I found the Xrd peak intensities for ZnS thin films increase with the film's thickness, but what should be the exact explanation of this phenomenon. Would anyone please provide me some valid references?
In XRD patterns usually intensity of the substrate peaks should decrease with increase in the film thickness because of more reflections from the film than the substrate but in my case substrate peaks intensity increases(intensity of the film peaks also increases) with film (poly crystalline film) thickness
Dear Md. Abu Sayeed ,
the reason is simple;
with increasing x-ray path length in the film the number of atoms which contribute to the x-ray diffraction processes increase, and thus the peak height of a diffraction peak increases.
You can derive the peak height versus thickness relation by yourself.
For an infinitesimal increase dx of the length x of x-ray path way in the film the increase dI of the intensity I of a peak is being proportional to dx:
dI ~ dx
However we have to consider the x-ray attenuation along the primary path way length x as well as the attenuation of the diffracted beam having the same path length inside the film:
dI ~ exp(-2µ*x)*dx; µ being the linear x-ray attenuation coefficient
In order to have the intensity I you have to integrate this relation:
I ~ integral[exp(-2µ*x)*dx] = -1/(2µ) *exp(-2µ*x) from x=0 to x
So we have:
I(x) ~ 1/(2µ)*(1 - exp(-2µ*x))
For Imax at x=infinity we will have: Imax= 1/(2µ)
So we will have the following ratio:
I(x)/Imax = 1-exp(-2µ*x).
The last step is to relate the path way length x to the depth d in the sample or the thickness d of the film:
we have: sin(theta)= d/x and thus x= d/sin(theta);
The final result is : I(d)/Imax = 1-exp(-2µ*d/sin(theta))
That's it....
Please make a sketch and recalculate the formulae...
Good luck
In general, the intensity of the diffracted beam from thin films increases as the thickness increases, as you could observe. To practical effects, this fact is mainly associated with the amount of material in the film, being the number of intensity counts directly proportional to the film volume. Specifically, the diffracted intensity is proportional to the effective volume of the crystalline material in the sample (see the exact equation on page 3 in XRD_2.pdf). Thus, it is expected that increasing the film thickness (keeping the same area) produces more intensity counts. More details about the physical fundamentals of the diffracted intensity nature can be found in XRD.pdf. On the other hand, I do not know if exists a fixed relation between XRD peak intensity and thickness. From my experience, I have never found any generalized model or theory that relates both parameters.
I hope this information can help you. Good luck!
You may also need a text book in this case, see this one:
Book Thin Film Analysis by X-Ray Scattering
Because, by increasing the film’s thickness, the flatness of film is also increased, so electrons of more surface’s atoms experience the X-rays. In this way, electrons (of each participating ring) of surface’s atoms deal with the reflection at same angle (point) printing the intensity (at the position of detector) to construct their dedicated peaks (in the pattern). Each constructed peak in the pattern is under greater count per second (relative intensity).
At fixed process parameters, you may deposit the thin films of same material. To deposit the films of different thicknesses, you need to vary the time of deposition. You will note in the patterns of X-ray analysis of your samples that by increasing the film’s thickness, resulted peaks in a pattern become sharp having lessened FWHM along with relative intensity.
You may consider it as a fixed relation between X-ray reflection (XRR) peak intensity and thickness. Please note that an X-ray diffraction is, in fact, an X-ray reflection. You may explain this important question further and by taking the help of the article given at link (https://www.researchgate.net/publication/352830671or http://dx.doi.org/10.13140/RG.2.2.27720.65287). Good luck!
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