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• #### 2. Procedures by Element 114-272

= (A0 / λ) [1− exp(−λk1t1/2)]

= (A0t1/2 / ln 2) [1 − exp(−k1 ln 2)] .

(11)

The relative efficiency η is given by

η = Nb / Nc = A0t1/2{1 − exp[−k1(ln 2)]} / (ln 2)[A0t1/2(k1 + k2)]

= {1 − exp[−k1(ln 2)]} / [(ln 2)(k1 + k2)] .

(12)

The variable k2 is the time required for discarding the current sample and positioning the new one; hence, k2 will depend on the chemical separation being used, the transport time of the separated sample to the detector position, and other physical restrictions of the autobatch system. It is convenient to maximize η for a given k2.

dη/dk1 = 0

= ((ln 2){exp[−k1(ln 2)]} / (ln 2)(k1 + k2)) − {1 − exp[−k1(ln 2)]} / [(ln 2)(k1 + k2)2] .

(13)

The above equation can be solved for k2, or

k2 = {exp [(k1)opt(ln 2)] − 1 } / ln 2 − (k1)opt .

(14)

From the k2 equation, the optimum counting time k1 can be deduced. Figure 11 shows a plot of k1 as a function of k2. For any value of k2, the optimum counting time k1 can be obtained from the graph. For example, if the sample changing time k2 is 1 (one half-life), then the optimum counting time k1 is 1.35 half-lives; if k2 is 2, the value for k1 is 1.9. The optimum counting time k1 increases as k2 increases.

For a given value of k2 and the corresponding optimum k1 value, the maximum efficiency ηmax can be calculated:

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