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

Below breakdown in a uniform field, the steady current grows by primary ionization as

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

where $\alpha$ is Townsend's primary ionization coefficient (ionizing collisions per cm per electron). It depends only on $E/p$, so with the field and pressure constant, $\alpha$ and $I_0$ are the same for both gap lengths.

Given

• $d_1 = 0.4$ cm, $I_1 = 5.5\times10^{-8}$ A (at 8 kV, so $E = 20$ kV/cm)
• $d_2 = 0.1$ mm $= 0.01$ cm, $I_2 = 5.5\times10^{-9}$ A

Solution

$$\begin{aligned} \frac{I_1}{I_2} &= e^{\alpha (d_1 - d_2)} \\ \frac{5.5\times10^{-8}}{5.5\times10^{-9}} &= 10 = e^{\alpha (0.4 - 0.01)} \\ \alpha &= \frac{\ln 10}{0.39} = \frac{2.3026}{0.39} \\ &= 5.904\ \text{cm}^{-1} \end{aligned}$$

Check: $e^{5.904\times0.39} = e^{2.3026} = 10$, as required.

Initial current: $I_0 = I_2\,e^{-\alpha d_2} = 5.5\times10^{-9}\times e^{-0.059} = 5.18\times10^{-9}$ A.

Answer: Townsend's primary ionization coefficient $\alpha \approx$ 5.90 per cm (5.904 cm⁻¹).

Note: if the second gap were 0.1 cm instead of 0.1 mm, the same method gives $\alpha = \ln 10/0.3 = 7.68$ cm⁻¹. Secondary ionization is neglected since the current ratio fits pure exponential growth.