Calculate the sum of the kinetic energies, in MeV, shared among the three outgoing alpha particles. The initial state is a proton and a boron-11 nucleus at rest, and the final state is three alpha particles with rest energy plus some total kinetic energy shared among the three alpha particles. The proton mass m=1.6726219E-27 kg, the boron-11 mass is 1.82814E-26 kg, and the mass of an alpha particle is 6.64648E-27kg. What is the kinetic energy of the 3 alpha particles in MeV?I found that the proton’s initial kinetic energy is 2.5 MeV. A disadvantage of fission power is the unavoidable generation of large amounts of highly radioactive heavy element waste that mustbe stored safely. Fusion power does not produce radioactive heavy elements, but most fusion reactions do generate energeticneutrons that can make the walls of the reactor somewhat radioactive. There is interest in a cleaner process, the fusion of a protonwith boron-11, which mostly produces three alpha particles (the nucleus of helium-4, consisting of two protons and two neutrons),with a low probability of producing neutrons. In principle, the reaction could create a carbon-12 nucleus, but the system is in such anexcited state that it falls apart into three alpha particles.Unfortunately, the electric repulsion between the proton and the boron-11 nucleus, which contains 5 protons and 6 neutrons, meansthat the input proton's kinetic energy has to be large in order to contact the boron-11 nucleus, though this kinetic energy is less thanthe output kinetic energy of the alpha particles, so there is a net energy gain. Another advantage of the proton-boron-11 reaction isthat the electrically-charged alpha particles can produce electric current directly, which is much more efficient than using energeticparticles to boil water to run a generator, as is the case in fission reactors and most planned fusion reactors.Let's consider a simplified but informative analysis of this reaction. Suppose all the electrons have been stripped from a boron-11atom, so that the proton will encounter a "naked" boron-11 nucleus. The proton starts from a location very far from the boron-11 withenough kinetic energy to come to rest at the surface of the boron-11; contact is necessary for the short-range nuclear force to act tofuse the two particles. The center-to-center distance between the two touching particles can be approximated in the following way:The radius of a proton is r = 0.88 x 10-15 m m, and its volume is . The volume of the boron-11 nucleus is approximately4л3ΑπR3 = 1134元3so the radius of the boron-11 nucleus is approximately R = 11 r. which is 1.96 x 10-15 m. For the proton totouch the boron nucleus the center-to-center distance dR+r=2.84 x 10-15 m.

Question

Calculate the sum of the kinetic energies, in MeV, shared among the three outgoing alpha particles. The initial state is a proton and a boron-11 nucleus at rest, and the final state is three alpha particles with rest energy plus some total kinetic energy shared among the three alpha particles. The proton mass m=1.6726219E-27 kg, the boron-11 mass is 1.82814E-26 kg, and the mass of an alpha particle is 6.64648E-27kg. What is the kinetic energy of the 3 alpha particles in MeV?I found that the proton’s initial kinetic energy is 2.5 MeV. A disadvantage of fission power is the unavoidable generation of large amounts of highly radioactive heavy element waste that mustbe stored safely. Fusion power does not produce radioactive heavy elements, but most fusion reactions do generate energeticneutrons that can make the walls of the reactor somewhat radioactive. There is interest in a cleaner process, the fusion of a protonwith boron-11, which mostly produces three alpha particles (the nucleus of helium-4, consisting of two protons and two neutrons),with a low probability of producing neutrons. In principle, the reaction could create a carbon-12 nucleus, but the system is in such anexcited state that it falls apart into three alpha particles.Unfortunately, the electric repulsion between the proton and the boron-11 nucleus, which contains 5 protons and 6 neutrons, meansthat the input proton's kinetic energy has to be large in order to contact the boron-11 nucleus, though this kinetic energy is less thanthe output kinetic energy of the alpha particles, so there is a net energy gain. Another advantage of the proton-boron-11 reaction isthat the electrically-charged alpha particles can produce electric current directly, which is much more efficient than using energeticparticles to boil water to run a generator, as is the case in fission reactors and most planned fusion reactors.Let's consider a simplified but informative analysis of this reaction. Suppose all the electrons have been stripped from a boron-11atom, so that the proton will encounter a &#34;naked&#34; boron-11 nucleus. The proton starts from a location very far from the boron-11 withenough kinetic energy to come to rest at the surface of the boron-11; contact is necessary for the short-range nuclear force to act tofuse the two particles. The center-to-center distance between the two touching particles can be approximated in the following way:The radius of a proton is r = 0.88 x 10-15 m m, and its volume is . The volume of the boron-11 nucleus is approximately4л3ΑπR3 = 1134元3so the radius of the boron-11 nucleus is approximately R = 11 r. which is 1.96 x 10-15 m. For the proton totouch the boron nucleus the center-to-center distance dR+r=2.84 x 10-15 m.

Accepted Answer

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