San José State University
Thayer Watkins
Silicon Valley
& Tornado Alley

The Quantum Constant Associated
with Nuclear Force Phenomena
is not Planck's Constant

Planck's constant is generally presumed to be a universal constant. Here are the various ways that Planck's constant h or its reduced form h arose in physics:

Note that all of the above involve an electromagnetic field. Clearly black-body radiation and photons involve the electromagnetic field. In an atom the electrons are held in orbit by an electric field. The de Broglie relations are not derived but asserted by analogy to electromagnetic radiation. The attempts to derive Schrödinger equations start with Maxwell equations and hence also pertain to electromagnetic fields. The Uncertainty relations are derived from a Schrödinger equation and hence pertain to electromagnetic fields. Thus in all of the above Planck's constant is involved with electromagnetic fields.

Planck's constant is dimensional and its value is determined from the dimensionless coupling constant αEM for the electromagnetic field by the formula. (This coupling constant is usually referred to as the fine strucure constant.)

h = D/αEM
h = (D/2π)/αEM

where D is a constant that depends upon the system of units used. The value of αEM is 1/137.036.

This formula says that in general the quantum constant for a force field is inversely proportional to its coupling constant. Thus, if there is a quantum constant q for the nuclear force field its value would be given by

q = D/αN
q = (D/2π)/αN

where αN is the coupling constant for the nuclear force field. Its value is approximately 1.0. The constant D would be the same for the same system of units as that for Planck'sconstant. Therefore

q/h = q/h = αEMN ≅ 1/137

The Bohr-Mottelson Rule for Nuclear Rotation

In 1969 there came another role for h but in the realm of the field of the nuclear force.

It will be shown that Planck's constant cannot be the appropriate constant empirically for the Bohr-Mottelson formula for nuclear rotation.

When the Bohr-Mottelson formula for nuclear rotation is applied for the deuteron with I=1 the result is the increditably high value of 4.74×1021 revolutions per second. (The details of the computation are given in the Appendix.) This about five billion trillion revolutions per second, or equivalently, five thousand billion billion revolutions per second. Note that this is the minimum rate of rotation according to the Bohr-Mottelson formula. See Nuclear Rotation.

For corroboration of this computed value what is needed is an empirical measurement of the rotation rate of a deuteron. Such is available through the nuclear magnetic moment measurements.

Nuclear Magnetic Moments

When an electrostatic charge is moving a magnetic field is generated. The strength of the generated magnetic field is proportional to the magnitude of the charge and the velocity at which it is moving. For a rotating system of charges the tangential velocity is proportional to the product of the angular rate of rotation and the orbit radius. Thus the measurement of the magnitude of the magnetic field generated by a rotating nucleus can give a value for its rate of rotation.

Consider a deuteron, a proton and neutron rotating about their center of mass. The magnetic moment of a deuteron includes that which due to the rotations of the proton and neutron about their own axes. The following diagram depicts the situation.

The net sum of the magnetic moments of a neutron and a proton is 0.87980 magnetons, intriguingly close to the magnetic moment of a deuteron, which is 0.8574. The difference of 0.02237 magnetons due to the deuteron rotation about its center of mass.

The value of the nuclear magnetron μN in SI units is

μN = eh/(2mp) = 5.051×10−27 Joules per second

where e and mp are the charge and mass of a proton, respectively.

This means that the magnetic moment due to the rotation of the deuteron is 1.1466×10−28 Joules per second.

Since μ is equal to (e/2)ωr²

ω = 2μ/(er²) = 2(0.02237)eh/(2mp)/[(e/2)r²]
which reduces to
ω = 2(0.02237)h/[mpr²]
ω = (0.04474)(1.055×10−34)/[1.6727×10−27(2.252×10−15)²]
and hence
ω = 5.56×1020 radians/sec
ν = 8.85×1019 rotations/sec

or roughly 88.5 million trillion rotations per second, or equivalently, 88.5 billion billion rotations per second.

In the Guide to the Nuclear Science Wall Chart there is this statement

[…] scientists can create nuclei which have very high angular momentum. Nuclei respond to this rotation, which can be as fast as a hundred billion billion revolutions per second[…]

The figure cited is roughly the same as the figure computed.

This is far less than the minimum rotation rate given by the Bohr-Mottelson formula with the constant being h. The rotation rate of a deuteron can be explained by the constant in the formula being q=h/137 and the integer I being 4.

The magnetic moments for other small nuclides are higher than for the deuteron but not so high as the values that result from a strict application of the Bohr-Mottelson formula. Their values can be explained by higher values for the integer I along with q=h/137.

Thus the constant in the Bohr-Mottelson formula cannot be h and should not be h because nuclear rotation is a phenomenon of the nuclear force whose quantum constant qh/137.


What appears to be the case is that Planck's constant applies for phenomena concerning the electromagnetic field but another constant applies in phenomena concerning the force involved in nuclei. This nuclear constant is about 1/137 of the magnitude of Planck's constant.

This ratio is justified in that a quantum constant like Planck's constant is inversely proportional to the coupling constant of its field. The coupling constant for the electromagnetic field, which is usually called the fine structure constant, is approximately 1/137. The coupling constant for the nuclear force is approximately 1.0. Thus the quantum constant for the nuclear force field should be 1/137 times the quantum constant of the electromagnetic field.

Thus the Bohr-Mottelson formula should be

Jω = q(I(I+1))½

where q is the quantum constant for the nuclear force field and qh/137.


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. (One fermi equals 10−15 meters.) 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
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.

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