When the emanation is removed from a radium compound by solution or heating, the activity _measured by the_ β _rays_ falls almost to zero, but increases in the course of a month to its original value. The curve showing the rise of β and γ rays with time is practically identical with the curve, Fig. 85, showing the recovery of the lost activity of radium measured by the α rays. The explanation of this result lies in the fact that the β and γ rays from radium only arise from the active deposit, and that the non-separable activity of radium gives out only α rays. On removal of the emanation, the activity of the active deposit decays nearly to zero, and in consequence the β and γ rays almost disappear. When the radium is allowed to stand, the emanation begins to accumulate, and produces in turn the active deposit, which gives rise to β and γ rays. The amount of β and γ rays (allowing for a period of retardation of a few hours) will then increase at the same rate as the activity of the emanation, which is continuously produced from the radium. =216. Effect of escape of emanation.= If the radium allows some of the emanation produced to escape into the air, the curve of recovery will be different from that shown in Fig. 85. For example, suppose that the radium compound allows a constant fraction α of the amount of emanation, present in the compound at any time, to escape per second. If _n_ is the number of emanation particles present in the compound at the time _t_, the number of emanation particles changing in the time _dt_ is λ_ndt_, where λ is the constant of decay of activity of the emanation. If _q_ is the rate of production of emanation particles per second, the increase of the number _dn_ in the time _dt_ is given by _dn_ = _qdt_ − λ_ndt_ − α_ndt_, or _dn_ ----- = _q_ − (λ + α)_n_. _dt_ The same equation is obtained when no emanation escapes, with the difference that the constant λ + α is replaced by λ. When a steady state is reached, _dn_/_dt_ is zero, and the maximum value of _n_ is equal to _q_/(λ + α). If no escape takes place, the maximum value of _n_ is equal to _q_/λ. The escape of emanation will thus lower the amount of activity recovered in the proportion λ/(λ + α). If _n₀_ is the final number of emanation particles stored up in the compound, the integration of the above equation gives $$ \frac {n} {n₀} = 1 − e^{-(λ + \alpha) t} $$ . The curve of recovery of activity is thus of the same general form as the curve when no emanation escapes, but the constant λ is replaced by λ + α. For example, if α = λ = ¹⁄₄₆₃₀₀₀, the equation of rise of activity is given by $$ \frac {n} {n₀} = 1 − e^{−2λ t} $$, and, in consequence, the increase of activity to the maximum will be far more rapid than in the case of no escape of emanation.