Tutorial 1 Kimball Tutorial 2 Tutorial 3 ⇒

Required:

Content: Using G.E. Kimball’s equation, introduce Kimball’s model, discuss energy minimum, and mechanical equilibrium, compute one electron systems and the first molecule: ES 16 June 2017//2002//1982

Some properties of the Kimball H atom, equilibrium of forces

Kimball’s equation for the H atom in its stationary ground state:

R is the expectation value of the electron-proton distance of the H1s state

Kinetic energy Eh:

Potential energy Eh:

Following Kimball, we invent a sphere of radius R, and fill it with one electron. We assume that its kinetic energy allows the electron to roam freely about, such, that it may be found equally likely at every point inside that sphere. Of course, such a sphere would immediately evaporate unless we keep it existing by a positive charge. For a start, this will be one proton at a distance r somewhere outside the sphere. The attraction of the electrostatic force leads to a race of the proton towards the cloud until the proton arrives at the clouds imagined rim and then drops further down into it to finally rest at the clouds center with x=y=z=0. C.F.Gauss has shown, that this will be an equilibrium position because in spherical symmetry the force on the proton exerted by a volume element of negative charge at (x,y,z) will be exactly compensated by an equivalent element at (-x,-y,-z). Let’s compute the energy of the proton-cloud system on that trajectory of the proton: Outside the sphere the proton sees the whole charge of -1 in the center of the sphere. As soon as r < R the spherical shell outside r of thickness R-r has no influence anymore because of Gauss’ argument just given. So, the attraction diminishes with decreasing r until it is zero for r=0. Ok, now the simple mathematics for this:

First part: proton-electron distance r arbitrarily large, -e electron, e proton charge, Coulomb’s law:

Second part: proton from r=R to r=0; -

Total potential energy change:

In atomic units, that is:

exactly the same value we have found in Tutorial 1 and reproduced above. Let’s check whether this is an energy minimum by setting the derivative dE/dr = 0:

No wonder! We have done the same kind of integration as in Tutorial1: In a spherically symmetric system the integral over r=0 to ∞ does not depend on the density distribution given that it adds up to same total charge and is maintained by the equilibrium between kinetic and potential energy, see below. Hence, we get the same value for R=x, whether we use from the H atom with a pronounced cusp at the origin or Kimball’s constant density sphere with radius R.

This now is **Kimball’s model**, distinct from the exact Kimball equation: The potential energy term of the Kimball equation for the H atom can be interpreted as the electrostatic interaction of a proton with an electron, “filling” a sphere of radius R=3/2 a0 with constant density and having the kinetic energy . We will investigate further down and in the following tutorials, whether this is true for non spherical systems, i.e. molecular chemistry, or only for the physics of the 1-electron H and homologous ions, like C(+5).

What are the forces, keeping the equilibrium? Increase the clouds radius by dr from R and get the forces for bringing it back to the equilibrium by again deriving dE/dr: In Mathematica’s notation this application of the Principle of d’Alembert becomes:

and in vector notation, unit vector:

These two forces counterbalance each other for r=3/2=R (solve the equation!), as they should. The right term is the Coulomb force which responds to the perturbation of the equilibrium by an infinitesimal energy increase, the left is the derivative of the kinetic energy, dT/dr, which reacts by an energy decrease. It is a measure of the “electron pressure” tending to blow the sphere up while the electrostatic force tends to deflate it. This is a typical quantum mechanical effect. In classical physics the force derived from a kinetic energy of the electron does not exist! Together with the Coulomb force it is the force maintaining the stability of matter in the universe we know by preventing the negative and positive charges to collapse onto each other (we do not address the stability of nuclei).

Kimball’s sphere has no wall! Its surface is just the locus where the proton or other nucleus in the center stabilizes the average distance R of the electron moving back and forth. This radial motion does not dissipate energy and is not thermal! Therefore, this kinetic energy is named “zero point (kinetic) energy”, a general property of the microscopic particles of matter if they suffer from a boundary limiting their free roaming. The boundary is caused by the Coulomb attraction, not a solid cage! Hence, it is also valid for the proton which is forced by Coulomb to sit (very near) to the center of the electron cloud. It forms its own probability cloud! But this is so tiny compared to the electron probability cloud that we usually neglect it, an aspect of the Born-Oppenheimer approximation (see tutorial6a). Its radius is almost 2000 times smaller, the ratio of proton mass/electron mass.

It is astounding how often we unconsciously (?) use the image of spheres with solid surfaces when contemplating microscopic phenomena. Think about the innumerable models of crystal lattices, densest packed metals, ionic lattices, models of molecules a.s.o. with spherical particles having an "atomic", "ionic" or "van der Waals" radius, often thought as impenetrable (“tangent sphere model”, a caricature of the model we discuss here). They are equally imaginary as Kimball's spheres and have precisely the same physical background of (mental) existence, the kinetic energy of “clamped” particles. In fact, it is probably not possible to talk chemistry, condensed matter physics or mineralogy/crystallography without this firmly established archetypical model, unless you apply abstract mathematical description to avoid it. All macroscopic wire and styrofoam models of molecules or crystal lattices are not adequate description of what they should show. They can only give an idea of symmetry. The rest is fairy tale.

Homologues of the H atom

The potential energy of an atomic ion with nuclear charge Z and one electron becomes Z times larger than for H:

This would have been clear in Tutorial 1 already, had we used the complete 1s wavefunction, which is:

The kinetic energy of the electron adjusts to the charge Z with a new cloud radius. Kimball’s equation is now:

The cloud shrinks with and the energy is now, as above with the wavefunction:

For He(+): Etot = -2.0 Eh, R=3/4 a0, Ionization energy 2*27.2114 = 54.42 eV (exp. 54.42)

and e.g.

for C(+5): Etot = -18.0 Eh, R=1/4 a0, Ionization energy 18*27.2114 = 489.8 eV (exp. 489.9)

Two nuclei, one electron: First molecule!

This is the potential energy of a proton in the cloud, but at a distance r from its center. V2 is changed, V1 keeps its former value:

In atomic units (ConditionalExpression valid!):

You discover immediately that a proton at r from the center is not in a stable position. It will move to the center, unless a force keeps it at r. We put a second proton into the cloud, exactly opposite to the first.They repel each other. We expect that they settle at a point where the attraction by the fraction of the electron charge in e*/3 compensates the repulsion, so that a stationary location exists. The following expression for the total energy contains all the terms:

Clear[R,r,E2];

the last term being the electrostatic repulsion energy of two protons at a distance of 2r. We now have two variables, r,R and need to minimize the energy of the system. We could easily solve this with pencil and paper, but we use the splendid minimization function of Mathematica:

A stable system! The protons have a distance of

Equilibrium requires the following forces to vanish:

which they do! ( )*e is the electron charge of the sphere between 0 and r.

/. t[[2]] is Mathematica’s substitution command, meaning that the values for R and r are to be taken from the result of the minimization in the second element of the list t. The first element is the energy of the minimum determined:

This is lower than the energy of the H atom, i.e. the chemical reaction just completed

H + → ΔE = -0.7276+0.5 Eh

is exothermic with an energy change of -0.2276 = -142 kcal/mol at zero Kelvin, the bonding energy. The bond distance, compiled above, is 1.567*52.9177 = 82.9 pm. The experimental values are -64.4 kcal/mol and 105.8 pm!

The bond energy is more than twice the experimental value, while the bond distance is 22% too small, a failure of the model! We have analyzed the reason for this discrepancy in protag.htm. The main problem is the assumption of a homogeneous density within the Kimball sphere. As we have seen in Tutorial 1, there exists a sharp cusp of density at or near every nucleus. In the H atom this is at the center of the cloud and does not influence the stability. In we have two cusps off-center. Therefore, the accumulation of charge near the nuclei disturbs the assumed homogeneous density. It diminishes the electron density between the two nuclei - in the bonding region -, and simultaneously screens the positive charges more than assumed. Both effects lead to a smaller binding energy and a larger bond distance. One can correct for this by lowering the density of the cloud near the origin and thus approximate the diminished value in the “real” molecular ion. This is shown in the figure with comments below it:

disks mark the proton locations in the vibration-less molecule ion:

blue: Kimball result, above, with dashed circle and open proton “disks”,

green: experimental position of protons

red: corrected Kimball result for effect of cusps, formula below

E2corr = -0.6023 is now very near the experimental ground state energy of of -0.6026 Eh. However, a lower density between the two protons can only be realized by increasing the size of Kimball’s sphere using a correction factor 1.208 for the kinetic energy. There is no law against such a correction, because for a one electron sphere the kinetic energy 9/8R^{2} is a lower limit defined by Heisenberg’s uncertainty relation.

In this simple, original model of Kimball, there is no tool foreseen for the density to react to the details of the electrical field inside the sphere, nor for its deformation to an often more appropriate ellipsoid. The “Floating Spherical Gaussian Orbital” model, FSGO, of A.A.Frost (1967) models the cusps by a Gaussian bell shape, and thus is a further development of Kimball’s ideas. We do not go into this here, because it leads to a complete sacrifice of the simplicity of Kimball’s model. This means, that we have to live with this correction, which pertains almost entirely to bonds with H atoms, e.g. C-H, N-H, O-H, can be tabled and is determined once and for all from experiment. It is of similar size, about 1.21 to 1.28 for all X-H bond clouds.

Take note of the fact, that the second proton in the one electron cloud makes it shrink from a radius of 1.5 in H to 1.243.. a0 in . The additional electric field, created by the second proton, "glues" the electron to a smaller average distance, hence a cloud with a 17% smaller expected radius. By chance the corrected red circle is almost the same size as that of the (original) H atom. The dashed blue circle and the red circle show approximately the shrinking for the uncorrected .