Question B 05
In the Chernobyl reactor explosion, which occurred on April 26, 1986, substantial amounts of the isotope strontium-90 (90Sr) contaminated the area around the nuclear plant. 90Sr decays at a rate proportional to its quantity. 90Sr has a half-life of 29 years; that is, it takes 29 years for a quantity of 90Sr to decrease by half. What is the proportion of 90Sr originally released which remains on April 26, 2012?
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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.
Recall that radioactive decay is an example of exponential decay, described by the formula , where is the percentage of material left after t years, and C is the initial amount.
Before you can calculate the current amount remaining, you will have to use the information about half-life to find k.
With a half life of 29 year, expect more than 50% of the material to be remaining at 26 years.
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
As stated in the second hint, we will first use the information about half-life to solve for the rate of decay, or k in the equation:
If is the percentage remaining after years, we know that of 90Sr remains after 29 years. At t = 0 the amount is 100 percent, hence choose C = 100. Plugging this into the formula gives:
Solving for , we get
Plugging this back into our formula for , we can now solve for the percentage remaining on April 26, 2012, when .
Simplifying will give:
or 53% remaining.