# Energy ionization of degenerated matter

Distance between electron and atomic nucleus in degeneracy matter is lower than non-degenerated matter. Electron energy binding with the nucleus must be higher. Whether temperature needed to degenerated matter become plasma is higher than temperature to ionization ordinary matter? For example, the white dwarf has a very high density a few tonnes per cubic centimeter. That means a few million tonnes per cubic meter. So the radius of atoms decreases about 100 times. Will the ionization energy and temperature increase 100 times? The Coulomb potential changes like 1/R. The ionization in this question is a transition of ordinary matter to plasma condition where the electrons and nucleus start moving freely between them.

I'm not familiar with condensed matter physics terminology, but what I could understand from your question is that degenerate matter becomes plasma at high temperatures, while non-degenerate matter becomes an ionized gas. So correct me if I'm wrong: in a Fermi-Dirac gas, where Pauli exclusion principle applies to electrons, there should be no way to tell whether matter is degenerate or not, the only thing being that there is a continuum of densities which increases degeneracy with density. In that case, both ionization and 'plasmification' of a Fermi gas should be described by the same equation. Indeed, I think that what you're looking for is Saha equation:

$$\frac{n_{i+1}n_e}{n_i} = \frac{2}{\lambda^3} \frac{g_{i+1}}{g_{i}} \exp \left[- \frac{\left(\epsilon_{i+1} - \epsilon_i\right)}{k_B T}\right]$$

Where, for a finite volume, $$n_e$$ is the total electron density (bound and unbound electrons), $$n_j$$ is the density of atoms with $$j$$ electrons removed, $$g_j$$ is the corresponding number of quantum states that accomodate this density (think of $$n_j$$ as specifying a macrostate, while $$g_j$$ is the number of microstates that correspond to the macrostate specified by $$n_j$$), $$\epsilon_{i+1} - \epsilon_i$$ is the binding energy of the $$i-$$th electron to the atom and $$\lambda = \sqrt{h^2/2 \pi m_e k_B T}$$ is the thermal de Broglie wavelength of the electron.

Binding energy seems to follow the relation that you guessed (i.e. if density increases as to decrease atoms' radii $$n-$$fold, then binding energy increases $$n-$$fold), but I'm not completely sure if nonlinearities in density-dependence of the partition function would propagate to energy levels (I don't even know how to obtain a partition function in this case). However, I feel that the ionization/'plasmification' temperature depends particularly on the initial and final states (as its clear from the Saha equation), as well as on density, such that it seems not to be the case that ionization temperature will increase $$100$$ times if radii shrink by this amount.

Indeed, if you suppose that all number densities get divided by the same factor $$f^3$$ (i.e. increase) as the consequence of multiplying the radius of a spherical white dwarf by a factor $$0 < f < 1$$ (i.e. decreasing the radius), the degeneracy equation gives:

$$\frac{n_e}{n^{\prime}_e} = \frac{\lambda^{\prime 3}}{\lambda^3} \Rightarrow \lambda^{\prime} = f \lambda \Rightarrow T^{\prime} = f^{-2} T$$

So, decreasing atoms' radii $$100$$ times makes temperature increase $$10000$$ times (assuming densities scale equally). Take a look at Alexander Fridman's "Plasma Chemistry", p. 94 for better clarity.

• Plasma is an ionized matter where electrons and the nucleus move freely between them. There are a few manners to ionization. I mean the ionization by the increased temperature to appropriate value. Whether temperature to matter becomes a plasma increase when the atomic radius decrease due to applied very high pressure. Mar 5, 2022 at 11:36