Thomas–Fermi Model - Kinetic Energy

Kinetic Energy

For a small volume element ΔV, and for the atom in its ground state, we can fill out a spherical momentum space volume Vf up to the Fermi momentum pf, and thus,

where is a point in ΔV.

The corresponding phase space volume is

The electrons in ΔVph are distributed uniformly with two electrons per h3 of this phase space volume, where h is Planck's constant. Then the number of electrons in ΔVph is

The number of electrons in ΔV is

where is the electron density.

Equating the number of electrons in ΔV to that in ΔVph gives,

The fraction of electrons at that have momentum between p and p+dp is,

\begin{align} F_\vec{r} (p) dp & = \frac{4 \pi p^2 dp} {\frac{4}{3} \pi p_f^3(\vec{r})} \qquad \qquad p \le p_f(\vec{r}) \\ & = 0 \qquad \qquad \qquad \quad \text{otherwise} \\ \end{align}

Using the classical expression for the kinetic energy of an electron with mass me, the kinetic energy per unit volume at for the electrons of the atom is,

\begin{align} t(\vec{r}) & = \int \frac{p^2}{2m_e} \ n(\vec{r}) \ F_\vec{r} (p) \ dp \\ & = n(\vec{r}) \int_{0}^{p_f(\vec{r})} \frac{p^2}{2m_e} \ \ \frac{4 \pi p^2 } {\frac{4}{3} \pi p_f^3(\vec{r})} \ dp \\ & = C_F \ ^{5/3} \end{align}

where a previous expression relating to has been used and,

Integrating the kinetic energy per unit volume over all space, results in the total kinetic energy of the electrons,

This result shows that the total kinetic energy of the electrons can be expressed in terms of only the spatially varying electron density according to the Thomas–Fermi model. As such, they were able to calculate the energy of an atom using this expression for the kinetic energy combined with the classical expressions for the nuclear-electron and electron-electron interactions (which can both also be represented in terms of the electron density).

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