Irodov 3.215: Ions Fading by Pairs – Amit & Deepali Sinha Foundation

Problem Irodov 3.215

An ionizer that makes $\dot n_i = 3.5\times10^{9}$ ion pairs per cm $^3$ of air per second has run long enough for the ion concentration to settle, and is then switched off. The ions are lost only by recombination, with coefficient $r = 1.67\times10^{-6}$ cm $^3$ /s. How long after the switch-off does the ion concentration fall by a factor $\eta = 2.0$ ?

Given

$\dot n_i = 3.5\times10^{9}$ cm $^{-3}$ s $^{-1}$
$r = 1.67\times10^{-6}$ cm $^3$ /s
$\eta = 2.0$

Find

the time

t

for the concentration to fall

\eta

times

Solution

Step 1The starting concentration

Having run long enough, the ionizer has brought the air to equilibrium, where the pairs it makes are lost by recombination at the same rate, $r n_0^2$ per unit volume per second:

\dot n_i = r n_0^2

So the concentration at the moment of switching off is

n_0 = \sqrt{\frac{\dot n_i}{r}} \tag{1}

Step 2The decay by recombination

After the switch-off nothing is produced. Positive and negative ions have the same concentration $n$ and recombine in pairs, so

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

Separating the variables and integrating from the switch-off, where the concentration is $n_0$ :

\int_{n_0}^{n}\frac{dn'}{n'^2} = -r\int_0^t dt' \quad\Longrightarrow\quad \frac{1}{n} - \frac{1}{n_0} = r t

Solving for the concentration:

n(t) = \frac{n_0}{1 + r n_0 t} \tag{2}

The ion concentration after the switch-off (solid): it falls to n_0/2 at t = τ, with τ = 1/(r n_0). Dashed: an exponential with the same initial slope, which the recombination decay lags behind. — The ion concentration after the switch-off (solid): it falls to $n_0/2$ at $t = \tau$ , with $\tau = 1/(r n_0)$ . Dashed: an exponential with the same initial slope, which the recombination decay lags behind.

Step 3The time to fall

\eta

times

Setting $n = n_0/\eta$ in (2):

1 + r n_0 t = \eta \quad\Longrightarrow\quad t = \frac{\eta - 1}{r n_0}

Substituting $n_0$ from (1) into the product $r n_0$ :

r n_0 = r\sqrt{\frac{\dot n_i}{r}} = \sqrt{r\dot n_i}

Numerically, with $\eta = 2.0$ :

t = \frac{2.0 - 1}{\sqrt{1.67\times10^{-6}\times 3.5\times10^{9}}}\ \text{s}

t = \frac{\eta - 1}{\sqrt{r\dot n_i}} = 13\ \text{ms}

(3)

Check

Units: $\text{cm}^3\text{s}^{-1}\cdot\text{cm}^{-3}\text{s}^{-1}$ is $\text{s}^{-2}$ , whose root is a rate, so $t$ is a time. Limits: $\eta = 1$ gives $t = 0$ , as it must; and the time grows only linearly with $\eta$ , so a hundredfold fall takes $99/\sqrt{r\dot n_i}$ , about $1.3$ s, about a hundred times the halving time, where an exponential decay would need fewer than seven half lives.

Remark

Recombination is a two-body loss, so the rate per ion, $r n$ , drops as the ions thin out: each halving takes twice as long as the one before. There is no fixed half life, unlike a radioactive decay or an $RC$ discharge, and the slow tail in the graph is why a gas stays weakly conducting long after its ionizer stops.