The reason is simle. X-ray peak is derived from the lattice sum over the entire crystal atoms. In ideal crystals, the sum is over infinite number which generate delta function on coherent spectral wavelength. Nano particles have much less atoms hence the lattice sum is not able to converge to a diffraction line but

bronden out. The smaller the particle the broad the diffraction peak. Therefore the brodening can be used to measure the particle size.

The method is described in any X-ray crystallography (see Azaroff or Warren's book)

However, in calculating the particle size, intrisic instrumental width must be deducted from the measurements by root square difference.

I have measured BaTiO3 particle by this method down to 50 angstrom 40 years ago.

my suggestion is to have a look e.g. at the great book of Warren (X-ray diffraction.. look for that on google) or at some of the papers you find in my list ;)

Though the exact reason may differ from one spectral technique to another, but broadly it may be explained on the basis of Heisenberg's Uncertainty Principle:

 x . p ≈ ħ

The nano particles, with infinitesimally small size, may be assumed to have very less mass ; meaning thereby that p value will be very small to make x very large i. e, the probability of its being present over a large space will be very high and thus the

line broadening is expected.

.. I think this is going way off.. it is easier to notice that the diffraction pattern of the object is the Fourier transform of it.

If you just consider the 1D case and you consider the object as a box, then it's easier to see that the FT of a box of width D is a sinc function having a width related to 1/D

I think it can simply explain by diffraction under non-ideal conditions:

Ref. BD Cullity "Elements of X-ray Diffraction"

The reason is simle. X-ray peak is derived from the lattice sum over the entire crystal atoms. In ideal crystals, the sum is over infinite number which generate delta function on coherent spectral wavelength. Nano particles have much less atoms hence the lattice sum is not able to converge to a diffraction line but

bronden out. The smaller the particle the broad the diffraction peak. Therefore the brodening can be used to measure the particle size.

The method is described in any X-ray crystallography (see Azaroff or Warren's book)

However, in calculating the particle size, intrisic instrumental width must be deducted from the measurements by root square difference.

I have measured BaTiO3 particle by this method down to 50 angstrom 40 years ago.

The reason given by Jin Shown Shie is correct and I agree with his answer.

For some easy literature, have a look at Scherrer's Equation (and related shape factors that are directly applied to nano-scale specimens measured by XRD). It is widely used as an estimation and comparison tool for particle sizes in journal articles alongside measurements made by SEM and TEM.

Comparison to "know" standards is the best way to use XRD.

In otherwords, ID & quantify the "instrumental factors" using "known standards".

Works well in 2D Real Time Bragg XRD Microscopy!

Scott A Speakman, Ph.D.; [email protected];

http://prism.mit.edu/xray

I refer to this a lot. Thanks S.N. Piramanayagam!

The challenge with conventional X-ray diffractometry is the deconvolution of the "particle size" & "strain effect" on the FWHM. Using "known standards" is helpful.

.. the real challenge is to try convincing people that some advancement occurred in X-ray powder diffraction from 1918 (Scherrer formula) to now... unfortunately the fact that it is "widely used" seems to justify a further quantitative use

I agree Matteo. I'm not sure if it is lethargy or fear of change? I see the same viscosity of thought when it comes to using 2D real time diffraction profiles as opposed to the conventional integrated linear "smudge".

Sorry! I forgot to add how powerful a tool the "smudge" conventional linear diffractogram is. The nano-structural information derivable from one of these is still phenomenal. Hence, its continued use over the past century.

Many orders of magnitude too slow compared with its 2D real time modern version.

Going back to the original question: “reasons for peak broadening in the case of Nano particles compared to the bulk"

Here is a specific scenario of Ni foil simulating the "bulk":

http://www.flickr.com/photos/85210325@N04/8008714677/in/set-72157632728981912

Is the question, why a nano powder of Ni of the same grain/particle/"diffracting domain" size as in the bulk Ni foil would show larger FWHM for Bragg peak using the conventional diffractogram (powder)?

Why would the mathematics or diffraction result change whether the diffracting domains were attached to each other in the "bulk" foil form or detached in the "Nano powder" form? Why would the "delta function" change other than the preferred orientation effects in the case of the foil?

2D Bragg Micrographs showing preferred orientation in foils:

http://www.flickr.com/photos/85210325@N04/8012716332/in/set-72157632728981912/

Every spot on the Debye-Scherrer Arc/Ring is a topograph of the diffracting domain. As the sample is rotated the film integrates the entire diffracting volume from each diffracting domain. Here is an extreme real time example with particle sizes larger than beam size:

http://www.flickr.com/photos/85210325@N04/7988731098/in/set-72157632728981912/

Real time Debye-Scherrer type data with a LaB6 powder standard from Beam Line X14A Brookhaven National Lab, NY:

http://www.youtube.com/watch?v=acgFl5M7P-I&list=PL7032E2DAF1F3941F

My objective is only to clarify my own understanding of the matter and not to question anyone else's knowledge. Thanks for your help.

Ravi

[email protected]

I read with interest the debate on this issue. However, for me the question is unclear. Author asks to inform about the peak broadening which occurs when the particle size falls in nano range. What is the nature of the peak remains unclear. I can think that it is a broadening of the photoluminescence peaks ...

Yuriy!

My understanding of the initial question: Would there be a difference in the FWHM as a result of "particle" or "grain size" if the sample were in Nano-powder form or in bulk form (e.g. sintered)? Why?

NO would be my answer! I’m assuming no net change in Nano-grain size or phase changes or preferred orientation in the two morphologies (loose powder & bulk). Pertaining to the frequently mentioned "Scherrer formula".

If the density in the bulk was greater than the loose powder, then the integrated intensity in a transmission/reflection experiment would increase proportionally. Factoring absorption events.

Key observation - “What is the nature of the peak?". Shape, shoulder profile, tail, etc.

FWHM of XRD peaks is just relared to crystalite size, not particle or grain!!!!!!!!

To get more information about "crystallite size, particle or grain size" please review the attached RG links.

https://www.researchgate.net/post/What_is_the_difference_between_crystallite_size_grain_size_and_particle_size

https://www.researchgate.net/search.Search.html?type=question&query=particle+size%2C+grain+size%2C+crystallite+size