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From Wikipedia

Helium atom

Helium is an element and the next simplest atom to solve after the hydrogen atom. Helium is composed of two electrons in orbit around a nucleus containing two protons along with either one or two neutrons, depending on the isotope. The hydrogen atom is used extensively to aid in solving the helium atom. The Niels Bohr model of the atom gave a very accurate explanation of the hydrogen spectrum, but when it came to helium it collapsed. Werner Heisenberg developed a modification of Bohr's analysis but it involved half-integral values for the quantum numbers. Thomas-Fermi theory, also known as density functional theory, is used to obtain the ground state energy levels of the helium atom along with the Hartree-Fock method.

## Introduction

The Hamiltonian of helium is given by

H\psi(\vec{r}_1,\, \vec{r}_2) = \Bigg[\sum_{i=1,2}\Bigg(-\frac{\hbar^2}{2\mu} \nabla^2_{r_i} -\frac{Ze^2}{4\pi\epsilon_0 r_i}\Bigg) - \frac{\hbar^2}{M} \nabla_{r_1} \cdot \nabla_{r_2} + \frac{e^2}{4\pi\epsilon_0 r_{12}} \Bigg]\psi(\vec{r}_1,\, \vec{r}_2)

where \mu = \frac{mM}{m+M} is the reduced mass of an electron with respect to the nucleus and r_{12} = |\vec{r_1} - \vec{r_2}| . Consider M = \infty so that \mu = m and the mass polarization term \frac{\hbar^2}{M} \nabla_{r_1} \cdot \nabla_{r_2} disappear. The Hamiltonian in atomic units (a.u.) for simplicity is given by

H\psi(\vec{r}_1,\, \vec{r}_2) = \Bigg[-\frac{1}{2}\nabla^2_{r_1} - \frac{1}{2}\nabla^2_{r_2} - \frac{Z}{r_1} - \frac{Z}{r_2} + \frac{1}{r_{12}}\Bigg]\psi(\vec{r}_1,\, \vec{r}_2).

The presence of the electron-electron interaction term 1/r12, makes this equation non separable. This means that \psi_0(\vec{r}_1,\, \vec{r}_2) can't be written as a single product of one-electron wave functions. This means that the wave function is entangled. Measurements cannot be made on on one particle without affecting the other. This is dealt with in the Hartree-Fock and Thomas-Fermi approximations.

## Hartree-Fock Method

The Hartree-Fock method is used for a variety of atomic systems. However it is just an approximation, and there are more accurate and efficient methods used today to solve atomic systems. The "many-body problem" for helium and other few electron systems can be solved quite accurately. For example the ground state of helium is known to fifteen digits. In Hartree-Fock theory, the electrons are assumed to move in a potential created by the nucleus and the other electrons. The Hamiltonian for helium with 2 electrons can be written as a sum of the Hamiltonians for each electron:

H = \sum_{i=1}^2 h(i) = H_0 + H^'

where the zero-order unperturbed Hamiltonian is

H_0 = -\frac{1}{2} \nabla_{r_1}^2 - \frac{1}{2} \nabla_{r_2}^2 - \frac{Z}{r_1} - \frac{Z}{r_2}

while the perturbation term:

H' = \frac{1}{r_{12}}

is the electron-electron interaction. H0 is just the sum of the two hydrogenic Hamiltonians:

H_0 = \hat{h}_1 + \hat{h}_2

where

\hat{h}_i = -\frac{1}{2} \nabla_{r_i}^2 - \frac{Z}{r_i}, i=1,2

En1, the energy eigenvalues and \psi_{n,l,m}(\vec{r}_i) , the corresponding eigenfunctions of the hydrogenic Hamiltonian will denote the normalized energy eigenvalues and the normalized eigenfunctions. So:

\hat{h}_i \psi_{n,l,m}(\vec{r_i}) = E_{n_1} \psi_{n,l,m}(\vec{r_i})

where

E_{n_1} = - \frac{1}{2} \frac{Z^2}{n_i^2} \text{ in a.u.}

Neglecting the electron-electron repulsion term, the SchrÃ¶dinger equation for the spatial part of the two-electron wave function will reduce to the 'zero-order' equation

H_0\psi^{(0)}(\vec{r}_1, \vec{r}_2) = E^{(0)} \psi^{(0)}(\vec{r}_1, \vec{r}_2)

This equation is separable and the eigenfunctions can be written in the form of single products of hydrogenic wave functions:

\psi^{(0)}(\vec{r}_1, \vec{r}_2) = \psi_{n_1,l_1,m_1}(\vec{r}_1) \psi_{n_2,l_2,m_2}(\vec{r}_2)

The corresponding energies are (in a.u.):

E^{(0)}_{n_1,n_2} = E_{n_1} + E_{n_2} = - \frac{Z^2}{2} \Bigg[\frac{1}{n_1^2} + \frac{1}{n_2^2} \Bigg]

Note that the wave function

\psi^{(0)}(\vec{r}_2, \vec{r}_1) = \psi_{n_2,l_2,m_2}(\vec{r}_1) \psi_{n_1,l_1,m_1}(\vec{r}_2)

An exchange of electron labels corresponds to the same energy E^{(0)}_{n_1,n_2} . This particular case of degeneracy with respect to exchange of electron labels is called exchange degeneracy. The exact spatial wave functions of two-electron atoms must either be symmetric or antisymmetric with respect to the interchange of the coordinates \vec{r}_1 and \vec{r}_2 of the two electrons. The proper wave function then must be composed of the symmetric (+) and antisymmetric(-) linear combinations:

\psi^{(0)}_\pm(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}} [\psi_{n_1,l_1,m_1}(\vec{r}_1) \psi_{n_2,l_2,m_2}(\vec{r}_2) \pm \psi_{n_2,l_2,m_2}(\vec{r}_1) \psi_{n_1,l_1,m_1}(\vec{r}_2)]

This comes from Slater determinants.

The factor \frac{1}{\sqrt{2}} normalizes \psi^{(0)}_\pm . In order to get this wave function into a single product of one-particle wave functions, we use the fact that this is in the ground state. So n_1=n_2=1,\, l_1=l_2=0,\, m_1=m_2=0 . So the \psi^{(0)}_{-} will vanish, in agreement with the original formulation of the Pauli exclusion principle, in which two electrons cannot be in the same state. Therefor the wave function for helium can be written as

\psi^{(0)}_0(\vec{r}_1, \vec{r}_2) = \psi_1(\vec{r_1}) \psi_1(\vec{r_2}) = \frac{Z^3}{\pi} e^{-Z(r_1 + r_2)}

Where \psi_1 and \psi_2 use the wave functions for the hydrogen Hamiltonian. For helium, Z = 2 from

E^{(0)}_0 = E^{(0)}_{n_1=1,\, n_2=1} = -Z^2 \text{ a.u.}

where E^{(0)}_0 = -4 a.u. which is approximately -108.8 eV, which corresponds to an ionization potential V_P^{(0)} = 2 a.u. ( \simeq 54.4 eV). The experimental values are E _0 = -2.90 a.u. ( \simeq -79.0 eV) and V_p = .90 a.u. ( \simeq 24.6 eV).

The energy that we obtained is too low because the repulsion term between the electrons was ignored, whose affect is to raise the energy levels. As Z gets bigger, our approach should yield better results, since the electron-electron repulsion term will get smaller.

So far a very crude independent-particle approximation has been used, in which the electron-electron repulsion term is completely omitted. Splitting the Hamiltonian showed below w

Atomic weight

Atomic weight (symbol: A) is a dimensionlessphysical quantity, the ratio of the average mass of atoms of an element (from a given source) to 1/12 of the mass of an atom of carbon-12 (known as the unified atomic mass unit). The term is usually used, without further qualification, to refer to the standard atomic weights published at regular intervals by the International Union of Pure and Applied Chemistry (IUPAC) and which are intended to be applicable to normal laboratory materials. These standard atomic weights are reprinted in a wide variety of textbooks, commercial catalogues, wallcharts etc., and in the table below. The fact "relative atomic mass of the element" may also be used to describe this physical quantity, and indeed the continued use of the term "atomic weight" has attracted considerable controversy since at least the 1960s (see below).

Atomic weights, unlike atomic masses (the masses of individual atoms), are not physical constants and vary from sample to sample. Nevertheless, they are sufficiently constant in "normal" samples to be of fundamental importance in chemistry.

## Definition

The IUPAC definition of atomic weight is:

An atomic weight (relative atomic mass) of an element from a specified source is the ratio of the average mass per atom of the element to 1/12 of the mass of an atom of C.

The definition deliberately specifies "An atomic weightâ€¦", as an element will have different atomic weights depending on the source. For example, boron from Turkey has a lower atomic weight than boron from California, because of its different isotopic composition. Nevertheless, given the cost and difficulty of isotope analysis, it is usual to use the tabulated values of standard atomic weights which are ubiquitous in chemical laboratories.

## Naming controversy

The use of the name "atomic weight" has attracted a great deal of controversy among scientists. Objectors to the name usually prefer the term "relative atomic mass" (not to be confused with atomic mass). The basic objection is that atomic weight is not a weight, that is the force exerted on an object in a gravitational field, measured in units of force such as the newton.

In reply, supporters of the term "atomic weight" point out (among other arguments) that

• the name has been in continuous use for the same quantity since it was first conceptualized in 1808;
• for most of that time, atomic weights really were measured by weighing (that is by gravimetric analysis) and that the name of a physical quantity should not change simply because the method of its determination has changed;
• the term "relative atomic mass" should be reserved for the mass of a specific nuclide (or isotope), while "atomic weight" be used for the weighted mean of the atomic masses over all the atoms in the sample;
• it is not uncommon to have misleading names of physical quantities which are retained for historical reasons, such as

It could be added that atomic weight is often not truly "atomic" either, as it does not correspond to the property of any individual atom. The same argument could be made against "relative atomic mass" used in this sense.

## Determination of atomic weight

Modern atomic weights are calculated from measured values of atomic mass (for each nuclide) and isotopic composition. Highly accurate atomic masses are available for virtually all non-radioactive nuclides, but isotopic compositions are both harder to measure to high precision and more subject to variation between samples. For this reason, the atomic weights of the twenty-two mononuclidic elements are known to especially high accuracy â€“ an uncertainty of only one part in 38&nbsp;million in the case of fluorine, a precision which is greater than the current best value for the Avogadro constant (one part in 20&nbsp;million).

The calculation is exemplified for silicon, whose atomic weight is especially important in metrology. Silicon exists in nature as a mixture of three isotopes: Si, Si and Si. The atomic masses of these nuclides are known to a precision of one part in 14&nbsp;billion for Si and about one part in one billion for the others. However the range of natural abundance for the isotopes is such that the standard abundance can only be given to about Â±0.001% (see table). The calculation is

A(Si) = (27.97693 Ã— 0.922297) + (28.97649 Ã— 0.046832) + (29.97377 Ã— 0.030872) = 28.0854

The estimation of the uncertainty is complicated, especially as the sample distribution is not necessarily symmetrical: the IUPAC standard atomic weights are quoted with estimated symmetrical uncertainties, and the value for silicon is 28.0855(3). Th

From Encyclopedia

Atomic Structure Atomic Structure

Question:Is '6.02 times ten to the power of 23 chemical units of a pure substance' right, or is there some standard test question answer? Thanks - thought the other was a little complicated!

Answers:"A mole of a substance is the mass of a substance that has the same number of particles as there are atoms in 12g of carbon 12" Im doing AS chemistry and I was also confused with the definition. I asked my teacher and he said just think of it as the mass of any substance that has the same number of atoms relative to a scale where 12g of carbon is 12. You can write down Avogadro's number, but he said it wasn't necessary unless it specifies. Hope this makes things clearer.

Question:Explain how the existence of electron energy levels in atoms give rise to line spectra, which may be emission or absorption spectra? Why is there dark lines in the spectrum? Please explain the difference between absorption, emission and continuous spectrum. What leads to continuous and line spectra?

Answers:Because of the energy levels having definite energies, transitions from one to another are always the same energy. This means there are some energies that there are no transitions corresponding to them, thus there are colors that are not possible to emit from a given element. Keep in mind that the element can absorb or emit these definite energies--so when a continuous spectrum--that's ALL colors passes through a given gas, ONLY these energies are absorbed, thus there are black bands which correspond exactly to the emission spectra . . . the absorbtion and emmission spectra are identical, except for the first you are absorbing the given colors and in the second emitting the colors. Continuous spectra usually come from hot things, black body radiation, whereas line spectra come from the electron energy transitions already discussed.

Question:what does it say about the proton+ electron's combined energy level's Is it that there combined energy is greater than the last possible energy level, which on the question is E(7)= - 0.28eV? or E(infinity)= 0e V? cheers by continuum i mean its not in an energy level- its ionised.

Answers:If the electron is no longer in a bound state, where the energy levels of the atom are the discrete values E_n = E_1 / n^2 < 0 (E_1 = - alpha^2 m c^2 / 2, with alpha the fine structure constant, m the electron mass, and c the speed of light), then it is in a scattering state, where the energy can take on any positive value. So, the answer to your question is E > 0.

Question:the definition is: a structure above the cellular level. What is the word for this meaning?

Answers:the organization for complex organisms goes as follows: ecosystems, organisms, organs systems, organs, tissues, cells, molecules, and atoms. any of these, larger than a cell, would answer your question