Ionization, Recombination, and Attachment

An electrical discharge is produced when gas molecules or atoms are ionized by electron impact. In the absence of losses, the change in the electron density is given by:

$$\frac{dn_e}{dt}\Bigg\vert _i = \int_I^\infty Nv\sigma_i(\varepsilon)n_e(\varepsilon)\,d\varepsilon = \nu_{i}n_e\;,$$ (2.7)

where the subscript $i$ denotes that this is an electron density change due only to electron impact ionization, $\sigma_i$ is the ionization cross section by electrons with energy $\varepsilon$, $n_e(\varepsilon)$ is the free electron density as a function of energy, $I$ is the ionization potential energy, and $\nu_i$ is the ionization frequency (the average number of ionization events per electron per second). The ionization potentials for the two major constituents of air, N$_2$ and O$_2$, are 15.6 eV and 12.2 eV respectively.

Assuming that $\nu_{i}$ is constant, the solution to Equation 2.7 is

$$n_{e}(t)\bigg\vert _i = n_{o}e^{\nu_{i}t}\;,$$ (2.8)

where $n_{o}$ is the initial electron density. In the absence of losses, the electron density increases exponentially with time.

After an ionization event, a free electron can ``recombine'' with a positive ion. In the absence of other processes, the change in the electron density due to recombination is given by:

$$\frac{dn_e}{dt}\Bigg\vert _r = -{\beta}n_{e}n_{+}\;,$$ (2.9)

where $\beta$ is the electron-ion recombination coefficient, $n_{+}$ is the positive ion density, and the subscript $r$ indicates that this is an electron density change due only to recombination. If one assumes that the plasma is neutral with only single ionization, then $n_e\!=\!n_{+}$. Substituting into Equation 2.9 and solving for $n_{e}(t)$ gives:

$$n_{e}(t)\bigg\vert _r = \frac{n_o}{1+{\beta}n_{o}t}$$ (2.10)

In addition to recombining with positive ions, electrons can also attach to neutral molecules if the gas has electronegative properties. Air is an electronegative gas because of the electron affinity of $O_2$. Electrons can attach to oxygen either in triple collisions ( $e + O_2 + M \rightarrow O_{2}^{-} + M$) or via dissociative attachment reactions ( $e + O_2 + 3.6~eV \rightarrow O^{-} + O$). The latter form of attachment dominates at $E/N$ values near that of breakdown. The change in electron density due to attachment is:

$$\frac{dn_e}{dt}\Bigg\vert _a = -\nu_{a}n_e\;,$$ (2.11)

giving:

$$n_{e}(t)\bigg\vert _a = n_{o}e^{-\nu_{a}t}$$ (2.12)

where $\nu_a$ is the attachment frequency.

At $E/N$ values near breakdown, electron loss is due primarily to attachment processes and in particular to dissociative attachment (Bazelyan and Raizer, 1997, p. 24). At a given $E/N$ value, both the ionization and attachment frequencies will be proportional to the air number density, $N$, since the collision rate is proportional to $N$ (see Equation 2.3). Figure 2.1 shows the density-normalized ionization frequency ($\nu_i/N$) and attachment frequency ($\nu_a/N$) plotted as a function of $E/N$ in units of Townsends ($1$ Td $\,=\!10^{-21}~$V$\,$m$^2$). For any given N, the ionization frequency will exceed that of attachment when $E/N \gtrsim 123$ Td. This is the basis for the breakdown strength of air at standard temperature and pressure (STP). An $E/N$ of 123 Td corresponds to a breakdown field of $\simeq\!3.3$ MV/m at STP.

Figure 2.1: The ionization frequency ($\nu_i$) and attachment frequency ($\nu_a$) for air, divided by the air number density ($N$), are plotted against the electric field ($E$) divided by $N$. Electron avalanches can develop when $\nu_i\!>\!\nu_a$, which occurs for $E/N \gtrsim 123$ Td. (From Bazelyan and Raizer (1997)).