Diffraction of X Rays

In 1900, when the properties of X-rays were being studied, Haga and Wind tried to show the wave nature of X-rays as similar to light radiations by using very fine slits. In 1913, Freidrich, Knipping and Laue passed a small beam of X-rays through a zinc sulphide crystal and allowed the emerging beam to fall on a photographic plate.

The developed plate showed a central spot surrounded by a number of small spots. This was caused by the diffraction of X-rays through the crystals in which atoms are arranged in a regular three dimensional lattice. Laue suggested that the atoms in certain crystals are arranged in such a manner that the crystal acts as a diffraction grating for the X-ray beam. The spacing between these atoms or molecules is of the order of 10^-8 cm.

Bragg’s law: W.H. Bragg and W.L. Bragg studied the diffraction of X-rays in detail. W.L. Bragg considered that when monochromatic X-rays impinge upon the atoms in the crystal lattice, each atom acts as a source of scattered radiation of the same wavelength. The crystal acts as a series of parallel reflecting planes. Thus, X-rays would be reflected according to ordinary laws of reflection and the planes would reflect X-rays at all angles. X-rays that penetrate more deeply into the crystal are reflected from the lower planes and thus several beams of X-rays reflected from various planes are obtained. The intensity of the reflected beam at certain angles will be maximum where the two reflected waves from two different planes have a phase difference equal to an integral multiple of the wavelength of X-rays, while at some other angles, the intensity of the reflected X-ray beam will be maximum.

Consider a beam of monochromatic X-rays of wavelength λ incident on a crystal and after reflection from the planes Y and Z goes along BC and EF respectively.

Let the crystal lattice spacing between the two planes be d and θ the glancing angle. The path difference between the reflected waves along BC and EF

= PE + EQ

In the Δ BPE, sin θ = PE/BE, PE = BE sin θ = d sin θ

Similarly, QE = d sin θ

∴ Path difference = 2 d sin θ

If this path difference is an integral multiple of wavelength, then constructive interference will occur between the reflected beams and they will reinforces with each other. Therefore, for the reflected beam to be of maximum intensity, 2d sin θ = nλ, where n = 1, 2, 3, etc. This is Bragg’s equation and represents Bragg’s law.

If n = 1, 2d sin θ = λ. This beam gives the spectrum of the first order.

From the equation, it is clear that if the wavelength of X-rays, which produces intense maxima at a glancing angle θ is known, then the distance between atomic planes of a crystal can be calculated. Similarly, if d is known, λ can be calculated.

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