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Radiographic Science - Essay Example

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A law often referred to as the inverse squire law can best explain the intensity of a beam of x-ray. This law states that the intensity of radiation from any given source of a radiation is inversely proportional to the squire of the distance from that given source. …
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? Radiographic Science Practical Essay writing-up Experiment To investigate the effect of the inverse square law on a beam of x-ray photons. Introduction. A law often referred to as the inverse squire law can best explain the intensity of a beam of x-ray. This law states that the intensity of radiation from any given source of a radiation is inversely proportional to the squire of the distance from that given source. Mathematically, the inverse squire law can be written as I ? . In this case, I represent intensity, with d representing the distance. In the case of an x-ray tube, the x-ray photon is often produced from the anode, which is a large metal piece connecting to the positive side of an electric circuit. As widely cited, the anode does not only convert the electronic energy to the x-radiations but also dissipates heat that is given out during the process. This, therefore, implies that in order for the process work effectively, there is need to make sure the material chosen to serve the purpose is appropriate for the work. In order to have an ideal situation, a significant number of electrons have to create the x-ray photons but not heat. It is worth noting that the amount of electronic energy, which is changed into the x-ray is somewhat dependent upon two factors. These are the atomic number of the material on the anode, and the electron energy. The function of the cathode, on the other hand, entails expelling electrons out of the electrical circuits while focusing them to a well-defined anode beam. This experiment seeks to demonstrate the application of the inverse squire law and it is, therefore, focused on investigate the effect of the inverse square law on a beam of x-ray photons. Method. The dosemeter was placed on the x-ray table. This indicated the photon’s relative number in the beam of x-ray and was proportional towards the intensity. The integral tape measure was used to have the x-ray’s tube positioned at 25cm. The sensitive plate was collimated on the dosemeter. An exposure of 40kV 10mAs was used, and the reading of the dosemeter recorded. After this, display was reset. The whole procedure was repeated twice and the mean reading calculated. Next, the x-ray tube distance was increased in steps of 25cm. The same exposure was repeated, and the readings recorded to a 100cm distance above the dosemeter. The dosemeter reading was predicted at each step, using a form of the inverse square law that was given by I1/I2= d22/d12. In this relation, I1 is the predicted reading at a d1 distance and I2 is the mean of reading at d2 distance. The excel software was used in plotting a dependent variable (mean reading of intensity) graph, against the variable that was independent (distance). Whenever the intensity was directly proportional to the plotted distance, a straight line was plotted. This is displayed in appendix A. Results. Table 1. d (cm) 1/d2 (x 10 -4cm-2) Mean Dosemeter readings (Gy/mGy) Calculated readings (Gy/mGy) 25 16 0 780 50 4 0.23 0.25 75 1.8 0.9 0.99 100 1 1.1 1.21 125 0.64 1.4 1.54 150 0.44 1.53 1.68 175 0.33 2 2.2 200 0.25 2.4 2.64 225 0.20 2.45 2.7 250 0.16 2.7 2.97 275 0.13 3.1 3.41 300 0.11 3.2 3.52 In each distance, 1/d2 was calculated, and a graph of intensity against 1/d2 plotted. As can be observed from the graph, there is a none-liner relationship between intensity was 1/ d2. This implied that intensity versus 1/d2 graph would give a straight line. All the data collected was recorded in a table, in the results section. Table 2. Photon energy (keV) 1/photon energy wavelength x 10-13 m 25 0.04 497 100 0.01 124 1000 0.001 12.4 2000 0.0005 6.2 Absorbed dose (mGy) Light output (arbitrary units) 0 0 0.23 0.25 0.9 0.99 1.1 1.21 1.4 1.54 1.53 1.68 2 2.2 2.4 2.64 2.45 2.7 2.7 2.97 3.1 3.41 3.2 3.52 Discussion. According to the plotted graph in appendix A and the collected results, it is evidenced that the intensity of the x-ray beam reduces as the length from the target increases due to the divergence of beam of x-ray. The reduction in intensity is, therefore, proportional to the distance squared from the target. A fall-off that is nonlinear in intensity with the measured distance is referred to as the inverse square law. This implies that changing the distance from the source of the x-ray from position A to B changes the x-ray beam intensity by (A/B)2. For instance, when the distance from the x-ray source is doubled, the x-ray beam intensity would be reduced four times. From the plotted graph, the intensity of a beam of x-ray would be given by the relation I ? , where d is the distance from the target to the measurement point. The x-ray quality is, therefore, dependent on the kVp, tube filtration and the waveform generator (Thayalan, 2003). Experiment 2: To investigate the attenuation with thickness of material of a beam of x-ray photons. Introduction. Attenuation is the x-ray photons removal from the beam through scattering or absorption. It occurs as the beam of x-ray goes through matter, and there exists a partial or complete loss of the x-radiation energy. Other x-ray photons may go through matter entirely without tissue interaction. Some of them may interact with an electron orbital, the nucleus or the entire atom. During interaction, there may be a complete loss of energy, which is referred to absorption or partial loss of energy referred to as scattering (Thayalan, 2003). After a complete absorption of photon energy and complete transfer to the tissue, the photon will fail to exist. A photon which interacts then scatters has enough energy to continue interacting with other atoms. In order to gain a deeper understanding about the attenuation of the x-ray photons, an experiment was set to find out the material thickness attenuation in a beam of x-ray photons. Method. This experimental set up was done procedurally. To start with, the x-ray tube was set at about 100 cm above the dosemeter on the table of x-ray and the collimate to the sensitive plate. With a view to make sure the collimation and distance remained in constant position, an exposure of 100mAs and 80kV while at the same time recording the readings. Then the entire procedure was redone followed by taking the transmission readings alongside increasing the aluminium thickness. Transmission, in this context, means the attenuation inverse. The measured transmission was considering, thus, the photons absorption from the beam of x-ray, alongside the photon removal from the beam of x-ray was achieved through scattering. In doing this, it was observed that whenever the aluminium was placed directly on the dosemeter, the photons that were scattered were detected, and their measurement were also taken especially in the cases where they were transmitted. Thereafter, the sheet of aluminium was placed far from the dosemeter, with a sole purpose of reducing the photons that were scattered and prevented from reaching the detector. By placing a 1mm Al sheet direct to the dosemeter, the scattering effect could be seen. This way, the 1mm aluminium sheet was carefully attached beneath the diaphram’s light beam by the use of the sticky tape. In this regard, similar factors of exposure were applicable while making sure the dosemeter reading are recorded. With a view to reduce on experimental errors, the entire process was repeated using the 4, 3, and 2 mm thicknesses of aluminium. Following this, the percentage transmission was calculated for each of the layers of 1mm, relative to the reading that was done original without the aluminium foil. It was then that analysis was done using Excel. This way, a graph of the dependent variable (percentage x-ray transmission versus the thickness of aluminium) was plotted. Results. The table below show the experimental results. Table 3. Al (mm) % transmission Loge transmission 0 100 4.60 1 68 4.22 2 54.5 4.00 3 45.5 3.82 Appendix A show a plot of intensity verses 1/d. Appendix B show a linear relationship existing between intensity of a photon and the distance of the photon from its source. Discussion. From the plotted analysed results, it is clear that with there is a linear relationship between intensity and the squire of the distance from the source of the radiation. The formula of exponential that relates intensity to materials thickness is given by; It = . or = The values I0 and It represent original and transmitted intensity respectively with E representing a constant of mathematics, while ? is the linear attenuation coefficient of a given material and x is the material thickness. Taking natural logarithm on both sides of the equation, the equation is reduced to – ?x. Basing on the equation of a straight line, it is evidenced that the natural logarithm graph of percentage transmission against x will result into a straight line whose gradient is –?. From the graph –=. = 11.5. This implies that the attenuation of the x-ray photon increased in those tissues, which were somewhat dense. More evidently, Attenuation in 1mm of aluminium was greater than that of 2mm. This is consistent with what has been found out in similar studies, which have indicated that the attenuation of metals with high atomic number is high as compared as other metals. The reason being that the atomic numbers of the density is extremely higher for the tissues in that are within the surrounding. As widely cited, those components that have a high degree of attenuation are referred to as radiopaque. This follows their ability to absorb the x-radiation photons easily. For other substances that have a low degree of attenuation, they are referred to as radiolucent given that the x-ray can easily penetrate through these substances. In this regard, Aluminium can be classified as a radiopaque substance because of their high power of attenuation (Thayalan, 2003). Clearly, from the graph, the aluminium thickness that would give 50% transmission approximates to 1.5mm. This value is the experimental HVL value of aluminium. However, the value slightly deviates from the theoretical one due to experimental errors. The HVL or the half-value layer involves the amount of the needed material to reduce the beam intensity of x-ray to one half of the initial or original value. Notably, the HVL can be affected by the quantity of filtration, as well as the kVp in the x-ray beam. In this case, if the thickness of the material is increased twice the amount, it may not give out the double attenuation since the mean energy of a beam of x-ray varies as it goes through the material. As the energy reduces, there would be a reduction in the number of the x-ray photons that are removed out from the beam. This implies that the beam that is in exit is more penetrating and is considered being of high energy compared to the entrance beam. When the beam removes the soft x-ray photons, the beam would be hardened hence increasing the potential of the x-ray photons that passes through the tissue. In soft tissues, the HVL is about 4cm. This means that the 4cm of a material decreases the intensity of x-ray to one-half of the original value. Therefore, HVL is responsible for the description of the penetration or the x-ray beam quality. Generally, it is certain that the x-ray beams that are more penetrating have a high HVL. Conclusion. In this experiment, it clear that all the objectives of the experiment were achieved. Though, experimental values were somewhat different from the theoretical, this happened following errors during the experiment. It was evidenced that the quality of beam described the x-ray beam penetration. The x-rays that have a higher kVp had a high penetration implying that they can go through into a tissue completely with no absorption. On the other hand, the x-ray beams with a high quality of the beam are known to give out images that have minimum noise (Thayalan, 2003). As the thickness of the tissue increased, the number of x-ray photons that were attenuated increased through scattering and absorption. This implied that a high number of photons would be attenuated in a tissue that is 3 mm compared to a tissue with 1mm. The factors of technical exposure should be altered to compensate the varying differences in the tissue. Whenever x-ray travel in the matter, they can be absorbed, scattered or transmitted. Attenuation occurs whenever the x-ray photons are removed due to the absorption of tissue and are thereby scattered from the beam of x-ray as it goes through matter. The x-ray photons may go through matter without being affected, they may be absorbed whereby they will transfer their energy towards the tissues, and they may be scattered. Whenever they are scattered their direction may change, and there may be a possibility of losing energy. In a diagnostic radiology photoelectric effect and Compton scatter are the two vital interactions of x-ray. Coherent scatter is vital in the diagnostic radiology. Research has shown out that the x-ray photons always move energy to the electrons in large quantities, and this is measured in kiloelectron volts. The electrons that are energetic, in turn lose their energy by interacting with the electrons orbital from the other different atoms thereby giving increased ionization. An electron that is energetic may give out many ion pairs that are additional. Reference. Thayalan, K., 2003. Basic Radiology physics. Oxford: Oxford publishers. APPENDIX APPENDIX A APPENDIX B. APPENDIX C. APPENDIX D. APPENDIX E. Read More
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