San José State University

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Thayer Watkins
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Nuclear Rotations

Aage Bohr and Ben Mottleson found that nuclear rotations obey the I(I+1) rule; i.e., have energy levels EI such that

#### EI = [h²/(2J)]I(I+1)

where I is an integer, J is the moment of interia of the rotating nuclear shell and h is Planck's constant divided by 2π. Note what this implies about the quantification of angular momentum. If ω is the rate of rotation then

#### EI = ½Jω² = [h²/(2J)]I(I+1) and hence ω² = [h²/J²]I(I+1) and therefore ω = {h/J][I(I+1)]½

Thus angular momentum L is given by

#### L = Jω = h[I(I+1)]½

This is in contrast to the Old Quantum Physics of Niels Bohr in which L would be hI.

Note that

#### ω = h[I(I+1)]½/J

and thus the smaller the moment of inertia the faster a structure rotates. However the angular momentum is always h[I(I+1)]]½ for any mode of rotation no matter what the moment of inertia is. Thus there is an equipartition of angular momenta among the various modes of rotation.

## Order of Magnitude Estimate of the Rate of Rotation of a Deuteron

A deuteron consists of a proton and a neutron. The neutron has slightly more mass than the proton. For purposes of this order of magnitude estimate the differences between the neutron and the proton will be ignored. It is thus the computation for a pseudo-deuteron.

The diameter of the deuteron is approximately 4.2 fermi. The charge radius of a proton is 0.877 fermi. Deducting two proton radii from the diameter of the deuteron gives a separation distance of the particle centers of 2.446 fermi and thus an orbit radius of 1.223 fermi.

The mass of a proton is 1.673×10-27 kg. Thus the moment of inertia of a deuteron for rotation about an axis perpendicular to its longitudinal axis is

#### J = 2(1.673×10-27)(1.223×10-15)² J = 5.005×10-57 kg m²

The minimum rate of rotation is thus

#### ω = √2(1.05457×10-34)/(5.005×10-57) = 2.98×1022 radians per second

The number of complete rotations per second is 4.74×1021.

There could also be rotation about the longitudinal axis. The moment of inertia of a solid sphere of mass M is (2/5)MR² where R is the radius of the sphere. The moment of inertia of the pseudo-deteron about the longitudinal axis is then

#### J = 2(2/5)(1.673×10-27(8.77×10-16)² = 1.287×10-56

And therefore the rate of rotation is

#### ω = √2h/J = 1.159×1021 radians per second and 1.845×1020 complete rotations per second

Thus the order of magnitude of the rotations of a deuteron is about 1020 per second.