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Atomic Energy Exercises - Assignment Example

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Atomic Energy Exercise Name Institution Response to question 7A Case 1 The quantity of energy needed to push particles close together for curve 1 is more compared to that for curve 2. This is so because of the combination of electrostatic repulsion force which extends over large distances and is combined with nuclear attraction force that extends over small distances…
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Atomic Energy Exercises
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Download file to see previous pages An approximation of the distance of separation for curve 2 appears to be bigger than that of curve 1 thus accounting for the smaller energy than that of curve 1. The maximum amount of force that is needed to push the particles together is greater in curve 1 than in curve 2. This is because the vector force that pulls particles towards each other is inversely proportional to the square of the separation distance and directly proportional to the products of the two charges. Case 2 There is immense proton repulsion between the protons within the nucleus because of the Coulomb’s law electrostatic interactions. In this regard, the force that is exerted by the center protons towards the outer protons is inversely proportion to the proton’s square of the separation distance. The separation distance squared comes about because the force field of an isolated proton charge is uniform towards all directions and becomes diluted with the separation distance just like the area of the centered sphere of the point charge which increases as its radius increases. Case 3 For the incoming protons, their potential energy is smaller compared to when they reach the bottom pit in the center. From Coulomb’s law (f=kq1q2/r2), potential energy is the product of force and distance which is given by PE= kq1q2/r. This means that as the protons move from outside they are far apart from each other hence remarkably little potential energy. When protons are in the center, the particles are close to each other leading to a small separation distance thus a greater potential energy compared to that of incoming protons. Case 4 Binding energy is the mechanically required energy to disassemble an atom into nucleus and free electrons (Jones International University, 2011). Helium has a lower binding energy compared to beryllium and iron. This is because helium has an atomic mass of 2, beryllium has an atomic mass of 4, whereas iron has an atomic mass of 26. In this case, iron has the greatest binding energy because of its atomic mass that is larger than that of beryllium and helium. This is so because, from the periodic table (of elements), the elements with greater atomic mass have increasing binding energy than those with a small atomic mass up to elements that are heavier than xenon which do not obey this trend. This is due to the increasing nucleon force in the nucleus as an extra nucleon gets attracted to other nucleons making the nucleus to be tightly bound. On the other hand, the amount of energy needed to pull the nuclei together for a tight bound is less for in an iron atom than the beryllium and helium atom. This is due to the fact that iron atom has a larger atomic mass than the helium and beryllium atoms. The larger the atomic mass, the more the energy levels of an atom and the less the energy needed to push the nucleus together for a tight bound. Case 5 Uranium 238 has 92 protons and 146 neutrons. Its binding energy can be given by: a - b /A1/3 - cZ2/A4/3- d (N-Z) 2 /A2 ± e/A7/4. Where a = 14.0, b = 13.0, c = 0.585, d= 19.3, and e = 33. A is the number of nucleons, Z is the number of protons in the nucleus, and N is the number of neutrons in the nucleus. B.E/A = 14-13/238^1/3- 0.585x92^2/238^4/3 – 19.3 (146-92) ^2/238^2± 33/238^7/4 = 0.1614-3.357-0.9936 ±0.002288 =-4.1892 ...Download file to see next pagesRead More
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