first, nearly compensates for the breaking up of C. The values of Q and R after several hours decrease exponentially, falling to half value in 28 minutes. 199. Case 3. Suppose that a primary source has supplied the matter A at a constant rate for any time T and is then suddenly removed. Required the amounts of A, B, C at any subsequent time.
Suppose that n_{0} particles of the matter A are deposited each second. After a time of exposure T, the number of particles P_{T} of the matter A present is given by
P_{T} = n_{0}[integral]_{0}^T e^{-λ_{1}t}dt = (n_{0}/λ_{1})(1 - e^{-λ_{1}T}.
At any time t, after removal of the source, the number of particles P of the matter A is given by
P = P_{T}e^{-λ_{1}t} = (n_{0}/λ_{1})(1 - e^{-λ_{1}T})e^{-λ_{1}t}.
Consider the number of particles n_{0}dt 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 = (n_{0}λ_{1})/(λ_{1} - λ_{2})(e^{-λ_{2}t} - e^{-λ_{1}t})dt = n_{0}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} = n_{0}[f(T)dt + f(T - dt)dt + . . . + f(0)dt]
= n_{0}[integral]_{0}^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 = n_{0}[integral]_{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).
