hydroxide is shown in Fig. 79. The initial drop in the curve is quite absent, and the curve, starting from the minimum, is practically identical with the curve shown in Fig. 48, which gives the recovery curve of thorium hydroxide after the first two days. This residual activity—about 25 per cent. of the maximum—is non-separable from the thorium by any chemical process that has been tried.
The initial rise of activity of Th X, after it has been separated, will now be considered. In all cases it was found that the activity of the separated Th X had increased about 15 per cent. at the end of 24 hours, and then steadily decayed, falling to half value in about four days.
This peculiarity of the Th X curve follows, of necessity, from the considerations already advanced to explain the drop in the recovery curve. As soon as the Th X is separated, it at once produces from itself the emanation, and this in turn produces thorium A and B. The activity due to B at first more than compensates for the decay of activity of the Th X itself. The total activity thus increases to a maximum, and then slowly decays to zero according to an exponential law with the time. The curve expressing the variation of the activity of the separated Th X with time can be deduced from the theory of successive changes already considered in chapter IX. In the present case there are four successive changes occurring at the same time, viz. the change of Th X into the emanation, of the emanation into thorium A, of A into B, and of B into an inactive product. Since, however, the change of the emanation into thorium A (about half changed in one minute) is far more rapid than the changes occurring in Th X or thorium A and B, for the purposes of calculation it may be assumed without serious error that the Th X changes at once into the active deposit. The 55 minute change will also be disregarded for the same reason.
Let λ_{1} and λ_{2} be the constants of decay of activity of Th X and of thorium A respectively. Since the activity of Th X and of thorium A falls to half value in 4 days and 11 hours respectively, the value of λ_{1} = ·0072 and of λ_{2} = ·063, where 1 hour is taken as the unit of time.
The problem reduces to the following: Given the matter A
