if the peak intensity of (h k l ) plane in XRD for thin film/ bulk be maximum ,what shape is its ring in the electron diffraction in the TEM ( Thinner or wider) ?
There should be no quqlitative difference but there will of couse be quantitative differences. The positions of diffraction peaks in any diffraction experiment (X-ray, electron or neutron) will give information about the lattice spacing and angles which should be the same for all diffractions albeit for the same wavelength of radiation. The intensities however will be quite different because the scattering power of atoms will be different with different radiations. The intensities are goverened by the structure factors which are dected by the atomic positions. The structure factors consists of geometrical part that are dependent of atomic positions. These depent on hkl values and the atomic positions and are there same for all radiations. But structure factors also contain form factors or scattering lengths (for neutrons); they are different for different radiations. Also there will be Debye Waller factors depending on the mean squre displacements. These are also the same for different radiations. The thin films may be epitexial or may consist of epitexial domains. The incident radiation may fall only on a single domain or may cover several domains with different orientations depending on the area covered by the radiation. So one has to be careful about the details.
As far as I recall the peak width directly related with 1/N where N is the number of cell along the diffraction column in the Warren theory. That means diffraction lines or the spots in the reciprocal lattice associated especially with high Braggs angles are most affected from the film thickness in showing broadening. Actually, in the extreme case three dimensional reciprocal lattice is converted to the two dimensional system because of the broadening of the lattice spots along the thickness of the film.
''The maximum peak intensity of X-ray diffraction pattern in the electron diffraction pattern will look like widen for a thin film''
One can even derive following expression to show the following inverse connection between the fraction broadening of electron-diffraction pattern and the number of planes in the thin film taking part in the diffraction process, namely:
Δθmax/θB = 1/3(N −1),
Where it was assumed that high-energy electrons 100 keV used, and lambda is about 0.037 A, Hence: Sin Theta= Theta; which gives Bragg's Law as :
2d ThetaB=Lamtha
I got a lot of reading to do here. I'll catch up and then, I'll be back :-)
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