What are the distributions of electron speeds in a star and in a planet?

Ideally I would like to have an $x$-$y$ graph of ($x$) speed relative to centre of mass of the body (star or planet) against ($y$) the number or percentage of electrons having that speed at a given moment in time. For simplicity I can ignore the small proportion of electrons in the atmosphere. Also please ignore effects on electron speed caused either by speed of the body rotating around its axis or by orbital revolution. I do not need high precision just a rough picture.

UPDATE 20160707 I have accepted CR Drost's answer as it is very helpful. But I am not confident that I have got the necessary understanding to answer my question.

A useful related question with a respected answer is here: how-fast-do-electrons-travel-in-an-atomic-orbital.

Originally I inquired about electron speeds in the Sun and the Earth but following input from Chris Drost's answer this seems too ambitious (for my high-school level physics ability). So I will focus on a simpler imaginary model comprising a white dwarf star and a planet consisting of just a metallic iron core.

Relevant Information

According to wikipedia (Quantum and classical regimes) the classical Maxwell-Boltzmann formula for particle speeds ($v$)

$$f(v) = \left( \frac{m}{2\pi kT} \right)^{3/2} 4\pi v^2 \mathrm{e}^{-\frac{mv^2}{2kT} }$$

applies if the concentration of particles corresponds to an average interparticle separation $\bar{R}$ that is much greater than the average de Broglie wavelength $\bar{\lambda}$ of the particles. Here $\bar{\lambda} = \frac{h}{\sqrt{3mkT}}$ with $h$ the Plank constant ($h=6.62606957\times 10^{-34} \,\mathrm{J\cdot s}$) and $m$ the particle mass ($m_e = 9.11 \times 10^{-31} \,\mathrm{kg}$ for an electron).

But when $\bar{R}$ is not "much greater" than $\bar{\lambda}$ the Fermi-Dirac equation for energy (applicable to Fermions e.g. electrons, protons, neutrons) applies: $$f(E) = (e^{(E-E_f)/kT}+1)^{-1}.$$ where $k$ is the Boltzmann constant ($k=1.3806488 \times 10^{-23} \,\mathrm{J/K}$), $T$ is temperature and $E_f$ is the Fermi Energy wikipedia. The section "Typical Fermi energies" gives example of calculations of the Fermi energy for (i) conduction electrons in metals ($\sim 10\,\mathrm{eV}$ for a density of $10^{28}$ to $10^{29}$ electrons$/m^3$), (ii) for degenerate electrons in white dwarfs (~ $3 \times 10^5 \,\mathrm{eV}$) and (iii) for nucleons in an atom.

To obtain the distribution of electron speeds I would simply apply the relation: $E_i = \frac{1}{2} m V_i^2$ to the electron energies.

White Dwarf

The temperature of a white dwarf is (ignoring its thin outer shell) fairly uniform at about $10^7 \,\mathrm{K}$. The Fermi energy for degenerate electrons in white dwarfs is $\sim 3 \times 10^5 \,\mathrm{eV}$. The Fermi-Dirac formula becomes (using MKS units): $$f(E) = (e^{(E-(3 \times 10^5 \,\mathrm{eV}))/(1.3806488 \times 10^{-23}\, \mathrm{J/K}\cdot 10^7 \,\mathrm{K})}+1)^{-1}.$$ $$f(V) = (e^{((\frac{1}{2}m V^2)-(3 \times 10^5\,\mathrm{eV} \cdot 1.6 \times 10^{-19}\,\mathrm{J/eV}))/(1.3806488 \times 10^{-23}\,\mathrm{J/K}\cdot 10^7\, \mathrm{K})}+1)^{-1}.$$

Initial calculations indicate that $F(V) =1$ for $V = 0$ to $320,000 \,\mathrm{km/s}$ and $F(V) = 0$ for $V \geq 330,000 \, \mathrm{km/s}$ using electron rest mass.

Using the relativistic electron mass instead ($m' = 1/\sqrt{(1-v^2/c^2)}$) indicates that $F(V) =1$ for $V = 0$ to $244,000\, \mathrm{km/s}$ and $F(V) = 0$ for $V = \geq 248,000\, \mathrm{km/s}$.

Presumably because the electrons are degenerate their speed does not depend on the speed of local nucleons. Is this true?

Metal Planet

The surface temperature of Mercury is between $100$ and $700\,\mathrm{K}$. The temperature of Earth's inner core is about $6000\,\mathrm{K}$.

Having assumed suitable core and outer temperatures for our imaginary metal planet, next we can sub-divide the planet into a nested sequence of spherical shells of some arbitrary thickness and apply a temperature gradient radially across the shells and calculate a speed distribution for the electrons in each shell based on shell temperature. The mass density of each shell would be required to calculate the number of electrons in each shell. And the elemental composition can be used to determine the proportion of conduction to bound electrons in each atom.

In the metal there are two sets of electrons to consider (a) conduction electrons and (b) electrons bound to the nucleus.

Can "non-thermal" motions of electrons in their orbitals be ignored?

The speeds of (a) conduction electrons can be obtained using Fermi-Dirac. The Fermi energy of $E_f \approx 10\,\mathrm{eV}$ quoted earlier applies to metals in lab conditions, so a different figure may be required according to pressure (function of depth of burial below planet surface).

For (b) bound electrons in the metal. I am unclear whether I need to give different treatments according to position of electron in the atom. I do not know how to obtain the Fermi-energies of these different electrons. Or can I just use Maxwell-Bolzman for these electrons?

Presumably I need to determine the speed distribution for the atomic nucleus at each depth and then convolve (smear) the nucleus velocity distribution onto the raw electron distribution in each shell. This presumably is required for bound electrons but not for conduction electrons. Am I right?

• Honestly, I'd use Fermi-Dirac whenever I can when talking about electrons that are only thermally excited. The difference in the statistics comes from the identicalness of the particles, not necessarily their density. However, it is totally 100% true that at low densities (and the high $T$ of the sun may effectively give it a "low" density), the Fermi-Dirac statistics will mirror Maxwell-Boltzmann statistics. – CR Drost Jul 10 '15 at 14:50