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It is said that if the Bragg peaks are sharp, the material is crystalline and if the peaks are broadened over a range of angle then it is amorphous. But how that happens? I mean how the peak sharpness decides the crystalline nature? Please someone explain me in brief

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  • $\begingroup$ Consider the diffraction pattern to be the Fourier transform of the atomic positions. If all the atoms are precisely on crystal lattice sites there will be a few discrete points. If the atoms are not precisely on lattice sites, this smears the intensity out, resulting in broader peaks. $\endgroup$ – Jon Custer Aug 19 '16 at 14:10
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In a crystalline material, the collective diffraction pattern is sharp because there are many atoms adding repetitively, and off-Bragg-angle scattering (such as a single atom would create) is not reinforced by the neighbors. In an amorphous material, there are still clusters of crystalline structure, but randomly interspersed with either another competing crystal structure or with defects, so that the long-range order is lacking. Because the local structure puts four or five atoms into a nearly crystalline cluster, you can get a four-atom reinforcement of the diffraction at a peak.

Just as a diffraction grating gives a sharp spectral line, and a two-slit diffraction experiment gives multiple broad peaks, so a crystalline material gives a sharp line and an amorphous one gives blurry lines.

Physically, the scattering that arises from N atoms gives rise to a peak wave function N times that of a single atom; the peak power density however, is N**2 times that produced by a single atom. Because the scattered energy is conserved, the amplitude rise due to large N occurs only in a narrow angular spread.

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