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Using Methods Derived from the Old Quantum Physics of Niels Bohr |
The deuteron, the nucleus of heavy hydrogen, is a very interesting, important special case. Its two nucleons, the proton and neutron, are held together by the nuclear force alone. This nuclear force is carried by the π meson. As with gravitation or the Coulomb force there is an inverse-square-of-the-distance dependence because force-carrying particles are spread over a spherical surface whose area is proportional to the square of the distance from the source.. However there is an essential difference for the nuclear force as compared to the electrostatic Coulomb force in that the particles carrying the electrostatic force, photons, do not decay whereas the π mesons of the nuclear force do decay quite rapidly with distance. The proportion of the mesons which survive is an exponential function of distance. Therefore the formula for the nuclear force would be of the form
where d is the distance between the nucleons, H* and λ* are constants with dimensions of kg·m³/s² and m−1, respectively. The material which follows is an attempt to work out the implications of this force formula for the quantum level phenomena of the deuteron.
Specifically the purpose of the analysis is to derive from the quantization of angular momentum quantizations of orbital radii and velocities. From these quantizations follows quantizations of kinetic energy, potential energy and total energy. The change in total energy when the nucleons move from one state to another corresponds to the energy of the gamma photon emitted or absorbed. From the measured energy of emitted photons an estimate of H* may then be obtained.
At this point it is convenient to switch notation. Let r stand for the radius of the orbit of a particle with respect to the center of mass of the particle pair. For the electron in an proton-electron pair r is essentially equal to the distance d between the particles. For the proton-neutron pair of a deuteron r is essentially equal to one half of the distance between the particles. This means the formula for the force experienced by each nucleon as a function of the radius r of their orbits with respect to their center of mass is
where H=H*/4 and λ=2λ*. It is conventient to express λ* as 1/d0; i.e., d0 = 1/λ*. This means that r0 = 1/(2λ*) = ½d0
Hideki Yukawa established that the mass of the force-carrying particles is related to the λ parameter. From the mass of the π mesons it is known that d0 is about 1.522 fermi (1.522×10−15 meters). Thus r0 is about 0.761 fermi.
In mechanics the force is conveniently represented as the negative of the derivative of a potential function V(r). This means that the potential function is given by the integration of the force function with with respect to distance. For the Coulomb force of −α/r² this gives a potential function of V(r)=−α/r. Yukawa hypothesized a potential function which is just this potential function multiplied by an exponential factor; i.e., V(r)=(−α/r)e−λr. However it is the force function that is multiplied by the exponential factor due to the decay of the force-carrying mesons; i.e.,
where d is the distance between the two nucleons and H* and λ* are constants characteristic of the nuclear force. Therefore the potential function for the nuclear force is really of the form
It is convenient to change the variable of integration from s to λ*s=z. Thus
The standard procedure in quantum mechanics is to solve Schroedinger's equation for the particular potential function which applies. This works beautifully for the Coulombic potential and about a half dozen other special cases but gives no analytic results for the cases that deviate from those special cases. The Schroedinger equation cannot be solved even for the Yukawa potential function. There is no way to solve the Schroedinger equation analytically for even more complicated potential functions. Physicists then rely upon purely numerical methods of solution based upon perturbation theory.
What does work analytically is the analysis of Niels Bohr's Old Quantum Theory. Using
Bohr's analysis it can be established that the angular momentum
is quantized for any potential function in exactly the same way that it is for the
Coulombic force. This means that angular momentum must be equal to an integral multiple of
h, Planck's constant divided by 2π. This is also true whether the kinetic
energy is of the Newtonian form ½mv², where m is the mass of the particle and
v is its velocity or the relativistic form m0c²[1/(1−β²)½ − 1 ]
where β is the velocity relative to the speed of light, v/c.
In a circular orbit the attractive force balances the centrifugal force; i.e.,
This means that orbital velocity v is a function of orbit radius r; i.e.,
Angular momentum is defined as pθ=mvr, which on the basis of the relation between v and r means
But previous analysis found angular momentum to be quantized. That is to say,
where l is an integer and h is Planck's constant divided by 2π.
This means that
Now let r/r0 be denoted as z.
The function ze−z rises from 0 at z=0 to a maximum of e−1 at z=1 and falls asymptotically to 0 as z goes to +∞, as shown below.
For sufficiently small values of angular momentum there will be two values of r but there is a maximum angular momentum for which there are any solutions for r.
The levels shown in green above correspond to the quantity h²/(Hmr0)
multiplied by the square of an integer l.
The value of this coefficient can be calculated. The value of
h² in SI units is 1.11×10−68. For the present the dependence of
mass m on the velocity will be ignored and its value taken to be the rest mass of proton, m0.
The value of m0
is 1.67×10−27 kg. As
indicated above the value of r0 is 7.61×10−16 m.
The maximum value that ze−z can attain is e−1=0.3679 and this is achieved at z=1. This means that if there is to be a solution to the equation
On the other hand, if there is to be no solution to the equation then it must be that
It is assumed at the lowest level of angular momentum is (l=1). The case of (l=0) is treated elsewhere.
Since in a circular orbit v = [He−r/r0/(mr)]1/2 and H ≥ 3.219×10-26 this means that at the lowest level the tangential velocity of a nucleon is such that
which is 68.7 percent of the speed of light. Thus relativistic effects should be considered.
The balance of the attractive nuclear force with the centrifugal force under Special Relativity is
This can be reduced to
Multiplying by the denominator on the left produces a quadratic equation in β²; i.e.,
where ζ² = [He−r/r0/(m0c²r)]². The solutions are
The negative solution must be discarded. Thus
For large values of ζ the solution is approximately β=1.
It is shown elsewhere that even in the relativistic case angular
momentum pθ=mvr is quantized in increments of h so
where l is an integer.
For a circular orbit
Therefore
Since
pθ = lh it follows that
where γ=H/(hc). This is not an entirely satisfactory
expression of quantization because r on the RHS is quantized in some as yet undetermined
manner, yet the similarity with the non-relativistic case makes it of interest.
From the analysis of the previous section it is known that
where ζ = [He−r/r0/(m0c²r0)], which is
the same as γ(hc/(m0c²r0).
In principle the two expressions for β can be equated the result solved for r as a function of l. This provides the quantization of r and subsequently that of β and the rest of the characteristics of the system.
The equating of the two expressions for β gives
After squaring and rearranging the above equation reduces to
Since 1/ζ=m0c²r/(He−r/r0) the previous equation is equivalent to
Previously the expression H/(hc) was defined as γ. Utilizing
this expression the previous equation can be put into the form
It is mathematically convenient to deal with ξ=2r/r0 as a variable so the above equation takes the form
To simplify matters still more for computation let
(m0c²r0/(2ch)) be denoted as μ. The equation to be solved for ξ is
then
Although the above equation is transcendental and cannot be solved analytically, approximations to any degree of accuracy can be readily be obtained numerically, provided of course the equation has a solution at all.
The parameter μ evaluates to:
Since RHS of the previous transcendental equation is a function of the angular momentum quantum number l the solutions for ξ should be labeled with ξ. Thus there is a solution set {ξl; l=1,2,…,n}.
This set generates the set of quantum levels of r and these and the equation
generates the quantum levels of β and v. The set of quantum levels of β generate the quantum levels of kinetic energies. The levels of r and the definition of potential energy generate the quantum levels of potential energy and total energy.
It has been known since 1935 that if a deuteron is hit by a sufficiently high energy photon the deuteron disintegrates into a free proton and a free neutron. The energy level required for this photodisintergration of the deuteron is 2.23 Mev. This is 3.57×10−13 joules. A convenient unit of energy for this analysis is the rest mass energy of a proton, which is 938.272 MeV. Thus the energy of the gamma photon involved in the formation and dissociation of a deuteron is equal to 0.002371 of the rest mass energy of a proton, or about a quarter of 1 percent.
The uncombined state of the nucleons can be considered to be one in which their kinetic energies are zero and the potential energy of their system is zero. Thus the energy levels of the deuteron are the same as the changes in the energy levels for the change from an uncombined to combined state or vice versa.
It might be thought to be reasonable identify this energy as the change in potential energy of each of the l=1 level nucleons. However on an atomic level the change in potential energy for an electron transition between levels is twice the magnitude of the photon enery. The other half of the change in potential energy goes into a change in the kinetic energy of the electron. A photon that knocks one nucleon out of its orbit disassociates the deuteron.
The procedure is to estimate H* by searching for a value that is consistent with a change in total energy equal to the dissociation energy of 2.22457 MeV. This is carried out as follows.
(The parameter μ² has a known value of 2.935.
These have to be doubled to account for the energies of the two nucleons.
Suppose γ is taken to have a value of 2.0 and l is equal to 1.0. The graph of the LHS and RHS of the transcendental equation and their difference looks like this.
In this case, l=1, there is a solution, ξ=6.08. There is also another solution for small ξ which cannot be discerned in the graph because of its scale. For the case l=2 there is no solution. The quadratic function of the RHS is below the exponential function on the LHS at the origin (ξ=0)and can never catch up as ξ increases.
The table below gives the solutions for a number of values of γ.
| γ | 1.0 | 0.8 | 0.8032 | |
| upper | lower | lower | lower | |
| ξ | 3.74 | 0.425 | 1.18 | 1.158 |
| r/r0 | 1.87 | 0.2125 | 0.59 | 0.579 |
| r (fermi) | 1.42307 | 0.16171 | 0.44899 | 0.44062 |
| d (fermi) | 2.8614 | 0.32343 | 0.88276 | 0.88124 |
| β | 0.154 | 0.80856 | 0.44346 | 0.45016 |
| E | 2.024186 | 3.39897 | 2.23141 | 2.23977 |
| ∫(e-z/s²)ds | -0.02045 | -3.16465 | -0.49731 | -0.51667 |
| V/(m0c²) | -0.01403 | -1.84678 | -0.23217 | -0.24217 |
| T | 2.010153 | 1.55219 | 1.99924 | 1.99780 |
| T-2 | 0.010153 | -0.44781 | -0.00076 | -0.00240 |
The figure to be matched on the bottom line is -0.002371, the energy of the gamma ray emitted or absorbed in the formation or dissociation of a deuteron. This value is achieved for γ equal to slightly less than 0.8032. For the upper solution for γ=1.0 the computations indicate that the total energy of the nucleon system in increased instead being decreased in the formation of the deuteron. This does not happen. Therefore the physically relevant solution is the lower solution value.
Since γ=H/hc, value of γ=0.8032 implies that
H=3.162336×10-26 kg·m³/s and therefore H*=4H is equal to
1.265×10-25 kg·m³/s. This is remarkably close to the critical value
of H* previously determined, 1.28765×10-25 which represented the dividing
line between there being a solution and there not being a solution.
One interesting aspect of the solution is that there is a loss in potential energy in the formation of the deuteron that is equal to 0.24127 times the rest mass energy of the proton and about 99 percent of this goes into an increase in kinetic energy of the deuteron, with only about 1 percent going for the energy of the emitted gamma ray.
The results imply a separation distance for the nucleons of 0.88 fermi. This is significantly smaller than other estimates of the size of the deuteron.
The model implies only one stable state for the deuteron and this is consistent with the empirical observation.
The general conclusion is that the model of the nuclear force being given by the formula
is logically coherent and consistent the empirical information available for the deuteron.
(To be continued.)
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