Taking into account the recoil of the daughter nucleus gives the kinetic energy of the alpha particle:Ĭlassically, the the measured 8.78 MeV for the alpha particle is not sufficient to pass the 26.4 MeV barrier. For the Polonium-212 alpha decay, the differences between the atomic masses are: Converting this mass difference to its E=mc 2 energy equivalent gives the net available kinetic energy given to the products since the effects of both the nuclear strong force attraction and the electromagnetic repulsion are factored into this atomic mass difference. In this way the masses of the electrons cancel out. To find this net energy yield, one can just take the difference between the atomic masses of the the constituents of the decay, using the atomic mass of helium for the alpha particle. If the mass energy of the parent is greater than the sum of the mass energies of the daughter nucleus and the emitted particles, that excess energy is given to the emitted particles plus the recoil energy of the daughter nuclide. To find the energy available to give to the particles that result from the alpha decay, you use the difference in binding energy between the parent and daughter nucleus. The measured energy of the emitted alpha particle is 8.78MeV with a half-life of 0.3 microseconds. Using the example of Polonium-212 alpha decay, the height of the electric potential at the nuclear radius is about 26.4 MeV. To explore this phenomenon further, the steps will be approximate and somewhat qualitative because they involve both classical and quantum ideas. It would be acted upon by the inverse square law Coulomb force, proportional to the slope of the potential energy curve. If an alpha particle is outside the nucleus, it would experience the electric repulsive force from the nuclear positive charge, having a large electric potential energy as shown. However, the strong nuclear force is an extremely short range force, and this negative potential energy essentially disappears at the radius of the nucleus as illustrated below. Within the nuclear interior, the strong force attraction between nucleons produces a deep well of potential energy, even though it must overcome the strong electromagnetic repulsion of the protons to do so. By examining the coupling constants for the fundamental forces, it is clear that the strong force would be expected to dominate for phenomena within the nucleus. The nuclear strong force and the electromagnetic force influence radioactivity in the form of alpha decay. Strong and Electromagnetic Forces in Alpha Decay Calculating the ratio of the wavefunction outside the barrier and inside and squaring that ratio gives the probability of alpha emission. Inside the barrier, the solution to the Schrodinger equation becomes a decaying exponential. To evaluate this probability, the alpha particle inside the nucleus is represented by a free-particle wavefunction subject to the nuclear potential. Quantum mechanical tunneling gives a small probability that the alpha can penetrate the barrier. The Coulomb barrier faced by an alpha particle with this energy is about 26 MeV, so by classical physics it cannot escape at all. The illustration represents an attempt to model the alpha decay characteristics of polonium-212, which emits an 8.78 MeV alpha particle with a half-life of 0.3 microseconds. This extraordinary dependence upon kinetic energy suggests an exponential process, and is modeled by quantum mechanical tunneling through the Coulomb barrier. This half-life range depends strongly on the observed alpha kinetic energy which varies only about a factor of two from about 4 to 9 MeV. The half-lives of heavy elements which emit alpha particles varies over 20 orders of magnitude, from about a tenth of a microsecond to 10 billion years. Alpha Particle Tunneling Alpha Halflife vs Kinetic Energy
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