Secret Behind X-rays scattering by electrons - atoms - unit cells (Crystal) || XRD analysis

Structure Factor - Scattering of x-rays from electron ⟹ atom ⟹ unit cell - Contribution in XRD Peak intensity ✍🥀
Bragg’s equation (nλ= 2dsinθ) tells us about the peaks position (2θ). However, tells nothing about the peak intensities.
The intensities of the XRD peaks depend on many factors but here we only discuss the “Structure factor”
When x-rays shine on a crystal can scatter in three ways: As crystal is made up of unit cells ⟹ unit cell is made up of atoms ⟹ atom is made up of electrons
✅ 1) - Scattering from electron : Polarization factor
✅ 2) - Scattering from an atom : Atomic scattering factor (f)
✅ 3) - Scattering from a unit cell : Structure factor (F)
This is the structure factor formula:
F = ∑f exp2πi(hx+ky+lz)
To establish an exact relation between atom position and intensity, first considering how x-rays are scattered by a single electron, followed by an atom, and finally by all the atoms in the unit cell.
✅ 1- Scattering of x-rays by an electron - Polarization factor ✍🥀
X-rays are EM wave characterized by an E-field whose strength varies sinusoidally with time. Since E-field exerts a force on charged objects (in this case electron) so, the oscillating E-field of the incident x-ray sets any electron it encounters into oscillatory motion about its mean position. When any charged object (electron) is accelerating or decelerating, it emits an EM waves (i.e., x-rays). In this scenario, an electron is said to scatter x-rays. The scattered beam shows exactly the same λ and frequency as the incident x-rays and is said to be coherent with it. Here, the intensity of the scattered beam is just small fraction of the incident x-rays!
Moreover, the x-rays scatter from the ONLY tightly bounded electrons NOT from the loose bounded electrons ❗
✅ 2- Scattering by of x-rays by an atom (atomic scattering factor, f)✍🥀
As atom made up of electrons (as shown in the attached image) so the wave scattered by an atom is simply the sum of the waves scattered by its component electrons, right? Or more concisely, an atom containing Z (atomic number) electrons, the amplitude of the wave scattered is Z times the amplitude of the wave scattered by a single electron. This simply means that the scattered wave is proportional to the Z in the atom. A quantity f, shown in the “Structure factor” formula is called the atomic scattering factor, and defined as a ratio of amplitudes
🥀f = (Amplitude of the x-rays scattered by atom)/(Amplitude of the x-rays scattered by single electron)🥀
✅ 3- Scattering by a unit cell (Structure factor, F) ✍🥀
We must now restrict ourselves to consideration of coherent scattering from all atoms of the unit cell called the structure factor, F. It is obtained by simply adding together all the waves scattered by the individual atoms.
F= (Amplitude of the x-rays scattered by all atoms in a unit cell)/(Amplitude of the rays scattered by single electron)
Finally, the intensity diffracted x-rays by all atoms in a unit cell is given by
F = ∑f exp2πi(hx+ky+lz)
or
I ∝ |F|^2

[@Rahelukumarakalva]

Superb explaination Sir can you explain How to identify clay minerals near peak points in XRD graph

[@qamarwali]

Thank you for the nice comment. I don't understand the question very well. However, let me answer generally. If you are looking for something else, please do comment. Identifying clay minerals or any other minerals by the XRD is simply involves analyzing the peak positions (2θ) and peak intensities of the XRD pattern, then we compare them with the known standard reference patterns for the clay minerals or other minerals. The standard databases are like PDF, ICDD (commercially available), download CIF file, or just look into the literature for the similar clay minerals to find the JCPDS #, ...... Other ways also help like using Simulation software or Rietveld refinement method..... Generally, if we get SHARP XRD peaks, the sample will be CRYSTALLINE If we get Sharp Peaks, the crystallite size must be LARGE... If we get BROAD XRD peaks, the sample may be Amorphous If we get BROAD XRD peaks, the sample may be made up of nanomaterials (size less than 50 nm) If we get BROAD peaks, the crystallite size must be SMALL... The *problem with XRD analysis* is, we can not differentiate the XRD pattern for Amorphous & Nanomaterials as *Both produces Broad XRD peaks!*

[@AnwarAli-ee3ch]

Thumb up sir your way of explaining is unique ❤❤ keep it up sir

[@qamarwali]

Thanks 😊. I am happy you like it.