AP Chemistry Question 160: Answer and Explanation
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2. Approximately how many half-lives will it take to reduce 10.0 kg of a radioactive substance to 1.0 microgram of that substance?
- A. 8
- B. 13
- C. 29
- D. 34
Correct Answer: D
(D) Mathematically, the fraction left is 1 microgram/10 kilograms, which is a ratio of (10-6)/(10 × 103) or 1/10-10. For half-lives, the fraction remaining is (1/2)n, where n is the number of half-lives. By equating (1/2)n = (1/10-10), we can take the logarithm of both sides of the equation to get n log (1/2) = -10. We can invert the fraction within the log term to get n log (2) = 10. (Note that this operation changes the sign.) Finally, n = 10/(log 2) = 33.2. This needs to be rounded to the next highest half-life, 34. This may be solved without a calculator by recalling that log (2) = 0.3.