**Bragg’s law** was first proposed by Sir William Bragg and his son Sir Lawrence Bragg. They studied about the diffraction of X-rays and devised a method for determining the wavelength of X-rays.

## Bragg’s Law Statement

When a monochromatic X-rays are incident upon a crystal, atoms in different layers acts as a source of scattering radiation of same wavelength. The intensity of the reflected beam will be maximum at certain incident angle when the path difference between two reflected wave from two different planes is an integral multiple of wavelength of X-rays.

i.e For maximum intensity, Path difference = nλ where n = 1,2,3,... and λ = wavelength of X-rays

Let us consider a parallel beam of monochromatic X-rays *AB* and *DE* of wavelength *λ* is incident on parallel crystal planes *X* and *Y* respectively, both with a equal glancing angle *θ*. Ray *AB* strikes an atom at *B* and is reflected along *BC*. Similarly, ray *DE* strikes an atom at *E* and is reflected along *EF*.

Now, draw *BG* and *BH* perpendicular to *DE* and *EF* respectively. From above figure, the path difference between the rays *DEF* and *ABC* is

path difference = *GE* + *EH*

From triangle BGE and BEH, EH = GE = BEsinθ or, EH = GE = dsinθ where d = interplaner spacing So, path difference = 2dsinθ

We know, for maximum intensity

Path difference = nλ or, 2dsinθ = nλ

This is called **Bragg’s Law**.

Here n gives the order of diffraction.

In general,

2dsinθ_{n}= nλ

## Numerical Problem for Bragg’s Law

X-rays of wavelength 3.6 X 10^{-11}m undergoes first order reflection at a glancing angle of 4.8*°* from a crystal. Find the spacing of the atomic planes in the crystal.

**Solution:**

Wavelength of X-ray(λ) = 3.6 X 10^{-11}m

Glancing angle(θ) = 4.8*°*

Order of diffraction (n) = 1

Let the spacing of atomic planes in the crystal be d. Then using Bragg’s law, we have

Hence the spacing of the atomic planes is 2.5 X 10^{-10}m.