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Economics Dynamics Problems 64

Economics Dynamics Problems 64 - − t The half-life of a...

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48 Economic Dynamics where F [ x ( t )] = f ( x ) dx Using this equation we can solve for x = x ( t ) which gives the general solution to equation (2.19). Example 2.13 Radioactive decay and half-life Equation (2.9) specified the differential equation that represented the radioactive decay of atomic particles. We can employ the feature of separability to solve this equation. Thus, if dn dt = − λ n λ > 0 then we can re-write this equation dn n = − λ dt Integrating both sides, and letting c 0 denote the coefficient of integration, then dn n = − λ dt + c 0 ln n = − λ t + c 0 n = e λ t + c 0 = ce λ t c = e c 0 At t = t 0 , n = n 0 . From this initial condition we can establish the value of c n 0 = ce λ t 0 c = n 0 e λ t 0 n = n 0 e λ t e λ t 0 = n 0 e λ ( t
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Unformatted text preview: − t ) The half-life of a radioactive substance is the time necessary for the number of nuclei to reduce to half the original level. Since n denotes the original level then half this number is n / 2. The point in time when this occurs we denote t 1 / 2 . Hence n 2 = n e − λ ( t 1 / 2 − t ) 1 2 = e − λ ( t 1 / 2 − t ) − ln 2 = − λ ( t 1 / 2 − t ) . . . t 1 / 2 = t + ln 2 λ = t + . 693 λ Usually, t = 0 and so t 1 / 2 = . 693 λ These results are illustrated in Fgure 2.11....
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