| d²X² | = 8πe(q − αn1n2)( | 1 | + | 1 | ) (7). |
| dx² | k1 | k2 |
an equation which is very useful, because it enables us, if we know the distribution of X², to find whether at any point in the gas the ionization is greater or less than the recombination of the ions. We see that q − αn1n2, which is the excess of ionization over recombination, is proportional to d²X²/dx². Thus when the ionization exceeds the recombination, i.e. when q − αn1n2 is positive, the curve for X² is convex to the axis of x, while when the recombination exceeds the ionization the curve for X² will be concave to the axis of x. Thus, for example, fig. 11 represents the curve for X² observed by Graham (Wied. Ann. 64, p. 49) in a tube through which a steady current is passing. Interpreting it by equation (7), we infer that ionization was much in excess of recombination at A and B, slightly so along C, while along D the recombination exceeded the ionization. Substituting in equation (7) the values of n1, n2 given in (3), (4), we get
| d²X² | 8πe[q - | α | (I + | k2 | dX² | )(I - | k2 | dX² | )]( | 1 | + | 1 | ) (8). | ||
| dx² | e²X²(k1 + k2)² | 8π | dx | 8π | dx | k1 | k2 |
| Fig. 11. |
This equation can be solved (see Thomson, Phil. Mag. xlvii. P. 253), when q is constant and k1 = k2. From the solution it appears that if X1 be the value of x close to one of the plates, and X0 the value midway between them,
| X1/X0 = | 1 |
| β2 − 2/β |
where β = 8πek1/α.
Since e = 4×10-10, α = 2×10-6, and k1 for air at atmospheric pressure = 450, β is about 2.3 for air at atmospheric pressure and it becomes much greater at lower pressures.
Thus X1/X0 is always greater than unity, and the value of the ratio increases from unity to infinity as β increases from zero to infinity. As β does not involve either q or I, the ratio of X1 to X0 is independent of the strength of the current and of the intensity of the ionization.