Elementary Differential Equations and Boundary Value Problems 9th Edition

Published by Wiley
ISBN 10: 0-47038-334-8
ISBN 13: 978-0-47038-334-6

Chapter 1 - Introduction - 1.2 Solutions of Some Differential Equations - Problems - Page 17: 12

Answer

(a) $r=0.02828/day$ (b) $Q(t) = 100e^{-.02828t}$ (c) $t\approx 24.5$ days.

Work Step by Step

1.2.12. In general, a radioactive material decays at a rate proportional to the amount present: If $Q(t)$ denotes the amount of material at time $t$, $dQ/dt = -rQ$, where $r>0$ is the decay rate. (a) If 100 mg. of $Th^{234}$ decays to 82.04 mg. in 1 wk. (= 7 days), find r for $Th^{234}$. $\textit{Solution}$: We have (1) $\frac{dQ}{dt} = -rQ$, (2) $Q_{0} = Q(0) = 100$ mg., (3) $Q_{1} = Q(7) = 82.04$ mg., time, $t$, measured in days. We assume $r \ne 0$ and $Q\ne 0$. Thus, rewriting (1), (4) $\frac{dQ/dt}{Q} = -r$, By integrating both sides of (4), $\ln(Q) = -rt + C$, where $C$ is a constant of integration, $Q=ce^{-rt}$, where $c = e^{C}$. Now, when $t=0$, $Q=100$ (using (2)), $c = 100$. Then, by (3), $82.04 = 100e^{-r\cdot 7}$, $\ln(82.04) = \ln 100 -7r$, $7r = \ln(100/82.04)$, $r = .02828/day$. (b) Find an expression for the amount of $Th^{234}$ at any time, $t$. $\textit{Solution}$: We use the general solution found in (a): $Q = ce^{-rt}$, and use the particular values found for $c$ and $r$ $Q(t) = 100e^{-.02828t}$ (c) Find the time required for the $Th^{234}$ to decay to one half the original amount. $\textit{Solution}$: This requires us to solve the following equation for $t$. $50 = 100e^{-.02828t}$, Proceeding, $\frac{1}{2} = e^{-.02828t}$ $t = -\ln(.5)/.02828$ $\approx24.5$ days.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.