Dear Pinky,

your calculated crystallite size is around 1 nm . This is about only  five atomic distances or two to three lattice plane distances. Do you believe that is ok? 

Please subtract your instrumental broadening contribution from your peak widths. After that you can go into the Scherrer equation.

Good luck

There's also strain broadening as well as instrumental.  Some research on Williamson-Hall is merited and looking up the vast number of RG discussions on the Scherrer equation is worthwhile.

Dear Gerhard,

Good day!

Thank you for sharing your thought. The reason why I tried to compute the Scherrer equation is to compare the cyrstallite size of the adsorbent we synthesize with the crystallite size in the literature. According to the journal J.H. Zhu et a. (2001) the crystallite size of HZSM5 that adsorb NDMA is 6.8 µm, however they did not discuss which lattice plane distances they were referring. So I do not know the answer to your question if it is ok. May I ask, what would be a good way to compare it with other journals?

I will try to subtract the instrumental broadening contribution from the peak widths. And post the answer here again.

Dear Alan, 

Good day!

Thank you for sharing your thought. 

Dear Pinky,

my question on whether you think that your calculated crystallite size is ok  and my other remarks on crystallite size aim at the assumption that your calculated size number seems to be very small; although you deal with NPs.

In your reply you mentioned a crystallite size of 6,8µm which is far larger than your value.

When considering the instrumental broadening your calulated crystallite size will increase. 

 You may also apply Williamsen-Hall evaluation ( as Alan proposed); so you can get  a little bit rid of strain contributions to peak width; but you have to have enough peaks for that.

But do not stress your crystallite size value too much. There are a lot a of contraints and assumptions dealing with crystallite size calculation/estimation. 

Please see for example  a discussion on that topic a few days ago.

https://www.researchgate.net/post/What_is_the_difference_between_Scherrer_and_Williamson-Hall_equation

Dear Friend,

For investigating the crystallographic of your materials as recently published and report released by some scientist you can be used "Modified Scherrer Equation" which is suitable for analysis and measuring the materials crystallinity and crystalline structure.

Dear Pinky,

for completeness:

for 'modified Scherrer equation' please have a look at the link.

 This crystallite size evaluation is derived from Scherrer equation and is based on the use of  a series of XRD peaks and performing some type of least square fit on the individual results.

Please have a look; interesting approach.

Final remark: the application of the modified equation will not significantly change the magnitude of your calculated crystallite size. There may be another fault.   

Article Modified Scherrer Equation to Estimate More Accurately Nano-...

Dear Pinky, I have got the solution.

For the  value of your FWHM please use the radian unit of that angle and not the degree unit. (Peak widths can be seen from your video to be around 0,1° in delta2theta; this value is similar to your FWHMs in your EXCEL sheet).

 Please convert degree of FMHM value to radian by multiplying by pi/180 ~ 0,0175.

So the value of your crystallite size will be increased by a factor of 180/pi ~ 57,3.

 But neverthess: please subtract instrumental broadening first.

 Good luck 

Dear Pinky,

I checked your excel sheet. Unfortunately you did not give the original values of the FWHM of the peaks.So please try to update the excel sheet. So that I can give my final opinion.

Thank you

.

Dear Kadhim,

the FWHM values are at column B of the excel sheet. But they are in degree and have not been converted into radian in Pinky's following calculations. 

Dear Martens,

Thank you for your clarifications. The FWHM has to be in radian scale.

But the FWHM in 2theta scale. What is about the 1theta scale?

Dear Kadhim,

please have a look at the link.

 Here the ß is the FWHM and is denoted as delta(2theta) in the 2theta scale. Therefore no necessity to divide by 2.

@ Pinky:

Please have also a look at the link.

 ß = FWHM  has to be in radians.

https://en.wikipedia.org/wiki/Scherrer_equation

Dear Martin,

Thank you very much for your clarifications.

Regards