Uses of X-rays, X-ray Diffraction and Bragg's Law

Uses of X-ray

There are various uses of X-rays of which some of them are given below:

1. X-rays are used in surgery to detect the fracture of bones, diseased organic and foreign bodies and growth in the human body. A widely used improved X-ray technique is computed tomography, the corresponding instrument is called a CT scanner.
2. X-rays are used to destroy malignant tumours and to cure intractable skin diseases soft X-rays are used if the affected parts are superficial while hard X-rays are used to destroy tumours very deep inside the body.
3. In industries, X-rays are used to detect any defect in radio values, tennis balls, rubber tyres and a presence of pearl in oysters. They are used to test the uniformity of the insulating materials and the quality of oil paintings.
4. The X-rays are used in detecting flaws in heavy girders, flaws, cracks and air bubbles in certain goods. They are used to test the quality of weldings, moulds and metal castings.
5. Property of penetration of X-rays is utilized in making the detection of the presence of smuggled gold, at airports, sea ports etc. They are used in detective departments for detection of explosive, opium and other contraband goods. They are also used in mints, where coins are made.
6. X-rays are widely used for research and to investigate the structure of crystals, arrangement of atoms and molecules in the matter and their behavior on different materials.

X-Ray Diffraction

The diffraction effects can be observed only if the spacing between the lines ruled on the grating is of the order of magnitude of the wavelength of light used. The grating element for the usual grating is of the order of 10^-6 m in the diffraction of a beam of light. A transmission diffraction grating has about 6000 lines per cm on the piece of glass. The width of the slit between two lines is of the order of a wavelength of light and hence is used to study the diffraction of ordinary light. The wavelength of X-ray in order of 1/1000^th of that of the grating used in the visible region of the wavelength of the visible light. Therefore, to observe diffraction with X-rays, we need a grating having a spacing 1/1000^th the visible region. We have suggested that crystals can be used for the diffraction of X-rays. In a crystal gratings, the atoms arranged in a regular pattern correspond to the grating lines and the distance between two atoms to the grating element.

T is a source of X-rays which gives X-ray beam of a continuous range of wavelengths. A narrow beam of X-rays from the X-ray tube is collimated by two slits S₁ and S₂ and is passed through a zinc sulphide crystal and allowed the emerging beam to fall on a photographic plate as shown in the figure. After an exposure of several hours and on developing the plate, many faint but regularly arranged spots around the central bright spot were observed on the plates as shown in the figure.

From this experiment, it can be concluded that

1. The atoms of a crystal are arranged in a regular three-dimensional lattice.
2. X-rays are electromagnetic waves.

Bragg’s Law

Consider a set of parallel atomic planes separated by a distance d. A narrow monochromatic X-ray beam of wavelength λ is allowed to fall on the crystal lattice at a glancing angle θand after reflection from the planes X and Y goes along QR and Q’R’ respectively. Since X-rays are much more penetrating than ordinary light, there is only partial reflection at each plane. The complete absorption takes place only after penetrating several layers. Consider two parallel rays PQR and P’Q’R’ in the beams, which are reflected by two atoms Q and Q’. Q’ is vertically below Q. the ray P’Q’R’ has a longer path than the ray PQR. To calculate the path difference, drop QT and QS perpendicular to P’Q’ and Q’R’ respectively.

\begin{align*} \text {Then, path difference} &= TQ’ + Q’S \\ \text {From the right angled triangle QTQ’ and QSQ’, we have} \\ \therefore \text {Path difference} &= d\sin \theta + d\sin \theta \\ &= 2d\sin \theta \\ \end{align*}

Since for maximum intensity, the path difference between two reflected rays should be integral multiple of λ,

Hence, \(2d\sin \theta = n\lambda \) where n = 1,2,3,….

This equation is known as Bragg’s law or Bragg’s equation. The integer n gives the order of the scattered beam. For example, for first order reflection n= 1. If θ₁ is the corresponding glancing angle, then from Bragg’s equation \(2d\sin \theta_1 = \lambda \)

Similarly, for second order reflection n= 2. If θ₂ is the corresponding glancing angle, then \(2d\sin \theta_2 = 2\lambda \)

Reference

Manu Kumar Khatry, Manoj Kumar Thapa, Bhesha Raj Adhikari, Arjun Kumar Gautam, Parashu Ram Poudel. Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.

S.K. Gautam, J.M. Pradhan. A text Book of Physics. Kathmandu: Surya Publication, 2003.