Science:Math Exam Resources/Courses/MATH110/April 2011/Question 05

MATH110 April 2011

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[hide] Question 05
In the 1986 Chernobyl reactor explosion, substantial amounts of the isotope strontium-90 (⁹⁰Sr) contaminated the area around the nuclear plant. ⁹⁰Sr undergoes exponential radioactive decay with a half-life of 29 years. Write down a ``calculator-ready`` expression for the percentage of the origina ⁹⁰Sr remaining today (use the year the exam was written).

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[show] Hint
The formula for exponential radioactive decay is $S(t)=S_{0}e^{kt}$ , where S₀ is the initial quantity and k is a rate of decay. How can you use the information about the half-life of ⁹⁰Sr to find k?

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[show] Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We know that the amount of ⁹⁰Sr remaining at a time $t$ is modeled by the equation $S(t)=S_{0}e^{kt}$ , where $S_{0}$ is the initial quantity and k is a rate of decay. Our first task is to find k. Using the information that the half-life of ⁹⁰Sr is 29 years, we know that when t = 29, S(29) = S₀/2. Plugging this into our equation of exponential decay, we get: ${\begin{aligned}S_{0}e^{k\cdot 29}&={\frac {S_{0}}{2}}\\e^{k\cdot 29}&={\frac {1}{2}}\\k\cdot 29&=\ln \left({\frac {1}{2}}\right)\\k&={\frac {\ln \left({\frac {1}{2}}\right)}{29}}\end{aligned}}$ Note that k < 0 as expected: the amount of ⁹⁰Sr declines over time. Putting k back into the formula $S(t)=S_{0}e^{kt}$ , we get $\displaystyle S(t)=S_{0}e^{(\ln(1/2)/29)t}$ You can leave this expression as is, or simplify it to get $\displaystyle S(t)=S_{0}\left(1/2\right)^{t/29}$ To find the percentage of ⁹⁰Sr remaining today, we simply plug in the amount of time that has elapsed since the initial explosion in 1986; in 2011, that is 25 years. So $\displaystyle S(25)=S_{0}(1/2)^{25/29}$ The percentage of $S_{0}$ remaining is thus ${\frac {S(25)}{S(0)}}={\frac {S_{0}(1/2)^{25/29}}{S_{0}}}=(1/2)^{25/29}$