A disadvantage of fission power is the unavoidable generation of large amounts of highly radioactive heavy element waste that must be stored safely. Fusion power does not produce radioactive heavy elements, but most fusion reactions do generate energetic neutrons that can make the walls of the reactor somewhat radioactive. There is interest in a cleaner process, the fusion of a proton with 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 an excited 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, means that the input proton's kinetic energy has to be large in order to contact the boron-11 nucleus, though this kinetic energy is less than the output kinetic energy of the alpha particles, so there is a net energy gain. Another advantage of the proton-boron-11 reaction is that the electrically-charged alpha particles can produce electric current directly, which is much more efficient than using energetic particles 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-11 atom, so that the proton will encounter a "naked" boron-11 nucleus. The proton starts from a location very far from the boron-11 with enough kinetic energy to come to rest at the surface of the boron-11; contact is necessary for the short-range nuclear force to act to fuse 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 approximately 4л 3 Απ R3 = 11 3 4元 3 so the radius of the boron-11 nucleus is approximately R = 11 r. which is 1.96 x 10-15 m. For the proton to touch the boron nucleus the center-to-center distance d R+r=2.84 x 10-15 m.

A disadvantage of fission power is the unavoidable generation of large amounts of highly radioactive heavy element waste that must be stored safely. Fusion power does not produce radioactive heavy elements, but most fusion reactions do generate energetic neutrons that can make the walls of the reactor somewhat radioactive. There is interest in a cleaner process, the fusion of a proton with 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 an excited 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, means that the input proton's kinetic energy has to be large in order to contact the boron-11 nucleus, though this kinetic energy is less than the output kinetic energy of the alpha particles, so there is a net energy gain. Another advantage of the proton-boron-11 reaction is that the electrically-charged alpha particles can produce electric current directly, which is much more efficient than using energetic particles 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-11 atom, so that the proton will encounter a "naked" boron-11 nucleus. The proton starts from a location very far from the boron-11 with enough kinetic energy to come to rest at the surface of the boron-11; contact is necessary for the short-range nuclear force to act to fuse 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 approximately 4л 3 Απ R3 = 11 3 4元 3 so the radius of the boron-11 nucleus is approximately R = 11 r. which is 1.96 x 10-15 m. For the proton to touch the boron nucleus the center-to-center distance d R+r=2.84 x 10-15 m.