Diffraction of X-rays through crystal lattice: (Bragg’s law)
X-rays are used to identify the structure of a crystal. The property of diffraction of x-rays are used in this procedure.
The wave length of ordinary light is very large comparing with the value of inter planar spacing.
Hence ordinary light cannot penetrate trough the lattice planes.
The wave length of ordinary light is very large comparing with the value of inter planar spacing.
Hence ordinary light cannot penetrate trough the lattice planes.
But X-rays penetrate through the lattice planes and they are scattered by the atoms of the crystal lattice. Each atom acts as an opaque and the distance between two atoms acts as narrow slit. Hence the crystal lattice is considered as diffraction grating.
When x-rays are scattered by the successive parallel planes they diffract with each other and produce diffraction pattern.by studying the diffraction pattern we can find different sets of planes located in the lattice at different directions.
Bragg used X-rays to study the structure of crystals.
Bragg used X-rays to study the structure of crystals.
He derived the condition of maxima of diffraction pattern which is
called Bragg’s law.
In the above diagram two x-rays R1 and R2 are scattered by the two parallel and successive planes of a crystal lattice.
In the diagram the atoms are denoted by red colored dots and the crystal planes passing through them are denoted by dotted lines.
The inter planar distance is equal to d.
The two X-rays R1 and R2 incident on the atom at the angle Ɵ and also scattered at the same angle
The X-rays scattered at equal angle diffract and produce diffraction pattern.
The diffraction pattern consists maximum intensity for the condition, the path difference between the two X- rays is equal to integral multiples of wavelength of X-rays.
Path difference = n λ, n = 0, 1, 2...
travel equal distances before AC and beyond AC’. Also it is observed that the second ray R2 travels
an extra path equal to BC and BC’.
Hence the path difference = BC + BC’
It is clear from the diagram that BC= BC’
Hence the path difference = 2BC
From the triangle ACB, sinƟ = BC / AB
AB = inter planar distance d
Therefore sinƟ = BC / d
And BC = d sinƟ
Hence the path difference = 2BC = 2d sinƟ
Hence the condition for maximum intensity is 2d sinƟ =nλ
It is the Bragg’s law for diffraction of X- rays through the crystals.
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In the diagram the atoms are denoted by red colored dots and the crystal planes passing through them are denoted by dotted lines.
The inter planar distance is equal to d.
The two X-rays R1 and R2 incident on the atom at the angle Ɵ and also scattered at the same angle
The X-rays scattered at equal angle diffract and produce diffraction pattern.
The diffraction pattern consists maximum intensity for the condition, the path difference between the two X- rays is equal to integral multiples of wavelength of X-rays.
Path difference = n λ, n = 0, 1, 2...
Calculating the path difference:
Draw two perpendicular lines AC and AC’ from the point A. it can be observed that the two X-raystravel equal distances before AC and beyond AC’. Also it is observed that the second ray R2 travels
an extra path equal to BC and BC’.
Hence the path difference = BC + BC’
It is clear from the diagram that BC= BC’
Hence the path difference = 2BC
From the triangle ACB, sinƟ = BC / AB
AB = inter planar distance d
Therefore sinƟ = BC / d
And BC = d sinƟ
Hence the path difference = 2BC = 2d sinƟ
Hence the condition for maximum intensity is 2d sinƟ =nλ
It is the Bragg’s law for diffraction of X- rays through the crystals.
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