Hi Pushkar,
It sounds to me like you have the right idea.
Using this procedure with free nanoparticles tends to give sizes that correlate to TEM sizes, but are about ten times smaller, just as you found. This is shown in more detail in this later paper of mine:
“Comparison of extended X-ray absorption fine structure and Scherrer analysis of x-ray diffraction as methods for determining mean sizes of polydisperse nanoparticles,” S. Calvin, S. X. Luo, C. C. Broadbridge, J. K. McGuinness, E. Anderson, A. Lehman, K. H. Wee, S. A. Morrison, and L. K. Kurihara, Appl. Phys. Lett. 87, 233102 (2005).
I’m still not completely convinced as to why the EXAFS size for free nanoparticles is often so much smaller than the TEM and XRD sizes. One possibility, raised by Anatoly Frenkel, is that there are often lots of molecular-scale fragments that TEM and XRD don’t pick up, but EXAFS does. Another possibility, explored in some of my papers, is that the size distribution is broad enough so that, for XRD in particular, the signal is dominated by the largest crystallites. Another possibility, not well explored, is that the amplitude reduction factor S02 has a weak dependence on R, or that there’s some other R-dependent effect (e.g. a surface effect) that is systematically reducing the sizes given by EXAFS.
Overall, then, my advice is that EXAFS-determined sizes can be compared to other EXAFS-determined sizes for similar systems. For example, if you have two sets of nanoparticles of the same composition, EXAFS can be used to get the relative size of those two samples. EXAFS-determined sizes, however, can not be compared directly to TEM and XRD determined sizes (or, for that matter, to other methods), particularly for polydispersed samples,in part because each technique places different weight on the low and high end of the actual size distribution.
—Scott Calvin
Sarah Lawrence College
On Jan 21, 2016, at 4:01 AM, pushkar shejwalkar mailto:pshejwalkar2004@gmail.com> wrote:
Dear All,
I sent a related question in the month of Dec but it seems that the question did not appeared on list because of mailbox being full, as we all have received an email. Since mailing list is started again I am sending this question once more.
My question is actually regarding finding out the nanoparticle size. In this respect, I am referring to Harris's work in which by using a simple equation we can roughly estimate the nanoparticle size
J. Appl. Phys., Vol. 94, No. 1, 1 July 2003
Now the question is that in this paper and another paper I found online as well, it refers to Nnano and Nbulk. I am guessing that this is basically coordination number of nanoparticle and bulk. But my first question is how do I calculate it?? e.g. I am working on Pd K-edge. So should I fit Pd foil with Pd crystal data and get the CN? but that would be 12 anyways. So would Nbulk be 12 in that case? and what will be Nnano then?
In another context, CN is proportional to amplitude, so in that case is it directly proportional? as in, can I simply replaced in formula, Nbulk/Nnano with Amplitudebluk/amplitudenano?
Third thing is that, if above is true then how should/can I calculated the amplitude? isnt that amplitude is always calculated by Feff? if so could someone help me with the procedure to do so? How will I know amplitude for nanoparticles, because I cannot have FEFF for nanoparticles, right?
I went through lot of material, however, so far my calculations give me a solution that is vaguely correlated to TEM images. The TEM value is about 2nm (rouhgly 20 angstrom). my current values I am getting by solving the equation by my current understanding is only 2Ao. about 10 times less. What I am doing is simply plotting the crude data in athena to get x(R) Vs radial distance graph, calculate A-3 for nanoparticles and bulk and using this as amplitudes. It is possible that I may be doing something totally irrelevant, so please help me in this aspect.
Thank you
Sincerely
Pushkar
--
Best Regards,
Pushkar Shejwalkar.
Post-doctoral -Researcher,
Tokyo Engineering University,
Tokyo-to
Japan
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