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The Distribution of Energies
of Leptons Expelled from Nuclei

The distributions of energies of electrons and positrons expelled from the radio-active isotope of copper, 64Cu, are known.

These measured distribution of two particles differ but that difference can readily be accounted for from the positrons being repelled by the nuclei after expulsion whereas the electrons are subject to the electrostatic attraction. The distributions of energies of the two particles could be the same at expulsion but the distribution of energies for positrons shifted to the right where the energies are observed and the distribution of energies for electrons shifted to the left. For more on this matter see beta decay.

The following material is an attempt to account for that distribution of expulsion energies based upon a few simple semiclassical assumptions. The most important assumption is that leptons and hadrons are subject to a repulsion. It is often asserted that leptons are not subject to the strong force, now generally called the nuclear force, that holds nuclei together. However usually what is meant is that the leptons are not subject to the attraction of the nuclear force. This leaves open the possibility that leptons are repelled by the nucleons. Such a repulsion would explain the expulsion of electrons and positrons from nuclei when they are created there.

The assumptions of the model are then:

It is generally believed that the nuclear force involves, in addition to other terms, an exponential factor which is a function of distance. The fourth assumption is based upon the notion that at the distances involved within a nucleus the exponential function changes so little that it may be considered constant.

Based upon the second assumption the distribution of the radii r0 within the nucleus where leptons are created is of the form

n(r) = αr0²
for 0≤r0≤R

where R is the effective radius of the nucleus.

The force on a lepton at a distance r from the center of the nuclei is the force due to the nuclear matter contained within a sphere of radius r. The nuclear material in the shell beyond r has no effect. If the nuclear force is of the form

F = HN/d²

where H is a constant, N is the amount of nucleonic matter and d is the distance.

For a lepton at a distance r from the center of the nucleus N is (4/3)πr³ρ, where is ρ is the density of nucleons. Thus the force on a lepton at r is

F = H(4/3)πr³ρ/r² = Gr

where G=H(4/3)πρ.

The non-relativistic equation of motion for the lepton is then

ml(d²r/dt²) = Gr

where ml is the mass of the lepton.

Multiplying both sides of the above equation by (dr/dt) and integrating from t=0 where r=r0 to the time at which the lepton leaves the nucleus where r=R, the effective radius of the nucleus, gives

½mlv(R)² − ½mlv(r0)² = ½G[R² − r0²]

The LHS of the above is the gain in kinetic energy by the lepton (ignoring relativistic effects). Thus the kinetic energy K(R) of the lepton as it leaves the nucleus is given by

K(R) = K(r0) + ½G[R² − r0²]

If the kinetic energy of the lepton at the instant of creation is zero then a lepton with kinetic energy K is one that was created at a distance from the center of the nucleus r0 given by

r0 = (R² − 2K/G)½

The number of these leptons is given by αr0² so the number leptons of kinetic energy K is

n(K) = α(R² − 2K/G)
for 0≤K≤GR²/2

This is a triangular distribution which is roughly comparable to the observed distribution

For the triangular distribution Kmax is equal to GR²/2 so the value of G would be 2Kmax/R². This value is 2(0.615×106)(1.6×10−19)/(1.6×10−14)² which is 7.688×1014. The volume of the 64Cu nucleus is (4/3)π(1.6×10−14)³ = 1.7157×10−41. Therefore the density of nucleons is 64/1.7157×10−41=3.73×1042. Since G=(4/3)πHρ this means that H is equal to (3/4)G/(πρ) which is 4.92×10−29 kg-m4/s2.

The distribution of energy would be modified if any of the following conditions prevailed:

Although the shape of the energy spectra for beta-decay may be accounted for by the model the model is incompatible with the magnitudes of the energies involved. If a positron were created in the  64Cu nucleus and physically moved to the surface it would have at the surface a potential energy of almost 6 million electron volts (Mev). This potential energy would be converted in kinetic energy as positron is repelled from the positive nucleus. However the maximum kinetic energy observed for the positrons of  64Cu decay is less than ).7 of 1 Mev. The resolution of this discrepancy requires a quantum mechanical view in which what is created in the decay process is not particles per se but instead probability density functions for the particles. The particles then appear in a region of an allowable potential energy without passing through the region of excessively high potential energy.

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