Then λ1 = 1·75 × 10-5 (sec)-1.
Since the maximum activity is reached after an interval T = 220 minutes (see [Fig. 65]), substituting the values of λ1 and T in the equation, the value of λ2 comes out to be
λ2 = 2·08 × 10-4 (sec)-1.
This value of λ2 corresponds to a change in which half the matter is transformed in 55 minutes.
Substituting now the values of λ1, λ2, T, the equation reduces to
The agreement between the results of the theoretical equation and the observed values is shown in the following table:
| Time in minutes | Theoretical value of It/IT | Observed value of It/IT |
|---|---|---|
| 15 | ·22 | ·23 |
| 30 | ·38 | ·37 |
| 60 | ·64 | ·63 |
| 120 | ·90 | ·91 |
| 220 | 1·00 | 1·00 |
| 305 | ·97 | ·96 |
After 5 hours the activity decreased nearly exponentially with the time, falling to half value in 11 hours.
It is thus seen that the curve of rise of activity for a short exposure is explained very satisfactorily on the supposition that two changes occur in the deposited matter, of which the first is a rayless change.