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The Magnetic Dipole Moment and the Rotational Kinetic Energy of a Deuteron |
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The binding energy of a nuclide is its mass deficit expressed in energy units via the Einstein equation E=mc². Its mass deficit is the difference between the total mass of its constituent nucleons and its measured mass,
The mass of a particle is measured by injecting it into a cloud chamber pervaded by a magnetic field. The particle moves in a helix which can be observed because the particles create a track in the cloud chamber. The initial velocity of the particle can be determined from the length of the cloud track it forms. The radius of the helix can be measured. From the radius of the path, the velocity of the particle and the strength of the magnetic field the mass of the particle can be calculated from an equation based upon a balance of the force imposed upon the particle by the magnetic field and centrifugal force. The centrifugal force is mv²/r where m is the mass of the particle, v is its velocity perpendicular to the direction of the magnetic field and r the radius of path. The force exerted upon the particle by the magnetic field is e(Bv), where e is the charge of the particle, B the magnetic field intensity and v the velocity of the particle.
One measured quantity that might depend directly upon the rate of rotation of a nucleus is its magnetic dipole moment. That of a deuteron is 0.8574 magnetons which in joules per second is 4.3307346×10^{-27}. The net sum of the magnetic dipole moments of the proton and neutron is 0.86242 magnetons or 4.4437070×10^{-27} J/T.
This might mean that the moment due to the rotation of the deutron is 0.112972×10^{-27} J/T or rounded off 1.13×10^{-28} joules per second.
Since the magnetic dipole moment M is given by the formula
where Q is charge. The magnetic dipole moment due to the rotation of the deuteron would depend only on the motion of the proton since the neutron is electrically neutral. Thus M is sensitive to the radius of the protons orbit.
The radius of the proton's orbit can be obtained from the separation distnce s of the ceners of the two nucleons. Let r_{p} and r_{n} be the radii of the proton and neutron, respectively and m_{p} and m_{n} their masses. Then
Therefore
The separation distance of the centers is obtained by subtracting the radii of the nucleons from the diameter of the deuteron. The measured radius of the deuteron is 2.1424×10^{-15} m and hence its diameter is 4.2824×10^{-15} m. The particle radius of the proton is 0.877551×10^{-15} m. The radius of the neutron is less certain but the figure of 1.0×10^{-15} m is a reasonable value. This gives the separation distance of the centers as 2.4072×10^{-15} m. This puts the value of r_{p} at approximately 1.2×10^{-15} meters. This gives the rate of rotation ω as
The tangential velocity of the nucleons would then be
This is slightly less than 4 percent of the speed of light so non-relativistic analysis is appropriate.
The rotational kinetic energy of the deuteron is then
This is an insignificant amount compared to the energy of the gamma ray created with the formation of the deuteron.
The reason for there being no amount of the magnetic dipole moment attributable to the rotation of the deuteron is that the direction of the nuclear rotation is random with respect to the direction of the spin rotation. Thus there would be equal numbers of the following.
The equal numbers cancel out the effect of the rotation. The slight difference between the magnetic dipole moment of the deuteron and the net value for the proton and neutron is probably due to some slight interaction of the nucleon spins and has nothing to due with the rotation of the deuteron.
(To be continued.)
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