San Josť State University

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The Quantization of Rotational Kinetic Energy
as a Result of the Quantization of Angular Momentum

The purpose of this webpage is to show how the quantization of angular momentum in two-body rotating system leads generally to the quantization of rotational kinetic energy. In the case of the linear momentum p and kinetic energy E of a body it is simple to express their relationship as

E = p²/m

where m is the mass of the body. In the case of angular momentum and rotational kinetic energy the analysis can be carried one step further because angular momentum is quantized.

Consider two bodies of masses m and M with their centers separated by a distance s. The system rotates about the center of mass and the distances from that center of mass are given by:

mrm = MrM
and hence
rm/rM = M/m

The separation distance s is given by

s = rm + rM = rM[rm/rM + 1]
or, equivalently
s = rM[M/m + 1] = rMM[1/m + 1/M]

The expression [1/m+ 1/M] is the reciprocal of the reduced mass μ of the two bodies. Thus

s = rMM/μ
or, equivalently
μs = MrM

Thus, very neatly

rMM = rmm = sμ

Angular Momentum

If the system is rotating at a rate ω then its angular momentum L is given by

L = mωrm² + MωrM² = mrmωrm + MrMωrM
and, since MrM=mrm
L = mωrm[rm+rM]
and hence
L = mωrms

But mrm is equal to μs so

L = ωμs²

The angular momentum is quantized; i.e.,

L = nh
and thus
ωμs² = nh

where n is a positive integer (known as the principal quantum number) and h is Planck's constant divided by 2π.

This means that

ω = nh/(μs²)

Kinetic Energy

The rotational kinetic energies of the two bodies are

Km = ½mω²rm²
and
KM = ½Mω²rM²

These can be expressed as

Km = ½(mωrm)²/m
and
KM = ½(MωrM)²/M

Since both mωrm and MωrM are equal to ωμs the total kinetic energy K is given by

K = ½(ωμs)²/m + ½ (ωμs)²/M
= ½(ωμs)²(1/m + 1/M)
= ½(ωμs)²/μ
and hence
K = ½ω²μs²

Since the angular momentum L is equal to μωs² and it is quantized as nh

K = ½Lω = ½nhω

But it was previously found that ω is equal to nh/(μs²) so

K = (nh)²/(2μs²)

This formula can be examined for particular cases. Consider first the case of the hydrogen atom. The mass of the proton is about 1836 times greater than the mass of the electron. Therefore the reduced mass of the system is essentially the same as the mass of the electron, which is 9.109383×10-31 kg. However, precisely the reduced mass of the proton-electron system is 9.10442×10-31 kg. The radius of the first Bohr orbit is 5.29177×10-11 meters. Thus

K = 1.112122×10-68/(2*9.10442×10-31*28.0028×10-22)
K = 2.18106×10-18 joules
K = 13.6 electron volts (eV)

This is the Rydberg constant (13.6 eV) and thus the validity of the formula is verified.

Now consider the case of the deuteron.

K = (nh)²/(2μs²)

Twice the reduced mass for the neutron and proton in a deuteron is 1.67374921×10-27 kilograms. The separation distance of the centers of the nucleons in a deuteron is 2.252×10-15 meters. Planck's constant divided by 2π in the MKS system is 1.054571×10-34 joule-sec and squared is 1.112122×10-68. Thus for n=1

K = 1.112122×10-68/(1.67374921×10-27*(5.071504×10-30))
= 1.31016295×10-12 joules
= 8.177390 million electron volts (MeV).

When a deuteron is formed there is an emission of a gamma ray of energy 2.224573 MeV. This means that when a deuteron is formed there is a loss of 10.401963 MeV of potential energy, 8.17739 MeV of which goes into its rotational kinetic energy and 2.224573 MeV of which goes into the emission of a gamma photon.

The energies involved in the formation of a deuteron are very important because the mass of the neutron is deduced from the masses of the deuteron and the proton and an estimate of the mass deficit of the deuteron. The conventional assumption is that the mass deficit of the deuteron is equal to the energy of the gamma photon involved in its formation, 2.224573 MeV. The above indicates that the mass deficit of the deuteron is 8.17739 MeV greater than is presumed and thus the mass of the neutron is correspondingly 8.17739 MeV greater. This is an increase of less than 0.1 of 1 percent, but conceptually quite important because the binding energies of all nuclides would be in error by their number of neutrons times 8.17739 MeV.

(To be continued.)