I just performed a XRD analysis on a metallic glass particle and got an amorphous halo around 2(theta)=44 degree. How can I justify why the peak appears around 44 degrees?
Dear Mohammad Reza Rezaei ,
the amorphous halo or the broad 'humps' in the XRD pattern of amorphous samples are due to the superposition of the scattered x-ray waves ejected from of the nearest neighbour atoms in the amorphous sample.
Here the radial distribution function(RDF) and/or the pair distribution functions (PDF) of the arrangements of the atoms the sample play a role.
Your 44° is a consequence of the local geometrical arrangement of your metallic glass atoms. Please 'google' for 'metallic glass XRD' and you can see some XRD pattern peaking aroundabout 40 to 44° (so your measurements seem to be OK).
For water such hump shows up at about 25° (in 2theta) and for glass of various types around 20 to about 30°.
Some more details and links with respect to the evaluation of amorphous XRD pattern please see my answer at a recent question on this topic (I will not repeate all the text here again...):
https://www.researchgate.net/post/How_would_you_qualitatively_quantitively_analyse_the_amorphous_phases_of_glass_beyond_SEM-EDX_analysis
Best regards
G.M.
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