Helium Coulomb repulsion

Another example of the difficulty is offered in figure B3.1.5. Flere we display on the ordinate, for helium s (Is ) state, the probability of finding an electron whose distance from the Fie nucleus is 0.13 A (tlie peak of the Is orbital s density) and whose angular coordinate relative to that of the other electron is plotted on the abscissa. The Fie nucleus is at the origin and the second electron also has a radial coordinate of 0.13 A. As the relative angular coordinate varies away from 0°, the electrons move apart near 0°, the electrons approach one another. Since both electrons have opposite spin in this state, their mutual Coulomb repulsion alone acts to keep them apart. 

I have included the arbitrary parameter X in order to keep track of orders of magnitude. I will later set it to unity. In the case of the helium problem above, the perturbation would be just the Coulomb repulsion between the electrons. 

The intensity of a peak in RBS is determined by the cross section o for scattering. At MeV energies, the helium ion penetrates deeply into the atom and approaches the nucleus of the target atom to within 10 4 nm, i.e. well within the radius of the K-electron shell. This means that the scattering event depends only on the Coulomb repulsion between the two nuclei, whereas screening by the electrons (which is important in LETS) plays no role. Thus the scattering cross section is a... 

Following the core helium flash, the star quickly ( 106 years) moves to the Horizontal Branch (HB), where it burns 4He in a convective core, and hydrogen in a shell (that provides most of the luminosity). This corresponds to points 10-13 in Figure 3. The coulomb repulsion is larger for He than for H,... 

The traditional explanation of Hund s rule is as follows Electrons with the same spin tend to keep out of each other s way (recall the idea of Fermi holes), thereby minimizing the Coulombic repulsion between them. The term that has the greatest number of parallel spins (that is, the greatest value of 5) will therefore be lowest in energy. For example, the 5 term of the helium ls25 configuration has an antisymmetric spatial function that vanishes when the spatial coordinates of the two electrons are equal hence the 5 term is lower than the 5 term. 

The mutual avoidance of electrons in the helium atom or in the hydrogen molecule is caused by Coulombic repulsion of electrons (described in the previous subsection). As we have shown in this chapter, in the Haitree-Fock method the Coulomb hole is absent, whereas methods that account for electron correlation generate sueh a hole. However, electrons avoid each other also for reasons other than their charge. The Pauli principle is another reason this occurs. One of the consequences is the fact that electrons with the same spin coordinate cannot reside in the same place see p. 34. The continuity of the wave function implies that the probability density of them staying in the vicinity of each other is small i.e.. 

In Section 7.5.3 we saw that the ls 2s configuration of the helium atom can exist in two forms one where the electron spins are parallel, and the other where the spins are paired. The state with parallel spins was found to have the lower energy because the antisymmetric spatial wavefunction required the electrons to stay further apart, thereby reducing the coulombic repulsion energy. 

What is the correction to the energy of the helium atom, assuming that the perturbation can be approximated as a coulombic repulsion of the two electrons ... 

The helium atom consists of a system with two electrons around a nucleus. This model can be applied to any two-electron system with an atomic number Z including H, Li", and Be. The Hamiltonian includes kinetic energy operators for the two electrons, the Coulombic repulsion potential between the electrons, and a Coulombic attraction between each electron and the nucleus. This is shown schematically in Figure 84. 

Now lei us turn to the problem of how the composition of a nucleus affects its stability. The forces that exist between the particles in the nucleus are very large. The most familiar of ihe intranuclear forces is the coulomb force of repulsion which the protons must exert on one another. In order to appreciate the magnitude of this repulsive force, let us compare the force between two protons when they are separated by 10 8 cm, as they are in the hydrogen molecule, with the force between two protons separated by 10-18 cm, as they are in a helium nucleus. In the first case we have... 

Note that the other electrons do not block the influence of the nucleus they simply provide additional repulsive coulombic interactions that partly counteract the pull of the nucleus. For example, the pull of the nucleus on an electron in the helium atom is less than its charge of +2e would exert but greater than the net charge of +e that we would expect if each electron balanced one positive charge exactly. 

To be sure, the experiments of Nicholson and Merton on the radii of the atomic orbits in mixtures of hydrogen and helium relative to the distance between molecules and of Ramsauer on the long free paths of electrons in argon and other gases do not favor Coulomb s law. It may be that the repulsion between nuclei and the attraction between nucleus and electron follow different laws. 

The potential energy of the helium atom has three parts, all coulombic in nature there is an attraction between electron 1 and the nucleus, an attraction between electron 2 and the nucleus, and a repulsion between electron 1 and electron 2 (because they are both negatively charged). Each part depends on the distance between the particles involved the distances are labeled r, r2, and ri2 as illustrated in Figure 12.3. Respectively, the potential energy part of the Hamiltonian is thus... 

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