Question

If you were able to produce nanocrystalline powders of FeTi of 10 nm crystallite size, for incorporating into the ferrite, what would be the peak width for the 100% peak in the X-ray diffraction pattern of this compound?

Answer 1

X-ray diffraction spectrum for the nanocrystalline Fe₅₀Ti₅₀ (a) before and (b) after
oxidation in the thermobalance up to 550°C. (,,) Fe₅₀Ti₅₀, (V) Fe, (4)) Fe₂TiO₅,
(v) TiO₂, (o) Fe₂0₃. Two peaks (x) have not yet been identified.

So, let's now discuss

How to calculate the Width of the peaks of XRD Spectra

We need to calculate FWHM , which stands for Full With at Half Maximum,that is applied at any bell-shaped curve, such as a Gaussian, Lorentian, triangle, rectangle

We can use Origin software or fullprof software to calculate the FWHM of peaks

then to calculate width of the crystal size , after collectting all the data of experiments

use this tool

https://www.mtixtl.com › tech-articles › xrd_scherrer_calculator

Here you can copy & paste the data of

Peak position 2θ (°)

FWHM Bsize (°)

Dp (nm)

you will find the value of crystal size of diameter using this equation

Crystallite size Dp = K / (B cos )

crystal Size = (0.9 x λ)/ (d cosθ)

K = 0.9

λ = 1.54060 Å (in the case of CuKa1) so, 0.9 x λ = 1.38654

Θ = 2θ/2 (in the example = 20/2)

d = the full width at half maximum intensity of the peak (in Rad) – you can calculate it using Origin software.

To convert from angle to rad

Rad = (22 x angle) / (7 x 180) = angle x 0.01746

Example: if d = 0.5 angle (θ)

= (22 x 0.5)/ (7x 180) = 0.00873 rad