|San José State University|
& Tornado Alley
Deuteron with a Nuclear Strong Force
that is an Exponentially Weighted
Inverse Distance Squared Law
Consider two particles of equal mass m with their centers separated by a distance s. Let ω be the rate of rotation of the particles about their center of mass. The particles will travel in circular orbits of radius r equal to s/2. The angular momentum of the system is then 2m(ωr)r, which is equal to mωs²/2. Angular momentum is quantized so
where n is an integer, called the principal quantum number, and
h is Planck's constant
divided by 2π.
Thus the angular rate of rotation is given by
The kinetic energy K of the system is 2(½m(ωr)²) which reduces to mω²s²/4. If the expression above for ω² is substituted into this formular the result is
This formula will be used later.
For circular orbits of radius r the centrifugal force on each particle is mω²r which is equivalent to mω²s/2. Thus a balance requires
Equating the two expressions found above for ω² gives
where λ =
The function σ*
exp(−σ) has the form
This is the quantization condition for σ which leads to the quantization of ω and the other variables.
The separation distance s1 in the physical deuteron is about 2.252 fermi (2.252×10-15 m), Thus s1/s0 is equal to 1.4796, a pure number. When this value is substituted into the LHS of the above equation the result is
This is essentially at the maximum of the function σ*exp(−σ) so n can only possibly have the value of 1.
Values for all quantities in the quantization condition are known except for the constant H.
If s1 is the actual separation distance then
The potential energy function V(s) is given for s1 by
If the above expression for H is substituted into this formula the result is
The function exp(s1/s0) may be taken inside the integral to give
The variable of integration may be changed from z to y=z-s1 to give
Another change of the variable of integration to p=y/s0 gives
The integral in the above equation can be approximated by numerical integration. For s1/s0 equal to 1.4796 its value is 0.24683. This makes the ratio V(s1)/K(s1) equal to −0.73042.
Since the magnitude of V(s1) is less than the magnitude of K(s1) it is physically impossible for the two particles to come together with a principlal quantum number of 1 without an input of energy. The base state has to be one of zero angular momentum. From this base state of principal quantum number of zero a state with a high principal quantum number would require an input of energy.
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