## Short Note on Bragg’s Law:

(1) Consider a series of parallel atomic planes which after the scattering of X-rays produce intense spots.

(2) Let the planes be separated by a distance ‘d’.

(3) Consider a narrow beam of X-rays from source be incident on these planes at an angle ‘θ’.

(4) The beam will be scattered in all directions by atoms in various atomic planes.

(5) Let us consider the portion of the beam which is scattered at an angle ‘θ’ to the atomic planes, so that angle of reflection is equal to the angle of incidence.

(6) Consider a ray PA reflected at ‘A’ in a direction AR, from 1^{st} atomic layer and another ray QB reflected at another atom C, in the direction CS from 2^{nd} layer.

(7) From A draw AB and AD perpendicular to QC and CS respectively.

(8) It is clear that rays QC and CS travel a longer path PA and AR and the path difference is BC + CD.

(9) In figure to understand the concept, the atomic spacing is greatly enlarged, actually the reflected rays interfere with each other and produces a single impression on a photographic plate.

(10) When a path difference is a whole wavelength or a multiple of whole wavelength nλ, then the two waves AR and CS will reinforce each other and produce an intense spot (i.e. constructive interference).

∴ Path Difference = BC + CD = nλ ……….(i) |

(11) But from the figure,

BC = CD = d sin θ ……….(ii) ∴ d sin θ + d sin θ = nλ or 2d sin θ = nλ ……….(iii) Where n = 1, 2, 3,……. |

**Equation (iii) is known as Bragg’s Law**, where ‘n’ is always an integer which stands for the order of reflection.

Thus by knowing interplanar distance ‘d’, the wavelength of X-ray can be calculated.

## Comments (No)