Irodov 3.212: Counting Ions Through Their Drift – Amit & Deepali Sinha Foundation

Problem Irodov 3.212

X-rays ionize the air between two parallel plates, each of area $S = 500$ cm $^2$ , a distance $d = 20$ mm apart. With a voltage $V = 100$ V across the plates, a current $I = 3.0$ $\mu$ A flows, far below the saturation current. The mobilities of the air ions are $u_0^+ = 1.37$ cm $^2$ /(V $\cdot$ s) for the positive ions and $u_0^- = 1.91$ cm $^2$ /(V $\cdot$ s) for the negative ones. Find the concentration of the positive ions.

Two parallel plates a distance d apart with ionized air between them. — Two parallel plates a distance $d$ apart with ionized air between them.

Given

$d = 20$ mm
$S = 500$ cm $^2$
$V = 100$ V
$I = 3.0$ $\mu$ A
$u_0^+ = 1.37$ cm $^2$ /(V $\cdot$ s)
$u_0^- = 1.91$ cm $^2$ /(V $\cdot$ s)
uniform field, space charge negligible (assumed)

Find

the positive ion concentration

n

Solution

Step 1Why the concentration is uniform

Far below saturation, almost every ion pair the X-rays make is lost by recombination in the air, and only a small fraction is carried to the plates. Production balances recombination locally, so the positive and negative ions have the same concentration $n$ throughout the gap. The field is taken uniform, its space charge negligible (assumed):

E = \frac{V}{d}

Step 2The current carried by both kinds of ion

The positive ions drift along the field with the speed $u_0^+E$ and the negative ions against it with $u_0^-E$ . Both motions carry current the same way, so their contributions add:

j = e n\,u_0^+E + e n\,u_0^-E = e n\,(u_0^+ + u_0^-)\,\frac{V}{d}

The current through the plate area is $I = jS$ :

I = \frac{e n\,(u_0^+ + u_0^-)\,V S}{d}

Step 3The concentration

Solving for $n$ , with lengths in cm throughout ( $d = 2.0$ cm, $u_0^+ + u_0^- = 3.28$ cm $^2$ /(V $\cdot$ s)) so that $n$ comes out per cm $^3$ :

n = \frac{3.0\times10^{-6}\times 2.0}{1.60\times10^{-19}\times 3.28\times 100\times 500}\ \text{cm}^{-3}

n = \frac{I d}{e\,(u_0^+ + u_0^-)\,V S} = 2.3\times10^{8}\ \text{cm}^{-3}

(1)

Check

Units: $\text{A}\cdot\text{m}$ over $\text{C}\cdot(\text{m}^2/(\text{V}\cdot\text{s}))\cdot\text{V}\cdot\text{m}^2$ is $\text{A}\cdot\text{s}/(\text{C}\cdot\text{m}^3) = \text{m}^{-3}$ , a concentration. Size: air at standard conditions holds about $2.7\times10^{19}$ molecules per cm $^3$ , so only about one molecule in $10^{11}$ is ionized. The gas is a very weak conductor made of rare ions drifting through neutral air, the picture the solution assumed, in which the ions barely disturb the field.

Remark

Below saturation the gap obeys Ohm’s law: (1) says $I \propto V$ at fixed $n$ , with conductivity $e n(u_0^+ + u_0^-)$ . This holds only while the drift removes few ions; raising $V$ sweeps them out before they recombine, the concentration falls, and the current levels off at the saturation value $e\dot n_i S d$ ( $\dot n_i$ the pairs produced per unit volume per second), which collects every pair produced.