Irodov 3.213: Ions That Must Cross in Half a Period – Amit & Deepali Sinha Foundation

Problem Irodov 3.213

Two plane parallel electrodes, $1$ and $2$ , are a distance $l$ apart. The gas is ionized only in a thin layer next to the surface of electrode $1$ . An alternating voltage $V = V_0\sin\omega t$ is applied between the electrodes, and a galvanometer $G$ is connected in series on the side of electrode $2$ . As the frequency $\omega$ is lowered, the galvanometer shows a current only once $\omega < \omega_0$ , where $\omega_0$ is a cut-off frequency. Find the mobility of the ions that reach electrode $2$ .

Electrodes 1 and 2 edge on, the thin ionized layer (dashed) on the inner face of electrode 1, the galvanometer G on the electrode 2 side, and the alternating voltage V applied at the terminals. — Electrodes $1$ and $2$ edge on, the thin ionized layer (dashed) on the inner face of electrode $1$ , the galvanometer $G$ on the electrode $2$ side, and the alternating voltage $V$ applied at the terminals.

Given

gap $l$
$V = V_0\sin\omega t$
current only for $\omega < \omega_0$
uniform field, space charge negligible (assumed)

Find

the ion mobility

u_0

Solution

Step 1How far an ion gets in one half period

The field between the plane electrodes is taken uniform (assumed), so it follows the voltage:

E(t) = \frac{V_0\sin\omega t}{l}

An ion of mobility $u_0$ drifts with the speed $u_0E$ . Ions of one sign leave the layer whenever the field drives them toward electrode $2$ ; take the half period $0 < t < \pi/\omega$ for them. An ion leaving the layer at $t_0$ has covered, by the time $t$ of that half period, the distance

x(t) = \int_{t_0}^{t}\frac{u_0V_0}{l}\,\sin\omega t'\,dt' = \frac{u_0V_0}{\omega l}\,\left(\cos\omega t_0 - \cos\omega t\right)

Step 2The ion that goes furthest

The ion moves forward until $t = \pi/\omega$ , where the voltage reverses and drives it back toward electrode $1$ . Its furthest point is therefore

x_{\mathrm{max}} = x\!\left(\frac{\pi}{\omega}\right) = \frac{u_0V_0}{\omega l}\,(1 + \cos\omega t_0)

The largest of these belongs to the ion that sets out at the start of the half period, $t_0 = 0$ :

x_{\mathrm{max}} = \frac{2u_0V_0}{\omega l} \tag{1}

Step 3The cut-off and the mobility

The path of the ion that leaves the layer at t = 0. At the cut-off frequency (solid) it reaches electrode 2, at x = l, exactly at t = π/ω_0. Dashed: at ω = 1.5 ω_0 the voltage reverses first and the ion turns back. — The path of the ion that leaves the layer at $t = 0$ . At the cut-off frequency (solid) it reaches electrode $2$ , at $x = l$ , exactly at $t = \pi/\omega_0$ . Dashed: at $\omega = 1.5\,\omega_0$ the voltage reverses first and the ion turns back.

A current flows only if some ions reach electrode $2$ , that is if (1) is at least $l$ :

\frac{2u_0V_0}{\omega l} \geq l \quad\Longleftrightarrow\quad \omega \leq \frac{2u_0V_0}{l^2}

The cut-off frequency is the bound itself, $\omega_0 = 2u_0V_0/l^2$ . Solving for the mobility:

u_0 = \frac{\omega_0 l^2}{2V_0}

(2)

Check

Units: $\text{s}^{-1}\,\text{m}^2/\text{V}$ is $\text{m}^2/(\text{V}\cdot\text{s})$ , a mobility. Trend: a more mobile ion crosses faster, so it keeps getting through at higher frequencies, $\omega_0 \propto u_0$ ; a wider gap needs a slower voltage, $\omega_0 \propto 1/l^2$ , one factor of $l$ for the longer path and one for the weaker field.

Remark

This is a time of flight measurement with no clock: the voltage itself times the ions. With different ions in the gas, each kind has its own cut-off by (2), and the current first appears, as $\omega$ falls, for the most mobile ones, so the highest cut-off picks out the fastest ion present.