Page:Radio-activity.djvu/406

 long exposure, measured by the β rays. The equation expressing the decay of activity, measured by the α rays, differs considerably from this, especially in the early part of the curve. Several hours after removal the activity decays according to an exponential law with the time, decreasing to half value in 28 minutes. This fixes the value of λ_{3}. The constant α and the value of λ_{2} are deduced from the experimental curve by trial. Now we have already shown (section 207) that in the case of the active deposit from thorium, where there are two changes of constants λ_{2} and λ_{3}, in which only the second change gives rise to a radiation, the intensity of the radiation is given by

I_{t}/I_{0} = (λ_{2}/(λ_{2} - λ_{3}))e^{-λ_{3}t} - (λ_{3}/(λ_{2} - λ_{3}))e^{-λ_{2}t}

for a long time of exposure (see equation 8, section 198). This is an equation of the same form as that found experimentally by Curie and Danne. On substituting the values λ_{2}, λ_{3} found by them,

λ_{2}/(λ_{2} - λ_{3}) = 4·3, and λ_{1}/(λ_{1} - λ_{3}) = 3·3.

Thus the theoretical equation agrees in form with that deduced from observation, and the values of the numerical constants are also closely concordant. If the first as well as the second change gave rise to a radiation, the equation would be of the same general form, but the value of the numerical constants would be different, the values depending upon the ratio of the ionization in the first and second changes. If, for example, it is supposed that both changes give out β rays in equal amounts, it can readily be calculated that the equation of decay would be

I_{t}/I_{0} = (·5λ_{2}/(λ_{2} - λ_{3}))e^{-λ_{3}t} - ·5(λ_{3}/(λ_{2} - λ_{3}) - 1)e^{-λ_{2}t}.

Taking the values of λ_{2} and λ_{3} found by Curie, the numerical factor e^{-λ_{2}t} becomes 2·15 instead of 4·3 and _{3}t} ?]1·15 instead of 3·3. The theoretical curve of decay in this case would be readily distinguishable from the observed curve of decay. The fact that the equation of decay found by Curie and Danne involves the necessity of an initial rayless change can be shown as follows:—