![]() ![]() The answer was supplied in 1928 by the Russian physicist George Gamow (1904–1968). The α α particle inside the nucleus does not have enough kinetic energy to get over the rim, yet it does manage to get out by quantum mechanical tunneling. An α α particle outside the range of the nuclear force feels the repulsive Coulomb force. So how does the α α particle get out?įigure 31.27 Nucleons within an atomic nucleus are bound or trapped by the attractive nuclear force, as shown in this simplified potential energy curve. Yet the protons and neutrons do not have enough kinetic energy to get over the rim. ![]() In α α decay, two protons and two neutrons spontaneously break away as a 4 He 4 He unit. The slope of the hill outside the bowl is analogous to the repulsive Coulomb potential for a nucleus, such as for an α α particle outside a positive nucleus. That is, they are bound by an average of 8 MeV per nucleon. Protons and neutrons have kinetic energy, but it is about 8 MeV less than that needed to get out (see Figure 31.27). In a nucleus, the attractive nuclear potential is analogous to the bowl at the top of a volcano (where the “volcano” refers only to the shape). If it could find a tunnel through the barrier, it would escape, roll downhill, and gain kinetic energy. Figure 31.26 The marble in this semicircular bowl at the top of a volcano has enough kinetic energy to get to the altitude of the dashed line, but not enough to get over the rim, so that it is trapped forever.
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