Electromagnetic research Essay

We can relate the assiduity to the strewing of irradiation in the roentgen ray region by using the view attenuation coefficient, /? , and the mass vitality-absorption coefficient, en/?. In the X-ray region (10-100 KeV), we will figure that as the electron density ontogenys, the x-ray signal also increases1. X-ray scattering is considered to be due to the electron density of the atoms in a sample2. J. J. Thomson derived a formula relating the intensity of coherent scattering from a section. If the incident radiation is not polarized, then his relation takes the form I(2? ) = Io (e4)/(r2 m2 c4) (1 + cos2 2? )/2where e is the charge of the particle, r is the distance from the scatterer, m is the mass of the scatterer, c is the speed of light, and (1 + cos22? )/2 represents the partial polarization of the scattered photon. An electron is 2000 times lighter than a proton. Most nuclei be made up of many nucleons that have an purge greater mass than the mass of a proton. Becaus e the scattered intensity is inversely comparative to the square of the mass of the particle emitting the X-ray photon, scattering of X rays from the nucleus is considered negligible compared to the scattering from the electrons of an atoms.We can conclude that X-ray scattering is due to the density of the electron. The prospect for coherent scattering subsides with increasing photon elan vital and increases with increasing nuclear number (increasing number of interfering electrons). As a fraction of the fundamental mass attenuation coefficient, ? coh/? is maximal at atomic numbers around Z=10 and photon energies in the interval 30-50 keV. At higher atomic numbers, the relative fraction decreases due to the strong increase of photoelectric absorption with increasing atomic number3. In silicon, for example, the relative probability for coherent scattering is 14% at 50 keV.Let us refer to table 1, where the densities of elements could be found, and figures 2. 1, 2. 3 and 2. 4, the Absorption of Electromagnetic radiation sickness by Gold, Silicon, and Iron respectively. The density of silicon is 2. 330, iron is 7. 874 and gold is 19. 32. In the graphs, as the radiation increases we will notice that the absorption decreases. For gold, this will expand until it reaches 200 eV where a discontinuity takes place. In terms of the densities of the elements, as the density of the element decreases, the coherent scattering also decreases.For example, gold, which has an atomic number of 79, contains a higher electron density than an element such as silicon and iron (of lower atomic number 14 and 26 respectively) then the photon-absorption mathematical processes are stronger (higher ) making gold suitable as a radiation-shielding material. However, the decrease in () with increasing photon energy (below 1 MeV) means that relatively deep-chested sheets are needed to absorb hard (short-wavelength) x-rays or gamma -rays, compared to well-to-do (long-wavelength) x-r ays for example.Photon absorption in the visible region of the spectrum seems on the atomic sight of the atoms and their bonding. Pure silicon (Si) is strongly absorbing but silicon combine with oxygen is transparent. For the energetic photons in the x-ray regime, photon absorption is much easier to forecast and is independent of the details of atomic arrangement. It depends primarily on the electron parsimoniousness per unit volume. Since the concentration of atoms per unit volume only differ by computes of 2 or 3 from each other, the electron concentration in 2 materials can be estimated from the atomic number, Z.Gold (Z=79) absorbs x-rays much much efficiently than silicon (Z=14) or iron (Z=26). X-ray absorption does depend on the energy of the x-rays and decreases with increasing x-ray energy, E. Absorption decreases nearly proportional to the cube of the energy (i. e. absorption proportional to (1/E3). The eventual rise in indicates that a third process occurs at high photon energy this is mate production, in which a pair of elementary particles (a particle and its antiparticle of the same mass but opposite electrostatic charge) is created from the energy (hf) of the original photon4.In this case, the two particles are an electron and an anti-electron (more commonly known as a positron, whose embossment mass m0 is the same as that of an electron but whose charge is +e). equalise production can be represented by an equation, which represents the conservation of total energy (or mass-energy) hf = 2(m0 c2) + K(-e) + K(+e) Here, (m0 c2) = 0. 511 MeV is the rest energy of an electron, which is equal to that of the positron, so the factor of 2 represents the fact that two particles of identical rest mass are created. K(-e) and K(+e) represent the energizing energy of the electron and positron, immediately after their creation.If the photon energy were exactly 2m0c2 = 1. 02 MeV, the two particles would be created at rest (with zero kinetic energy) a nd this would be an example of the complete conversion of energy into mass. For photon energies below 2m0c2, the process cannot occur in other words, 1. 02 MeV is the threshold energy for pair production. For photon energies to a higher place the threshold, a photon has more than enough energy to create a particle pair and the surplus energy appears as kinetic energy of the two particles.BIBLIOGRAPHYGiacovazzo, Carmelo. Crystallography. Retrieved 9 June 2008, . McAlister, B. C. and Grady, B. P. The Use of Monte-Carlo Simulations to Calculate Small-Angle Scattering Patterns. Macromolecular Symposia, 2003. The American physical Society. X-Ray Radiation from Non-linear Thomson Scattering. Vol. 91, No. 19, 13 November 2003. Retrieved 9 June 2008, . Weidner, R. T. and Sells, R. L. Pair ware and Annihilation. Retrieved 9 June 2008, .