Townsend's primary ionization coefficient (0.4 cm to 0.1 cm)

In a uniform field below breakdown, the steady current grows with gap length by Townsend's primary ionization only:

$$I = I_0\, e^{\alpha d}$$

where $\alpha$ is Townsend's first (primary) ionization coefficient, i.e. the number of ionizing collisions made by one electron per cm of travel in the field direction. $\alpha$ depends only on $E/p$; here the field is kept constant ($E = 8/0.4 = 20$ kV/cm), so $\alpha$ is the same for both gaps and $I_0$ is the same.

Given

• $d_1 = 0.4$ cm, $I_1 = 5.5\times10^{-8}$ A
• $d_2 = 0.1$ cm, $I_2 = 5.5\times10^{-9}$ A

Solution

Divide the two current equations so that $I_0$ cancels:

$$\begin{aligned} \frac{I_1}{I_2} &= \frac{I_0 e^{\alpha d_1}}{I_0 e^{\alpha d_2}} = e^{\alpha (d_1 - d_2)} \\ \frac{5.5\times10^{-8}}{5.5\times10^{-9}} &= 10 = e^{\alpha (0.4 - 0.1)} \\ \alpha &= \frac{\ln 10}{0.3} = \frac{2.3026}{0.3} \\ &= 7.675\ \text{cm}^{-1} \end{aligned}$$

Check: $e^{7.675\times0.3} = e^{2.3026} = 10$.

The initial (cathode) current is then $I_0 = I_2 e^{-\alpha d_2} = 5.5\times10^{-9}\times e^{-0.7675} = 2.55\times10^{-9}$ A.

Answer: Townsend's primary ionization coefficient $\alpha \approx$ 7.68 ionizing collisions per cm (7.675 cm⁻¹), at $E = 20$ kV/cm.

Note: the gap is short and the current ratio is exactly exponential, so secondary ionization ($\gamma$) is neglected; it would show up only as an upward deviation from the straight line of $\ln I$ against $d$ at larger gaps.