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Electron and Positron Beta-Decay |
The energy spectra of electron and positron beta-decay look quite different as, for example, the ones shown below for the decay of ^{64}Cu nuclei.
However this observed difference can be accounted for by the fact that the electron is attracted by the nucleus once it is created whereas the positron is repelled. However this selective retardation or acceleration prevails only until the particles reach the outer shell of the electrons of the atom. Beyond that point the atom is electrostatically neutral.
The equation of motion for the decay particles has to be relativistically valid because the measured energies imply speeds up to 75% of the speed of light. The proper dynamic equation is therefore
where p is the momentum of the particle, r is the distance for the center of the nucleus, e is the charge of the proton and Z is the atomic number of the nucleus. The charge of the particle is represented as q; q=+e for the positron and q=−e for the electron. Thus q=±e.
The momentum p is m_{0}v/(1−β²)^{½} where v is velocity, β is velocity relative to the speed of light (v/c) and m_{0} is the rest mass of the particle.
To solve the equation for the dynamics of the particle, both sides of the equation are multiplied by v=(dr/dt) to give:
Letting v/c=β, p/(m_{0}c=α and kZeq=γ this last equation takes the form of
This gives on the RHS an expression in terms of r and dr/dt which can be integrated from t=t_{0}, the time at which the particle is outside of the nucleus, to t=t_{1}, the time at which the particle reaches the outer shell of the electrons. The effective Z changes between the time the particle reaches the inner shell of electrons and when passes beyond the outer shell, but this complication will be neglected.
For integration the LHS requires the expression of β as a function of α. This is achieved as follows.
Both sides of the definition of momentum are divided by c to give
The LHS is α² so the solution for β² is given by
The dynamic equation now takes the form
The results of the integration of this equation from t=t_{0} to t=t_{1} are
where r_{1}=r(t_{1}) and r_{0}=r(t_{0}).
Although t_{1} might be different for electrons and positrons the distance r_{1} would be the same. Therefore the terms on the RHS of the above equation except for q are the same for electrons and positrons. Since β²=α²/(1+α²) it follows that
Since the kinetic energy K is given by m_{0}c²[1/(1−β²)^{½} − 1] this means that the integration of the dynamic equation is equivalent to
Since the LHS is just the change in kinetic energy K the equation can be represented as
where Γ=kZe²(1/r_{0} − 1/r_{1}), a positive number. The positive value of ±Γ applies to the positrons and the negative value to the electrons.
Thus the distribution of kinetic energies of the positrons at the point of measurement is just the distribution at the point of expulsion shifted Γ energy units to the right. The distribution of kinetic energies of the electrons at the point of measurement is the distribution at the point of expulsion shifted to left, but with the provision that any electron with a kinetic energy less than Γ will not make it to the point of observation. The initial distribution of the positron energies would be the observed distribution shifted backwards an amount of energy equal to one half of the difference in energy levels for a corresponding part of the electron and positron energy distribution, say the levels at which the distribution curves cross the energy axis. The initial distribution of electron energies would be the observed distribution shifted the same amount to the right, with the provision that the distribution for energies less than Γ the distribution is unknown. For the electrons and positrons expelled for the ^{64}Cu nuclei the distribution would like something like the following:
The value of Γ is readily calculable for comparison with the observed value. The value of ke² is 2.309×10^{−28} and Z is 64. If it is assumed that the particles are ejected at the surface of the nucleus then r_{0} is 1.6×10^{−14}. The distance of the outer shell of electrons r_{1} is 1.17×10^{−10} is large compared to this value r_{0} that its reciprocal is small compared to that of r_{0}. For these values of the parameters Γ equals 9.235×10^{−13} joules, which is the same as 5.7645 MeV. However this is two orders of magnitude larger than the measured value of 0.045 MeV.
This is indeed a puzzle. If a positron is created in a ^{64}Cu nucleus and then ejected there is no denying that at the surface of the nucleus the positron has a potential energy of 5.77 MeV and most of this would be converted into kinetic energy by the time it passes through the electron shells of the Cu atom. However the maximum energy observed in the positron energy distribution is less than 0.7 MeV. The answer seems to be the quantum mechanical one that there is an energy barrier in the immediate vicinity of the nucleus surface which precludes the creation of positron there. What is created in the nuclear processes is not a positron or electron but instead a probability density function which has a value of zero in the region of high potential energy but is nonzero elsewhere. Sometimes the positron appears in the allowable region beyond the high potential energy barrier and is repelled beyond the atom. In effect there is a tunneling into the region of achievable potential energy. This argument presumes that the potential energy within the nucleus is comparable to half the difference in kinetic energy observed for positron and electron emissions; i.e., about 0.045 MeV. That would be at a distance of about 2×10^{−12} m, which is more than 100 times the radius of the nucleus but only about 1.75% of the distance to the outer shell of electrons of the Cu atom.
The conclusion of this analysis is that the beta decay spectra require the quantum mechanics perspective of the fundamental reality being probability density functions instead of particles per se.
(To be continued.)
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