Irodov 3.214: Saturation Counts, Recombination Balances

Problem Irodov 3.214

Ultraviolet radiation ionizes the air between two closely spaced plates uniformly. The volume of air between the plates is $V = 500$ cm $^3$ , and the saturation current is $I_{sat} = 0.48$ $\mu$ A. Find:

the number of ion pairs produced per unit volume per unit time;
the equilibrium concentration of ion pairs, given the recombination coefficient of air ions $r = 1.67\times10^{-6}$ cm $^3$ /s.

Two closely spaced plates with air between them.

Given

$V = 500$ cm $^3$
$I_{sat} = 0.48$ $\mu$ A
$r = 1.67\times10^{-6}$ cm $^3$ /s
in (b), ions lost by recombination only (assumed)

Find

(a)

\dot n_i

; (b) the equilibrium concentration

n

Solution

Step 1(a) Every pair counted at saturation

At saturation the field sweeps every ion to a plate before it can recombine. Each pair delivers the charge $e$ round the circuit: its positive ion to one plate and its negative ion to the other. With $\dot n_i$ pairs made per unit volume per second in the volume $V$ :

I_{sat} = e\,\dot n_i V

Solving for the production rate, with $V$ in cm $^3$ to give it per cm $^3$ :

\dot n_i = \frac{0.48\times10^{-6}}{1.60\times10^{-19}\times 500}\ \text{cm}^{-3}\,\text{s}^{-1}

\dot n_i = \frac{I_{sat}}{eV} = 6.0\times10^{9}\ \text{cm}^{-3}\,\text{s}^{-1}

(1)

Step 2(b) Production against recombination

With no current drawn off (assumed), ions disappear only by recombining. A positive ion meets negative ones at a rate proportional to their concentration, so recombination removes $r n^+n^-$ pairs per unit volume per second, and the two concentrations are equal, both $n$ . The rate of change of the concentration is

\frac{dn}{dt} = \dot n_i - r n^2 \tag{2}

In equilibrium the right side vanishes, production and recombination balance:

\dot n_i = r n^2

Solving for $n$ and substituting the rate from (1):

n = \sqrt{\frac{6.0\times10^{9}}{1.67\times10^{-6}}}\ \text{cm}^{-3}

n = \sqrt{\dot n_i/r} = 6.0\times10^{7}\ \text{cm}^{-3}

(3)

Check

Units: (a) A over $\text{C}\cdot\text{cm}^3$ is pairs per cm $^3$ per second; (b) $\text{cm}^{-3}\text{s}^{-1}$ over $\text{cm}^3\,\text{s}^{-1}$ is $\text{cm}^{-6}$ , whose root is a concentration. Stability: by (2) a concentration above the equilibrium value recombines faster than it is made and one below it grows, so the system settles there.

Remark

The equilibrium concentration grows only as the square root of the production rate, because recombination is a two-body process: doubling the radiation raises the ion density by just $\sqrt2$ . A current drawn from the gap competes with recombination for the same ions, which is why the current saturates at $e\dot n_i V$ however strong the field: beyond that point there are no more ions to collect.