Braggs Law StatementBragg EquationBraggs Law Applications

## What is Bragg’s Law?

Bragg’s law is a special case of Laue diffraction which determines the angles of coherent and incoherent scattering from a crystal lattice. When X-rays are incident on a particular atom, they make an electronic cloud move just like an electromagnetic wave. The movement of these charges radiates waves again with similar frequency, slightly blurred due to different effects, and this phenomenon is known as Rayleigh scattering.

The same process takes place upon scattering neutron waves via nuclei or by a coherent spin interaction with an isolated electron. These wavefields which are re-emitted interfere with each other either destructively or constructively, creating a diffraction pattern on a film or detector. The basis of diffraction analysis is the resulting wave interference, and this analysis is known as Bragg diffraction.

### Bragg Equation

According to Bragg Equation:

**nλ = 2d sinΘ**

Therefore, according to the derivation of Bragg’s Law:

- The equation explains why the faces of crystals reflect X-ray beams at particular angles of incidence (Θ, λ).
- The variable
*d*indicates the distance between the atomic layers, and the variable Lambda specifies the wavelength of the incident X-ray beam. - n as an integer.

This observation illustrates the X-ray wave interface, which is called X-ray diffraction (XRD) and proof for the atomic structure of crystals.

Bragg was also awarded Nobel Prize in Physics in identifying crystal structures starting with NaCl, ZnS, and diamond. Diffraction has been developed to understand the structure of every state of matter by any beam e.g, ions, protons, electrons, neutrons with a wavelength similar to the length between the molecular structures.

### Derivation of Bragg’s Law

Consider the following figure of beams in which the phases of the beams coincide when the incident angle is equal to the reflecting angle. The incident beams are parallel to each other until they reach the point z. When they are at the point z, they strike the surface and travel upwards. At point B the second beam scatters. AB + BC is the distance traveled by the second beam. The extra distance is known as the integral multiple of the wavelength.

nλ = AB + BC

We also know that AB = BC

nλ = 2AB (equation 1)

d is the hypotenuse of the right triangle Abz. Ab is the opposite of the angle θ.

AB = d sinθ (equation 2)

Substituting equation 2 in equation 1

nλ = 2d sinθ

The above equation is Bragg’s law expression.

Following is the table explaining the other **Physics related laws**:

## Applications of Bragg’s Law

There are numerous applications of Bragg’s law in the field of science. Some common applications are given in the points below.

- In the case of XRF (X-ray fluorescence spectroscopy) or WDS (Wavelength Dispersive Spectrometry), crystals of known d-spacings are used as analyzing crystals in the spectrometer.
- In XRD (X-ray diffraction) the inter-planar spacing or d-spacing of a crystal is used for characterization and identification purposes.

### Bragg’s Diffraction

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.

### Solved Examples

**Example 1:**

The wavelength of the X-rays is 0.071 nm which is diffracted by a plane of salt with 0.28 nm as the lattice constant. Determine the glancing angle for the second-order diffraction. Assume the value of the plane of salt to be 110 and the given salt is rock salt.

**Solution:**

Given:

Wavelength of the X-rays = 0.071 nm

Lattice constant = 0.28 nm

Plane = 110

Order of diffraction = 2

Glancing angle =?

Using Bragg’s law:

2d sin Ө = nλ

Rock salt has FCC, therefore, \(d=\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}\)

Substituting the values,

\(d=\frac{0.28\times 10^{-9}}{\sqrt{1^{2}+1^{2}+0^{2}}}=\frac{0.28\times 10^{-9}}{\sqrt{2}}m\)

Substituting in the Bragg’s equation,

Ө = 21°

**Example 2:**

The wavelengths of first-order X-rays are 2.20A0 at 2708’. Find the distance between the adjacent Miller planes.

**Solution:**

Using Bragg’s law,

2d sin Ө = nλ

Where,

n = 1

λ = 2.29 A°

Ө = 27°8’

Substituting the values, we get

d = 2.51 A°

## Bragg’s law Conclusion

The concluding ideas from Bragg’s law are:

- The diffraction has three parameters i.e, the wavelength of X rays,λ
- The crystal orientation defined by the angle θ
- The spacing of the crystal planes, d.

The diffraction can be conspired to occur for a given wavelength and set of planes. For instance, changing the orientation continuously i.e, changing theta until Bragg’s Law is satisfied.

## Frequently Asked Questions

### State if the given statement is true or false: Bragg’s law is not enough to explain the diffraction by crystalline solids.

The given statement is true because the atoms present at the non-corner positions might result in scattering at Bragg angles which is out-of-phase.

### What is the minimum interplanar spacing that is required for Bragg’s diffraction to occur?

### State if the given statement is true or false: The electronic clouds move when the X-ray is incident on an atom.

The given statement is true. Since X-ray is an electromagnetic wave, it makes the electronic cloud around the atom to move.

### Bragg’s law experiment is based on which scattering of waves?

Bragg’s law experiment is based on the Rayleigh scattering in which the charges are scattered without a change in their wavelength.

### Calculate the wavelength for the first-order spectrum if the angle f incidence is 30 degrees.

From the Bragg’s equation, we know that \(n\lambda =2d\;sin\theta\)

By substituting the values for

n=1

\(sin\theta\)=30 degrees = 1/2

We get, \(\lambda =d\).

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