/caption Helium is the second lightest element in the known universe. It is also the second most abundant. According to some estimates helium accounts for as much as 24 percent of the Universe. This book is divided into three chapters. The first chapter discusses the experimental results and compares with the equations used to generate the tables. These equations are supplemented by a vapor pressure equation, which represents the 1958 He-4 scale of temperature that is defined in terms of the vapor pressure of helium-4.
Next:Hydrogen Molecule Ion Up:Variational Methods Previous:Variational PrincipleA helium atom consists of a nucleus of charge surroundedby two electrons. Let us attempt to calculate its ground-state energy.
Let the nucleus lie at the origin of our coordinatesystem, and let the position vectors of the two electrons be and , respectively. The Hamiltonian of the system thustakes the form
Gas Constant Of Helium
where we have neglected any reduced mass effects.The terms in the above expression represent the kinetic energy of the firstelectron, the kinetic energy of the second electron, the electrostaticattraction between the nucleus and the first electron, the electrostaticattraction between the nucleus and the second electron, and theelectrostatic repulsion between the two electrons, respectively.It is the final term which causes all of the difficulties. Indeed, if thisterm is neglected then we can write| (1181) |
where
In other words, the Hamiltonian just becomes the sum of separate Hamiltonians for each electron. In this case, we would expect thewavefunction to be separable: i.e.,
| (1183) |
Hence, Schrödinger's equation
reduces to
| (1185) |
where
Of course, Eq. (1185) is the Schrödinger equation of a hydrogen atom whosenuclear charge is , instead of . It follows, from Sect. 9.4 (making the substitution ), that if both electrons are in their lowest energystates then
| (1187) |
| (1188) |
where
Here, is the Bohr radius [see Eq. (679)]. Note that is properly normalized. Furthermore,
| (1190) |
where is the hydrogen ground-stateenergy [see Eq. (678)]. Thus, our crude estimatefor the ground-state energy of helium becomes
Unfortunately, this estimate is significantly different from the experimentallydetermined value, which is . This factdemonstrates that the neglected electron-electron repulsion term makes alarge contribution to the helium ground-state energy.Fortunately, however, we can use the variational principle to estimate this contribution.
Let us employ the separable wavefunction discussed above as our trialsolution. Thus,
| (1192) |
R Constant For Helium
The expectation value of the Hamiltonian (1180) thus becomeswhere
| (1194) |
The variation principle only guarantees that (1193) yields anupper bound on the ground-state energy. In reality, we hopethat it will give a reasonably accurate estimate of this energy.
It follows from Eqs. (678), (1192) and (1194) that
| (1196) |
where is the angle subtended between vectors and .If we perform the integral in space before that in space then
where
| (1198) |
Our first task is to evaluate the function . Let be a set of spherical polar coordinates in space whose axis of symmetry runs in the direction of . It followsthat . Hence,
| (1200) |
Making the substitution , we can see that
Now,
| (1202) |
giving
But,
| (1204) |
| (1205) |
yielding
Since the function only depends on the magnitude of ,the integral (1197) reduces to
| (1207) |
which yields
Hence, from (1193), our estimate for the ground-stateenergy of helium is
| (1209) |
This is remarkably close to the correct result.
We can actually refine our estimate further. The trial wavefunction (1192) essentially treats the two electrons as non-interacting particles. Inreality, we would expect one electron to partially shield the nuclearcharge from the other, and vice versa. Hence, a bettertrial wavefunction might be
We can rewrite the expression (1180) for the Hamiltonianof the helium atom in the form
| (1211) |
where
is the Hamiltonian of a hydrogen atom with nuclear charge ,
| (1213) |
is the electron-electron repulsion term, and
It follows that
| (1215) |
where is the ground-state energy of a hydrogenatom with nuclear charge , is the value of the electron-electron repulsion term whenrecalculated with the wavefunction (1210) [actually, all weneed to do is to make the substitution ], and
Here, is the expectation value of calculatedfor a hydrogen atom with nuclear charge . It follows fromEq. (695) [with , and making the substitution ] that
| (1217) |
Hence,
since . Collecting the various terms, our new expression for the expectationvalue of the Hamiltonian becomes| (1219) |
The value of which minimizes this expression is the root of
It follows that
| (1221) |
The fact that confirms our earlier conjecture that the electrons partiallyshield the nuclear charge from one another. Our new estimatefor the ground-state energy of helium is
This is clearly an improvement on our previous estimate (1209) [recall that thecorrect result is eV].
Obviously, we could get even closer to the correct value of thehelium ground-state energy by using amore complicated trial wavefunction with more adjustable parameters.
Note, finally, that since the two electrons in a helium atom are indistinguishable fermions, the overall wavefunction must be anti-symmetric with respect to exchange of particles (see Sect. 6).Now, the overall wavefunction is the product of the spatial wavefunctionand the spinor representing the spin-state. Our spatial wavefunction (1210) is obviously symmetric with respect to exchange ofparticles. This means that the spinor must be anti-symmetric. It is clear, from Sect. 11.4, that if the spin-state ofan system consisting of two spin one-half particles (i.e., two electrons)is anti-symmetric with respect to interchange of particles then the system isin the so-called singlet state with overall spin zero. Hence,the ground-state of helium has overall electron spin zero.
Next:Hydrogen Molecule Ion Up:Variational Methods Previous:Variational PrincipleRichard Fitzpatrick2010-07-20