Consider the number of particles _n₀dt_ of the matter _A_ produced during the interval _dt_. At any later time _t_, the number of particles _dQ_ of the matter _B_, which result from the change in _A_, is given (see equation 4) by $$ dQ = \frac {n₀ λ_1} {λ_1 − λ_2} (e^{–λ_2 t} − e^{–λ_1 t}) dt = n₀ f(t) dt $$ .... (10). After a time of exposure _T_, the number of particles _Q_{T}_ of the matter _B_ present is readily seen to be given by $$ Q_T = n₀ (f(T) dt + f(T − dt) dt + ... + f(0) dt) = n₀ \int₀^T f(t) dt $$ . If the body is removed from the emanation after an exposure _T_, at any later time _t_ the number of particles of _B_ is in the same way given by $$ Q = n₀ \int_t^{T+t} f(t) dt $$ . It will be noted that the method of deduction of _Q_{T}_ and _Q_ is independent of the particular form of the function _f_(_t_). Substituting the particular value of _f_(_t_) given in equation (10) and integrating, it can readily be deduced that $$ \frac {Q} {Q_T} = \frac {ae^{–λ_2 t} − be^{–λ_1 t}} {a - b} $$ .... (11), where $$ a = \frac {(1 − e^{–λ_2 T} {λ_2} $$, $$ b = \frac {(1 − e^{–λ_1 T} {λ_1} $$, In a similar way, the number of particles _R_ of the matter _C_ present at any time can be deduced by substitution of the value of _f_(_t_) in equation (5). These equations are, however, too complicated in form for simple application to experiment, and will not be considered here. =200.= CASE 4. _The matter A is supplied at a constant rate from a primary source. Required to find the number of particles of A, B, C at any subsequent time t, when initially A, B, C are absent._ The solution can be simply obtained in the following way. Suppose that the conditions of Case 2 are fulfilled. The products _A_, _B_, _C_ are in radio-active equilibrium and let _P₀_, _Q₀_, _R₀_ be the number of particles of each present. Suppose the source is removed. The values of _P_, _Q_, _R_ at any subsequent time are given by equations (7), (8) and (9) respectively. Now suppose the source, which has been removed, still continues to supply _A_ at the same constant rate and let _P₁_, _Q₁_, _R₁_ be the number of particles of _A_, _B_, _C_ again present with the source at any subsequent time. Now we have seen, that the rate of change of any individual product, considered by itself, is independent of conditions and is the same whether the matter is mixed with the parent substance or removed from it. Since the values of _P₀_, _Q₀_, _R₀_ represent a steady state where the rate of supply of each kind of matter is equal to its rate of change, the sum of the number of particles _A_, _B_, _C_ present at any time with the source, and in the matter from which it was removed, must at all times be equal to _P₀_, _Q₀_, _R₀_, ..., that is _P₁_ + _P_ = _P₀_, _Q₁_ + _Q_ = _Q₀_, _R₁_ + _R_ = _R₀_.