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Energy Relationships in a Model of a
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

mωs²/2 = nh

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

ω = 2nh/(ms²)
and, for later use,
ω² = 4n²h²/(m²s4)

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

K = m[4n²h²/(m²s4)]s²/4
which reduces to
K = n²h²/(ms²)

This formula will be used later.

The Balance of Centrifugal Force by the Central Force

For circular orbits of radius r the centrifugal force on each particle is mω²r which is equivalent to mω²s/2. Thus a balance requires

mω²s/2 = H*exp(−s/s0)/s²
which means that
ω² = 2H*exp(−s/s0)/(ms3)

The Quantization Conditions

Equating the two expressions found above for ω² gives

2H*exp(−s/s0)/(ms3) = 4n²h²/(m²s4)
which reduces to
s*exp(−s/s0) = n²h²/(mH)
or, upon dividing both sides by s0
(s/s0)exp(−s/s0) = n²h²/(mHs0)
which, letting σ denote s/s0,
can be expressed as
σ*exp(−σ) = λn²

where λ = h²/(mHs0). 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 Physical Dimensions of the Deuteron

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

σ*exp(−σ) = 0.33695

This is essentially at the maximum of the function σ*exp(−σ) so n can only possibly have the value of 1.

Determination of the Force Constant H

Values for all quantities in the quantization condition are known except for the constant H.

If s1 is the actual separation distance then

H =(2n²h²)/[ms1*exp(−s1/s0)]
or, equivalently
H =(2n²h²*exp(s1/s0)/(ms1)

The Potential Energy Function

The potential energy function V(s) is given for s1 by

V(s1) = −H∫s1[exp(−z/s0)/z²]dz

If the above expression for H is substituted into this formula the result is

V(s1) = −[2n²h²/(ms1)]exp(s1/s0)∫s1[exp(−z/s0)/z²]dz

The function exp(s1/s0) may be taken inside the integral to give

V(s1) = −[2n²h²/(ms1)]∫s1[exp(−(z-s1)/s0)/z²]dz

The variable of integration may be changed from z to y=z-s1 to give

V(s1) = −[2n²h²/(ms1)]∫0[exp(−y/s0)/(y+s1)²]dy

Another change of the variable of integration to p=y/s0 gives

V(s1) = −[2n²h²/(ms1)]∫0[exp(−p)/(s0p+s1)²](s0dp)
which reduces to
V(s1) = −[2n²h²/(ms1s0)]∫0[exp(−p)/(p+s1/s0)²]dp

Since K(s1)=n²h²/(ms1²)

V(s1)/K(s1) = −2(s1/s0)∫0[exp(−p)/(p+s1/s0)²]dp

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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