Electromagnetic research Essay Example for Free

Electromagnetic research EssayWe can relate the density to the sprinkling of radiation in the X-ray region by victimization the passel attenuation coefficient, /? , and the mass energy-absorption coefficient, en/?. In the X-ray region (10-100 KeV), we will find that as the negatron density increases, the x-ray signal in like manner increases1. X-ray scattering is considered to be due to the negatron density of the atoms in a sample2. J. J. Thomson derived a formula relating the intensity of coherent scattering from a particle. If the hap 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 velocity of light, and (1 + cos22? )/2 represents the uncomplete polarization of the scattered photon. An electron is cc0 times lighter than a proton. Most nuclei are made up of many nucleons that have an even greater m ass than the mass of a proton. Because the scattered intensity is inversely proportional 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 probability for coherent scattering decreases with increasing photon energy and increases with increasing atomic government issue (increasing number of interfering electrons). As a fraction of the total 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 congeneric 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 fo und, and figures 2. 1, 2. 3 and 2. 4, the Absorption of Electromagnetic irradiation by Gold, Silicon, and Iron respectively. The density of silicon is 2. 330, iron is 7. 874 and currency is 19. 32. In the graphs, as the radiation increases we will notice that the absorption decreases. For gold, this will continue 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) therefore the photon-absorption runes 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 thick sheets are needed to absorb sticky (short-wavelength) x-rays or gamma -rays, compared to soft (long-wa velength) x-rays for example.Photon absorption in the visible region of the spectrum depends on the atomic arrangement of the atoms and their bonding. Pure silicon (Si) is strongly absorbing plainly silicon combined with oxygen is transparent. For the energetic photons in the x-ray regime, photon absorption is much easier to predict and is independent of the details of atomic arrangement. It depends primarily on the electron concentration per unit volume. Since the concentration of atoms per unit volume only differ by factors of 2 or 3 from each other, the electron concentration in two materials can be estimated from the atomic number, Z.Gold (Z=79) absorbs x-rays much more 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 terce proces s 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 rest mass m0 is the same as that of an electron but whose charge is +e). Pair 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 lucifer 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 kinetic energy of the electron and positron, immediately afterwards their creation.If the photon energy were exactly 2m0c2 = 1. 02 MeV, the two particles would be created at rest (with zero kineti c energy) and 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 above 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, http//xrayweb. chem. ou. edu/notes/crystallography. html. 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 ray from Non-linear Thomson Scattering. Vol. 91, No. 19, 13 November 2003. Retrieved 9 June 2008, http//www. eecs. umich. edu/USL-HFS/TaPhouc_prl_03. pdf. Weidner, R. T. and Sells, R. L. Pair Production and Annihilation. Retrieved 9 June 2008, http//physics. pdx. edu/egertonr/ph311-12/pair-p a. htm.