Townsend ionisation coefficient
How fast does ionisation multiply along the field at this pressure and field strength?
- α
- Ionisation coefficient, /cm
- A
- Gas constant A
- B
- Gas constant B
- p
- Pressure, Torr
- E
- Electric field, V/cm
LaTeX
α = A \cdot p \cdot \exp\left(-\frac{B \cdot p}{E}\right)
Result
α
Ionisation coefficient
0.007405 /cm
Ion pairs generated per centimetre of path an electron travels along the field.
LaTeX
0.007405\,\mathrm{/cm} = 12 \cdot 5\,\mathrm{Torr} \cdot \exp\left(-\frac{180 \cdot 5\,\mathrm{Torr}}{100\,\mathrm{V/cm}}\right)
This result is a link — the address bar holds your numbers, so it can be pasted into a post and opened to the same answer.
What this looks like
The formula behind the curve
- α
- Ionisation coefficient, /cm
- A
- Gas constant A
- B
- Gas constant B
- p
- Pressure, Torr
- E
- Electric field, V/cm
LaTeX
α = A \cdot p \cdot \exp\left(-\frac{B \cdot p}{E}\right)
The formula behind the curve
- α
- Ionisation coefficient, /cm
- A
- Gas constant A
- B
- Gas constant B
- p
- Pressure, Torr
- E
- Electric field, V/cm
LaTeX
α = A \cdot p \cdot \exp\left(-\frac{B \cdot p}{E}\right)
Method
- Look up the gas's empirical constants A and B — fitted to measured ionisation rates for that gas, in torr and volts per centimetre.
- Multiply A by the pressure.
- Divide B times the pressure by the field, and take the negative exponential of that ratio. This term collapses toward zero when the field is weak relative to the pressure, which is why a low field at high pressure ionises almost nothing.
- Multiply the two together to get α — ion pairs per centimetre of travel.
Assumptions
- A and B are constants over the pressure and field range entered. The real coefficients drift outside the range they were fitted over — roughly tens to a few hundred V/(cm·torr) for these three gases — and this does not know where that range ends.
- The gas is pure. A mixture, or contamination from the electrode material or the tube surface, shifts the effective A and B away from the pure-gas figures used here.
- The field is uniform across the gap, matching the electric field calculation this feeds from.