X-Rays

Introduction X-ray were discovered by Roentgen in 1985. The scientist, Barkla, in 1906 proved that X-ray are electromagnetic radiation of very small.

Properties of X-Rays X-Rays travel in straight lines and are not deflected by electric and magnetic fields X-Rays is electromagnetic waves with wavelength range 0.01-10 nm. X-Rays ionize the medium through which they pass X-Rays is highly penetrating. They can penetrate 2-3cm in Wood and 1.15 cm in aluminum. They can penetrate through hands, feet, arms and legs. X-Rays Cause the phosphorescent substances to glow and affect the photographic plates.

Production Of X-rays Fig. X-ray Tube The Cathode is heated by filament F through which a current (i) is passed. It emits electron by thermionic emission. Voltage (V) accelerates the electrons, When then fall on a target which is inset within the face of the anode. The Face of the target is inclined, at about 600 to the incident beam of electrons, so that the X-rays that leave the target pass through the side of the tube. The Tube is evacuated to prevent any obstruction in the path of electrons.

Experimental Observations Fig.2 a)Continiouus spectrum of X-rays from a tungsten target. b) Spectrum of X-rays from Mo And W at V= 35 KeV If V Increases, then the intensity of X-rays increases and higher penetrating power are produced (shown in Fig.2 a) If the Current i through the filament is increased, The Intensity of X-Rays increases. Spectra of X-rays with different targets are different. (shown in Fig.2 b) As the Voltage (V) increases, wavelength is deceases. (shown in Fig.2 a) For A same voltage, λmin is same for all element producing X-rays, i.e. λmin is independent of the targets. (shown in Fig.2 b)

As the Voltage (V) increases, wavelength is deceases. Fig. Continuous spectrum of X-rays from a tungsten target If V Increases, then the intensity of X-rays increases and higher penetrating power are produced If the Current i through the filament is increased, The Intensity of X-Rays increases. As the Voltage (V) increases, wavelength is deceases.

Spectra of X-rays with different targets are different. Spectrum of X-rays from Mo And W at V= 35 KeV Spectra of X-rays with different targets are different. For a same voltage, λ min is same for all element producing X-rays, i.e. λ min is independent of the targets.

X-RAY Diffraction If we use to a gratings for study of diffraction of X-Rays, it must have more than 108 lines, it is impossible to make such a grating, Alternately solutions were suggested by laues and Braggs.

Fig. Diffraction of X-rays Laues’s Diffraction Fig. Diffraction of X-rays Laue’s was aware about the fact that a crystal is made up of the regularly spaced atom with interionic spacing of the order of an armstrong (A0). The Wavelength of X-rays are also of the order of Armstrong. So laue’s suggested that the possibility of the crystal as a diffraction grating. In this experiment, the X-Rays passed through a fine pencil aperture in the sheets of lead at S1, S2 and S3 trough a crystal and then on photographic plate. After an exposure, the photographic plate is developed. It exhibited a central spot due to the direct beam of X-Rays. Around the central spot were distributed regularly arranged spot, fainter than central spot. These spot were called Laue’s spot.

Conclusions X- Rays are electromagnetic wave X-rays could be diffracted and their wavelength could be measured. The wavelength of X-rays was shown around 0.1- 0.4 A0. Atoms in crystals were arranged in definite geometrical pattern.

Bragg’s Law of X-ray Diffraction Fig. Bragg’s Diffraction Consider one set of Bragg planes. Let a parallel beam of monochromatic X- Rays of wavelength (λ) be incident on it, making angle (θ) with a plane. Ray AB reflected along BC and ray A1B1 along B1D. Angle of reflection is equal to the angle of incident in each case. IF BE is normal to A1B1and BF is normal to the B1D, Then EB1 and B1F will be the path difference between the reflected beams. In ∆ BB1E, sin θ =B1E/ BB1, but BB1 = d Hence B1E = d sin θ Similarly, In In ∆ BB1F, B1F = d sin θ

2d sinθ= nλ where n is an integer. If the path difference in an integral multiple of λ, then constructive interference will occur between the reflected beams. There will consequently be a maximum intensity in the reflected beam only when 2d sinθ= nλ where n is an integer.

Bragg’s X-ray Single Crystal Spectrometer Fig. X-ray Spectrometer