Question:

H H K4 The Qvalue is 4 0 MeV and so this nuclear reaction

Last updated: 6/30/2023

H H K4 The Qvalue is 4 0 MeV and so this nuclear reaction

H H K4 The Qvalue is 4 0 MeV and so this nuclear reaction liberates about 1 MeV per nucleon roughly the same as the fission reaction This reaction can be performed in the laboratory by accelerating a beam of deuterons on to a deuterium target In order to observe the reaction we must get the incident and target deuterons clased enough that the nuclear force can produce the reaction that is we must overcome the mutual Coulomb repulsion of the two particles We can estimate this Coulomb repulsion by calculating the electrostatic repulsion of two deuterons when they are just touching The radius of a deuteron is about 1 5 fm and the electrostatic potential energy of the two charges separated by about 3 fm is about 0 5 MeV We can produce such a beam of deuterons in many of the accelerators available in nuclear physics laboratories The beam currents of such accelerators are typically of the order of microamperes If every particle in the beam produced a reaction hardly a reasonable assumption the total power produced would be 4W An output of 4W assuming we could extract every bit of the energy liberated in the reaction which appears as the kinetic energies of the products H and H hardly makes this device a useful power source According to the passage the fusion reaction initiated by accelerators in laboratories is not useful source of energy because A Q value per reaction is very low C Radioactive daughter nuclei are produced B The energy losses are high D The power output is low For initiating the reaction the energy of incident deuterons should be of the order of B 1 0 MeV A 0 5 MeV C 4 0 MeV D 4 5 MeV What is the mass of tritium atom if mass of hydrogen atom 1 0078 amu Mass of Deuterium atom 2 0141 amu 931 5 MeV c A 3 0161 amu B 3 0204 amu C 3 0247 amu D 3 0118 amu