(thorium X) all of one kind, which changes into B (thorium B), find the activity of A and B together at any subsequent time. This corresponds to Case I. (section 197). The amount Q of B at any time T is given by
Q = (λ_{1}n_{0}/(λ_{1} - λ_{2}))(e^{-λ_{2}t} - e^{-λ_{1}t}),
and the activity I at any time of the two together is proportional to λ_{1}P + Kλ_{2}Q, where K is the ratio of the ionization of B compared with that of A.
Then I_{t}/I_{0} = (λ_{1}P + Kλ_{2}Q)/(λ_{1}n_{0}) = e^{-λ_{1}t}[1 + (Kλ_{2}/(λ_{2} - λ_{1}))(1 - e^{-(λ_{2} - λ_{1})t})],
where I_{0} is the initial activity due to n_{0} particles of Th X.
By comparison of this equation with the curve of variation of the activity of Th X with time, shown in Fig. 47, it is found that K is almost ·44. It must be remembered that the activity of the emanation and Th X are included together, so that the activity of thorium B is about half of the activity of the two preceding products.
The calculated values of I_{t}/I_{0} for different values of t are shown in the second column of the following table, and the observed values in the third column.
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| Time |Theoretical|Observed|
| | value | value |
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| 0 | 1·00 | 1·00 |
| ·25 days| 1·09 | — |
| ·5 " | 1·16 | — |
| 1 " | 1·15 | 1·17 |
| 1·5 " | 1·11 | — |
| 2 " | 1·04 | — |
| 3 " | ·875 | ·88 |
| 4 " | ·75 | ·72 |
| 6 " | ·53 | ·53 |
| 9 " | ·315 | ·295 |
|13 " | ·157 | ·152 |
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