Bragg’s Law – Solved Examples
Example 1: A beam of X-rays of wavelength 0.071 nm is diffracted by 110 nm2 plane of rock salt with lattice constant of 0.28 nm. Find the glancing angle for the second-order diffraction.
Solution:
Given,
- Wavelength (λ) of X-rays = 0.071 nm
- Lattice constant (a) = 0.28 nm
- Plane = 110 nm2
- Order of diffraction = 2
Glancing Angle θ = ?
Bragg’s Law is 2d sin θ = nλ
d = a / √ (h2 + k2 + l2) , because rock salt is FCC
d = 0.28 x 10-9/ √(12 + 12 + 02) = 0.28 x 10-9/√2 m
Substitute in Bragg’s Equation
2 × 0.28 × 10-9/ √2 sinθ = 2 × 0.071 × 10-9
sin θ = √2 × 0.071/ 0.28
sin θ = 0.3586
θ = sin-1 (0.3586) = 21.01° ~ 21°
Example 2: Wavelengths of first-order X-rays are 2.20 Å at 27°8′. Find the distance between the adjacent Miller planes.
Solution:
Using Bragg’s law,
2d sin Ө = nλ
where,
- n = 1
- λ = 2.20 Å
- Ө = 27°8’
Substituting the values, we get
d = 2.20 Å
Example 3: X-rays of 173 pm wavelength are reflected by the (111) plane of a cubic primitive crystal at θ = 30°. The unit cell length (in pm) is closest to,
- 173
- 300
- 346
- 600
- None of the Above/More than One of Above
Solution:
Substituting this into Bragg’s law and solving for a, we get:
a = d x √3/ sin theta
a = (λ / 2 x sinθ) x √3
a = (173pm/ 2 x sin30° ) x √3
a ≈ 299.636 pm
Therefore, the length of the unit cell is closest to 300 pm Option (2) is correct
Bragg’s Law
Bragg’s Law is a law that helps in understanding coherent and incoherent scattering from a crystal lattice. In this article, we will see Bragg’s Law, its equation, derivation, application, etc.
In this article, we will learn about how to study Bragg’s Law.
Table of Content
- What is Bragg’s Law?
- Bragg’s Equation
- Derivation of Bragg’s Law
- Bragg’s Diffraction
- Bragg’s Spectrometer
- Applications of Bragg’s Law