Page:Radio-active substances.djvu/98

from which it proceeds; when the solution is evaporated, the solid salt obtained is but slightly active, because it contains only a small amount of emanation. Gradually the emanation is accumulated in the salt, the activity of which rises to a limiting value, which is reached when the production of the emanation by the radium compensates the loss by external emission and by local transformation into Becquerel rays.

When a radium salt is heated, the external emission of the emanation is greatly increased, and the phenomena of induced radio-activity are more intense than when the salt is at the ordinary temperature. But when the salt returns to the ordinary temperature it is exhausted, as is the case after being dissolved, and contains but a small amount of emanation, its activity having become greatly reduced. Gradually the emanation accumulates afresh in the solid salt, and the radiation increases.

It may be said that radium gives rise to a constant generation of any emanation—part of which escapes to the exterior, the remainder being transformed in the radium itself into Becquerel rays. When radium is raised to a red heat, it loses the greater part of its capacity to cause the induction of activity; otherwise stated, the evolution of the emanation is lessened. Consequently, the proportion of the emanation utilised in the radium itself should be greater, and the substance attains a higher limit of radio-activity. We will endeavour to establish theoretically the law of rise of activity of a solid radium salt which has been dissolved or has been heated. We will assume that the intensity of radiation of radium is, at each instant, proportional to the quantity of emanation, $q$, present in the radium. We know that the emanation is spontaneously destroyed according to a law such that, at each instant—

I,

$q=q_{0}e^{-{\frac {t}{\theta }}}$ . . . . . .

$q_{0}$ being the amount of the emanation at the moment of starting the observation, and $\theta$ the time constant, equal to 4·97 × 10⁵ secs.

Now let $\Delta$ be the evolution of the emanation by radium, a quantity which I will assume constant. Let us consider what would occur if no emanation were escaping to the exterior. The emanation generated would then be completely utilised by the radium for the production of the radiation. We have from Formula 1—

${\frac {dq}{dt}}=-{\frac {q_{0}}{\theta }}e-{\frac {t}{\theta }}=-{\frac {q}{\theta }}$;