Why do we set the number of degrees of freedom of an electron in a gas to 2 rather than 3? I was reading up on basic properties of plasmas and came across this description in the Wikipedia on how to convert between eV and Kelvin units to describe the amount of energy of the electrons:

The SI unit of temperature is the kelvin (K), but using the above relation the electron temperature is often expressed in terms of the energy unit electronvolt (eV). Each kelvin (1 K) corresponds to 8.617 333 262...×10−5 eV; this factor is the ratio of the Boltzmann constant to the elementary charge.[5] Each eV is equivalent to 11,605 kelvins, which can be calculated by the relation $⟨ E ⟩ = k_{\text{B}} T$.

This last equation has me confused. Why do we not write instead
$$⟨ E ⟩ = \frac32 k_{\text{B}} T$$
after all an electron has three translational degrees of freedom? What am I missing?
 A: You are mixing two things:

*

*What is the average kinetic energy of the system in $n$-dimensions? Thermodynamics states  that
$$\bar E = \frac{f}{2}k_B T$$
where $f$ is the number of degree of freedom.

*How do we convert between the energy unit "Joule" and the temperature unit "Kelvin"? Since $k_B T$ has the dimension of an energy, we use
$T = E/k_B$ to express an energy in terms of Kelvin, see here.

Using this convention, we do not have to think about the number of degrees of freedom. Furthermore, we are even allowed to associate a temperature with a system, which does not have a temperature. E.g. consider a single electron, which obviously has no temperature. It's perfectly fine to state that the electron has a kinetic energy of  $1K \approx 1.38e-23 J$. This is analog to expressing an energy in terms of $eV$ -- even if the particle is not electrically charged.
A: 
Why do we set the number of degrees of freedom of an electron in a gas to 2 rather than 3?

In a magnetized plasma, most electrons are said to be gyrotropic -- their velocities can be decomposed into parallel and perpendicular (with respect to the background, quasi-static magnetic field), where the latter is assumed to be azimuthally symmetric.  Thus, there is really only two degrees of translational freedom.

Temperature as an energy instead of the thermodynamic quantity we all know and love.

Plasmas are almost never in equilibrium so temperature will not have the same meaning as a thermodynamic system.  Regardless of this, it's often the case that when discussing temperature folks will keep $k_{B}$ (= 1.38064852 x 10-23 J K-1) and $T$ together (or even absorb $k_{B}$ into $T$) so that what is called temperature is really an energy.
Generally one finds the second velocity moment of the velocity distribution function (e.g., see https://physics.stackexchange.com/a/218643/59023) of the particle species of interest, multiplied by some constants, to get the pressure tensor.  Then one takes one-third the trace divided by the number density to get what most plasma physicists will call the temperature of the plasma.

Conversion between electron volts and kelvin.

We know that 1 eV = 1.6021766208 x 10-19 J and we know that J and K are just a constant apart.  So it's not difficult to show that 1 eV ~ 11,604.5 K.
