Helium lowest energy

VV e now wish to establish the general functional form of possible wavefunctions for the two electrons in this pseudo helium atom. We will do so by considering first the spatial part of the u a efunction. We will show how to derive functional forms for the wavefunction in which the i change of electrons is independent of the electron labels and does not affect the electron density. The simplest approach is to assume that each wavefunction for the helium atom is the product of the individual one-electron solutions. As we have just seen, this implies that the total energy is equal to the sum of the one-electron orbital energies, which is not correct as ii ignores electron-electron repulsion. Nevertheless, it is a useful illustrative model. The wavefunction of the lowest energy state then has each of the two electrons in a Is orbital ... 

It can now be seen that there is a direct and simple correspondence between this description of electronic structure and the form of the periodic table. Hydrogen, with 1 proton and 1 electron, is the first element, and, in the ground state (i.e. the state of lowest energy) it has the electronic configuration ls with zero orbital angular momentum. Helium, 2 = 2, has the configuration Is, and this completes the first period since no... 

With these two assumptions, we can propose the electronic arrangement of lowest energy for each atom. We do so by mentally placing electrons successively in the empty orbitals of lowest energy. The electron orbital of lowest energy is the Is orbital. The single electron of the hydrogen atom can occupy this orbital. In the helium... 

The helium atom has the highest ionization potential of any atom. It has a nuclear charge of + 2, and the electrons reside in the lowest energy level close to the nucleus. 

The outer shell of the helium atom is full and complete the shell can only accept two electrons and, indeed, is occupied by two electrons. Similarly, argon has a complete octet of electrons in its outer shell. Further reaction would increase the number of electrons if argon were to undergo a covalent bond or become an anion, or would decrease the number of electrons below the perfect eight if a cation were to form. There is no impetus for reaction because the monatomic argon is already at its position of lowest energy, and we recall that bonds form in order to decrease the energy. 

Helium Helium has two electrons, both of which fit into the lowest-energy, Is orbital. The two electrons have opposite spins. 

Electrons fill the orbitals in the lowest energy level first, and then proceed to fill up the orbitals in other energy levels. If an atom has only two electrons, such as the element helium, those two electrons fill the lowest energy level, and the atom is stable. A helium atom does not easily gain, lose, or share electrons because its only orbital is full. 

Early on, Niels Bohr had speculated that electrons were particles circling an atoms nucleus in quantum shells with fixed energies. Helium, he knew, has two electrons. Because it is a very stable atom, one that refuses to gain or lose electrons under most conditions, Bohr concluded that two electrons filled the lowest energy shell, which he called n= 1. Bohr offered no reason why two electrons would completely fill that energy shell he simply based his conclusion on the known properties of helium. 

The Pauli exclusion principle states that no two electrons in an atom can have the same set of quantum numbers. Each electron exists in a different quantum state. Consequently, none of the electrons in an atom can have the same energy. The Is orbital has the following set of allowable numbers n= 1, t=0, m =0, m=+1/2 or -1/2. All of these numbers can have only one value except for spin, which has two possible states. Thus, the exclusion principle restricts the Is orbital to two electrons with opposite spins. A third electron in the Is orbital would have to have a set of quantum numbers identical to that of one of the electrons already in the orbital. So, the third electron needed for lithium must go into the next higher energy shell, which is a 2s orbital. The question about the Bohr atom that had so vexed scientists—why two electrons completely fill the lowest energy shell in helium—was now answered. There are only two electrons in the lowest energy shell because the quantum numbers derived from Schrodinger s equation and Paulis principle mandate it. 

A helium atom has two electrons, so we need two sets of quantum numbers. To represent the atom in its lowest energy state, we want each electron to have the lowest energy possible. If we let the first electron have the value of 1 for its principal quantum number, its set of quantum numbers will be the same as one of those given previously for the one electron of hydrogen. The other electron of helium can then have the other set of quantum numbers. 

As described in Ref. [25], the Hartree approach has been applied to get energies and density probability distributions of Br2(X) 4He clusters. The lowest energies were obtained for the value A = 0 of the projection of the orbital angular momentum onto the molecular axis, and the symmetric /V-boson wavefunction, i.e. the Eg state in which all the He atoms occupy the same orbital (in contrast to the case of fermions). It stressed that both energetics and helium distributions on small clusters (N < 18) showed very good agreement with those obtained in exact DMC computations [24],... 

The main exception to the rule is helium, which is at lowest energy when it has two electrons in its valence shell. 

---