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

MATH110 April 2011

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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).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

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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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 ${\frac {S(25)}{S(0)}}={\frac {S_{0}(1/2)^{25/29}}{S_{0}}}=(1/2)^{25/29}$

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