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of Two-Particle Systems |
A system has kinetic energy and potential energy. Let these quantities be denoted by K and V, respectively. When the properties of a system are quantized there is a principal quantum number n. The allowable levels of K and V are a function of n, denoted as K_{n} and V_{n}. When the principal quantum number n changes from a higher level to a lower level the kinetic and potential energy change. The potential energy decreases and the kinetic energy increases. These changes do not necessarily preserve energy and the difference in energy is emitted as a photon. Let γ be the energy of the emitted photon. Then
If there exists two coefficients, α and β, such that
where n is a positive integer, called the principal quantum number, then
This is the scheme that was found to explain the spectrum of the hydrogen atom to a high degree of accuracy. See the Bohr Model.
Two specific subcases will be analyzed first to acquaint the reader with the analysis before the general case is examined. Those two cases are: 1. Positronium 2. The Hydrogen Atom.
Positronium is an electron and a positron rotating about their center of mass. Let m be the common mass of an electron and positron. Let s be their separation distance and r their orbit radius, with r=s/2. Let ω be the rate of rotation of the system.
The electrostatic attraction experienced by the particles is -k/s², where k is the constant of the electrostatic force. A balance of centrifugal force and the electrostatic force for maintaining a circular orbit requires:
(By limiting the analysis to circular orbits the special case of zero angular momentum is missed. This is dealt with elsewhere.)
The angular momentum of each particle is m(ωr)r=mωr²=mωs²/4. The total angular momentum is then 2mωs²/4=mωs²/2.
Angular momentum is quantized; i.e.,
where n is a positive integer and h is Planck's constant divided by 2π.
For a background on the quantization of angular momentum see Bohr's Condition.
This quantization equation squared is
Setting equal the two expressions derived for ω² yields
The expression (h²/(mk)) arises often enough to justify giving it a special
label. However, it is appropriate to first introduce the concept of the reduced mass for
two particles. The reduced mass μ of masses m_{1} and m_{2}
is given by
Thus for a system of two equal masses of m the reduced mass is m/2. Therefore the previous equation
Let
It is a natural unit of length scale. It is twice the Bohr radius for electrons and thus the diameter of the innermost electron orbit in a hydrogen atom. Its value is 1.05835442×10^{-10} meters or 105,835.442 fermi.
Thus the quantization condition for the separation distance for positronium is
This means the spatial dimensions of positronium are the same as those of the hydrogen atom.
Now consider the kinetic and potential energies.
The kinetic energy for the two particles is given by
The potential energy is given by
It is immediately clear that K=−V/2. Thus the total energy of the system is K+V=V/2. The spectrum is given by
This means that
The frequency ν of the photon is given by
Before going on to the next case there are some observations to be made about the scale length σ.
The constant for the electrostatic force k can be expressed as αhc, where α is
the fine structure constant 1/137.06. Thus σ can also be expressed as
This means that
The energy of the γ ray emitted upon the formation of positronium corresponds to a change in 1/n² from 0 to 1. Thus
Since the rest mass energy of an electron is 0.511 MeV the energy of the γ ray emitted on the formation of positronium should be, according to the model, 6.8 eV. The increase in kinetic energy of the system then should also be 6.8 eV and the decrease in potential energy 13.6 eV. This is the same as the ionization energy of electrons for a hydrogen atom.
The analyis is basically the same as for positronium but now r=s. The mass of proton is about 1836 times that of the electron. The reduced mass of the system is virtually the same as the mass of the electron. In the standard treatment the center of mass is taken to be at the center of the proton and the angular momentum of the system is entirely in the electron and is equal to m(ωr)r, which is the same as mωs². Thus
The dynamic balance condition for a circular orbit, is now
Therefore
The potential energy is then
The kinetic energy of the electron is ½mω²s² and thus
Again K=−½V. Thus, the spectrum is given by
The value of σ for this case is one half of its value for positronium because the the reduced mass of the system is twice that for positronium. The value of γ is then 13.6 eV, the ionization energy. The increase in kinetic energy is then also 13.6 eV and the decrease in potential energy is 27.2 eV.
The hydrogen atom model presumes that the proton is infinitely massive compared to the electron. Although the model gives excellent empirical results even though the mass ratio is only 1836 instead of infinity there are other exotic combinations such as a muon and proton where the finite mass ratio must be taken into account.
Let m_{1} and m_{2} the masses of two particles subject to elecrostatic attraction. The radii of their orbits about their center of mass are given by
where s is their separation distance (r_{1}+r_{2}).
The electrostatic force on each particle is −k/s². Thus dynamic balance for circular orbits requires
The expression m_{1}m_{2}/(m_{1}+m_{2}) is just the reduced mass of the system, which is denoted as μ. Thus the previous equation is
The quantization of the angular momentum of the system requires
The expression (m_{1}²r_{1}²+m_{2}r_{2}²) can be greatly simplified.
Thus the quantization of angular momentum condition can be expressed as
Hence
The potential energy of the system is then
The kinetic energy is then
Once again K=−½V. Thus the spectrum of the system, in energy terms, is
The relevant factor for this case is that mass is given by
where m_{0} is the rest mass of the electron.
The two equations which were previously derived for the hydrogen atom are:
The second equation can be put into a form which matches the LHS of the first equation; i.e.,
This means that
The quantity ωs is the tangential velocity of an electron. The above equation is thus a quantization condition for the tangential velocity of the electron. This velocity relative to the speed of light c is then
The equation
The result is
This is the quantization condition for the separation distance s.
The potential energy V is again given by
The kinetic energy in relativistic terms is
It is completely determined by the value of β, whose quantized values have already been
determined to be k/(nhc).
The first order approximation of K is K* = ½m(ωs)². This reduces to
This reveals that K is approximately equal to −½V. (The potential energy V is negative so −½V is positive.) Thus the energy spectrum of the system is given approximately by
To obtain a higher degree of approximation V and K can be expressed as functions of x=(1/n²) and ΔV and ΔK approximated as (∂V/∂x)Δx and (∂K/∂x)Δx. First let 1/(1−β²)^{½} be denoted as B. Then
Thus
Since B = (1−(k/hc)²x)^{−½}
Therefore
Thus the spectrum is given approximately by
A general attractive central force between two particles can be represented in the form
where s is the separation distance between the centers of the two particles, H is a positive constant and f(s) is a function such that f(0)=1. The major relevant cases for f(s) are: 1. The electrostatic case in which f(s)=1 for all s; 2. The nuclear force case in which f(s)=exp(−λs) where λ is a positive constant reflecting the rate of decay of the force carrying particles, the π mesons. The parameter λ can be expressed as 1/r_{0} so that f(s)=exp(−s/r_{0}).
Consider the case in which the two particles have an equal mass of m. These is roughly the case of the deuteron.
The quantization of angular momentum equation is then
Dynamic balance for circular orbits requires
Equating the two expressions for ω² yields
Now is the time to note that the reduced mass μ of the system is m/2. Thus the previous equation is
This gives the allowable values of the separation distance s. For f(s)=1 this is just s=σn². For f(s)=exp(−λs) the function sf(s) is one that rises from 0 at s=0 to a maximum and thereafter falls asymptotically toward zero. There would be a maximum value n that would have a solution.
From the quantized values of s the quantized values of ω are determined.
The kinetic energy is given by
The quantity ω²s² can be represented as
This means the kinetic energy can be respresented as
The quantity ω²s² can also be represented as
Thus kinetic energy can be expressed as
If s is approximately proportional to n² then K is approximately inversely proportional to n². For a further investigation of the dependences of s and K on n see n dependence.
Another way of representing the kinetic energy is to note that if sf(s)=σn² then s²=σ²n^{4}/f²(s). Thus
Since σ=h²/(μH) and H can be expressed as ζhc,
where ζ is the structure constant for the nuclear force,
Thus
An estimate of ζ as 1.58 is given in Force Constants.
The increase in kinetic energy from the case of n=+∞ (1/n²=0) to n=1 (1/n²=1) is then
The potential energy V is given by
Note that it was found above that one representation of kinetic energy was as ½H(f(s)/s). Thus, when s goes from ∞ where V=0 to s_{0}
This means that if γ and s_{0} are known, along with f(s), then an estimate of H can be obtained as
Some invesigation of empirical values is appropriate at this point. Estimates of the parameters of = -H*exp(-s/r_{0})/s² are available. The values are H=1.92570×10^{−25} kg·m³/s² and r_{0}=1.522×10^{-15}=1.522 fermi. The reduced mass of a proton and neutron is 8.368746×10^{−28} kg.
Since h²=1.112121×10^{−68}
m^{4} kg^{2} / s^{2} this means
The charge radius of the deuteron as determined by scattering has been found to be 2 fermi ±0.1 fermi. Thus the separation distance s is about 4 fermi.
The change in kinetic energy when the separation distance s goes from s=∞ to s=4 fermi=4×10^{-15} m is then
Thus, according to the model and the parameter estimates, the mass deficit of a deuteron would be 2.59198+2.22457=4.81655 MeV rather than 2.22457 MeV. For a separation distance of 4.2 fermi the value of ΔK would be 2.35100 fermi. This makes the energy of the gamma ray and the increase in kinetic energy roughly equal, which is what occurs for the electrostatic force.
If the separation distance were 5 fermi instead of 4 the value of ΔK would be 1.6888 MeV and thus the mass deficit of a deuteron would be 3.88344 MeV rather than 2.22457 MeV, as is conventionally assumed.
The mass of the neutron is not directly measurable; instead it is estimated from the measured masses of the deuteron and the proton and the energy of the gamma ray involved in the formation or disassociation of a deuteron. The traditional analysis derives the mass of a neutron by presuming that the energy of the gamma ray is exactly equal to the mass deficit of the deuteron. If the conversion of mass goes partially into the energy of the gamma ray and the rest goes into an increase in the kinetic energies of the proton and neutron then the accepted estimate of the mass of the neutron in error by being less than the actual mass. A mass deficit for a deuteron of 3.88344 MeV, corresponding to a separation distance in the deuteron of 5 fermi, would mean the mass of the neutron is underestimated 1.6888 MeV; i.e., it is 941.2548 MeV rather than 939.566 MeV.
Using the conventional estimate of the mass of the neutron, the beryllium isotope having four protons and one neutron has a negative mass deficit (mass surplus) of 0.75 MeV. The mass deficit is also called the binding energy. This means that the Be^{5} supposedly has a negative binding energy. This value may just be the result of an underestimate of the mass of the neutron. This lends plausibility to the above estimates.
In the preceding investigation an empirical estimate of the separation distance of the proton and neutron was used. The model itself implies a value for the separation distance. It is the solution to the equation
The estimate of r_{0} as 1.522 fermi is available in Nuclear Force Formula.
Since the estimate of σ is 0.069 the transcendental equation to be solved is
The solution is z=0.0475487 and thus s_{0}=1.522*0.0475487=0.07237 fermi. This is too small, indicating that the value of H used to compute σ was too large.
If s_{0}=4.2 fermi is taken as the correct solution then
Thus the value of H is given by
which is about one fourth the magnitude of the estimate of H used previously.
The value of the increase in kinetic energy is given by
This means the mass deficit of the deuteron is 2.22457+2.351=4.57557 MeV rather than 2.22457 MeV. The decrease in potential energy is divided 51.4% into an increase in kinetic energy and 48.6% into the energy of the emitted gamma ray. It also means the mass of the neutron is 941.917 MeV instead of 939.566 MeV. This also means that the mass deficit (binding energy) of the Be^{5} nuclide is 1.6 MeV instead of −0.75 MeV.
The kinetic energy of the system in terms of the tangential velocity v is
This means that relativistic adjustments will amount to only about 1/8 of 1 percent.
The relativistic form of the model is not needed for numerical significance but would be worthwhile for further theoretical analysis.
Here relativistic means only that mass is given by
The conditions for the quantization of angular momentum and the dynamic balance can be reduced to
where μ is the reduced mass of the particles.
From these it can deduced that
The tangential velocity of the particles is then ½ωs. Thus
The equation
Noting that μ=μ_{0}/(1-β²)^{½}
and H=ζhc the previous formula may be expressed
as
This may be further reduced to
If s is known for a specific value of n, say n=1, then the above formula can be used to estimate ζ.
For the nuclear force formula f(s)=exp(−s/r_{0}) where r_{0}=1.522 fermi.
Using the conventional masses of the proton and neutron
μhc = 3.1637145×10^{-26} joules-meters (m³ kg/s²).
Using s=4.2 fermi for n=1 then gives
From these values the increase in kinetic energy created from the formation of the deuteron is
The mass of the neutron is then 941.910 MeV rather than the conventional 939.566 MeV.
Since the conventional mass of the neutron was used in computing the reduced mass of the system the computation must be repeated until a solution is found for H, ζ, ΔK and m_{n}.
After a small number of iterations the solutions for s=4.2 fermi converge to:
The mass deficit of the deuteron is then 4.5749 MeV, which means that the mass of the neutron is
This would mean that the mass of the neutron is greater than than of the proton by 7.136 electron masses instead of the conventional 2.532. It means that the mass deficit or binding energy of the He4 (alpha particle) would be 33 MeV instead of 28.3 MeV. This value corresponds very closely with the figure of 32.94963 MeV that was found in one statistical analysis of the binding energies of 2932 nuclides. For more on this investigation see The Enigma of Nuclear Mass Deficits.
The value of s=4.2 may correspond to the limits of the charge distance rather than the separation of the centers of the nucleons. The diameters of the proton and neutron are thought to be about 1 fermi. Thus the separation of the centers would be 4.2-0.5-0.5=3.2 fermi. For a value of 3.2 fermi the solutions values are:
This would mean the mass deficit of the deuteron would be 6.2789 MeV and hence the mass of the neutron would be
This exceeds the mass of the proton by 9.89 electron masses. For this value of neutron mass the mass deficit (binding energy) of the alpha particle would be 36.4 MeV.
The deuteron has one nucleonic bond wherea the alpha particle has six. The binding energies per bond of the deuteron and alpha particle are then 6.2789 MeV dne 6.05 MeV, respectively.
where n is the principal quantum number of the system.
where s is the separation distance of the nucleons. The values of the parameters are r_{0}=1.522 fermi and H is in the range of 3×10^{-26} to 5×10^{-26} m²kg/s.
where γ is the energy of the photon emitted upon its formation and ΔK is the increase in kinetic energy of the nucleons resulting from its formation.
(To be continued.)
APPENDIX
The dependence of s on n can be elucidated by considering d(ln(s))/d(ln(n)); i.e.,
Since ds/d(ln(s)) is equal to s
Thus, of course, if f'(s)=0 then d(ln(s))/d(ln(n))=2. If f(s)=exp(−λs) then d(ln(s))/d(ln(n))=2/(1−λs). Thus For values of λs which are small relative to 1, d(ln(s))/d(ln(n)) is only slightly larger than 2. This means that the kinetic energy is approximately of the form
where ρ is a constant.
However, λ can be considered to be the reciprocal of a scale length s_{0}; i.e. λ=1/s_{0}. The previous analysis indicated that for s<<s_{0} s is approximately proportional to n², but for s>s_{0} the dependence is drastically altered since
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