Problem_1.8soln

# Problem_1.8soln - First Order ODEs Problem_1.8 The...

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First Order ODEs Problem_1.8 The radioactive decay of a substance is described by the following ODE (1) Y ÅÅÅÅÅÅÅ t = - kY where k is the decay constant ( k > 0 ). If the amount of material at t=0 is Y 0 , determine an expression for the amount of material Y(t) at any other time t > 0. Next consider two arbitrary times t 1 and t 2 , where t 1 < t 2 . If t 2 is selected such that there is half as much radioactive material as there was at time t 1 , show that (2) t 2 - t 1 = Ln H 2 L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ k T ª t 2 - t 1 is called the half-life of the radioactive material. If carbon 14 has a half-life of 5568 years, what is the decay constant for carbon 14. Solution First we rearrange the equation as follows (3) Y ÅÅÅÅÅÅÅ Y = - k t Integrating once gives (4) 1 ÅÅÅÅ Y Y = - k t The result is (5) Ln

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Unformatted text preview: C This result can be reorganized as (6) Y H t L = C 1 ‰-k t Finally we use the initial condition to determine C 1 where Y H L = Y . The result is (7) Y H t L = Y ‰-k t The amounts of radioactive material at times at t 1 and t 2 are (8) Y H t 1 L = Y ‰-k t 1 , Y H t 2 L = Y ‰-k t 2 Then in order that the Y H t 2 L is half that at Y H t 1 L we have (9) Y ‰-k t 2 = 1 ÅÅÅÅ 2 Y ‰-k t 1 Simplifying this expression gives (10) ‰-k H t 2-t 1 L = 1 ÅÅÅÅ 2 Solving for t 2-t 1 gives (11) T = H t 2-t 1 L = Ln H 2 L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ k If T = 5568 years, then k is (12) k = Ln H 2 L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 5568 = 0.000124488 yrs-1 2 Problem_1.8.nb...
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