Townsend's first and second coefficients from data

In a uniform field with constant $E/p$, Townsend's theory gives the current

$$I = \frac{I_0\,e^{\alpha d}}{1 - \gamma\left(e^{\alpha d} - 1\right)}$$

For small gaps $\gamma(e^{\alpha d} - 1) \ll 1$ and $I \approx I_0 e^{\alpha d}$, so a plot of $\ln I$ against $d$ is a straight line of slope $\alpha$ (first coefficient). At larger gaps the current rises faster than this line; the upward deviation gives $\gamma$ (second coefficient).

Step 1: Tabulate $\ln I$

From 1.0 mm to 2.0 mm the current doubles every 0.5 mm, a clean exponential (straight-line) region. From 3.0 mm onwards the current rises far faster: this is the secondary-ionization region.

Step 2: First ionization coefficient $\alpha$

From the straight-line region (1.0 to 2.0 mm):

$$\begin{aligned} \alpha &= \frac{\ln I_{2.0} - \ln I_{1.0}}{d_{2.0} - d_{1.0}} = \frac{\ln(8\times10^{-13}/2\times10^{-13})}{0.2 - 0.1\ \text{cm}} \\ &= \frac{\ln 4}{0.1} = 13.86\ \text{cm}^{-1} \end{aligned}$$

Step 3: Second ionization coefficient $\gamma$

Taking $I_0 = 6.5\times10^{-14}$ A (the minimum current, as stated) and rearranging the current equation:

$$\gamma = \frac{1 - \dfrac{I_0\,e^{\alpha d}}{I}}{e^{\alpha d} - 1}$$

At $d = 3.0$ mm $= 0.3$ cm:

$$\begin{aligned} e^{\alpha d} &= e^{13.86\times0.3} = e^{4.159} = 64.0 \\ I_0 e^{\alpha d} &= 6.5\times10^{-14}\times64 = 4.16\times10^{-12}\ \text{A} \\ \gamma &= \frac{1 - 4.16\times10^{-12}/6.5\times10^{-12}}{64 - 1} = \frac{1 - 0.64}{63} \\ &= 5.71\times10^{-3} \end{aligned}$$

Repeating at the other points in the upturn region:

At 5.0 mm the gap is practically at breakdown ($\gamma e^{\alpha d} \approx 1$), so that point is not used. Average of 3.0–4.0 mm: $\gamma = (5.71 + 6.87 + 3.76)/3\times10^{-3} = 5.45\times10^{-3}$.

Answer: Townsend's first ionization coefficient $\alpha \approx$ 13.9 per cm; second ionization coefficient $\gamma \approx$ 5.5 × 10⁻³ (about $5.7\times10^{-3}$ from the first upturn point at 3 mm).

Check of breakdown: with $\gamma = 5.45\times10^{-3}$, the condition $\gamma(e^{\alpha d} - 1) = 1$ gives $e^{\alpha d} \approx 184$, i.e. $d \approx \ln 184/13.86 = 0.38$ cm. This matches the very steep rise of current beyond 3.5 mm. The measured currents at 4 and 5 mm are still finite, so the data do not fit a single $\gamma$ exactly; this scatter comes from the experimental data, and a graphical fit gives values of the same order.