Irodov 3.217: An Avalanche Grows Exponentially – Amit & Deepali Sinha Foundation

Problem Irodov 3.217

A gas fills the gap $d$ between the two plane plates of a capacitor. One plate emits $\nu_0$ electrons per second. Driven by the electric field, the electrons ionize the gas molecules, each electron producing $\alpha$ new electrons (and as many ions) per unit length of its path. Ionization by the ions themselves is negligible. Find the electron current arriving at the opposite plate.

The gas-filled capacitor, gap d: the bottom plate emits ν_0 electrons per second, and x is the distance from it. — The gas-filled capacitor, gap $d$ : the bottom plate emits $\nu_0$ electrons per second, and $x$ is the distance from it.

Given

gap $d$
$\nu_0$ electrons per second from one plate
$\alpha$ new electrons per unit path length
no ionization by ions
steady state, no recombination or capture of electrons (assumed)

Find

the electron current

I

at the opposite plate

Solution

Step 1The growth of the electron flow

In the steady state, let $\nu(x)$ electrons per second cross the plane at the distance $x$ from the emitting plate. All of them move toward the opposite plate, and in the layer between $x$ and $x + dx$ each one makes $\alpha\,dx$ new electrons, which join the flow (assumed: none recombine or are captured):

d\nu = \alpha\,\nu\,dx

Step 2Integrating across the gap

Separating the variables and integrating from the emitting plate, where the flow is $\nu_0$ :

\int_{\nu_0}^{\nu}\frac{d\nu'}{\nu'} = \alpha\int_0^x dx' \quad\Longrightarrow\quad \ln\frac{\nu}{\nu_0} = \alpha x

So the flow grows exponentially across the gap:

\nu(x) = \nu_0\,e^{\alpha x} \tag{1}

The electron flow across the gap, drawn for α d = 1.2: it starts at ν_0 on the emitting plate and grows exponentially to ν_0e^α d at the opposite plate. — The electron flow across the gap, drawn for $\alpha d = 1.2$ : it starts at $\nu_0$ on the emitting plate and grows exponentially to $\nu_0e^{\alpha d}$ at the opposite plate.

Step 3The current at the opposite plate

Each electron carries the charge $e$ , so the current at $x = d$ is $e$ times (1) there:

I = e\,\nu_0\,e^{\alpha d}

(2)

Check

Units: $\alpha$ is per unit length, so $\alpha d$ is a pure number and $e\nu_0$ is a current. Limit: with no ionization, $\alpha = 0$ , the current is $e\nu_0$ , just the emitted electrons; for a weak avalanche, $\alpha d \ll 1$ , it is $e\nu_0(1 + \alpha d)$ , the emitted electrons plus the $\alpha d$ each makes on one crossing.

Remark

Every electron started at the cathode arrives as a swarm of $e^{\alpha d}$ , so the gap is an amplifier whose gain is exponential in $\alpha d$ ; this is the principle of the proportional counter. The current still vanishes when the emission stops, because the ions, neglected here, create no new electrons: once they do, by striking the cathode or ionizing in turn, the discharge can sustain itself, and that is the onset of breakdown.