unfortunately our budget is already maxed out, so we cannot afford replacement software for the foreseeable future (this would be TOPAS P from Bruker-Axs).

However, luckily the XRD Control software (XRD Commander) and the XRD plot viewing software (Eva) are unaffected, and as such we can still perform XRD analysis (admittedly of a somewhat limited fashion).

This was fine until now, where I have had a request to obtain crystallite size for a sample. I now need a critical eye and input/suggestions on my manual calculation method.

I have obtained a 30-day trial of software called PeakFit, with this I have managed to use the extracted raw data (see attached .txt files), which when converted it into an Excel format can be imported into PeakFit.

View attachment 12827
View attachment 12828

Now PeakFit isn’t strictly speaking XRD software, however, it has allowed me to do a baseline subtraction, FFT Filter smoothing of the raw data and do a Voigt(amplitude) peak fit.

All this has enabled me to obtain a FWHM value for the Reflections of interest i.e. those at 2-Theta 7.0, 10.5 and 18.5, the FWHM values (from PeakFit not raw data):

7.0 = 0.202
10.5 = 0.230
18.5 = 0.265

Using the same approach above on an NaCl Std, to take account of instrumental broadening etc, the NaCl FWHM = 0.153.

Other parameters are:
x-axis is 2-Theta scanned from 3 to 35
y-axis is counts (scintillation detector)
λ = 1.54
k = 0.9

Now, as the crystallite size isn’t in the μm range I have went with the following version of the Sherrer Equation:

d = 0.9λ
...BcosΘ

Where B = bsample-bstandard


Please see attached Excel spreadsheet for details. (please rename XRD Crystallite Size - Sample Calc.txt to XRD Crystallite Size - Sample Calc.xls)

View attachment 12829

We expect the sample to have an average primary crystal diameter of between 40 and 120nm.
Crystallite diameter perpendicular to β (001) plane of between 22 and 33nm
Crystallite diameter perpendicular to monoclinic β (200) plane of between 17 and 27nm.
Crystallite diameter perpendicular to β (010) plane of between 10 and 20nm.

Although it looks like I have obtained an answer within the ranges above I really need to know; Am I fudging it or does my method stand up to scrutiny?

All suggestions, comments and questions welcome.

 

[lastlegend]

hi friends I need pdf card , could you please send mail. My mail furkanalan@gmail.com