Page:Radio-activity.djvu/375

 This fixes the value of the constants of one of the changes. Let us assume for the moment that this gives the value of λ_{1}.

Then     λ_{1} = 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 λ_{1} and T in the equation, the value of λ_{2} comes out to be

λ_{2} = 2·08 × 10^{-4} (sec)^{-1}.

This value of λ_{2} corresponds to a change in which half the matter is transformed in 55 minutes.

Substituting now the values of λ_{1}, λ_{2}, T, the equation reduces to

I_{t}/I_{T} = 1·37(e^{-λ_{2}t} - e^{-λ_{1}t}).

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| Observed value of |
 * | I_{t}/I_{T}   |  I_{t}/I_{T}  |
 * 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.

Further data are required in order to fix which of the time constants of the changes refers to the first change. In order to settle this point, it is necessary to isolate one of the products of the changes and to examine the variation of its activity with time. If, for example, a product can be separated whose activity decays to half value in 55 minutes, it would show that the second change is the more rapid of the two. Now Pegram has examined the radio-active products obtained by electrolysis of thorium solutions.