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Bragg's Law describes the relationship between the angles of incidence and reflection of electromagnetic waves as they interact with a crystal lattice structure. The law is based on the principles of constructive interference and diffraction.
- Bragg’s Law is a special case of Laue diffraction to understand the angles of coherent and incoherent scattering from a crystal lattice.
- When the X-rays are incident on an atom, it moves the electronic cloud just like an electromagnetic wave does.
- The derivation of Bragg's Law involves the phases of beams that coincide when the incident angle is equal to the angle of reflection.
Bragg’s law states that “When a beam of x-ray is incident on a crystal at an angle θ, the reflected beam of x-ray also has the same angle of scattering θ. Constructive interference will occur when path difference ‘d’ is equal to the whole number ‘n’ of the wavelength.” Mathematically, Bragg’s law equation can be represented as,
nλ = 2d sinθ
Where:
- n is an integer representing the order of reflection.
- λ is the wavelength of the incident X-ray beam.
- d is the distance between the atomic layers in the crystal lattice.
- θ is the angle of incidence and scattering.
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Key Terms: Braggs Law, Bragg’s law equation, Constructive Interference, Path Difference, Wavelength, Scattering, Reflection
Bragg’s Law
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Bragg's law in Physics states that when an X-ray is an incident on a crystal surface, its angle of incidence, θ reflects back with the same angle of scattering θ.
- When X-rays are incident on an atom, they form an electronic cloud moving like an electromagnetic wave.
- The movement of the charges releases waves of the same frequency as X-rays but are slightly blurred due to various effects.
- This phenomenon is called Rayleigh scattering.
- Bragg’s law basically explains the relation between an X-ray light shooting and its reflection from a crystal surface.
- Bragg’s Law helps to determine the angles formed by coherent and incoherent scattering from a crystal lattice.
The video below explains this:
Bragg’s Law Detailed Video Explanation:
State Bragg's law
Bragg's equation for the diffraction of an X-ray is written as follows –
When the X-rays are incident on a crystal surface, the angle of incidence (θ), will reflect back with the same angle of scattering, θ. When the path difference equals a whole number, n, of wavelength, λ, constructive interference will occur.
- The same process occurs when neutron waves are scattered via nuclei or a coherent spin interaction along with an isolated electron.
- The fields of wave that are re-emitted interfere with each other constructively and destructively, creating a pattern of diffraction on a detector or a film.
- The analysis of diffraction and the interference of the resulting wave is referred to as Bragg diffraction.
- The diffraction analysis is the resulting wave interference, and this analysis is known as Bragg diffraction.
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Bragg’s Equation
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The Braggs equation can be represented as –
nλ = 2d sinθ
Therefore, according to Bragg’s law of diffraction,
- The equation gives the reason as to why the crystal lattices reflect X-ray light beams at specific angles of incidence, that is, (Θ, λ).
- n, in the equation, denotes an integer.
- ‘d’ is a variable that indicates the distance existing between the atomic layers.
- λ (lambda) is a variable that denotes the wavelength of the x-ray wave that was incident upon the surface.
Equation of Bragg’s Law
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Solved ExamplesExample 1: Justify the Following: Braggs equation has no solution if n = 2 and λ > d. (2 Marks) Reason: Bragg's equation is given by nλ=2dsinθ. Solution: Braggs equation is nλ=2dsinθ. If n = 2 and λ > d, then there is no solution because the sine of an angle can't be more than 1. Example 2: In Bragg's equation for diffraction of X-rays, n represents? (2 Marks) Solution: W.L. Bragg and W.H. Bragg worked out and gave a mathematical relation to determining the inter-atomic distance from the X-ray diffraction pattern. Braggs equation is nλ=2dsinθ where n is an integer i.e. 1,2,3,4 etc. which represents the order of reflection. |
Bragg’s Law Derivation
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Braggs law Statement and Derivation – Bragg's Law states that when X-rays or other electromagnetic radiation waves interact with a crystal lattice, the waves undergo constructive interference if the following condition is met: the path difference between waves reflected from adjacent crystal lattice planes is equal to an integer multiple of the wavelength of the incident radiation.
To derive Bragg's equation, let us consider a figure to observe the phases of beams that coincide when the incident angle is equivalent to the angle of reflection.
- The incident beams are parallel to each other before they reach point, z.
- When they eventually get to point z, they travel upwards after striking the surface.
- The second beam scatters when it reaches point B.
- AB + BC can be considered the distance travelled by the second beam.
- The extra distance it travelled is referred to as the integral multiple of the wavelength.
Derivation of Bragg’s Law
Here is the derivation of Bragg’s equation,
nλ = AB + BC
Now, we know that, AB = BC
Thus, nλ = 2AB ----- 1)
Now, considering d as the hypotenuse of the right-angled triangle, ABz, it is clear that AB is the opposite of the angle θ.
Therefore, AB = d sinθ ----- 2)
Substituting equation 2 in equation 1, we get, nλ = 2d sinθ
Applications of Bragg’s Law
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There are various applications of Braggs Law of diffraction such as –
- Bragg’s Law is useful for conducting the measurements of wavelengths.
- Alongside that, it is also used to determine the lattice spacings of crystals.
- In Wavelength Dispersive Spectrometry (WDS) or X-ray fluorescence spectroscopy (XRS), the crystals of known d-spacings are used to analyze crystals in the spectrometer.
- It plays a huge part in Crystallography.
- In X-ray diffraction (XRD), the d-spacing or inter-planar spacing of a crystal is used for identification purposes.
Solved ExampleGive a detailed explanation of how Bragg’s Law is used in X-ray diffraction. Solution: The application is explained in detail below:
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Bragg’s Diffraction
As per Bragg’s law, diffraction has three parameters
- The wavelength of X-rays, λ
- Crystal orientation, θ
- Spacing of the crystal planes, d.
The diffraction occurs for a given wavelength and set of planes. For example, changing the orientation continuously, i.e., changing θ until Bragg’s Law is satisfied.
Limitations of Bragg’s Law
Bragg's Law provides valuable insights into crystal structures and diffraction phenomena but it also has certain limitations which are as follows –
- It assumes a perfect crystal lattice structure, which may not always be the case in real materials.
- Deviations from ideal crystal structures can introduce errors in the measurements of lattice spacings.
- Bragg's Law is based on the assumption of monochromatic radiation, but in practice, X-ray sources have a broad range of wavelengths.
- This can lead to overlapping diffraction patterns and difficulty in accurately determining the angles and intensities of diffracted beams.
- The law also assumes that the crystal is static during the diffraction process, while in reality, the vibrations and thermal motion of atoms can affect the scattering patterns.
Things to Remember
- Bragg’s law states that an X-ray incident on a crystal at an angle θ results in a reflected beam with the same angle of scattering θ.
- Bragg’s law is a special case of Laue Diffraction.
- Constructive interference will occur when path difference ‘d’ is equal to the whole number ‘n’ of the wavelength.
- The equation of Bragg’s law: nλ = 2d sinΘ; where ‘n’ is an integer, ‘d’ is the distance, and Lambda is the wavelength.
- It determines the respective angles given out by coherent and incoherent scattering of crystal lattice surfaces.
- Bragg's law has applications in X-ray crystallography to determine crystal structures and for analyzing material properties.
- The extra distance the second beam travels in Bragg’s law is known as the integral multiple of the wavelength.
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Previous Year Questions
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Sample Questions
Ques 1. The wavelengths of the first-order X-rays are given as 2.20A0 at 2708’. Calculate the distance between the adjacent Miller planes. (3 marks)
Ans. Using Bragg’s law,
2d sin Ө = nλ
We know that, n = 1
λ = 2.29 A°
Ө = 27°8’
Substituting the values in the equation, we get
d = 2.51 A°
Ques 2. What is the difference between Interference and Diffraction? (2 marks)
Ans. Interference is a property that is formed from the waves of two distinct coherent sources. The secondary wavelets, however, are formed from the same wave but from different parts to produce a phenomenon known as Diffraction.
Ques 3. The wavelength of the X-rays is given as 0.071 nm. It is diffracted by a plane of salt with a lattice constant of 0.28 nm. Calculate the glancing angle for the second-order diffraction. Assume the salt plane value is 110, and the given salt is rock salt. (5 marks)
Ans. Given: Wavelength of X-rays = 0.071 nm
Lattice constant = 0.28 nm; Salt Plane = 110, and Order of diffraction = 2
Glancing angle = ?
Using Bragg’s law: 2d sin Ө = nλ
Rock salt has FCC, therefore,
\(\begin{array}{l}d=\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}\end{array}\)
Putting the value in the equation,
\(\begin{array}{l}d=\frac{0.28\times 10^{-9}}{\sqrt{1^{2}+1^{2}+0^{2}}}=\frac{0.28\times 10^{-9}}{\sqrt{2}}m\end{array}\)
Substituting the value in Bragg’s Equation, we get
Ө = 21o
Ques 4. Define Diffraction. (2 marks)
Ans. Diffraction is the phenomenon that takes place when a wave visibly runs into an obstacle or an aperture. It is simply a noticeable bending of light when passing across the corners of an object. In general, the amount of bending of the light is based on the relative size of the light’s wavelength to the size of the opening. It is the scattering of waves, be it of light, when passing around or through an object.
Ques 5. Consider that an X-ray of wavelength 100 \(\dot{A}\) tends to be incident on an atom that originally has an angle of 90°. Given the statement, determine the value of d for the first-order spectrum. (3 marks)
Ans. As per the given equation,
The wavelength is given as = 100 \(\dot{A}\)
The angle as mentioned in the question is = 90 degrees
Now, since we already know that Bragg’s law states = nλ = 2dsinθ.
Therefore, by substituting the equation with values as mentioned in the question, we get
-> λ = 10-8 m, n = 1, sinθ = 1.
With that said, 10-8 = 2 x d x 1
Hence, d = 50 \(\dot{A}\).
Ques 6. What is the order of diffraction in Bragg’s Law? (2 marks)
Ans. Bragg’s law, in general, outlines a condition for a plane of wave that is witnessed to be diffracted by a group of lattice planes. Bragg’s law claims the equation, nλ = 2d sinθ, wherein ‘d’ signifies the interplanar spacing or the d-spacing, θ usually refers to the angle between the individual atomic planes. Lambda, alongside the former, represents the wavelength and ‘n’ is an integer. The whole representation is known to be the order of diffraction.
Ques 7. Derive the calculation of d-spacing as per Bragg’s Law. (2 marks)
Ans. D-spacing, in accordance with Bragg’s law, can be calculated as,
\(\lambda\) = 2 d sin, wherein, \(\lambda\) is considered as the wavelength of the X-ray beam which can be denoted by 0.154 nm (it appears when the first incident upon the surface of the crystal lattice).
On the other hand, the letter ‘d’ mainly signifies the distance amassed between the individual atomic planes (the GO layers or the sheets), and θ dominantly represents the diffraction of the angle.
Ques 8. What is Bragg’s diffraction? (2 marks)
Ans. Bragg’s diffraction was first proposed by William Henry Bragg and William Lawrence Bragg, in 1913. Bragg’s diffraction occurs when a subatomic particle or electromagnetic radiation waves have wavelengths that are comparable to atomic spacing in a crystal lattice.
Ques 9. What will the path difference be equal to when the peaks of scattered intensity are observed? (2 marks)
Ans. In Bragg’s law experiment, the x-ray radiations are scattered by the crystal lattice. The peaks of scattered intensity are observed when the path difference is equal to the integral multiple of the wavelength. Also, the incident angle should be equal to the reflecting angle.
Ques 10. Do X-rays have larger wavelengths than gamma rays? (2 marks)
Ans. Yes. X-rays have larger wavelengths than gamma rays because they have lesser energy than gamma rays.
Ques 11. Why is X-ray diffraction used for studying the crystal structure of solids? (2 marks)
Ans. X-ray diffraction is used for studying the crystal structure of solids because, for diffraction to occur, the obstacle size should be comparable to that of the incident ray wavelength.
Ques 12. What should be the path difference between two waves if destructive interference is to occur? (2 marks)
Ans. For destructive interference to occur, the path difference between two waves should be \(n\lambda\).
Ques 13. What should be the path difference between two waves if constructive interference is to occur? (2 marks)
Ans. For constructive interference to occur, the path difference between two waves should be \(2\pi\).
Ques 14. Bragg’s law is not a sufficient condition for diffraction to occur. True or False? (1 mark)
Ans. True. Atoms at non-corner positions may result in out-of-phase scattering at Bragg angles.
Ques 15. The miller indices h, k, and l of parallel planes in a BCC lattice should satisfy which of the following x-ray diffraction reflection rules? (1 mark)
a.h+l+k should be even
b.h, k, l should either be even or odd
c.h,k,l should form Pythagorean triplet
d. All planes should allow reflection
Ans. a. h+k+l should be even.
If the sum of miller indices becomes odd, then destructive interference occurs.
Ques 16. What is the minimum interplanar spacing required for Bragg’s diffraction? (2 marks)
Ans. The minimum interplanar spacing required for Bragg’s diffraction is \(\frac{\lambda}{2}\)
Ques 17. How is reflection different from diffraction? (2 marks)
Ans. Reflection is a surface phenomenon where large portions of the incident waves are reflected. Reflection can also occur at any incident angles whereas diffraction can only occur at Bragg angle.
Ques 18. Why do K-beta x-rays have shorter wavelengths than K-alpha waves? (2 marks)
Ans. K-beta x-rays have shorter wavelengths than K-alpha waves because K-alpha waves are formed when there is an electron transition from the L shell to the K shell. K beta x-rays are formed when there is an electron transition from the M shell to the K shell. Thus K-beta x-rays have more energy and shorter wavelengths than K-alpha x-rays.
Ques 19. Does the electron cloud move when an x-ray is incident on an atom? (2 marks)
Ans. Since X-ray is an electromagnetic wave, it makes the electron cloud around the atom move when it is incident on an atom.
Ques 19. Why do waves scatter in Bragg’s law experiment? (2 marks)
Ans. The waves undergo a phenomenon called Rayleigh scattering in Bragg’s law experiment. Here, the re-radiated waves have the same frequency as that of the incident wave.
Ques 20. Why is Bragg’s law a special case of Laue diffraction? (2 marks)
Ans. Bragg’s law is a special case of Laue diffraction because in Laue diffraction if the incident angle is equal to the scattering angle and the wavelength is integer times the path difference, then satisfies Bragg’s law.
Ques 21. If the angle of incidence is 300, the wavelength of the first-order spectrum is equal to? (2 marks)
Ans. We know that,
\(n\lambda = 2dsin\theta\)
\(\theta\) = 300
n=1
we get,
\(\lambda = d\)
Ques 22. If an X-ray of wavelength 100 A0 is incident on an atom at an angle of 900, then what should be the value of d for the first order spectrum? (2 marks)
Ans. We know that,
\(n\lambda = 2dsin\theta\)
\(\lambda\)= 10-8m
n=1
\(sin\theta\)=1
Therefore,
10-8 = 2d
d= 50 A0
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