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The Magnetic Dipole Moment and
Rotational Kinetic Energy of a Deuteron

The Magnetic Dipole Moment of a Deuteron

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

M = (Q/2)ωr²
ω = 2M/(Qr²)

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 rp and rn be the radii of the proton and neutron, respectively and mp and mn their masses. Then

mprp = mnrn
and
rp + rn = s

Therefore

rn/rp = mp/mn
and
rp(1+mp/mn) = s
and hence
rp = [mn/(mp + mn)]s

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 rp at approximately 1.2×10-15 meters. This gives the rate of rotation ω as

ω = 2(1.13×10-28)/[(1.6×10-19)(1.2×10-15)²)] = 9.8×1020 radians per second

The tangential velocity of the nucleons would then be

v = ωrp = 1.177×105 meters per second

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

K = ½(mp + mn)v²
K = (0.5)(3.348×10-27)(1.177×105
K= 2.3135×10-17 joules

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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